calculus ii for engineers - faculty.uaeu.ac.aefaculty.uaeu.ac.ae/jaelee/class/1120/ln.pdfcalculus ii...

Calculus II for Engineers

My Students1

Fall 2010

1It is based mostly on the textbook, Smith and Minton’s Calculus Early Transcendental Functions, 3rd Ed and ithas been reorganized and retyped by Jae Lee.

Calculus II for Engineers Fall, 2010

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CONTENTS

9 Parametric Equations and Polar Coordinates 79.1 Plane Curves and Parametric Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79.2 Calculus and Parametric Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159.3 Arc Length and Surface Area in Parametric Equations . . . . . . . . . . . . . . . . . . . . . 159.4 Polar Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

9.4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 169.4.2 Conversion from Rectangular to Polar Coordinates . . . . . . . . . . . . . . . . . . 179.4.3 Conversion from Polar to Rectangular Coordinates . . . . . . . . . . . . . . . . . . 189.4.4 Graph of Polar Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

9.5 Calculus and Polar Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 229.6 Conic Sections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 229.7 Conic Sections in Polar Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

10 Vectors and the Geometry of Space 2510.1 Vectors in the Plane . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

10.1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2510.1.2 Representation of Vector in terms of Component . . . . . . . . . . . . . . . . . . . 2610.1.3 Representation of Vector in terms of Standard Basis . . . . . . . . . . . . . . . . . . 2810.1.4 Representation of Vector in Polar Form . . . . . . . . . . . . . . . . . . . . . . . . 2910.1.5 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

10.2 Vectors in Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3110.2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3110.2.2 Vectors in R3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3310.2.3 Some Problems from Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

10.3 The Dot Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3510.3.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3510.3.2 Geometric Interpretation of Dot Product . . . . . . . . . . . . . . . . . . . . . . . . 3510.3.3 Components and Projections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3610.3.4 Application: Work Done . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

10.4 The Cross Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3910.4.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3910.4.2 Application: Area, Distance, Volume, Torque . . . . . . . . . . . . . . . . . . . . . 40

10.5 Lines and Planes in Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4410.5.1 Lines in R3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4410.5.2 Planes in R3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

10.6 Surfaces in Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

3


11 Vector–Valued Functions 4711.1 Vector–Valued Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

11.1.1 Vector–Valued Functions in Plane . . . . . . . . . . . . . . . . . . . . . . . . . . . 4711.1.2 Vector–Valued Functions in Space . . . . . . . . . . . . . . . . . . . . . . . . . . . 4811.1.3 Arc Length . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

11.2 The Calculus of Vector–Valued Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . 5011.2.1 Limit, Continuity and Derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5011.2.2 Interpretation of Derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5111.2.3 Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

11.3 Motion In Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5211.4 Curvature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

11.4.1 Arc Length Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5311.4.2 Curvature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

11.5 Tangent and Normal Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5511.6 Parametric Surfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

12 Functions of Several Variables and Partial Differentiation 5712.1 Functions of Several Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5712.2 Limits nd Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5712.3 Partial Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

12.3.1 Partial Derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5712.3.2 Higher–Order Partial Derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58

12.4 Tangent Planes And Linear Approximations . . . . . . . . . . . . . . . . . . . . . . . . . . 5912.5 The Chain Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

12.5.1 Chain Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6012.5.2 Implicit Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62

12.6 Gradient and Directional Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6412.6.1 Functions of Two Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6412.6.2 Functions of Three Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71

12.7 Extrema of Functions of Several Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . 7512.7.1 Definition and Second Derivatives Test . . . . . . . . . . . . . . . . . . . . . . . . 7512.7.2 Linear Regression: Method of Least Squares . . . . . . . . . . . . . . . . . . . . . 8012.7.3 Steepest Ascent Algorithm: Estimating Local Extrema without Critical Points . . . . 8012.7.4 Absolute Extrema . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80

12.8 Constrained Optimization and Lagrange Multipliers . . . . . . . . . . . . . . . . . . . . . . 81

13 Multiple Integrals 8313.1 Double Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83

13.1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8313.1.2 Double Integral over a Rectangle . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8313.1.3 Geometrical Application . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8413.1.4 Double Integral over a General Region . . . . . . . . . . . . . . . . . . . . . . . . 8413.1.5 Changing Order of Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8613.1.6 Linearity of Double Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87

13.2 Area, Volume, and Center of Mass . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8813.2.1 Area . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8813.2.2 Volume . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8813.2.3 Moments and Center of Mass . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9113.2.4 Moments of Inertia . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92

13.3 Double Integrals in Polar Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9313.3.1 Polar Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93

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13.3.2 Double Integrals in Polar Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . 9313.4 Surface Area . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9713.5 Triple Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98

13.5.1 Formulas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9813.5.2 Applications: Volume, Mass and Center of Mass . . . . . . . . . . . . . . . . . . . 99

13.6 Cylindrical Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10213.6.1 Cylindrical Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10213.6.2 Evaluating Triple Integrals with Cylindrical Coordinates . . . . . . . . . . . . . . . 103

13.7 Spherical Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10613.7.1 Spherical Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10613.7.2 Triple Integrals in Spherical Coordinates . . . . . . . . . . . . . . . . . . . . . . . 108

13.8 Change of Variables in Multiple Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . 110

14 Vector Calculus 11114.1 Vector Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11114.2 Line Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111

14.2.1 Line Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11114.2.2 Application: Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112

14.3 Independence of Path and Conservative Vector Fields . . . . . . . . . . . . . . . . . . . . . 11314.4 Green’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11314.5 Curl and Divergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114

14.5.1 Curl . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11414.5.2 Divergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115

14.6 Surface Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11614.7 The Divergence Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11614.8 Stokes’ Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11614.9 Applications of Vector Calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116

15 Appendix: Distances between Various Objects 11715.1 Two–Dimensional Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117

15.1.1 Two Points P(x0,y0) and Q(x1,y1) . . . . . . . . . . . . . . . . . . . . . . . . . . . 11715.1.2 Line L : y = mx+b and Point P(x0,y0) not lying on L . . . . . . . . . . . . . . . . . 11715.1.3 Parallel Lines L : y = mx+b and M : y = mx+ c . . . . . . . . . . . . . . . . . . . 118

15.2 Three–Dimensional Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11915.2.1 Two Points P(x0,y0,z0) and Q(x1,y1,z1) . . . . . . . . . . . . . . . . . . . . . . . . 11915.2.2 Line L and Point Q(x0,y0,z0) not lying on L:

L : x = x1 +at, y = y1 +bt, z = z1 + ct . . . . . . . . . . . . . . . . . . . . . . . . 11915.2.3 Parallel Lines L and M:

L : x = x0 +at,y = y0 +bt,z = z0 + ct and M : x = x1 +a′s,y = y1 +b′s,z = z1 + c′s . 12015.2.4 Skew Lines L and M (Version 1):

L : x = xL +aLt, y = yL +bLt, z = zL + cLt andM : x = xM +aMs, y = yM +bMs, z = zM + cMs . . . . . . . . . . . . . . . . . . . . 121

15.2.5 Plane Π : ax+by+ cz = d and Point P(x0,y0,z0) not lying on Π . . . . . . . . . . . 12215.2.6 Plane Π : ax+ by+ cz = d and Line L : x = x0 + a′t,y = y0 + b′t,z = z0 + c′t not

intersecting with Π . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12315.2.7 Parallel Planes Π0 : ax+by+ cz = d and Π1 : a′x+b′y+ c′z = e . . . . . . . . . . . 12415.2.8 Appendix: Skew Lines L and M (Version 2):

L : x = xL +aLt, y = yL +bLt, z = zL + cLt andM : x = xM +aMs, y = yM +bMs, z = zM + cMs . . . . . . . . . . . . . . . . . . . . 125

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Chapter 9

Parametric Equations and Polar Coordinates

§9.1 Plane Curves and Parametric Equations.Given any pair of functions x(t) and y(t) defined on the same domain D, the equations

x = x(t), y = y(t)

are called parametric equations and t is called the parameter.For each choice of t, the parametric equations specify a point (x,y) = (x(t),y(t)) in the xy–plane. Thecollection of all such points is called the graph of the parametric equations. In the case where x(t) and y(t)are continuous functions and D is an interval of the real line, the graph of the parametric equations is a curvein the xy–plane and it is called the plane curve.

Example 9.1.1. Sketch the plane curve defined by the parametric equations x= 6−t2, y= t/2 for −2≤ t ≤ 4.

ANSWER 1 (NAIVE METHOD). We make a table and based on the table we plot the points.

t x y

−2 2 −1

−1 5 −1/2

0 6 0

1 5 1/2

2 2 1

3 −3 3/2

4 −10 2

Table 9.1: Graph of x = 6− t2, y = t/2, −2 ≤ t ≤ 4□

As t increases from −2 to 4, the point (x,y) moves from (2,−1) to (−10,2) along the plane curve. Thisdirection is called the orientation of the plane curve.

ANSWER 2 (EQUATION OF x AND y). As an easier way, we may consider eliminating the parameter t andgetting the equation of x and y.

y = t/2 −→ t = 2y −→ x = 6− t2 = 6− (2y)2 = 6−4y2 −→ x = 6−4y2.

Moreover, since −2 ≤ t ≤ 4, so we have −1 ≤ y = t/2 ≤ 2. Thus, we have the following equation with theinterval of y:

x = 6−4y2, −1 ≤ y ≤ 2.

If we sketch its graph, we have exactly same graph as the one in Answer 1 above. □

Example 9.1.2. Sketch the plane curve defined by the parametric equations x = 2cos t, y = 2sin t, for (1)0 ≤ t ≤ 2π and (2) 0 ≤ t ≤ π and (3) −π/2 ≤ t ≤ π/2.

7


Figure 9.1: Plane curves of x = 2cos t, y = 2sin t with various intervals of t

ANSWER. We eliminate t by using the identity cos2 t + sin2 t = 1 and get the equation of x and y and sketchits graph.

cos t =x2, sin t =

y2

−→ x2

22 +y2

22 = cos2 t + sin2 t = 1 −→ x2 + y2 = 22,

which is a circle centered at the origin of radius 2.(1) 0 ≤ t ≤ 2π: In this case, we have

−2 ≤ x = 2cos t ≤ 2, −2 ≤ y = 2sin t ≤ 2,

which means the plane curve is traversed one whole circle on the xy–plane. Also we observe

1. as t moves from 0 to π/2, (x,y) moves from (2,0) to (0,2),2. as t moves from π/2 to π , (x,y) moves from (0,2) to (−2,0),3. as t moves from π to 3π/2, (x,y) moves from (−2,0) to (0,−2),4. as t moves from 3π/2 to 2π , (x,y) moves from (0,−2) to (2,0).

It implies that the plane curve has the counterclockwise orientation. See the figure 9.1.(2) 0 ≤ t ≤ π: In this case, we have

−2 ≤ x = 2cos t ≤ 2, 0 ≤ y = 2sin t ≤ 2,

which means the plane curve is traversed top semicircle on the xy–plane. Also we observe

1. as t moves from 0 to π/2, (x,y) moves from (2,0) to (0,2),2. as t moves from π/2 to π , (x,y) moves from (0,2) to (−2,0).

It implies that the plane curve has the counterclockwise orientation. See the figure 9.1.(3) −π/2 ≤ t ≤ π/2: In this case, we have

0 ≤ x = 2cos t ≤ 2, −2 ≤ y = 2sin t ≤ 2,

which means the plane curve is traversed right semicircle on the xy–plane. Also we observe

1. as t moves from −π/2 to 0, (x,y) moves from (0,−2) to (2,0),2. as t moves from 0 to π/2, (x,y) moves from (2,0) to (0,2).

It implies that the plane curve has the counterclockwise orientation. See the figure 9.1. □

Exercise 9.1.3. Sketch the plane curve defined by the parametric equations x = 2sin t, y = 2cos t, for 0 ≤ t ≤2π .

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Example 9.1.4. Sketch the plane curve defined by the parametric equations.

1. x = 2cos t, y = 3sin t, for 0 ≤ t ≤ 2π2. x = 2+4cos t, y = 3+4sin t, for 0 ≤ t ≤ 2π3. x = 3cos(2t), y = 3sin(2t), for 0 ≤ t ≤ 2π4. x = 3cos(t/2), y = 3sin(t/2), for 0 ≤ t ≤ 2π

Figure 9.2: Plane curves of Example 9.1.4

ANSWER. (1) x = 2cos t, y = 3sin t, for 0 ≤ t ≤ 2π: By eliminating t, we have

cos t =x2, sin t =

y3

−→ x2

22 +y2

32 = cos2 t + sin2 t = 1 −→ x2

22 +y2

32 = 1,

which is a ellipse passing through the points (±2,0) and (0,±3). In fact, the plane curve is a whole ellipseand it has the counterclockwise orientation. See the figure 9.2.(2) x = 2+4cos t, y = 3+4sin t, for 0 ≤ t ≤ 2π: By eliminating t, we have

cos t =x−2

4, sin t =

y−34

−→ (x−2)2

42 +(y−3)2

42 = cos2 t + sin2 t = 1

−→ (x−2)2 +(y−3)2 = 42,

which is a circle centered at (2,3) of radius 4. In fact, the plane curve is a whole circle and it has thecounterclockwise orientation. See the figure 9.2.(3) x = 3cos(2t), y = 3sin(2t), for 0 ≤ t ≤ 2π: By eliminating t, we have

cos(2t) =x3, sin(2t) =

y3

−→ x2

32 +y2

32 = cos2(2t)+ sin2(2t) = 1 −→ x2 + y2 = 32,

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which is a circle centered at the origin of radius 3. In fact, the plane curve is a whole circle and it has thecounterclockwise orientation. But, the circle is traced out twice: first time for 0 ≤ t ≤ π and second time forπ ≤ t ≤ 2π . It is because of 2t in the parametric equations. See the figure 9.2.(4) x = 3cos(t/2), y = 3sin(t/2), for 0 ≤ t ≤ 2π: By eliminating t, we have

cos(t/2) =x3, sin(t/2) =

y3

−→ x2

32 +y2

32 = cos2(t/2)+ sin2(t/2) = 1 −→ x2 + y2 = 32,

which is a circle centered at the origin of radius 3. In fact, the plane curve has the counterclockwise orientationand it is a top semicircle. It is because of t/2 in the parametric equations. See the figure 9.2. □

Exercise 9.1.5. Sketch the plane curve defined by the parametric equations.

1. x = cos(3t), y = sin(3t), for 0 ≤ t ≤ 2π2. x = cos(4t), y = sin(4t), for 0 ≤ t ≤ 2π3. x = cos(t/3), y = sin(t/3), for 0 ≤ t ≤ 2π4. x = cos(t/4), y = sin(t/4), for 0 ≤ t ≤ 2π

Example 9.1.6 (CHALLENGEABLE). Sketch the plane curve defined by the parametric equations: (1) x =2tan t, y = 2sin t and (2) x = 2cos t, y = 3sin(2t).

Figure 9.3: Plane curves of x = 2tan t, y = 5sin t and x = 2cos t, y = 3sin(2t)

ANSWER. (1) x = 2tan t, y = 5sin t: A simple observation gives

sin t =y5,

sin tcos t

= tan t =x2

−→ y5= sin t =

x2

cos t −→ 2y = 5xcos t.

The given equation sin t = y/5 implies the following triangle.

..√

52 − y2

.5

.y

.t

It also implies cos t =√

52 − y2 /5 and thus we have

2y = 5x

√52 − y2

5−→ 4y2 = x2(52 − y2) −→ y2 =

5x2

x2 +4−→ y =±

√5x2

x2 +4.

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See the figure 9.3.(2) x = 2cos t, y = 3sin(2t): A simple formula says sin(2t) = 2sin t cos t and so we have

cos t =x2, y = 3sin(2t) = 6sin t cos t = 6sin t

x2= 3xsin t −→ y = 3xsin t.

The given equation cos t = x/2 implies the following triangle.

..x

.2

.√

22 − x2

.t

It also implies sin t =√

22 − x2 /2 and thus we have

y = 3xsin t = 3x

√22 − x2

2−→ y =

32

x√

22 − x2 .

See the figure 9.3. □

In the challengeable problems as above, we need to use the triangle technique and the basic formulas:

sin(A±B) = sinAcosB± cosAsinB cos(A±B) = cosAcosB∓ sinAsinB

sin(2A) = 2sinAcosA cos(2A) = 2cos2 A−1 = 1−2sin2 A

sin2(

A2

)=

1− cosA2

cos2(

A2

)=

1+ cosA2

tan2 A+1 = sec2 A cos2 t + sin2 t = 1

So far when parametric equations are given, we have sketched its plane curve via two ways: naive method(table) and elimination method (finding the equation of x and y by eliminating t). Now let us consider theother way. Suppose we have a graph of x and y or the equation of x and y. Then how can find parametricequations having the given graph as its plane curve?

Example 9.1.7. Find parametric equations for the line segment joining the point A(1,2) (initial point) andthe point B(4,7) (terminal point).

ANSWER 1 (USING EQUATION OF x AND y). The easiest way may be using the equation of x and y satisfy-ing the conditions given in the problem, if we can find it. In this problem, it is easy to see that the equationof the line passing through A and B is found by

y =53(x−1)+2 −→ y =

53

x+13.

• When we set x = t, we can easily have y = 53t + 1

3 , i.e., we have the parametric equations

x = t, y =53

t +13.

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Since the orientation of the line segment is from A to B, so as t increases, the point (x,y) on the line shouldmove from A to B. In our parametric equations, x = t and it should start at A(1,2). Thus t must start at 1.Also the line segment should end at B(4,7) and so t must end at 4. That is, the given line segment should berepresented by the parametric equations,

x = t, y =53

t +13, 1 ≤ t ≤ 4.

• When we set y = t, the equation x = 3y−15 implies x = 3t−1

5 and so we have

x =3t −1

5, y = t.

Since the orientation of the line segment is from A to B, so as t increases, the point (x,y) on the line shouldmove from A to B. In our parametric equations, y = t and it should start at A(1,2). Thus t must start at 2.Also the line segment should end at B(4,7) and so t must end at 7. That is, the given line segment should berepresented by the parametric equations,

x =3t −1

5, y = t, 2 ≤ t ≤ 7.

There are many other parametric equations for the given line segment. □

ANSWER 2 (USING FORM x = a+bt , y = c+dt). Since we are asked to find the parametric equations forthe line segment, we set up

x = a+bt, y = c+dt, α ≤ t ≤ β ,

where a, b, c, d and α and β are all constants to be determined. (Each equation on x and y represents a lineon the tx–plane and ty–plane, respectively.)For the simplest case, let us choose α = 0 and β = 1 and find other constants. Since A(1,2) is the startingpoint of the line segment, x = a+bt should be 1 at t = α , i.e., at t = 0. So we have

1 = x = a+b(0) = a −→ a = 1.

Similarly, since A(1,2) is the starting point of the line segment, y = c+dt should be 2 at t = α , i.e., at t = 0.So we have

2 = y = c+d(0) = c −→ c = 2.

By the same argument, since B(4,7) is the end point of the line segment, x = a+bt and y = c+dt should be4 and 7, respectively, at t = β , i.e., t = 1. So we have

4 = x = a+b(1) = a+b = 1+b −→ b = 37 = y = c+d(1) = c+d = 2+d −→ d = 5.

Therefore, we find parametric equations

x = 1+3t, y = 2+5t, 0 ≤ t ≤ 1.

Clearly there are many other parametric equations representing the given line segment. □

Remark 9.1.8. In the Answers above, we have found three parametric equations for the line segment:

1. x = t, y = 5t+13 , for 1 ≤ t ≤ 4

2. x = 3t−15 , y = t, for 2 ≤ t ≤ 7

3. x = 1+3t, y = 2+5t, for 0 ≤ t ≤ 1

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Figure 9.4: Portion of parabola y = x2 from A(−1,1) to B(3,9)

Are they same? They must be same, because they represent the same line segment. But then how can weprove that those three parametric representations are exactly same?

Example 9.1.9. Find parametric equations for the portion of the parabola y = x2 from A(−1,1) to B(3,9).

ANSWER. See the figure 9.4. Since the equation of x and y is given, we just set x = t or y = t.• Let us set x = t. Then it is easy to see y = t2 and t moves from t =−1 to t = 3, i.e.,

x = t, y = t2, −1 ≤ t ≤ 3.

• Let us set y = t. Then it is easy to see x2 = t and x =±√

t . In this case, we need to be careful. We observethat t = 0 corresponds to the origin (x,y) = (0,0). But t = 1 corresponds to (−1,1) and/or (1,1).If t = 1 corresponds to the point (1,1), then there is no way to trace out the portion between the origin andthe point A(−1,1).If t = 1 corresponds to the point A(−1,1), then it means the graph is traversed from the origin (t = 0) to thepoint A (t = 1) as t moves from 0 to 1. But then in order for the graph to pass by the point B(3,9), it should betraced out the portion between the origin and the point A(−1,1) again, i.e., it is traced out the portion twice,which is not acceptable. For this reason, the parametric equations y = t and x =±

√t does not represent the

given portion and so it is not acceptable. □

Exercise 9.1.10. Find parametric equations for the portion of the parabola y = −x2 + 4 from A(−1,3) toB(3,−5).

Example 9.1.11. Sketch the plane curve defined by the parametric equations x = t2 −2, y = t3 − t.

Figure 9.5: Plane curve of x = t2 −2, y = t3 − t

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ANSWER. The equation of x and y corresponding to the parametric equations is obtained by

t2 = x+2 −→ t =±√

x+2 −→ y = t3 − t =±(x+2)3/2 ∓ (x+2)

of which graph is not easy to sketch. So we use the naive method – table. We plot the x–intercepts andy–intercepts and connect those points.• x–intercept: A x–intercept is a point whose y–component is 0. So we solve y = 0 for t:

y = t3 − t = t(t2 −1) = t(t −1)(t +1) = 0 at t = 0,±1.

At t = 0, we have (x,y) = (−2,0). At t =±1, we have (x,y) = (−1,0).• y–intercept: A y–intercept is a point whose x–component is 0. So we solve x = 0 for t:

x = t2 −2 = (t −√

2 )(t +√

2 ) = 0 at t =±√

2 .

At t =−√

2 , we have (x,y) = (0,−√

2 ). At t =√

2 , we have (x,y) = (0,√

2 ).• Signs of x and y: x = t2 −2 is negative for t ∈ (−

√2 ,

√2 ) and positive for t ∈R− [−

√2 ,

√2 ]. (R means

the set of all real numbers.) y= t3−t is negative for t ∈ (−∞,−1)∪(0,1) and positive for t ∈ (−1,0)∪(1,∞).Combining all pieces of the information above, we have the following table.

t −√

2 −1 0 1√

2

x + 0 − −1 − −2 − −1 − 0 +

y − −√

2 − 0 + 0 − 0 +√

2 +

Based on the table, we have the figure 9.5. □Exercise 9.1.12. Sketch the plane curve defined by the parametric equations x = t3 − t, y = t4 −5t2 +4.

Example 9.1.13. Suppose that a missile is fired toward your location from 500 miles away and follows aflight path given by the parametric equations

x = 100t, y = 80t −16t2, for 0 ≤ t ≤ 5. (9.1.1)

Two minutes later, you fire an interceptor missile from your location following the flight path

x = 500−200(t −2), y = 80(t −2)−16(t −2)2, for 2 ≤ t ≤ 7. (9.1.2)

Determine whether the interceptor missile hits its target.

Figure 9.6: Plane curve of Attacking Missile (Red) and Interceptor (Blue)

ANSWER. When two paths meet, they should have the same x and same y component at the same time t. Sowe solve the equations for t:

100t = x = 500−200(t −2) −→ 100t = 500−200(t −2) −→ t = 3

80t −16t2 = y = 80(t −2)−16(t −2)2 −→ t = 3.5

It implies when t = 3, two paths have the same x but different y’s (difference in y is 32). When t = 3.5, theyhave the same y but different x’s (difference in x is 150). Therefore, they do not meet at all. See the figure 9.6.(The figure in the far left–hand side shows the intersection. But in fact they do not meet as in the secondfigure.) □

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Remark 9.1.14. For the attacking missile following the path (9.1.1), the given interceptor (9.1.2) is useless.How can we modify (9.1.2) so that the modified interceptor can intercept the attacking missile? The modifiedinterceptor should follow the path as given in the third one in figure 9.6.

§9.2 Calculus and Parametric Equations.Skip. Please read the textbook.

§9.3 Arc Length and Surface Area in Parametric Equations.Skip. Please read the textbook.

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§9.4 Polar Coordinates.

□ 9.4.1 Introduction.A point on the plane can be described in terms of the horizontal and vertical distances from the origin. Thissystem is referred as the system of rectangular coordinates. See the figure 9.7.

..x

.y

.y

.x.(x,y)

.O

Figure 9.7: Rectangular Coordinates.

..x

.y

.r

.(r,θ)

.θ

.O

Figure 9.8: Polar Coordinates.

As an alternative description of a point on the plane, we use (r,θ), where r is the distance from the originto the point and θ is the angle (in radians) measured from the positive x–axis counterclockwise to the rayconnecting the origin and the point. Here r and θ are called the polar coordinates for the point and this systemis referred as the system of polar coordinates. See the figure 9.8.By the definition,

1. any point on the positive x-axis has θ = 0,2. any point on the positive y-axis has θ = π/2,3. any point on the negative x-axis has θ = π ,4. any point on the negative y-axis has θ = 3π/2.

See the figure 9.9.

..θ = 0.θ = π

.θ = π/2

.θ = 3π/2

.O

Figure 9.9: Angles θ = 0, θ = π/2, θ = π , θ = 3π/2

..x

.y

.O .A(2,0).D(2,π)

.B(3,π/2)

.C(−3,π/2)

Figure 9.10: Plots of points in Example 9.4.1

Example 9.4.1. Plot the points with the indicated polar coordinates (r,θ) and determine the correspondingrectangular coordinates (x,y) for (1) (r,θ) = A(2,0), (2) (r,θ) = B(3,π/2), (3) (r,θ) =C(−3,π/2) and (4)(r,θ) = (2,π).

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ANSWER. (1) (r,θ) = A(2,0): Since θ = 0, the point A should be on the positive x–axis and by r = 2, itshould be located 2 units away from the origin. Thus, it corresponds to (x,y) = (2,0).(2) (r,θ) = B(3,π/2): Since θ = π/4, the point B should be on the positive y–axis and by r = 3, it should belocated 3 units away from the origin. Thus, it corresponds to (x,y) = (0,3).(3) (r,θ) = C(−3,π/2): The angle is the same as in (2), but a negative value of r indicates that the point islocated 3 united in the opposite direction. So, the point C should be on the line corresponding to θ =−π/2(which is same as θ = 3π/2) with the distance 3, i.e., it corresponds to (x,y) = (0,−3).(4) (r,θ) = D(2,π): Since θ = π , the point D should be on the negative x–axis and by r = 2, it should belocated 2 units away from the origin. Thus, it corresponds to (x,y) = (−2,0). See the figure 9.10. □Example 9.4.2. Find a polar coordinate representation (r,θ) of the rectangular point (x,y) = (1,1).

ANSWER. When we draw a line from the origin to the point (x,y) = (1,1), we can measure the lengthr =

√12 +12 =

√2 of the line segment and the angle θ = π/4 from the positive axis axis to the line segment.

Thus, the point (x,y) = (1,1) has the polar coordinate representation (r,θ) = (√

2 ,π/4). We notice that wecan specify the same point by using a negative value of r, r =−

√2 , with the angle θ = π/4+π = 5π/4. So

the point (x,y) = (1,1) has another polar coordinate representation (r,θ) = (−√

2 ,5π/4). In fact, all of thepolar points are

(r,θ) =(√

2 ,π4+2nπ

)and (r,θ) =

(−√

2 ,π4+π +2nπ

). □

Remark 9.4.3. Each point in the plane has infinitely many polar coordinates representations. For a givenangle θ , the angles θ ± 2π , θ ± 4π and so on, all correspond to the same ray. For convenience we use thenotation θ +2nπ (for any integer n) to represent all of these possible angles.

□ 9.4.2 Conversion from Rectangular to Polar Coordinates.Is there a better way to convert a point given in rectangular coordinates to the one in polar coordinates? Wemay use the following relationship between (x,y) and (r,θ):

r2 = x2 + y2, tanθ =yx.

Example 9.4.4. Find all polar coordinate representations (r,θ) for the rectangular points (1) (x,y) = (2,3)and (2) (x,y) = (−3,1).

ANSWER. (1) (x,y) = (2,3): By the formula, we have

r2 = x2 + y2 = 4+9 = 13 −→ r =±√

13 and tanθ =yx=

32

−→ θ ≈ 0.98.

Since (x,y) = (2,3) is in the first quadrant, so it is basically represented by (r,θ) = (√

13 ,0.98). Thus, allpolar representations are

(r,θ) =(√

13 ,0.98+2nπ)

and (r,θ) =(−√

13 ,0.98+π +2nπ),

where n is any integer.(2) (x,y) = (−3,1): By the formula, we have

r2 = x2 + y2 = 9+1 = 10 −→ r =±√

10 and tanθ =yx=

1−3

−→ θ ≈−0.32.

Since (x,y) = (−3,1) is in the second quadrant, so it is basically represented by (r,θ) = (−√

10 ,−0.32).Thus, all polar representations are

(r,θ) =(−√

10 ,−0.32+2nπ)

and (r,θ) =(√

10 ,−0.32+π +2nπ),

where n is any integer. □

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□ 9.4.3 Conversion from Polar to Rectangular Coordinates.Now conversely, for a point (r,θ) given in polar coordinates, how can we find its rectangular representation(x,y)? We use the following equations:

x = r cosθ , y = r sinθ ,

which is very important and worth to be memorized.

Example 9.4.5. Find the rectangular coordinates (x,y) for the polar points (1) (r,θ) = (3,π/6) and (2)(r,θ) = (−2,3).

ANSWER. (1) (r,θ) = (3,π/6): By the formula, we have

x = r cosθ = 3cosπ6=

3√

32

, y = r sinθ = 3sinπ6=

32.

Thus, the point corresponds to (x,y) = (3√

3/2,3/2).(2) (r,θ) = (−2,3): By the formula, we have

x = r cosθ =−2cos3 ≈ 1.98, y = r sinθ =−2sin3 ≈−0.28.

Thus, the point corresponds to (x,y) = (1.98,−0.28). □

□ 9.4.4 Graph of Polar Equation.Let us refresh our memory on a graph of an equation/a function. A graph means a set of all points satisfyingthe equation or the points obtained by the function. For instance, the graph G of the line y = mx+ b on theplane means the following set

G = { (x,y) : x and y satisfies y = mx+b }= { (x,mx+b) : x is any real number } .

Also the graph G of the function y = f (x) means the set

G = { (x,y) : x and y satisfies y = f (x) }= { (x, f (x)) : x is any real number } .

In this topic, we sketch the graphs of the polar equations r = constant, θ = constant and r = f (θ), somefunction of θ .

Example 9.4.6. Sketch the graphs of (1) r = 2 and (2) θ = π/3.

ANSWER 1 (USING DEFINITION). (1) r = 2: According to the definition of the polar coordinates, the graphG of r = 2 means the following set

G = { (r,θ) : r and θ satisfies r = 2 }= { (r,θ) : the point (r,θ) is 2 units away from the origin } .

Since there is no condition on θ , when we collect all points 2 units away from the origin, we have a circlecentered at the origin of radius 2. Thus, the graph of r = 2 is such a circle.(2) θ = π/3: According to the definition of the polar coordinates, the graph G of θ = π/3 means the set

G = { (r,θ) : r and θ satisfies θ = π/3 }= { (r,θ) : ray connecting the origin to point (r,θ) has the angle π/3 from the positive x–axis } .

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..x.−2 .2

.y

.−2

.2 .θ = π/3

.π/3

.r = 2

Figure 9.11: Graphs of r = 2 and θ = π/3.

Since there is no condition on r, when we collect all points satisfying the condition, the graph of θ = π/3becomes a line with the slope tan(π/3) =

√3 and passing through the origin, i.e., in rectangular coordinates,

it corresponds to y =√

3 x.We should be careful. The graph of θ = π/3 is in fact the same as the graph of θ = π/3+π , because thereis no condition on r. For example, (r,θ) = (−2,π/3) should be on the graph of θ = π/3 and as we know, itis also on the graph of θ = π/3+π . See the figure 9.11. □

ANSWER 2 (USING EQUATION OF x AND y). Let us use the formula to convert the polar equation to therectangular equation.(1) r = 2:

x2 + y2 = r2 = 22 −→ x2 + y2 = 22,

which is a circle centered at the origin of radius 2.(2) θ = π/3:

yx= tanθ = tan

π3=√

3 −→ y =√

3 x,

which is a straight line with the slope√

3 and passing through the origin. See the figure 9.11. □

As we can see, using the equation of x and y is much easier to sketch the graph of the given polar equations.So whenever the rectangular representation of the polar equation is available and it is easy to sketch, we willfollow this method.

Example 9.4.7. Find the polar equation(s) corresponding to the hyperbola x2 − y2 = 9.

Figure 9.12: Hyperbola x2 − y2 = 9

ANSWER. The formula implies

x = r cosθ , y = r sinθ

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−→ 9 = x2 − y2 = r2 cos2 θ − r2 sin2 θ = r2 (cos2 θ − sin2 θ)= r2 cos(2θ)

−→ r2 cos(2θ) = 9 −→ r2 = 9sec(2θ) −→ r =±3√

sec(2θ) .

The graph of the polar equations r =±3√

sec(2θ) is the hyperbola x2−y2 = 9. Specifically, r = 3√

sec(2θ)corresponds to the right branch of the hyperbola, because the points (r,θ) satisfying r = 3

√sec(2θ) should

correspond to positive x = r cosθ > 0. r = −3√

sec(2θ) corresponds to the left branch of the hyperbola,because the points (r,θ) satisfying r =−3

√sec(2θ) should correspond to negative x = r cosθ < 0. See the

figure 9.12. □

Example 9.4.8. Sketch the graph of the polar equation r = sinθ .

Figure 9.13: Graph of r = sinθ

ANSWER (USING EQUATION OF x AND y). By the formula, we have

r = sinθ −→ r2 = r sinθ −→ x2 + y2 = r2 = r sinθ = y −→ x2 + y2 − y = 0.

We recall the formula on the completing the square: a quadratic function f (x)= ax2+bx+c has the followingvertex form,

f (x) = ax2 +bx+ c = a(

x+ba

)2

− b2 −4ac4a

.

Using the form, we can rewrite the result

x2 + y2 − y = 0 −→ x2 +

(y− 1

2

)2

=122 ,

of which graph is a circle centered at the point (0,1/2) of radius 1/2. Therefore, the graph of r = sinθ is thecircle centered at the point (0,1/2) of radius 1/2. See the figure 9.13. □

Example 9.4.9 (Limacon1). Sketch the graph of the polar equation r = 3+2cosθ .

ANSWER 1 (USING RECTANGULAR EQUATION). Just as we did in Example 9.4.6, let us try to find thecorresponding rectangular equation of x and y. By the formulas, we have

r = 3+2cosθ −→ r2 = 3r+2r cosθ

1In mathematics, limacons (pronounced with a soft c), also known as limacons of Pascal, are heart–shaped mathematicalcurves. The cardioid is a special case, with a cusp. They arise in polar coordinates in the form r = a+bcosθ which in Cartesiancoordinates is (x2 + y2 −bx)2 = a2(x2 + y2).

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Figure 9.14: Graph of r = 3+2cosθ

−→ x2 + y2 = r2 = 3r+2r cosθ =±3√

x2 + y2 +2x

−→ x2 + y2 −2x =±3√

x2 + y2 −→(x2 + y2 −2x

)2= 9(x2 + y2)

of which graph is not easy to sketch. It is because of the constant term 3. Without 3, r = 2cosθ is easilyconverted to the rectangular equation, which is (x−1)2 + y2 = 1. For this reason, we do not follow this wayto sketch the graph. □

ANSWER 2 (NAIVE METHOD – TABLE). We make the following table and based on the table, we sketchthe graph.

Interval cosθ r = 3+2cosθ

[0,π/2] Decreases from 1 to 0 Decreases from 5 to 3

[π/2,π] Decreases from 0 to −1 Decreases from 3 to 1

[π,3π/2] Increases from −1 to 0 Increases from 1 to 3

[3π/2,2π] Increases from 0 to 1 Increases from 3 to 5


Example 9.4.10 (Cardioid). Sketch the graph of the polar equation r = 2−2sinθ .

ANSWER. We make the following table and based on the table, we sketch the graph.

Interval sinθ r = 2−2sinθ

[0,π/2] Increases from 0 to 1 Decreases from 2 to 0

[π/2,π] Decreases from 1 to 0 Increases from 0 to 2

[π,3π/2] Decreases from 0 to −1 Increases from 2 to 4

[3π/2,2π] Increases from −1 to 0 Decreases from 4 to 2


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Figure 9.15: Graph of r = 2−2sinθ

Exercise 9.4.11. Sketch the graph of the polar equation r = 1−2sinθ .

Example 9.4.12 (Four–Leaf Rose). Sketch the graph of the polar equation r = sin(2θ).ANSWER. We make the following table and based on the table, we sketch the graph. Since sin(2θ) = 0 at2θ = 0,π,2π,3π,4π , i.e., θ = 0,π/2,π,3π/2,2π . So the graph of sin(2θ) is given as in the figure 9.16. Thegraph shows 8 intervals as in the given table below.

Interval r = sin(2θ) Quadrant

[0,π/4] Increases from 0 to 1 First

[π/4,π/2] Decreases from 1 to 0 First

[π/2,3π/4] Decreases from 0 to −1 Fourth

[3π/4,π] Increases from −1 to 0 Fourth

[π,5π/4] Increases from 0 to 1 Third

[5π/4,3π/2] Decreases from 1 to 0 Third

[3π/2,7π/4] Decreases from 0 to −1 Second

[7π/4,2π] Increases from −1 to 0 SecondFigure 9.16: Graph of r = sin(2θ)


§9.5 Calculus and Polar Coordinates.Skip. Please read the textbook.

§9.6 Conic Sections.Skip. Please read the textbook.

§9.7 Conic Sections in Polar Coordinates.Skip. Please read the textbook.

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Figure 9.17: Graph of r = sin(2θ)

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Chapter 10

Vectors and the Geometry of Space

§10.1 Vectors in the Plane.

□ 10.1.1 Introduction.The term vector is used to indicate a quantity (such as displacement or velocity or force) that has both mag-nitude and direction. Vectors only impart magnitude and direction. Vectors don’t impart any informationabout where the quantity is applied. This is an important idea to always remember in the study of vectors.In a graphical sense vectors are represented by directed line segments. The length of the line segment is themagnitude of the vector and the direction of the line segment is the direction of the vector. However, becausevectors don’t impart any information about where the quantity is applied, any directed line segment withthe same length and direction will represent the same vector. We denote a vector by printing a letter inboldface (v) or by putting an arrow above the letter (−→v ). In the lecture notes, usually I will use the arrowedvector.The zero vector, denoted by 0, has length 0. It is the only vector with no specific direction.Suppose a particle moves along a line segment from point A to point B. The corresponding displacementvector has initial point A (the tail) and terminal point B (the tip) and we indicate this vector by writing

−→AB.

The length of the vector−→AB (referred as the magnitude of the vector

−→AB) is denoted by ∥−→AB∥.

When discussing vectors, we refer to real numbers as scalars. It is very important that you begin now tocarefully distinguish between vector and scalar quantities.(0) Equality: two vectors u and v are identical, i.e., u = v if and only if u and v have the same length andsame direction.(1) Addition: Consider two vectors u and v, where u has the initial point A and the terminal point B and vhas the initial point B and the terminal point C. Simply, u =

−→AB and v =

−→BC. Then the addition or the sum of

u and v is defined as follows:u+v =

−→AB+

−→BC =

−→AC.

In other words, if we add two vectors, the result is also a vector (called the resultant vector) and it is madewith the initial point of one vector and the terminal point of the other vector. Just like numbers, we can switchthe order, i.e.,

v+u =−→BC+

−→AB =

−→AB+

−→BC =

−→AC.

Figure 10.1: Sum of Two Vectors

We summarize as follows:

Definition 10.1.1. If u and v are vectors positioned so the initial point of u is at the terminal point of v, thenthe sum u+v is the vector from the initial point of u to the terminal point of v.

This definition is sometimes called the Triangle Law. See the figure 10.1.(2) Multiplication: We will study two important products on vectors, which are called the dot product andcross product. Since they are very important, we will study them later in Section 10.3 and 10.4. Here, we

25


discuss the scalar multiplication, i.e., multiplying a vector by a scalar (= a real number). If we multiply avector u by a scalar 3, then the result is 3u, which is also a vector. (We don’t write anything between 3 andu.) The vector 3u has the same direction as the vector u, but the length of the vector 3u is three time longerthan the length of the original vector u. However, if we multiply a vector v by a scalar −1, then the resultingvector −v has the opposite direction as the original vector v and same length as the vector v.

Figure 10.2: Difference of Two Vectors

(3) Subtraction: The subtraction u− v can be understood as the sum of u and −v, where −v is the vectormultiplied by a scalar −1. Therefore u−v can be handled by the addition and multiplication above. See thefigure 10.2.(4) Division: There is no division between vectors. We cannot divide a vector by a vector. We cannot dividea scalar (i.e., a real number) by a vector.

Figure 10.3: Regular Hexagon

Example 10.1.2. A regular hexagon1 is given as in the figure 10.3. When two vectors a and b are given,express other vectors in terms of a and b.

□ 10.1.2 Representation of Vector in terms of Component.Since the location of the initial point is irrelevant, we typically draw vectors with their initial point located atthe origin. Such a vector is called a position vector. For the position vector a with initial point at the originO(0,0) and terminal point at the point A(a1,a2), we denote the vector by

a =−→OA = ⟨a1,a2⟩ .

We call a1 and a2 the components of the vector a; a1 is the first component and a2 is the second component.Be careful to distinguish between the point (a1,a2) and the position vector ⟨a1,a2⟩. The magnitude of theposition vector a follows directly from the Pythagorean Theorem. We have

∥a∥= ∥⟨a1,a2⟩∥=√

a21 +a2

2 .

For two position vectors a = ⟨a1,a2⟩ and b = ⟨b1,b2⟩, the operations are defined as follows:

1. Zero Vector: The zero vector 0 is expressed as 0 = ⟨0,0⟩.1a hexagon with six sides of equal length

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2. Equality: a = b is equivalent to say that a1 = b1 and a2 = b2.3. Sum: a+b = ⟨a1,a2⟩+ ⟨b1,b2⟩= ⟨a1 +b1,a2 +b2⟩.4. Scalar Multiplication with scalar c (= real number): ca = c⟨a1,a2⟩= ⟨ca1,ca2⟩.5. Difference: a−b = ⟨a1,a2⟩−⟨b1,b2⟩= ⟨a1 −b1,a2 −b2⟩.

As we can see above, all the operations above are done componentwise.Observations:

1. If a vector a satisfies ∥a∥= 0, then the vector a should be the zero vector, i.e., a = 0.2. For a scalar c and a vector a = ⟨a1,a2⟩, we observe

∥ca∥= ∥c⟨a1,a2⟩∥= ∥⟨ca1,ca2⟩∥=√(ca1)2 +(ca2)2

=√

c2(a21 +a2

2) =√

c2√

a21 +a2

2 = |c|√

a21 +a2

2 = |c|∥a∥.

That is, simply we deduce∥ca∥= |c|∥a∥.

Please do not forget the absolute of c.3. When c =−1 in the observation 2 above, we get

∥−a∥= |−1|∥a∥= ∥a∥, i.e., ∥−a∥= ∥a∥.

Example 10.1.3. For vectors a = ⟨3,2⟩ and b = ⟨−2,5⟩, compute (1) a+b, (2) 4a, (3) 3a+2b, (4) 2a−3band (5) ∥2a−3b∥.

Definition 10.1.4. Two vectors having the same direction or opposite direction are called parallel.

For two vectors a and ca, where c is a nonzero scalar, they are parallel. When c is positive, a and ca havethe same direction. When c is negative, a and ca have the opposite direction. Hence, we deduce that two(nonzero) position vectors a and b are parallel if and only if b = ca, for some scalar c. In this event, we saythat b is a scalar multiple of a.

Example 10.1.5. Determine whether the given pair of vectors is parallel: (1) a = ⟨2,3⟩ and b = ⟨4,5⟩ and(2) a = ⟨2,3⟩ and b = ⟨−4,−6⟩.

Example 10.1.6. Find the constants a and b such that two vectors u = ⟨2a,a⟩ and v = ⟨b,3⟩ are parallel andthe magnitude of a is 5.

We denote that the set of all position vectors in two–dimensional space by

V2 = { ⟨x,y⟩ : x, y ∈ R } .

Theorem 10.1.7. For any vectors u, v and w in V2, and any scalars a and b in R, the following hold:

1. Commutativity: u+v = v+u2. Associativity: u+(v+w) = (u+v)+w3. Zero Vector: u+0 = u4. Additive Inverse: u+(−u) = 05. Distributive Law: a(u+v) = au+av6. Distributive Law: (a+b)u = au+bu7. Multiplication by 1: (1)u = u8. Multiplication by 0: (0)u = 0

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Figure 10.4: Position Vectors

For any two points A(a1,a2) and B(b1,b2), we can consider two position vectors−→OA = ⟨a1,a2⟩ and

−→OB =

⟨b1,b2⟩. By the Triangle Law or definition of sum of vectors 10.1.1, we have−→AB =

−→OB−−→

OA = ⟨b1,b2⟩−⟨a1,a2⟩= ⟨b1 −a1,b2 −a2⟩ ,

where the last equality comes by the component wise operations. That is, the vector−→AB is obtained by

−→AB = ⟨b1 −a1,b2 −a2⟩ .

See the figure 10.4.

Example 10.1.8. Find the vector represented by the directed line segment with initial point A(2,−3) andterminal point B(−2,1).

□ 10.1.3 Representation of Vector in terms of Standard Basis.We recall that any point (a,b) on the plane can be represented as

(a,b) = (a,0)+(0,b) = a(1,0)+b(0,1),

where (1,0) and (0,1) work as the base. Thanks to the Theorem 10.1.7 above, we can express a vectorv = ⟨a,b⟩ in the exactly same way:

v = ⟨a,b⟩= ⟨a,0⟩+ ⟨0,b⟩= a⟨1,0⟩+b⟨0,1⟩ ,where ⟨1,0⟩ and ⟨0,1⟩ are called the standard basis vectors. Since they are very important, we denote themas follows:

i = ⟨1,0⟩ , and j = ⟨0,1⟩ .Hence, the vector v = ⟨a,b⟩ is represented by

v = ⟨a,b⟩= a⟨1,0⟩+b⟨0,1⟩= ai+bj, i.e., v = ai+bj.

Here we call a and b the horizontal and vertical components of v, respectively.

Definition 10.1.9. Any vector u with the magnitude 1, i.e., ∥u∥= 1, is called a unit vector.

We observe that ∥i∥ = 1 and ∥j∥ = 1. That is, the standard basis vectors are unit vectors. Unit vectors arevery important and for any nonzero vectors, we can always find a unit vector with the same direction.

Theorem 10.1.10. For any nonzero position vectors v, a unit vector having the same direction as v is givenby

u =1

∥v∥v =

v∥v∥

.

Since 1/∥v∥ is a scalar, (1/∥v∥)v and v have the same direction and 1/∥v∥ makes the magnitude of (1/∥v∥)vto be 1. (A vector’s magnitude is sometimes called its norm.) The process of dividing a nonzero vector by itsmagnitude is sometimes called normalization. We will see this process again later in Chapter 12 Functionsof Several Variables and Partial Differentiation.

Example 10.1.11. Find the unit vector in the direction of the vector v = 2i− j.

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□ 10.1.4 Representation of Vector in Polar Form.Skip. Please read the textbook.

□ 10.1.5 Applications.Whenever two or more forces are acting on an object, the net force/resultant force acting on the object issimply the sum of all of the force vectors.

Example 10.1.12. At a certain point during a jump, there are two principal forces acting on a sky diver:gravity exerting a force of 180 pounds straight down and air resistance exerting a force of 180 pounds up and30 pounds to the right. What is the net force acting on the sky diver?

Example 10.1.13. An airplane has an airspeed of 400 mph. Suppose that the wind velocity is given by thevector w = ⟨20,30⟩. In what direction should the airplane head in order to fly due west (i.e., in the directionof the unit vector −i = ⟨−1,0⟩)?

Figure 10.5: Hanging Crate

Example 10.1.14. Two ropes are attached to a large crate. See the figure 10.5. Suppose that that rope Aexerts a force of ⟨−164,115⟩ pounds on the crate and rope B exerts a force of ⟨177,177⟩ pounds on the crate.If the crate weighs 275 pounds, what is the net force acting on the crate? Based on your answer, which waywill the crate move?

Example 10.1.15. The thrust of an airplane’s engines produces a speed of 600 mph in still air. The windvelocity is given by ⟨−30,60⟩. In what direction should the airplane head to fly due west?

Figure 10.6: Velocities Around Airplane

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ANSWER. See the figure 10.6. Let us define velocities (vectors).

1. a = ⟨x,y⟩: Velocity of Thrust2. w = ⟨−30,60⟩: Wind Velocity3. ⟨0,−c⟩: Net Force, where c is a positive real number.

Then, the net force comes from the sum of all forces:

⟨0,−c⟩= a+w = ⟨x,y⟩+ ⟨−30,60⟩= ⟨x−30,y+60⟩ −→ x = 30, y =−c−60.

By the given information, ∥a∥=√

x2 + y2 = 600:

6002 = x2 + y2 = 302 +(−c−60)2

c =−60±√

6002 −302 =−60±30√

399 = 539.25 and −659.25.

Since c should be positive, so we choose c =−60+30√

399 and thus

x = 30, y =−c−60 =−30√

399 −→ a =⟨

30,−30√

399⟩= 30

⟨1,−

√399

⟩tan−1

(yx

)= tan−1 −30

√399

30= tan−1

(−√

399)=−1.52078 rad =−87.134◦. □

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§10.2 Vectors in Space.

□ 10.2.1 Introduction.We now extend several ideas from the two–dimensional Euclidean space, R2 to the three–dimensional Eu-clidean space, R3. The three–dimensional Euclidean space is formed by adding an perpendicular2 axis tothe plane. The three coordinate planes in R3 (the xy–plane, the yz–plane and the xz–plane) divide space intoeight octants. The first octant is the one with x > 0, y > 0 and z > 0.

Example 10.2.1. Plot the points (1,2,3) and (3,−2,−6).

ANSWER. See the figure 10.7. □

Figure 10.7: Point (1,2,3)

In two–dimensional analytic geometry, the graph of an equation involving x and y is a curve in R2. Inthree–dimensional analytic geometry, an equation in x, y and z represents a surface in R3.

Example 10.2.2. What surfaces in R3 are represented by the following equations? (1) z = 3, (2) y = 5.

Figure 10.8: Planes z = 3 and y = 5

ANSWER. See the figure 10.8.(1) The surface of z = 3 is the set of points,

{ (x,y,3) : any x, y ∈ R } ,

which represents the plane parallel to the xy–plane and passing through the point (0,0,3).(2) The surface of y = 5 is the set of points,

{ (x,5,z) : any x, z ∈ R } ,

which represents the plane parallel to the xz–plane and passing through the point (0,5,0). □2perpendicular, orthogonal: intersecting at or forming right (= 90◦) angles

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When an equation is given, we must understand from the context whether it represents a curve in R2 or asurface in R3. For instance, y = 5 represents a plane in R3, but of course y = 5 can also represent a line inR2. In general in R3, if k is a constant, then x = k represents a plane parallel to the yz–plane, y = k is a planeparallel to the xz–plane, and z = k is a plane parallel to the xy–plane.

Example 10.2.3. Describe and sketch the surfaces in R3 represented by the equations (1) x− y = 0, (2)x+ y =−2.

Figure 10.9: Planes x− y = 0 and x+ y =−2

ANSWER. See the figure 10.9.(1) The surface of x− y = 0, i.e., y = x, is the set of points,

{ (x,x,z) : any x, z ∈ R } ,

which represents the plane parallel to the z–axis and passing through the point (0,0,0).(2) The surface of x+ y =−2, i.e., y =−x−2, is the set of points,

{ (x,−x−2,z) : any x, z ∈ R } ,

which represents the plane parallel to the z–axis and passing through the point (0,−2,0). □

The familiar formula for the distance between two points in a plane is easily extended to the following three–dimensional formula.

Theorem 10.2.4 (DISTANCE BETWEEN TWO POINTS IN THREE–DIMENSIONAL SPACE). The distanced(P1,P2) between the points P1(x1,y1,z1) and P2(x2,y2,z2) is

d(P1,P2) =√(x2 − x1)2 +(y2 − y1)2 +(z2 − z1)2 .

Example 10.2.5. Find the distance from the point P(2,−1,7) to the point Q(1,−3,5).

Example 10.2.6. Find an equation of a sphere with radius r and center C(a,b,c).

Hint. By definition, a sphere is the set of all points P(x,y,z) whose distance from C is r. Use the distanceformula above. □

Example 10.2.7. Show that x2 + y2 + z2 +4x−6y+2z+6 = 0 is the equation of a sphere and find its centerand radius.

Hint. Complete the squares in each variable. □

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□ 10.2.2 Vectors in R3.As in two dimensions, vectors in three–dimensional space have both direction and magnitude. We againvisualize vectors as directed line segments joining two points. A vector v is represented by any directedline segment with the appropriate magnitude and direction. The position vector a with terminal point atA(a1,a2,a3) (and initial point at the origin) is denoted by a = ⟨a1,a2,a3⟩.We denote the set of all three–dimensional position vectors by

V3 = { ⟨x,y,z⟩ : x, y, z ∈ R } .

� Magnitude: A position vector v = ⟨a,b,c⟩ has the magnitude

∥v∥= ∥⟨a,b,c⟩∥=√

a2 +b2 + c2 ,

by the distance formula in three dimensions.� The vector with initial point at P(a1,a2,a3) and terminal point at Q(b1,b2,b3) corresponds to the positionvector

−→PQ = ⟨b1 −a1,b2 −a2,b3 −a3⟩.� Arithmetic: for vectors u = ⟨u1,u2,u3⟩ and v = ⟨v1,v2,v3⟩, we have

1. u+v = ⟨u1,u2,u3⟩+ ⟨v1,v2,v3⟩= ⟨u1 + v1,u2 + v2,u3 + v3⟩.2. u−v = ⟨u1,u2,u3⟩−⟨v1,v2,v3⟩= ⟨u1 − v1,u2 − v2,u3 − v3⟩.3. cu = c⟨u1,u2,u3⟩= ⟨cu1,cu2,cu3⟩, where c is a scalar.

� For a scalar c and a vector v ∈V3, we have

∥cv∥= |c|∥v∥,

just like the case of the vectors in the plane.� The zero vector 0 in V3 is defined by 0 = ⟨0,0,0⟩.

Theorem 10.2.8. For any vectors u, v and w in V3, and any scalars a and b in R, the following hold:

1. Commutativity: u+v = v+u2. Associativity: u+(v+w) = (u+v)+w3. Zero Vector: u+0 = u4. Additive Inverse: u+(−u) = 05. Distributive Law: a(u+v) = au+av6. Distributive Law: (a+b)u = au+bu7. Multiplication by 1: (1)u = u8. Multiplication by 0: (0)u = 0

� Standard Basis Vectors: A vector v = ⟨a,b,c⟩ is expressed by

v = ⟨a,b,c⟩= ⟨a,0,0⟩+ ⟨0,b,0⟩+ ⟨0,0,c⟩= a⟨1,0,0⟩+b⟨0,1,0⟩+ c⟨0,0,1⟩ .

We denote the vectors ⟨1,0,0⟩, ⟨0,1,0⟩, and ⟨0,0,1⟩ by i, j and k, respectively, that is,

i = ⟨1,0,0⟩ , j = ⟨0,1,0⟩ , k = ⟨0,0,1⟩ .

We call them the standard basis vectors in V3. Using them, the vector v = ⟨a,b,c⟩ is expressed by

v = ⟨a,b,c⟩= ai+bj+ ck.

� Normalization: For a nonzero vector v ∈V3, a unit vector in the same direction as v is given by

u =1

∥v∥v.

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Example 10.2.9. Find the unit vector in the direction of the vector v = 2i− j−2k.

Example 10.2.10. Find a unit vector in the same direction as ⟨1,−2,3⟩ and write ⟨1,−2,3⟩ as the product ofits magnitude and a unit vector.

□ 10.2.3 Some Problems from Exercises.� Give an equation (e.g., z = 0) for the indicated figure: (1) xy–plane, (2) x–axis.� Find the displacement vectors

−→PQ and

−→QR and determine whether the points P(2,3,1), Q(0,4,2) and

R(4,1,4) are colinear3.� Use vectors to determine whether the points (2,1,0), (4,1,2) and (4,3,0) form an equilateral4 triangle.� Use vectors to determine whether the points (1,−2,1), (−2,−1,2), (2,0,2) and (−1,1,3) form a square.

Figure 10.10: Hanging Crate

� Two ropes are attached to a 500 pound crate. See the figure 10.10. Rope A exerts a force of ⟨10,−130,200⟩pounds on the crate, and rope B exerts a force of ⟨−20,180,160⟩ pounds on the crate. If no further ropes areadded, find the net force on the crate and the direction it will move. If a third rope C is added to balance thecrate, what force must this rope exert on the crate?� The thrust of an airplane’s engine produces a speed of 700 mph in still air. The plane is aimed in thedirection of ⟨6,−3,2⟩, but its velocity with respect to the ground is ⟨580,−330,160⟩ mph. Find the windvelocity.

3on the same line4having all sides or faces equal

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§10.3 The Dot Product.

□ 10.3.1 Definition.

Definition 10.3.1. The dot product of two vectors a = ⟨a1,a2,a3⟩ and b = ⟨b1,b2,b3⟩ in V3 is defined by

a ·b = ⟨a1,a2,a3⟩ · ⟨b1,b2,b3⟩= a1b1 +a2b2 +a3b3.

Likewise, the dot product of two vectors a = ⟨a1,a2⟩ and b = ⟨b1,b2⟩ in V2 is defined by

a ·b = ⟨a1,a2⟩ · ⟨b1,b2⟩= a1b1 +a2b2.

Be careful: The dot product gives a scalar, not a vector. That is, a ·b is a real number, not a vector.

Example 10.3.2. Compute the dot product a · b: (1) a = ⟨1,2,3⟩ and b = ⟨5,−3,4⟩, (2) a = 2i− 5j andb = 3i+6j.

It is easy to see that for the standard basis vectors i, j and k, we have

i · i = 1 = j · j = k ·k, and i · j = 0 = j ·k = k · i.

The dot product obeys many of the laws that hold for ordinary products of real numbers. These are stated inthe following theorem.

Theorem 10.3.3. If a, b, and c are vectors and s is a scalar, then

1. Commutativity: a ·b = b ·a2. Distributive Law: a · (b+ c) = a ·b+a · c3. (sa) ·b = s(a ·b) = a · (sb)4. 0 ·a = 05. a ·a = ∥a∥2

Example 10.3.4. For two vectors a and b, expand (a+b) · (a+b) and (a−b) · (a+b).

ANSWER. Using the Theorem 10.3.3, we have

∥a+b∥2 = (a+b) · (a+b) = (a+b) ·a+(a+b) ·b = a ·a+b ·a+a ·b+b ·b= ∥a∥2 +a ·b+a ·b+∥b∥2 = ∥a∥2 +2a ·b+∥b∥2

(a−b) · (a+b) = (a−b) ·a+(a−b) ·b = a ·a−b ·a+a ·b−b ·b= ∥a∥2 −a ·b+a ·b−∥b∥2 = ∥a∥2 −∥b∥2. □

□ 10.3.2 Geometric Interpretation of Dot Product.Let θ ∈ [0,π] be the angle between two nonzero vectors a and b. Then we have the following formula whichis used by physicists as the definition of the dot product.

Theorem 10.3.5.a ·b = ∥a∥∥b∥cosθ , cosθ =

a ·b∥a∥∥b∥

,

where 0 ≤ θ ≤ π .

Notice

1. a and b have the same direction if and only if θ = 0.2. a and b have opposite directions if and only if θ = π .

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3. a ·b = 0 if and only if θ = π/2. We say that a and b are orthogonal or perpendicular if θ = π/2.

We consider the zero vector 0 to be orthogonal to every vector.

Example 10.3.6. If the vectors a and b have lengths 4 and 6, respectively, and the angle between them isπ/3, find a ·b.

Example 10.3.7. Find the angle between the vectors: (1) a = ⟨2,1,−3⟩ and b = ⟨1,5,6⟩, (2) a = ⟨2,2,−1⟩and b = ⟨5,−3,2⟩.

Example 10.3.8. Determine whether the following pairs of vectors are orthogonal: (1) a = ⟨1,3,−5⟩ andb = ⟨2,3,10⟩ and (2) a = ⟨4,2,−1⟩ and b = ⟨2,3,14⟩.

Example 10.3.9. Show that 2i+2j−k is perpendicular to 5i−4j+2k.

The following two results provide us with some powerful tools for comparing the magnitudes of vectors.

Theorem 10.3.10 (CAUCHY–SCHWARTZ INEQUALITY). For any vectors a and b,

|a ·b| ≤ ∥a∥∥b∥

Theorem 10.3.11 (TRIANGLE INEQUALITY). For any vectors a and b,

∥a+b∥ ≤ ∥a∥+∥b∥

Exercise 10.3.12. For nonzero vectors, a and b, show that c = ∥a∥b+∥b∥a bisects the angle between a andb.

□ 10.3.3 Components and Projections.

Figure 10.11: Component and Projection

Consider two vectors−→PQ and

−→PR. Let S be the foot of the perpendicular line segment from R to the line

containing−→PQ. See the figure 10.11. Then, the length of the line segment PS is called the component of −→PR

onto−→PQ and denoted by Comp−→PQ

−→PR. (When we put a stick on the desert slantly, we see the shadow of the

stick on the desert. We can think of the line segment PS as the shadow of the directed stick−→PR.) Let θ be the

angle between−→PQ and

−→PR. Then since the line segment PS is the shadow of

−→PR onto

−→PQ, by the elementary

trigonometry, we observe

length of PS = ∥−→PR∥cosθ , i.e., Comp−→PQ−→PR = ∥−→PR∥cosθ .

Using the position vectors a =−→PR and b =

−→PQ, the component of a onto b is given by

Compb a = ∥a∥cosθ ,

where 0 ≤ θ ≤ π is the angle between a and b. The Theorem 10.3.5 implies

Compb a = ∥a∥cosθ = ∥a∥ a ·b∥a∥∥b∥

=a ·b∥b∥

, i.e., Compb a =a ·b∥b∥

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which is very important and worth to be memorized. Since ∥b∥ is a scalar, we can rewrite the equation as

Compb a = a ·(

b∥b∥

)Since b/∥b∥ is a unit vector in the direction of b, the component of a onto b can be understood as the dotproduct of a and the unit vector in the direction of b.Let us go back to the argument with

−→PQ and

−→PR. We assign a direction to the line segment PS. Since the

line segment is on the line containing the vector−→PQ, so we can consider the vector

−→PS, which is called the

projection of −→PR onto−→PQ and denoted by Proj−→PQ

−→PR, i.e.,

−→PS = Proj−→PQ

−→PR. Then we have

length of the line segment PS = ∥−→PS∥, i.e., ∥−→PS∥= Comp−→PQ−→PR.

Since−→PS is parallel to

−→PQ, so by multiplying the unit vector in the direction of

−→PQ , we get

−→PS =

(Comp−→PQ

−→PR) −→

PQ

∥−→PQ∥, Proj−→PQ

−→PR =

(−→PR ·−→PQ

∥−→PQ∥

) −→PQ

∥−→PQ∥, Proj−→PQ

−→PR =

−→PR ·−→PQ

∥−→PQ∥2

−→PQ.

Using the position vectors a =−→PR and b =

−→PQ, the projection of a onto b is given by

Projb a =a ·b∥b∥2 b.

In summary,

1. Component of a along b is a scalar given by

Compb a =a ·b∥b∥

.

2. Projection of a along b is a vector, which is parallel to b, and given by

Projb a =a ·b∥b∥2 b.

In some books, the component Compb a is called the scalar projection of a onto b. The projection Projb a iscalled the vector projection of a onto b.

Example 10.3.13. Find the component and projection of a along b: (1) a = ⟨2,3⟩ and b = ⟨−1,5⟩, (2)a = ⟨1,1,2⟩ and b = ⟨−2,3,1⟩.

Example 10.3.14. Show that, in general, Compb a = Compa b and Projb a = Proja b. When does the equalityhold?

□ 10.3.4 Application: Work Done.One use of projections occurs in physics in calculating work. The work done by a constant force F in movingan object through a distance d is defined as W = Fd, but this applies only when the force is directed alongthe line of motion of the object. Suppose, however, that the constant force is a vector F =

−→PR, pointing in

some other direction. If the force moves the object from P to Q, then the displacement vector is d =−→PQ.

The work W done by this force is defined to be the product of the component of the force F along d and thedistance moved:

W = (Compd F)∥d∥= F ·d∥d∥

∥d∥= F ·d, i.e., W = F ·d.

Thus the work W done by a constant force F is the dot product F ·d, where d is the displacement vector.

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Example 10.3.15. A wagon is pulled a distance of 100 m along a horizontal path by a constant force of 70N. The handle of the wagon is held at an angle of 35◦ above the horizontal. Find the work done by the force.

Example 10.3.16. You exert a constant force of 40 pounds in the direction of the handle of the wagon. Ifthe handle makes an angle of π/4 with the horizontal and you pull the wagon along a flat surface for 1 mile(5280 feet), find the work done.

Example 10.3.17. A force is given by a vector F = 3i+4j+5k and moves a particle from the point P(2,1,0)to the point Q(4,6,2). Find the work done.

Example 10.3.18. A car makes a turn on a banked road. If the road is banked at 10◦, show that a vectorparallel to the road is ⟨cos10◦,sin10◦⟩. If the car has weight 2000 pounds, find the component of the weightvector along the road vector. This component of weight provides a force that helps the car turn.

Example 10.3.19. In the sequel of the previous example, find the component of the weight vector along theroad vector for a 2500–pound car on a 15◦ bank.

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§10.4 The Cross Product.

□ 10.4.1 Definition.We start with the definition from the “Linear Algebra”:

Definition 10.4.1. (1) The determinant of 2×2 matrix of real numbers is a number defined by∣∣∣∣∣ a1 a2

b1 b2

∣∣∣∣∣= a1b2 −a2b1.

(2) The determinant of 3×3 matrix of real numbers is a number defined by∣∣∣∣∣∣∣∣a1 a2 a3

b1 b2 b3

c1 c2 c3

∣∣∣∣∣∣∣∣= a1

∣∣∣∣∣ b2 b3

c2 c3

∣∣∣∣∣−a2

∣∣∣∣∣ b1 b3

c1 c3

∣∣∣∣∣+a3

∣∣∣∣∣ b1 b2

c1 c2

∣∣∣∣∣= a1 (b2c3 −b3c2)−a2 (b1c3 −b3c1)+a3 (b1c2 −b2c1) .

Now we define the cross product of two vectors.

Definition 10.4.2. For two vectors a = ⟨a1,a2,a3⟩ and b = ⟨b1,b2,b3⟩ in V3, we define the cross product (orvector product) of a and b to be

a×b =

∣∣∣∣∣∣∣∣i j k

a1 a2 a3

b1 b2 b3

∣∣∣∣∣∣∣∣=∣∣∣∣∣ a2 a3

b2 b3

∣∣∣∣∣ i−∣∣∣∣∣ a1 a3

b1 b3

∣∣∣∣∣ j+∣∣∣∣∣ a1 a2

b1 b2

∣∣∣∣∣k= (a2b3 −a3b2) i− (a1b3 −a3b1) j+(a1b2 −a2b1)k.

Notice that a×b is also a vector in V3. To compute a×b, you must write the components of a in the secondrow and the components of b in the third row: the order is important! The cross product is defined only forvectors in V3. There is no corresponding operation for vectors in V2.

Example 10.4.3. Compute a×b: (1) a = ⟨1,3,4⟩ and b = ⟨2,7,−5⟩, (2) a = i+2j+3k and b = 4i+5j+6k.

Theorem 10.4.4. For any vector a in V3, we have

a×a = 0, and a×0 = 0.

Theorem 10.4.5. For any vectors a and b in V3, a×b is orthogonal to both a and b.

We recall the Remark 3 stating that two vectors u and v are orthogonal if and only if u · v = 0. A simplecomputation shows

(a×b) ·a = 0 = (a×b) ·b.

Hence, the Theorem 10.4.5 above is proven.Two different points determines one line, because there is only one line passing through two different points.Similarly, two different lines determines one plane, because there is only one plane containing those twodifferent lines. Given two vectors, we can consider two different lines containing each of the vectors andthose two lines determines one plane. Since the vector a×b is orthogonal to both a and b, so we can observethat the vector a×b is orthogonal to the plane determined by the vectors a and b. But, given a plane, outof which side of the plane does a×b point? A simple computation shows i× j = k, where i, j and k are thestandard basis vectors of V3. It illustrates the right–hand rule: If the fingers of your right hand curl in the

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Figure 10.12: Right–Hand Rule

direction of a rotation (through an angle less than 180◦) from a to b, then your thumb points in the directionof a×b. See the figure 10.12.By the right–hand rule or a simple computation, for the standard basis vectors we deduce

i× j = k, j×k = i, k× i = j,j× i =−k, k× j =−i, i×k =−j.

Since the cross product does not follow several of the rules you might expect a product to satisfy, you mightask what rules the cross product does satisfy. We summarize these in the Theorem below.

Theorem 10.4.6. For any vectors a, b and c in V3 and any scalar s, the following hold:

1. Anti–Commutativity: a×b =−(b×a)2. (sa)×b = s(a×b) = a× (sb)3. Distributive Law: a× (b+ c) = a×b+a× c4. Distributive Law: (a+b)× c = a× c+b× c5. Scalar Triple Product: a · (b× c) = (a×b) · c6. Vector Triple Product: a× (b× c) = (a · c)b− (a ·b)c

Now that we know the direction of the vector a×b, the remaining thing we need to complete its geometricdescription is its length ∥a×b∥. This is given by the following theorem.

Theorem 10.4.7. If 0 ≤ θ ≤ is the angle between a and b, then

∥a×b∥= ∥a∥∥b∥sinθ .

Since a vector is completely determined by its magnitude and direction, we can now say that a× b is thevector that is perpendicular to both a and b, whose orientation is determined by the right–hand rule, andwhose length is ∥a∥∥b∥sinθ . In fact, that is exactly how physicists define a×b.

Corollary 10.4.8. Two nonzero vectors a and b in V3 are parallel if and only if a×b = 0.

□ 10.4.2 Application: Area, Distance, Volume, Torque.

Area

If a and b are represented by directed line segments with the same initial point, then they determine a paral-lelogram5 with base ∥a∥, altitude ∥b∥sinθ . See the figure 10.13. The parallelogram has the area A obtainedby

Area A = (base)(altitude) = ∥a∥∥b∥sinθ = ∥a×b∥.Thus we have the following way of interpreting the magnitude of a cross product.

5a quadrilateral whose opposite sides are both parallel and equal in length

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Figure 10.13: Parallelogram formed by a and b

Theorem 10.4.9. The parallelogram formed by a and b has the area

A = ∥a×b∥.

Example 10.4.10. Find a vector perpendicular to the plane passing through the points P(1,4,6), Q(−2,5,−1)and R(1,−1,1).

Example 10.4.11. Find the area of the triangle with vertices P(1,4,6), Q(−2,5,−1) and R(1,−1,1).

Example 10.4.12. Find the area of the parallelogram with two adjacent sides formed by the vectors a =⟨1,2,3⟩ and b = ⟨4,5,6⟩.

Distance from a Point to a Line

Let d represent the distance from the point Q to the line through the points P and R. From elementarytrigonometry, we have that

d = ∥−→PQ∥sinθ

where θ is the angle between−→PQ and

−→PR. By the Theorem 10.4.7, we have

∥−→PQ×−→PR∥= ∥−→PQ∥−→PR∥sinθ = ∥−→PR∥d.

Solving this for d, we get the following formula.

Theorem 10.4.13. The distance from the point Q to the line through the points P and R is given by

d =∥−→PQ×−→

PR∥∥−→PR∥

.

Example 10.4.14. Find the distance from the point Q(1,2,1) to the line through the points P(2,1,−3) andR(2,−1,3).

Volume

Figure 10.14: Parallelepiped formed by a, b and c

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For any three non–coplanar vectors a, b and c, we consider the parallelepiped formed using the vectors asthree adjacent edges. See the figure 10.14. We choose a and b to form the base of the parallelogram. Thenthe area of the base is A = ∥a×b∥. See the topic 10.4.2. If θ is the angle between c and a×b, then the heightof the parallelepiped is h = ∥c∥|cosθ |. (We must use |cosθ | instead of cosθ in case θ > π/2.) Thereforethe volume V of the parallelepiped is

Volume V = (area of base)(height) = Ah = ∥a×b∥∥c∥|cosθ |= |c · (a×b) |.

Thus we have proven the formula.

Theorem 10.4.15. The volume of the parallelepiped determined by the vectors a, b, and c is the magnitudeof their scalar triple product:

V = |c · (a×b)| .

Since we can choose b and c to form the base of the parallelogram, by the same argument, the volume isobtained by

V = |a · (b× c)| .

That is, under the fixed products, we can rearrange the vectors:

V = |a · (b× c)|= |a · (c×b)|= |b · (c×a)|= |b · (a× c)|= |c · (a×b)|= |c · (b×a)| .

If we use the formula above and discover that the volume of the parallelepiped determined by a, b, and c is0, then the vectors must be coplanar6.� Fast Computation: For a = ⟨a1,a2,a3⟩, b = ⟨b1,b2,b3⟩ and c = ⟨c1,c2,c3⟩, we have

c · (a×b) =

∣∣∣∣∣∣∣∣c1 c2 c3

a1 a2 a3

b1 b2 b3

∣∣∣∣∣∣∣∣ .Example 10.4.16. Find the volume of the parallelepiped with three adjacent edges formed by the vectorsa = ⟨1,2,3⟩, b = ⟨4,5,6⟩ and c = ⟨7,8,0⟩.

Example 10.4.17. Use the scalar triple product to show that the vectors a = ⟨1,4,−7⟩, b = ⟨2,−1,4⟩, andc = ⟨0,−9,18⟩ are coplanar.

Torque

Figure 10.15: Wrench with Torque6lying in the same plane

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The idea of a cross product occurs often in physics. In particular, we consider a force F acting on a rigid bodyat a point given by a position vector r. For instance, if we tighten a bolt by applying a force to a wrench, weproduce a turning effect. See the figure 10.15. The torque τττ (relative to the origin) is defined to be the crossproduct of the position and force vectors:

τττ = r×F

and measures the tendency of the body to rotate about the origin. The direction of the torque vector indicatesthe axis of rotation. The magnitude of the torque vector is

∥τττ∥= ∥r×F∥= ∥r∥∥F∥sinθ ,

where θ is the angle between the position and force vectors. Observe that the only component of F that cancause a rotation is the one perpendicular to r, that is, ∥F∥sinθ . The magnitude of the torque is equal to thearea of the parallelogram determined by r and F.

Example 10.4.18. If you apply a force of magnitude 25 pounds at the end of a 15–inch–long wrench, at anangle of π/3 to the wrench, find the magnitude of the torque applied to the bolt. What is the maximum torquethat a force of 25 pounds applied at that point can produce?

Example 10.4.19. A bolt is tightened by applying a 40 N force to a 0.25 m wrench with an angle of 75◦

between the wrench and the force vector. Find the magnitude of the torque about the center of the bolt.

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§10.5 Lines and Planes in Space.In this section, we study the lines and planes in R3 and find the equations of a line and a plane.

□ 10.5.1 Lines in R3.We recall that a line in the xy–plane is determined when a point (a,b) on the line and the direction m of theline (its slope or angle of inclination) are given. The equation of the line can then be written using the point–slope form: y− b = m(x− a). Similarly, a line L in three–dimensional space is determined when we knowa point P1(x1,y1,z1) on L and the direction of L. In three dimensions the direction of a line is convenientlydescribed by a vector.Let us look for the line L that passes through the point P1(x1,y1,z1) and that is parallel to the position vectora = ⟨a1,a2,a3⟩. For any other point P(x,y,z) on L, we observe the vector

−→P1P is parallel to a. It implies

−→P1P = ta,

where t is some scalar. Since−→P1P = ⟨x− x1,y− y1,z− z1⟩, so the equation implies

⟨x− x1,y− y1,z− z1⟩= ⟨ta1, ta2, ta3⟩ −→ x = x1 + ta1, y = y1 + ta2, z = z1 + ta3.

We call the last equations parametric equations for the line L, where t is the parameter.Another way of describing the line L is to eliminate the parameter t from the parametric equations. If noneof a1, a2, or a3 is 0, we can solve each of these equations for t, equate the results, and obtain

x− x1

a1=

y− y1

a2=

z− z1

a3.

which is called symmetric equations of the line L.

Example 10.5.1. Find an equation of the line L through the point (1,5,2) and parallel to the vector ⟨4,3,7⟩.Also, determine where the line L intersects the yz–plane.

Example 10.5.2. Find an equation of the line L passing through the points P(1,2,−1) and Q(5,−3,4).

Example 10.5.3. (1) Find parametric equations and symmetric equations of the line that passes through thepoints P(2,4,−3) and Q(3,−1,1).(2) At what point does this line intersect the xy–plane?

Definition 10.5.4. Let L and M be two lines in R3, with parallel vectors vL and vM, respectively.

1. The lines L and M are parallel if and only if vL and vM are parallel.2. If L and M intersect, then

(1) the angle between lines L and M is same as the angle between parallel vectors vL and vM,(2) the lines L and M are orthogonal to each other if and only if the vectors vL and vM are orthogonal to

each other.

Example 10.5.5. Show that the lines L and M defined by

L : x−2 =−t, y−1 = 2t, z−5 = 2tM : x−1 = s, y−2 =−s, z−1 = 3s

are not parallel, yet do not intersect.

Definition 10.5.6. Nonparallel, nonintersecting lines are called skew lines.

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Note that it’s fairly easy to visualize skew lines. Draw two planes that are parallel and draw a line in eachplane (so that it lies completely in the plane). As long as the two lines are not parallel, these are skew lines.

Example 10.5.7. Show that the lines and with parametric equations

L : x = 1+ t, y =−2+3t, z = 4− tM : x = 2s, y = 3+ s, z =−3+4s

are skew lines, that is, they do not intersect and are not parallel (and therefore do not lie in the same plane).

□ 10.5.2 Planes in R3.Although a line in space is determined by a point and a direction, a plane in space is more difficult todescribe. A single vector parallel to a plane is not enough to convey the “direction” of the plane, but a vectorperpendicular to the plane does completely specify its direction. Thus a plane in space is determined by apoint in the plane and a vector that is orthogonal to the plane. This orthogonal vector is called a normalvector of the plane.Let us find an equation of the plane which passes through a point P1(x1,y1,z1) and orthogonal to the vectora = ⟨a,b,c⟩. Let P(x,y,z) be an arbitrary point in the plane. Then since P and P1 are both points in the plane,the vector

−→P1P = ⟨x− x1,y− y1,z− z1⟩ lies in the plane and so, must be orthogonal to a. By the formula on

the dot product, hence, we deduce

0 = a ·−→P1P = ⟨a,b,c⟩ · ⟨x− x1,y− y1,z− z1⟩= a(x− x1)+b(y− y1)+ c(z− z1) .

That is, the equation of the plane passing through the point (x1,y1,z1) with normal vector ⟨a,b,c⟩ is

a(x− x1)+b(y− y1)+ c(z− z1) = 0.

Example 10.5.8. Find an equation of the plane containing the point (1,2,3) with normal vector ⟨4,5,6⟩, andsketch the plane.

Example 10.5.9. Find an equation of the plane through the point (2,4,−1) with normal vector ⟨2,3,4⟩. Findthe intercepts and sketch the plane.

Note that if we expand out the equation of the plane, we get

0 = a(x− x1)+b(y− y1)+ c(z− z1) = ax+by+ cz− (ax1 +by1 + cz1) .

Since −(ax1 +by1 + cz1) is a constant, when we let d = −(ax1 +by1 + cz1), the equation above becomessimply

ax+by+ cz+d = 0.

where d =−(ax1 +by1 + cz1). We refer to this last equation as a linear equation in the three variables x, yand z. Conversely, it can be shown that if a, b, and c are not all 0, then the linear equation above represents aplane with normal vector ⟨a,b,c⟩.

Example 10.5.10. Find the plane containing the three points P(1,2,2), Q(2,−1,4) and R(3,5,−2).

Example 10.5.11. Find an equation of the plane that passes through the points P(1,3,2), Q(3,−1,6), andR(5,2,0).

In three dimensions, two planes are either parallel or they intersect in a straight line. Suppose that two planeshaving normal vectors a and b, respectively, intersect. Then the angle between the planes is the same as theangle between a and b. With this in mind, we say

1. the planes are parallel whenever their normal vectors are parallel and

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2. the planes are orthogonal whenever their normal vectors are orthogonal.

Example 10.5.12. Find an equation for the plane through the point (1,4,−5) and parallel to the plane definedby 2x−5y+7z = 12.

Example 10.5.13. Describe and draw the planes y = 3 and y = 8.

Example 10.5.14. (1) Find the angle between the planes x+ y+ z = 1 and x−2y+3z = 1.(2) Find symmetric equations for the line of intersection L of these two planes.

Example 10.5.15. Find the line of intersection of the planes x+2y+ z = 3 and x−4y+3z = 5.

Distance between Plane and Point

Let us find the distance between the plane ax+by+ cz+d = 0 and a point P0(x0,y0,z0) which is not on theplane. We choose a point P1(x1,y1,z1) lying on the plane. Then we observe the distance between P0 and theplane is simply the absolute value of the component of the vector

−→P1P along the normal vector a, i.e.,

distance between plane and P0 =∣∣∣Compa

−−→P1P0

∣∣∣ .Computing it, we get

Compa−−→P1P0 =

−−→P1P0 ·a∥a∥

=⟨x0 − x1,y0 − y1,z0 − z1⟩ · ⟨a,b,c⟩

∥⟨a,b,c⟩∥

=a(x0 − x1)+b(y0 − y1)+ c(z0 − z1)√

a2 +b2 + c2=

ax0 +by0 + cz0 − (ax1 +by1 + cz1)√a2 +b2 + c2

.

Since P1(x1,y1,z1) is on the plane ax+by+ cz+d = 0, we get

ax1 +by1 + cz1 +d = 0, i.e., d =−(ax1 +by1 + cz1) ,

which implies

Compa−−→P1P0 =

−−→P1P0 ·a∥a∥

=ax0 +by0 + cz0 − (ax1 +by1 + cz1)√

a2 +b2 + c2=

ax0 +by0 + cz0 +d√a2 +b2 + c2∣∣∣Compa

−−→P1P0

∣∣∣= |ax0 +by0 + cz0 +d|√a2 +b2 + c2

.

Thus, simply, the formula for the distance D between the plane ax+by+cz+d = 0 and a point P0(x0,y0,z0)which is not on the plane can be written as

D =|ax0 +by0 + cz0 +d|√

a2 +b2 + c2.

Example 10.5.16. Find the distance between the parallel planes:� Π1 : 2x−3y+ z = 6 and Π2 : 4x−6y+2z = 8.� Π1 : 10x+2y−2z = 5 and Π2 : 5x+ y− z = 1.

Example 10.5.17. The lines L : x = 1+ t, y =−2+3t, z = 4− t and M : x = 2s, y = 3+ s, z =−3+4s areskew. Find the distance between them.

§10.6 Surfaces in Space.Presentation via Mathematica and Webpage: Interactive Gallery of Quadric Surfaces(http://www.math.umn.edu/˜rogness/quadrics/index.shtml)Please read the textbook.

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Chapter 11

Vector–Valued Functions

§11.1 Vector–Valued Functions.We briefly discuss some examples of vector–valued functions.

□ 11.1.1 Vector–Valued Functions in Plane.

Definition 11.1.1. A vector–valued function r(t) in the two–dimensional space is a mapping from its domainD ⊂R to its range R ⊂V3, so that for each t ∈ D, r(t) = v for exactly one vector v ∈V2. We can always writea vector–valued function as

r(t) = f (t)i+g(t)j,

for some scalar functions f and g (called the component functions of r).

Example 11.1.2. Sketch a graph of the curve traced out by the endpoint of the two–dimensional vector–valued function

r(t) = (t +1)i+(t2 −2)j.

t r(t) =⟨t +1, t2 −2

⟩−3 ⟨−2,7⟩−2 ⟨−1,2⟩−1 ⟨0,−1⟩0 ⟨1,−2⟩1 ⟨2,2⟩2 ⟨3,7⟩3 ⟨4,14⟩4 ⟨5,23⟩ Figure 11.1: Graph of r(t) = (t +1)i+(t2 −2)j

ANSWER. We make the table as below and connect all the endpoints of the vectors in the table. Thenwe obtain the graph of the vector–valued function r(t). As t increases, the endpoints of the vector–valuedfunctions move from the second quadrant via the third and fourth quadrant to the first quadrant. This directionof increasing values of t is called the orientation of the curve. See the figure 11.1. □

ANSWER. Another way to get the graph is using x and y as follows: let x = t + 1 (first component of r(t))and y = t2 − 2 (second component of r(t)). Then, we get t = x− 1 and putting this into the equation of yimplies

y = t2 −2 = (x−1)2 −2, i.e., y = (x−1)2 −2,

of which graph is exactly same as the one in the figure 11.1. In this solution, it is not easy to determine theorientation of the curve compared to the solution above. □

Example 11.1.3. Sketch a graph of the curve traced out by the endpoint of the vector–valued function

r(t) = 4cos ti−3sin tj, t ∈ R.

47


Figure 11.2: Graph of r(t) = 4cos ti−3sin tj

ANSWER. One can use the table as the Example 11.1.2. But I prefer using x and y. The curve can be writtenparametrically as

x = 4cos t, y =−3sin t, t ∈ R.We deduce an equation of x and y by getting rid of the parameter t as follows:(x

4

)2+(y

3

)2= cos2 t + sin2 t = 1, i.e.,

(x4

)2+(y

3

)2= 1,

of which graph is an ellipse. See the figure 11.2. □

□ 11.1.2 Vector–Valued Functions in Space.

Definition 11.1.4. A vector–valued function r(t) in the three–dimensional space has the domain D ⊂ R andthe range R ⊂V3. We can write it as

r(t) = f (t)i+g(t)j+h(t)k,

for some scalar functions f , g and h (called the component functions of r).

Example 11.1.5. Plot the curve traced out by the vector–valued function r(t) = sin ti−3cos tj+2tk, t ≥ 0.

Figure 11.3: Graph of r(t) = sin ti−3cos tj+2tk

ANSWER. The curve is given parametrically by

x = sin t, y =−3cos t, z = 2t, t ≥ 0.

While most curves in three dimensions are difficult to recognize, you should notice that there is a relationshipbetween x and y here, namely,

x2 +(y

3

)2= sin2 t + cos2 t = 1.

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But since z = 2t, z will increase as t increases and so, the curve will wind its way up the cylinder, as tincreases. We call this curve an elliptical helix. See the figure 11.3. □Example 11.1.6. Plot the curve traced out by the vector–valued function

r(t) = ⟨3+2t,5−3t,2−4t⟩ , t ∈ R.

Figure 11.4: Graph of r(t) = ⟨3+2t,5−3t,2−4t⟩

ANSWER. Notice that the curve is given parametrically by

x = 3+2t, y = 5−3t, z = 2−4t, t ∈ R.

You should recognize these equations as parametric equations for the straight line parallel to the vector⟨2,−3,−4⟩ and passing through the point (3,5,2). See the figure 11.4. □

□ 11.1.3 Arc Length.� CALCULUS I FOR ENGINEERS:The curve of y = f (x) on [a,b] has the arc length

s =ˆ b

a

√1+[ f ′(x)]2 dx.

� ARC LENGTH OF VECTOR–VALUED FUNCTION IN PLANE:A vector–valued function r(t) = ⟨ f (t),g(t)⟩ for t ∈ [a,b] has the arc length

s =ˆ b

a

√[ f ′(t)]2 +[g′(t)]2 dt.

� ARC LENGTH OF VECTOR–VALUED FUNCTION IN SPACE:A vector–valued function r(t) = ⟨ f (t),g(t),h(t)⟩ for t ∈ [a,b] has the arc length

s =ˆ b

a

√[ f ′(t)]2 +[g′(t)]2 +[h′(t)]2 dt.

Example 11.1.7. Find the arc length of the curve traced out by the endpoint of the vector–valued function

r(t) =⟨2t, ln t, t2⟩ , 1 ≤ t ≤ e.

ANSWER. First, notice that for x(t) = 2t, y(t) = ln t and z(t) = t2, we have

x′(t) = 2, y′(t) =1t, z′(t) = 2t,

and the curve is traversed exactly once for 1 ≤ t ≤ e. (To see why, observe that x = 2t is an increasingfunction.) By the formula above, we have

s =ˆ 2

1

√22 +

(1t

)2

+(2t)2 dt = e2. □

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§11.2 The Calculus of Vector–Valued Functions.

□ 11.2.1 Limit, Continuity and Derivative.

Definition 11.2.1. For a vector–valued function r(t) = ⟨ f (t),g(t),h(t)⟩, the limit of r(t) as t approaches a isgiven by

limt→a

r(t) = limt→a

⟨ f (t),g(t),h(t)⟩=⟨

limt→a

f (t), limt→a

g(t), limt→a

h(t)⟩

provided all of the indicated limits exist. If any of the limits indicated on the right–hand side of the equationabove fail to exist, then limt→a r(t) does not exist.

Example 11.2.2. Find limt→0⟨t2 +1,5cos t,sin t

⟩.

Definition 11.2.3. The vector–valued function r(t) = ⟨ f (t),g(t),h(t)⟩ is continuous at t = a whenever

limt→a

r(t) = r(a)

(i.e., whenever the limit exists and equals the value of the vector–valued function).

Theorem 11.2.4. A vector–valued function r(t) = ⟨ f (t),g(t),h(t)⟩ is continuous at t = a if and only if all off , g and h are continuous at t = a.

Example 11.2.5. Determine for what values of t the vector–valued function r(t) =⟨e5t , ln(t +1),cos t

⟩is

continuous.

Definition 11.2.6. The derivative r′ of the vector–valued function r is defined by

r′(t) = limh→0

r(t +h)− r(t)h

,

for any values of t for which the limit exists. When the limit exists for t = a, we say that r is differentiable att = a.

Theorem 11.2.7. Let r(t) = ⟨ f (t),g(t),h(t)⟩ and suppose that the components f , g and h are all differentiablefor some value of t. Then r is also differentiable at that value of t and its derivative is given by

r′(t) =⟨

f ′(t),g′(t),h′(t)⟩

Example 11.2.8. Find the derivative of r(t) =⟨sin(t2),ecos t , t ln t

⟩.

Theorem 11.2.9. Suppose that r and s are differentiable vector–valued functions, f is a differentiable scalarfunction and a and b are any scalar constants. Then,

1. Linearity: [ar+bs]′ = ar′+bs′

2. Product Rule (Scalar Function Multiple): [ f r]′ = f ′r+ f r′

3. Product Rule (Dot Product): [r · s]′ = r′ · s+ r · s′4. Product Rule (Cross Product): [r× s]′ = r′× s+ r× s′

5. Chain Rule: [r( f )]′ = f ′r′( f )

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□ 11.2.2 Interpretation of Derivative.We say that the curve traced out by the vector–valued function r(t) = ⟨ f (t),g(t),h(t)⟩ on an interval I issmooth if r′ is continuous on I and r′(t) = 0, except possibly at any endpoints of I. Notice that this says thatthe curve is smooth provided f ′, g′ and h′ are all continuous on I and f ′(t), g′(t) and h′(t) are not all zero atthe same point in I.

Example 11.2.10. Is the plane curve traced out by the vector–valued function r(t) =⟨t3, t2⟩ smooth?

We refer to r′(a) as a tangent vector to the curve C at the point corresponding to t = a. Be sure to observethat r′(a) lies along the tangent line to the curve at t = a and points in the direction of the orientation of C.

Example 11.2.11. For r(t) = ⟨−cos2t,sin2t⟩, plot the curve traced out by the endpoint of r(t) and draw theposition vector and tangent vector at t = π/4.


Figure 11.5: Graph of r(t) = ⟨−cos2t,sin2t⟩

Theorem 11.2.12. ∥r(t)∥= constant if and only if r(t) and r′(t) are orthogonal, for all t.

The Theorem says that the path traced out by r(t) lies on a circle centered at the origin if and only if thetangent vector is orthogonal to the position vector at every point on the curve. Likewise, in three dimensions,it says that the curve traced out by r(t) lies on a sphere centered at the origin if and only if the tangent vectoris orthogonal to the position vector at every point on the curve.

□ 11.2.3 Integral.

Definition 11.2.13. The vector–valued function R(t) is an antiderivative of the vector–valued function r(t)whenever R′(t) = r(t).

Definition 11.2.14. If R(t) is any antiderivative of r(t), the indefinite integral of r(t) is defined to beˆ

r(t)dt = R(t)+ c,

where c is an arbitrary constant vector.

As in the scalar case, R(t)+ c is the most general antiderivative of r(t). (Why is that?) Noticeˆ

r(t)dt =ˆ

⟨ f (t),g(t),h(t)⟩ dt =⟨ˆ

f (t)dt,ˆ

g(t)dt,ˆ

h(t)dt⟩

That is, you integrate a vector–valued function by integrating each of the individual components.

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Example 11.2.15. Evaluate the indefinite integral´ ⟨

t2 +2,sin(2t),4tet2⟩

dt.

Definition 11.2.16. For the vector–valued function r(t) = ⟨ f (t),g(t),h(t)⟩, we define the definite integral ofr(t) on the interval [a,b] by

ˆ b

ar(t)dt =

ˆ b

a⟨ f (t),g(t),h(t)⟩ dt =

⟨ˆ b

af (t)dt,

ˆ b

ag(t)dt,

ˆ b

ah(t)dt

⟩

Theorem 11.2.17. Suppose that R(t) is an antiderivative of r(t) on the interval [a,b]. Then,

ˆ b

ar(t)dt = R(b)−R(a)

Example 11.2.18. Evaluate´ 1

0

⟨sin(πt),6t2 +4t

⟩dt.

§11.3 Motion In Space.Skip. Please read the textbook.

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§11.4 Curvature.

□ 11.4.1 Arc Length Function.Recall that the curve traced out by the endpoint of the vector–valued function r(t) = ⟨ f (t),g(t),h(t)⟩, fora ≤ t ≤ b, has the arc length

s =ˆ b

a

√[ f ′(t)]2 +[g′(t)]2 +[h′(t)]2 dt.

We define the arc length parameter s(t) to be the arc length of that portion of the curve from u = a up to u = t.That is,

s(t) =ˆ u=t

u=a

√[ f ′(u)]2 +[g′(u)]2 +[h′(u)]2 du =

ˆ u=t

u=a∥r′(u)∥du.

Example 11.4.1. Find an arc length parameterization of the circle of radius 4 centered at the origin.

ANSWER. Note that one parameterization of this circle is

C : x = f (t) = 4cos t, y = g(t) = 4sin t, 0 ≤ t ≤ 2π.

It gives the vector–valued function r(t) = ⟨4cos t,4sin t⟩ on [0,2π]. In this case, the arc length from u = 0 tou = t is given by

s(t) =ˆ t

0

√[ f ′(u)]2 +[g′(u)]2 du =

ˆ t

0

√[−4sinu]2 +[4cosu]2 du = 4

ˆ t

01dt = 4t.

That is, t = s/4, so that an arc length parameterization for C is

C : x = 4coss4, y = 4sin

s4, 0 ≤ s ≤ 8π. □

Exercise 11.4.2. Reparametrize the helix r(t) = ⟨cos t,sin t, t⟩ with respect to arc length measured from(1,0,0) in the direction of increasing t.

□ 11.4.2 Curvature.Consider the smooth curve C traced out by the endpoint of the vector–valued function r(t). Recall that foreach t, r′(t) can be thought of as both the velocity vector and a tangent vector, pointing in the direction ofmotion (i.e., the orientation of C). Notice that

T(t) =r′(t)

∥r′(t)∥

is also a tangent vector, but has length one (∥T(t)∥ = 1). We call T(t) the unit tangent vector to the curveC. That is, for each t, T(t) is a tangent vector of length one pointing in the direction of the orientation of C.

Figure 11.6: Graph of r(t) =⟨t2 +1, t

⟩ of 126


Example 11.4.3. Find the unit tangent vector to the curve determined by r(t) =⟨t2 +1, t

⟩.


From the figure 11.7, you can see that the tangent vector changes direction very slowly when C is fairlystraight, but it changes direction more quickly when C bends or twists more sharply.The curvature of C at a given point is a measure of how quickly the curve changes direction at that point.Specifically, we define it to be the magnitude of the rate of change of the unit tangent vector with respect toarc length. (We use arc length so that the curvature will be independent of the parametrization.)

Figure 11.7: Unit Tangent Vectors

Definition 11.4.4. The curvature κ of a curve is the scalar quantity

κ =

∥∥∥∥ dTds

∥∥∥∥where T represents the unit tangent vector and s represents the arc length parameter.

Note that, while the definition of curvature makes sense intuitively, it is not a simple matter to compute κdirectly from the definition. So we look for another form. The curvature is easier to compute if it is expressedin terms of the parameter t instead of s, so we use the Chain Rule to write

dTdt

=dTds

dsdt

−→ κ =

∥∥∥∥ dT/dtds/dt

∥∥∥∥= ∥T′(t)∥∥ds/dt∥

But dsdt = ∥r′(t)∥ from the definition of the arc length function, so finally we have

κ =∥T′(t)∥∥r′(t)∥

Example 11.4.5. Find the curvature of a straight line traced out by the vector–valued function

r(t) = ⟨at + x0,bt + y0,ct + z0⟩

for some constants a, b, c, x0, y0 and z0 (where at least one of a, c or e is nonzero).

Example 11.4.6. Show that the curvature of a circle of radius a is 1a .

Notice that the result of the example is consistent with your intuition. First, observe that you should be able todrive a car around a circular track while holding the steering wheel in a fixed position. (That is, the curvatureshould be constant.) Further, the smaller that the radius of a circular track is, the sharper you will need to turn(that is, the larger the curvature). On the other hand, on a circular track of very large radius, it would seem asif you were driving fairly straight (i.e., the curvature will be close to 0).Although the formula on curvature above can be used in all cases to compute the curvature, the formula givenby the following theorem is often more convenient to apply.

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Theorem 11.4.7. The curvature of the curve given by the vector–valued function r(t) is

κ =∥r′(t)× r′′(t)∥

∥r′(t)∥3

Example 11.4.8. Find the curvature of the helix traced out by r(t) = ⟨2sin t,2cos t,4t⟩.

Example 11.4.9. Find the curvature of the twisted cubic r(t) =⟨t, t2, t3⟩ at a general point and at (0,0,0).

In the case of a plane curve that is the graph of a function, y = f (x), we can derive a particularly simple for-mula for the curvature. Notice that such a curve is traced out by the vector–valued function r(t) = ⟨t, f (t),0⟩,where the third component is 0, since the curve lies completely in the xy–plane. Applying the formula to thisvector–valued function and putting the parameter t = x back, we obtain

κ =| f ′′(x)|

{1+[ f ′(x)]2}3/2

Example 11.4.10. Find the curvature of the parabola y = ax2 + bx+ c. Also, find the limiting value of thecurvature as x → ∞.

Example 11.4.11. Find the curvature of the parabola y = x2 at x = 0, x = 1 and x = 2.

§11.5 Tangent and Normal Vectors.Skip. Please read the textbook.

§11.6 Parametric Surfaces.Skip. Please read the textbook.

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Chapter 12

Functions of Several Variables and Partial Differentiation

§12.1 Functions of Several Variables.Skip. Please read the textbook.

§12.2 Limits nd Continuity.Skip. Please read the textbook.

§12.3 Partial Derivatives.

□ 12.3.1 Partial Derivative.In general, if is a function of two variables x and y, suppose we let only x vary while keeping y fixed, say y= b,where b is a constant. Then we are really considering a function of a single variable x, namely, g(x) = f (x,b).If g has a derivative at a, then we call it the partial derivative of f with respect to x at (a,b) and denote itby fx(a,b). Thus

fx(a,b) = g′(a), where g(x) = f (x,b).

By the definition of a derivative, we have

g′(a) = limh→0

g(a+h)−g(a)h

and so the equation above becomes

fx(a,b) = limh→0

f (a+h,b)− f (a,b)h

.

Similarly, the partial derivative of f with respect to y at (a,b), denoted by fy(a,b), is obtained by keepingx fixed (x = a) and finding the ordinary derivative at b of the function h(y) = f (a,y)

fy(a,b) = limh→0

f (a,b+h)− f (a,b)h

.

If we now let the point (a,b) vary, then fx and fy become functions of two variables.

Definition 12.3.1. If f is a function of two variables x and y, its partial derivatives are the functions fx andfy defined by

fx(x,y) = limh→0

f (x+h,y)− f (x,y)h

, fy(x,y) = limh→0

f (x,y+h)− f (x,y)h

.

Remark 12.3.2 (NOTATIONS FOR PARTIAL DERIVATIVES). If z = f (x,y), then we write

fx(x,y) = fx =∂ f∂x

=∂∂x

f (x,y) =∂ z∂x

= f1 = D1 f = Dx f .

fy(x,y) = fy =∂ f∂y

=∂∂y

f (x,y) =∂ z∂y

= f2 = D2 f = Dy f .

To compute partial derivatives, all we have to do is remember that the partial derivative with respect to x isjust the ordinary derivative of the function g of a single variable that we get by keeping y fixed. Thus we havethe following rule.

57


Remark 12.3.3 (RULE FOR FINDING PARTIAL DERIVATIVES OF z = f (x,y)).

1. To find fx, regard y as a constant and differentiate f (x,y) with respect to x.2. To find fy, regard x as a constant and differentiate f (x,y) with respect to y.

Example 12.3.4. If f (x,y) = x3 + x2y3 −2y2, find fx and fy and fx(1,2) and fy(1,2).

ANSWER.

fx =∂ f∂x

=∂∂x

x3 +∂∂x

(x2y3)−2∂∂x

y2 = 3x2 +2xy3 −2(0) = 3x2 +2xy3,

fy =∂ f∂y

=∂∂y

x3 +∂∂y

(x2y3)−2∂∂y

y2 = 0+3x2y2 −4y = 3x2y2 −4y,

fx(1,2) = 3(1)2 +2(1)(2)3 = 19, fy(1,2) = 3(1)2(2)2 −4(2) = 4 □

Example 12.3.5. For f (x,y) = 3x2 + x3y+4y2, find fx and fy and fx(1,0) and fy(2,−1).

Example 12.3.6. Find the partial derivatives fx and fy: (1) f (x,y) = exy + xy and (2) f (x,y) = sin

(x

1+y

).

Example 12.3.7. For a real gas, van der Waals’ equation states that(P+

n2aV 2

)(V −nb) = nRT.

Here, P is the pressure of the gas, V is the volume of the gas, T is the temperature (in degrees Kelvin), n isthe number of moles of gas, R is the universal gas constant and a and b are constants. Compute and interpret∂P∂V and ∂T

∂ P .

□ 12.3.2 Higher–Order Partial Derivative.If is a function of two variables, then its partial derivatives fx and fy are also functions of two variables, sowe can consider their partial derivatives ( fx)x, ( fx)y, ( fy)x, and ( fy)y, which are called the second partialderivatives of f . If z = f (x,y), we use the following notation:

Remark 12.3.8 (NOTATIONS FOR SECOND PARTIAL DERIVATIVES).

( fx)x = fxx = f11 =∂∂x

(∂ f∂x

)=

∂ 2 f∂x2 =

∂ 2z∂x2 .

( fx)y = fxy = f12 =∂∂y

(∂ f∂x

)=

∂ 2 f∂y∂x

=∂ 2z

∂y∂x.

( fy)x = fyx = f21 =∂∂x

(∂ f∂y

)=

∂ 2 f∂x∂y

=∂ 2z

∂x∂y.

( fy)y = fyy = f22 =∂∂y

(∂ f∂y

)=

∂ 2 f∂y2 =

∂ 2z∂y2 .

Thus the notation fxy or ∂ 2 f∂y∂x means that we first differentiate with respect to x and then with respect to y,

whereas fyx in computing the order is reversed.

Example 12.3.9. Find the second derivatives of f (x,y) = x3 + x2y3 −2y2.

ANSWER. From the Example 12.3.4 above, we have

fx = 3x2 +2xy3, fy = 3x2y2 −4y.

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Therefore,

fxx =∂∂x

(3x2 +2xy3)= 6x+2y3, fxy =

∂∂y

(3x2 +2xy3)= 6xy2

fyx =∂∂x

(3x2y2 −4y

)= 6xy2, fyy =

∂∂y

(3x2y2 −4y

)= 6x2y−4 □

Exercise 12.3.10. Find all second–order partial derivatives of f (x,y) = x2y− y3 + lnx.

Notice in the example above that fxy = 6xy2 = fyx. It turns out that this is true for most, but not all, of thefunctions that we will encounter. The following theorem discovered by Alexis Clairaut (1713 – 1765) givesconditions under which we can assert that fxy = fyx.

Theorem 12.3.11 (CLAIRAUT’S THEOREM). If fxy and fyx are continuous on an open set containing (a,b),then fxy(a,b) = fyx(a,b).

We can, of course, compute third–, fourth– or even higher–order partial derivatives.

Example 12.3.12. Compute fxxyz if f (x,y,z) = sin(3x+ yz).

ANSWER.

fx = 3cos(3x+ yz), fxx =−9sin(3x+ yz)fxxy =−9zcos(3x+ yz), fxxyz =−9 [cos(3x+ yz)− ysin(3x+ yz)] □

Exercise 12.3.13. For f (x,y) = cos(xy)− x3 + y4, compute fxyy and fxyyy.

Exercise 12.3.14. For f (x,y,z) =√

xy3z +4x2y, defined for x, y, z ≥ 0, compute fx, fxy and fxyz.

Exercise 12.3.15. The sag in a beam of length L, width w and height h is given by

S(L,w,h) = cL4

wh3

for some constant c. Show that

∂S∂L

=4L

S,∂S∂w

=− 1w

S,∂S∂h

=−3h

S.

Use this result to determine which variable has the greatest proportional effect on the sag.

§12.4 Tangent Planes And Linear Approximations.Skip. Please read the textbook.

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§12.5 The Chain Rule.

□ 12.5.1 Chain Rule.We recall from Calculus I that f (x) = (x2 +8)5 has the derivative

f ′(x) = 5(x2 +8)4(x2 +8)′ = 5(x2 +8)4(2x) = 10x(x2 +8)4.

The general form of the chain rule says that for differentiable functions h and g,

ddx

[h(g(x))] = h′(g(x))g′(x).

We now extend the chain rule to functions of several variables. This takes several slightly different forms,depending on the number of independent variables, but each is a variation of the already familiar chain rulefor functions of a single variable.For functions of more than one variable, the Chain Rule has several versions, each of them giving a rule fordifferentiating a composite function.

Chain Rule: Case 1

Theorem 12.5.1. Suppose that z = f (x,y) is a differentiable function of x and y, where x and y are bothdifferentiable functions of t. Then z is a differentiable function of t and

dzdt

=∂ f∂x

dxdt

+∂ f∂y

dydt

or z′(t) = fxx′(t)+ fyy′(t).

Example 12.5.2. If z = x2y+3xy4, where x = sin(2t) and y = cos t, find z′(t) = dzdt when t = 0.

ANSWER. For z = f (x,y) = x2y+3xy4, we compute

fx = 2xy+3y4, fy = x2 +12xy3, x′(t) = 2cos(2t), y′(t) =−sin t

z′(t) = fxx′+ fyy′ =(2xy+3y4)(2cos(2t))+

(x2 +12xy3)(−sin t)

=(2sin(2t)cos t +3cos4 t

)(2cos(2t))+

(sin2(2t)+12sin(2t)cos3 t

)(−sin t)

= 2cos(2t)(2sin(2t)cos t +3cos4 t

)− sin t

(sin2(2t)+12sin(2t)cos3 t

), z′(0) = 6. □

Exercise 12.5.3. For z = f (x,y) = x2ey, x(t) = t2 −1 and y(t) = sin t, find z′(t) = dzdt .

To remember the Chain Rule, it’s helpful to draw the tree diagram in the figure 12.1. We draw branches fromthe dependent variable z to the intermediate variables x and y to indicate that z is a function of x and y. Thenwe draw branches from x and y to the independent variables t. On each branch we write the correspondingpartial derivative. To find ∂ z

∂ t , we find the product of the partial derivatives along each path from z to t andthen add these products:

dzdt

=∂ f∂x

dxdt

+∂ f∂y

dydt

Example 12.5.4. Suppose the production of a firm is modeled by the Cobb–Douglas production functionP(k, l) = 20k1/4l3/4, where k measures capital (in millions of dollars) and l measures the labor force (inthousands of workers). Suppose that when l = 2 and k = 6, the labor force is decreasing at the rate of 20workers per year and capital is growing at the rate of $400000 per year. Determine the rate of change ofproduction.

We can easily extend the Theorem 12.5.1 above to the case of a function f (x,y), where x and y are bothfunctions of the two independent variables s and t, x = x(s, t) and y = y(s, t).

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Figure 12.1: Tree Diagrams, Case 1 and Case 2

Chain Rule: Case 2

Theorem 12.5.5. Suppose that z = f (x,y) is a differentiable function of x and y, where x = x(s, t) andy = y(s, t) are differentiable functions of s and t. Then

∂ z∂ s

=∂ f∂x

∂x∂ s

+∂ f∂y

∂y∂ s

, and∂ z∂ t

=∂ f∂x

∂x∂ t

+∂ f∂y

∂y∂ t

i.e., zs = fxxs + fyys, and zt = fxxt + fyyt .

Example 12.5.6. If z = f (x,y) = ex siny, where x = st2 and y = s2t, find ∂ z∂ s and ∂ z

∂ t .

ANSWER. For z = f (x,y) = ex siny, we compute

fx = ex siny, fy = ex cosy, xs = t2, xt = 2st, ys = 2st, yt = s2

zs = fxxs + fyys = ex siny(t2)+ ex cosy(2st) = tex (t siny+2scosy) = test2 (t sin(s2t)+2scos(s2t)

)zt = fxxt + fyyt = ex siny(2st)+ ex cosy(s2) = sex (2t siny+ scosy) = sest2 (

2t sin(s2t)+ scos(s2t)). □

Exercise 12.5.7. Suppose that z = f (x,y) = exy, x(u,v) = 3usinv and y(u,v) = 4uv2. Find the partial deriva-tives ∂ z

∂u and ∂ z∂v .

For the tree diagram in this Case of the Chain Rule, see the figure 12.1.

Exercise 12.5.8. Consider z = f (x,y), where x(r,θ) = r cosθ and y(r,θ) = ysinθ . Prove the followings.

1. fr = fx cosθ + fy sinθ .2. frr = fxx cos2 θ +2 fxy cosθ sinθ + fyy sin2 θ .3. fxx + fyy = frr +

frr + fθθ

r2 .

A slightly different use of a change of variables is demonstrated in the following example. An importantstrategy in solving some equations is to first rewrite and solve them in the most general form possible. Oneconvenient approach to this is to convert to dimensionless variables. As the name implies, these are typicallycombinations of variables such that all the units cancel. One example would be for an object with (one–dimensional) velocity v ft/s and initial velocity v(0) = v0 ft/s. The variable V = v

v0is dimensionless because

the units of V are ft/s divided by ft/s, leaving no units. Often, a change to dimensionless variables willsimplify an equation.

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Example 12.5.9. An object moves in two dimensions according to the equations of motion: x′′(t) = 0 andy′′(t) =−g, with initial velocity x′(0) = v0 cosθ and y′(0) = v0 sinθ and initial position x(0) = y(0) = 0.(1) Rewrite the equations and initial conditions in terms of the variables

X =gv2

0x, Y =

gv2

0y, T =

gv0

t.

(2) Show that the variables X , Y and T are dimensionless, assuming that x and y are given in feet and t inseconds.

ANSWER. (1) To transform the equations, we first need to rewrite the derivatives x′′ = d2xdt2 and y′′ = d2y

dt2 interms of X , Y and T . From the chain rule, we have

dxdt

=dxdT

dTdt

=d

dT

(v2

0Xg

)ddt

(gtv0

)=

v20

gdXdT

gv0

= v0dXdT

.

Again, we must be careful computing the second derivative. We have

d2xdt2 =

ddt

(dxdt

)=

ddt

(v0

dXdT

)=

ddT

(v0

dXdT

)dTdt

= v0d2XdT 2

ddt

(gtv0

)= v0

d2XdT 2

gv0

= gd2XdT 2 .

By the same argument, we havedydt

= v0dYdT

,d2ydt2 = g

d2YdT 2 .

Hence, the differential equations x′′(t) = 0 and y′′(t) =−g become, respectively,

gd2XdT 2 = 0, g

d2YdT 2 =−g, i.e.,

d2XdT 2 = 0,

d2YdT 2 =−1.

The initial conditions x′(0) = v0 cosθ and y′(0) = v0 sinθ become, respectively,

v0dXdT

∣∣∣∣T=0

= v0 cosθ , v0dYdT

∣∣∣∣T=0

= v0 sinθ , i.e.,dXdT

∣∣∣∣T=0

= cosθ ,dYdT

∣∣∣∣T=0

= sinθ .

The initial value problem is now

d2XdT 2 = 0,

d2YdT 2 =−1,

dXdT

∣∣∣∣T=0

= cosθ ,dYdT

∣∣∣∣T=0

= sinθ , X(0) = 0, Y (0) = 0.

(2) We observe X = gv2

0x has the units:

ft/s2

(ft/s)2 ft =ft2/s2

ft2/s2= 1.

Similarly Y and T have no units. □

□ 12.5.2 Implicit Differentiation.We suppose that an equation F(x,y) = 0 of the form defines y implicitly as a differentiable function of x, thatis, y = f (x), where F(x, f (x)) = 0 for all x in the domain of f . If F is differentiable, we can apply the ChainRule, Case 1, 12.5.1 to differentiate both sides of the equation F(x,y) = 0 with respect to x. Since both x andy are functions of x, we obtain

∂F∂x

dxdx

+∂F∂y

dydx

= 0, simply,dydx

=−

∂F∂x∂F∂y

=−Fx

Fy,

assuming Fy = 0.

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Theorem 12.5.10 (IMPLICIT DIFFERENTIATION).

1. For the equation F(x,y) = 0, we can find dydx as follows:

dydx

=−Fx

Fy

2. For the equation F(x,y,z) = 0, we can find ∂ z∂x and ∂ z

∂y as follows:

∂ z∂x

=−Fx

Fzand

∂ z∂y

=−Fy

Fz

Example 12.5.11. Find y′(x) = dy/dx if x3 + y3 = 6xy.

ANSWER. The given equation can be written as

x3 + y3 −6xy = 0.

Let F(x,y) = x3 + y3 −6xy. Then we have

F(x,y) = x3 + y3 −6xy = 0.

Using the formula above, we get

dydx

=−Fx

Fy=−3x2 −6y

3y2 −6x=−x2 −2y

y2 −2x. □

Exercise 12.5.12. Find ∂ z∂x and ∂ z

∂y if x3 + y3 + z3 +6xyz = 1.

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§12.6 Gradient and Directional Derivatives.For the animations or plots involved with the topics to be discussed here, please confer the following webpage:http://www.usd.edu/˜jflores/MultiCalc02/Contents.htmExplicitly, for the level curves, “Functions of Several Variables” in the webpage may be useful. For thegradient and directional derivatives, “Directional Derivatives and the Gradient Vector” in the webpage maybe helpful.

□ 12.6.1 Functions of Two Variables.In Calculus I, we learned the instantaneous rate of change f ′(a) of the function f (x) of one variable at x = a.It means the slope of the tangent line to the graph of f (x) at x = a.In Calculus II, we deal with the function f (x1,x2, . . . ,xn) of several variables. For such a function, how canwe define the instantaneous rate of change? What is the meaning of the rate of change?A graph of y = f (x) is given in the two–dimensional space. At each point (a, f (a)) on the graph, we candraw one tangent line with the slope f ′(a).The surface of z = f (x,y) is given in the three–dimensional space. At each point of the surface, we canfind one tangent plane. Since a plane contains infinitely many lines, at each point on the surface, we canthink of infinitely many tangent lines. We may say the instantaneous rate of change of f (x,y) at the pointP(a,b, f (a,b)) on the surface corresponds to the slope of the tangent line in the case of two–dimensionalspace. However, since there are so many tangent lines, we should choose only one by assigning the directionof the line.We give a special name to the instantaneous rate of change of f (x,y) at the point P(x,y) and in the directionof the unit vector u.

Definition 12.6.1. The directional derivative of f (x,y) at the point (a,b) and in the direction of the unitvector u = ⟨u1,u2⟩ is given by

Du f (a,b) = limh→0

f (a+hu1,b+hu2)− f (a,b)h

,

provided the limit exists.

When u = ⟨1,0⟩, the directional derivative becomes


f (a+h,b)− f (a,b)h

=∂ f (x,y)

∂x

∣∣∣∣(x,y)=(a,b)

= fx(a,b),

which is the value of the partial derivative of f (x,y) with respect to x at (x,y) = (a,b).When u = ⟨0,1⟩, the directional derivative becomes


f (a,b+h)− f (a,b)h

=∂ f (x,y)

∂y

∣∣∣∣(x,y)=(a,b)

= fy(a,b),

which is the value of the partial derivative of f (x,y) with respect to y at (x,y) = (a,b).In Calculus I, even if we learned the limit definition of the derivative, we did not use the limit definition tocompute the derivative. Rather than the limit definition, we applied for the formulas. By the same argument,we need formulas to get the directional derivatives easily.

Theorem 12.6.2. Suppose that f is differentiable at (a,b) and u = ⟨u1,u2⟩ is any unit vector. Then we canwrite

Du f (a,b) = fx(a,b)u1 + fy(a,b)u2.

The proof of this formula is based on the chain rule.

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Example 12.6.3. Compute the directional derivative at the given point in the direction of the indicated vector.(1) f (x,y) = x3y−4y2 at (2,−1) in the direction of u =

⟨1√2, 1√

2

⟩.

(2) g(x,y) =√

x2 + y2 at (3,−4) with u in the direction of ⟨3,−2⟩.(3) h(x,y) = sin(2x− y) at (π,0) with u in the direction from (π,0) to (2π,π).

ANSWER. We use the Theorem 12.6.2 above.(1) It is easy to get fx = 3x2y and fy = x3 −8y and so

Du f (2,−1) = fx(2,−1)1√2+ fy(2,−1)

1√2=

−12+16√2

= 2√

2 .

(2) The unit vector u in the direction of ⟨3,−2⟩ is obtained as follows:

u =⟨3,−2⟩

∥⟨3,−2⟩∥=

⟨3,−2⟩√13

.

Using this unit vector, we get the directional derivative:

gx(x,y) =x

g(x,y)=

x√x2 + y2

, gy(x,y) =y

g(x,y)=

y√x2 + y2

,

gx(3,−4) =35, gy(3,−4) =−4

5,

Dug(3,−4) = gx(3,−4)3√13

+gy(3,−4)−2√13

=9+85√

13=

175√

13.

(3) The vector in the direction from (π,0) to (2π,π) is ⟨2π −π,π −0⟩ = ⟨π,π⟩. The unit vector u in thedirection of ⟨π,π⟩= π ⟨1,1⟩ is obtained as follows:

u =π ⟨1,1⟩

∥π ⟨1,1⟩∥=

⟨1,1⟩√2

.

Using this unit vector, we get the directional derivative:

hx(x,y) = 2cos(2x− y), hy(x,y) =−cos(2x− y), hx(π,0) = 2, hy(π,0) =−1,

Duh(π,0) = hx(π,0)1√2+hy(π,0)

1√2=

2−1√2

=1√2. □

For convenience, we define the gradient of a function to be the vector–valued function whose componentsare the first–order partial derivatives of f . We denote the gradient of a function f by grad f or ∇ f (read “delf ”).

Definition 12.6.4. The gradient of f (x,y) is the vector–valued function

∇ f (x,y) =⟨

∂ f∂x

,∂ f∂y

⟩=

∂ f∂x

i+∂ f∂y

j, ∇ f (x,y) =⟨

fx, fy⟩= fxi+ fyj,

provided both partial derivatives exist.

Remark 12.6.5.

1. Recall from Section 11.2 The Calculus of VectorValued Functions: For the vector–valued functionr(t) = ⟨ f (t),g(t)⟩ of one variable t, we can get the derivative r′(t) = ⟨ f ′(t),g′(t)⟩, which is the tangentvector.

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2. The definition above is for the vector–valued function f (x,y) of two variables x and y:

∇ f (x,y) =⟨

fx, fy⟩.

When using the gradient, the Theorem 12.6.2 for the directional derivative can be simplified as the dot productof the gradient and the unit vector in the direction of interest. For any unit vector u = ⟨u1,u2⟩,

Du f (x,y) = fxu1 + fyu2 =⟨

fx, fy⟩· ⟨u1,u2⟩= ∇ f (x,y) ·u.

Thus we have the following theorem.

Theorem 12.6.6. If f is a differentiable function of x and y and u is any unit vector, then

Du f (x,y) = ∇ f (x,y) ·u.

Writing directional derivatives as a dot product has many important consequences.

Example 12.6.7. For f (x,y) = (x−y)2, compute the directional derivative of f (x,y) at the point (2,1) in thedirection of the indicated vector.(1) u in the direction of v = ⟨2,−1⟩.(2) u in the direction of w = ⟨−2,1⟩.

ANSWER. First, we compute the gradient of f :

∇ f (x,y) =⟨

fx, fy⟩= ⟨2(x− y),−2(x− y)⟩= 2(x− y)⟨1,−1⟩ .

So at the point (2,1), we have ∇ f (2,1) = 2⟨1,−1⟩.(1) The unit vector u in the direction of v = ⟨2,−1⟩ is obtained as follows:

u =v

∥v∥=

⟨2,−1⟩√5

.

By the Theorem 12.6.6,

Du f (2,1) = 2⟨1,−1⟩ · ⟨2,−1⟩√5

=6√5.

(2) The unit vector u in the direction of w = ⟨−2,1⟩ is obtained as follows:

u =w∥w∥

=⟨−2,1⟩√

5.

By the Theorem 12.6.6,

Du f (2,1) = 2⟨1,−1⟩ · ⟨−2,1⟩√5

=− 6√5. □

What is the graphical interpretation of the directional derivative? What does it mean by Du f (2,1) = 6√5

?Simply speaking, it represents the slope of the surface (i.e., slope of the tangent line to the surface) at thepoint.The point (2,1,1) is on the surface of z = f (x,y) = (x−y)2. In the case of (1) of the example above, the unitvector u is u = ⟨−2,1⟩/

√5 . Based on these two pieces of information, we consider a plane S such that (i)

it contains the point (2,1,1), (ii) it is perpendicular to the xy–plane and finally (iii) it is parallel to the unitvector u = ⟨−2,1⟩/

√5 . In this plane S, we introduce a new coordinate system having the direction of u as

the positive horizontal axis and (2,1) as the origin.Then, we have the two–dimensional graph of z = f (x,y) on this plane S with the coordinate system. Theslope of the tangent line to this image graph at the origin is Du f (2,1) = 6√

5. For the explicit explanation,

please confer the textbook.We can use a contour plot to estimate the value of a directional derivative. We refresh our memory on thelevel curve and the contour plot of f (x,y).

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Definition 12.6.8. A level curve or contour curve of f (x,y) is the (two–dimensional) graph of the equationf (x,y) = c, for some constant c. A contour plot of f (x,y) is a graph of numerous level curves f (x,y) = c,for representative values of c.

Example 12.6.9. The level curves of the function f (x,y) =√

9− x2 − y2 for c = 0,1,2,3 are√9− x2 − y2 = c i.e., x2 + y2 = 9− c2,

which is a family of concentric circles with center (0,0) and radius√

9− c2 . Since c = 0,1,2,3, we havetotal four level curves (circles) with radius 3,

√8 ,

√5 ,0, respectively.

For more information on the level curve and the contour curve, please read Section 12.1 Functions of SeveralVariables.

Example 12.6.10. Use a contour plot of z = f (x,y) = x2 + y2 to estimate Du f (1,−1) for u = ⟨−3,4⟩/√

5 .

Figure 12.2: Contour Plot of z = x2 + y2

Answer. The contour plot of f (x,y) = x2 + y2 for c = 0.2,1,2,3 is the set of circles with center (0,0) andradius

√0.2 ,1,

√2 ,

√3 , respectively. See the figure 12.2. Using the graph, we approximate the directional

derivative by estimating △z△u , where △u is the distance traveled along the unit vector u.

(1) When △u = 1, the terminal point of u with the initial point P(1,−1) is Q(2

5 ,−15

). (Why?

−→PQ =

⟨−3,4⟩/5, which is in the same direction as u and ∥−→PQ∥= 1.)The terminal point Q

(25 ,−

15

)is on the level curve z = x2+y2 = 0.2, because (2/5)2+(−1/5)2 = 1/5 = 0.2.

Thus, when traveling along the vector u as much as △u = 1, the vector appears to extend from z = 2 levelcurve (where the initial point P lies on) to z = 0.2 level curve (where the terminal point Q lies on). That is,

△z△u

=0.2−2

1=−1.8.

(2) We observe that the point R(

2(2+3√

6 )25 , 3−8

√6

25

)= R(0.7479,−0.6638) is on the z = 1 level curve and

the vector−→PR is in the same direction as u. Here,

−→PR =

⟨3(−7+3

√6)

25,−4(−7+2

√6)

25

⟩= ⟨0.0418,0.3361⟩ .

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So, when the vector extends from z = 2 level curve (where the initial point P lies on) to z = 1 level curve(where the initial point R lies on), the distance traveled along u is ∥−→PR∥ = 7−2

√6

5 ≈ 0.4202. That is, in thiscase, we have

△z△u

=1−2

0.4202≈−2.3798.

(3) If we use the Theorem 12.6.6, we have Du f (1,1) = ∇ f (1,1) ·u and so Du f (1,1) =−14/5 =−2.8. □

Keep in mind that a directional derivative gives the rate of change of a function in a given direction.

Remark 12.6.11. We raise some questions.

1. Calculus I: at what point does a given function have the horizontal tangent line and the vertical tangentline? The answer is involved with the critical numbers and the local extrema.

2. Calculus II: in what direction does a given function have its maximum or minimum rate of increase?

Let us answer to the Question 2. Combining the Theorem 10.3.5 in Section 10.3 with the Theorem 12.6.6,we deduce

Du f (a,b) = ∇ f (a,b) ·u = ∥∇ f (a,b)∥∥u∥cosθ = ∥∇ f (a,b)∥cosθ −→ Du f (a,b) = ∥∇ f (a,b)∥cosθ ,

where θ is the angle between ∇ f (a,b) and the directional unit vector u. Now we can say the followings:

1. The maximum value ∥∇ f (a,b)∥ of Du f (a,b) occurs when cosθ = 1, i.e., θ = 0, i.e., ∇ f (a,b) is inthe same direction as u. It implies

u =∇ f (a,b)

∥∇ f (a,b)∥.

2. The minimum value −∥∇ f (a,b)∥ of Du f (a,b) occurs when cosθ =−1, i.e., θ = π , i.e., ∇ f (a,b) isin the opposite direction as u. It implies

u =− ∇ f (a,b)∥∇ f (a,b)∥

.

3. The zero value Du f (a,b) = 0 occurs when cosθ = 0, i.e., θ = π/2, i.e., ∇ f (a,b) is orthogonal to u.4. Recall that the level curves are curves in the xy–plane on which f is constant. So at any point on thelevel curve, the directional derivative becomes zero. Why? For the level curve f (x,y) = c, c constant,we introduce a parameter t so that the variables x and y become functions of t and thus the equationturns to be f (x(t),y(t)) = c. If we differentiate the equation with respect to t, the Chain Rule impliesfxx′(t)+ fyy′(t) = 0, i.e.,

⟨fx, fy

⟩· ⟨x′(t),y′(t)⟩= 0. So we deduce

Du f (x,y) =⟨

fx, fy⟩·u = 0, where u =

⟨x′(t),y′(t)⟩∥⟨x′(t),y′(t)⟩∥

.

Thus, for this reason, the directional derivative becomes zero.Since ⟨x′(t),y′(t)⟩ is the derivative of ⟨x(t),y(t)⟩, so ⟨x′(t),y′(t)⟩ is in fact the tangent vector of the curve.That is, at any point (x,y), ∇ f (x,y) should be orthogonal to the tangent line to the level curve. (Aside:In Calculus I and II, “angle between a line/vector and a graph” means “angle between a line/vector andthe tangent line to the graph.” So “∇ f (a,b) is orthogonal to the tangent line to the level curve” can besimply said “∇ f (a,b) is orthogonal to the level curve”.) See the figure 12.3.

We summarize the answers in the following Theorem.

Theorem 12.6.12. Suppose that f is a differentiable function of x and y at the point (a,b). Then

1. the maximum rate of change of f at (a,b) is ∥∇ f (a,b)∥, occurring in the direction of the gradient,

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Figure 12.3: Level Curves of f (x,y) = 5+x2+2y, Yellow: Tangent Line at (1,−3), Blue: Gradient of f (x,y)at (1,−3)

2. the minimum rate of change of f at (a,b) is −∥∇ f (a,b)∥, occurring in the opposite direction of thegradient,

3. the rate of change of f at (a,b) is 0 in the direction orthogonal to ∇ f (a,b),4. the gradient ∇ f (a,b) is orthogonal to the level curve f (x,y) = c at the point (a,b), where c = f (a,b).

Remember that the directional derivative corresponds to the rate of change of the function f (x,y) in the givendirection.

Example 12.6.13. Find the directions of maximum and minimum change of f at the given point and thevalues of the maximum and minimum rates of change.(1) f (x,y) = y2e4x at (3,−1)(2) g(x,y) = xcos(3y) at (−2,π).

ANSWER. (1) We compute the gradient vector first:

∇ f (x,y) =⟨

fx, fy⟩=⟨4y2e4x,2ye4x⟩= 2ye4x ⟨2y,1⟩ , ∇ f (3,−1) =−2e12 ⟨−2,1⟩ .

By the Theorem 12.6.12, we deduce(i) the direction of maximum change is ∇ f (3,−1) = −2e12 ⟨−2,1⟩ and the maximum rate of change alongthis direction is ∥∇ f (3,−1)∥= ∥−2e12 ⟨−2,1⟩∥= 2e12

√5 ,

(ii) the direction of minimum change is −∇ f (3,−1) = 2e12 ⟨−2,1⟩ and the minimum rate of change alongthis direction is −∥∇ f (3,−1)∥=−2e12

√5 .

(2) We compute the gradient vector first:

∇g(x,y) =⟨gx,gy

⟩= ⟨cos(3y),−3xsin(3y)⟩ , ∇g(−2,π) = ⟨cos(3π),6sin(3π)⟩= ⟨−1,0⟩ .

By the Theorem 12.6.12, we deduce(i) the direction of maximum change is ∇g(−2,π) = ⟨−1,0⟩ and the maximum rate of change along thisdirection is ∥∇g(−2,π)∥= ∥⟨−1,0⟩∥= 1,(ii) the direction of minimum change is −∇g(−2,π) = ⟨1,0⟩ and the minimum rate of change along thisdirection is −∥∇g(−2,π)∥=−1. □

Note that the gradient vector ∇ f (a,b) gives the direction of fastest increase (i.e., maximum rate of change)of f .

Example 12.6.14. Find the direction of maximum increase of the function f (x,y) = 3x− x3 −3xy2 from thepoint A(0.6,−0.7) and sketch the path of steepest ascent.

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ANSWER. By the Theorem 12.6.12, the maximum increase occurs in the direction ∇ f (0.6,−0.7):

∇ f (x,y) =⟨

fx, fy⟩=⟨3−3x2 −3y2,−6xy

⟩= 3⟨1,0⟩−3

⟨x2 + y2,2xy

⟩,

∇ f (0.6,−0.7) = 3⟨1,0⟩−3⟨0.85,−0.84⟩= ⟨0.45,2.52⟩ .

The unit vector u in the direction ∇ f (0.6,−0.7) is

u =∇ f (0.6,−0.7)

∥∇ f (0.6,−0.7)∥=

⟨0.45,2.52⟩∥⟨0.45,2.52⟩∥

= ⟨0.1758,0.9844⟩ .

We observe the point A0(0.6,−0.7) is on the level curve z= f (x,y) = 0.702. A simple computation shows forthe point A1(0.6623,−0.5920) on the level curve z = f (x,y) = 1,

−−→A0A1 = ⟨0.0623,0.1080⟩ is in the direction

of u. We do the same argument with the new point A1 and find A2. Repeating the steps several times, wededuce the following table:

f (x,y) = c A0 A1 A2 A3 A4 A5

0.702 (0.6,−0.7)

1 (0.6623,−0.5920)

1.3 (0.7429,−0.4824)

1.6 (0.8437,−0.3614)

1.9 (1.0045,−0.1821)

2 (1,0)

Connecting each point on each level curve, we have the path of steepest ascent which remains perpendicularto each level curve through which it passes. Finding an equation for the path of steepest ascent is challenging.See the figure 12.4. □

Figure 12.4: At each point Ai, the gradient is perpendicular to each level curve

Remark 12.6.15 (ASIDE). In the Example above, we used the level curves between which the distance wasfixed 0.3. Due to the fixed distance, each vector

−−−−→AiAi+1 does not have the magnitude 1. Suppose that the

distance between given level curves L and M is more than 1. Then, how can we find the path of steepestascent between L and M? I will explain this in class. If you take a look at the exercises in the textbook, youwill see the problems asking to find the path of steepest ascent from the given contour plot (i.e., plot of levelcurves). From the starting point A0 on the level curve L0, we move to the direction of the gradient and findthe point A1 on the level curve L1. Keeping this argument, we can get the desired path.

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□ 12.6.2 Functions of Three Variables.Most of the results above extend easily to functions of any number of variables.

Definition 12.6.16. The directional derivative of f (x,y,z) at the point (a,b,c) and in the direction of theunit vector u = ⟨u1,u2,u3⟩ is given by

Du f (a,b,c) = limh→0

f (a+hu1,b+hu2,c+hu3)− f (a,b,c)h

,

provided the limit exists.The gradient of f (x,y,z) is the vector–valued function

∇ f (x,y,z) =⟨

∂ f∂x

,∂ f∂y

,∂ f∂ z

⟩=

∂ f∂x

i+∂ f∂y

j+∂ f∂ z

k,

provided all the partial derivatives are defined.

We extend the Theorem 12.6.6 on the directional derivative of functions of two variables as follows:

Theorem 12.6.17. If f is a differentiable function of x, y and z and u is any unit vector, then

Du f (x,y,z) = ∇ f (x,y,z) ·u.

As for the functions of two variables, we have

Du f (x,y,z) = ∇ f (x,y,z) ·u = ∥∇ f (x,y,z)∥∥u∥cosθ = ∥∇ f (x,y,z)∥cosθ ,

where θ is the angle between the vectors ∇ f (x,y,z) and u. Also by the same argument, the direction ofmaximum increase (i.e., fastest increase) at any point is given by the gradient at that point.

Example 12.6.18. The temperature at a point (x,y,z) is given by

T (x,y,z) = 200e−x2−3y2−9z2,

where T is measured in ◦C (centigrade) and x,y,z in meters.(1) Find the rate of change of temperature at the point P(2,−1,2) in the direction toward the point Q(3,−3,3).(2) In what direction does the temperature increase fastest at P(2,−1,2)?(3) Find the maximum rate of increase at P(2,−1,2).

ANSWER. Since the main idea to solve the problems is the gradient, we first compute the gradient of T :

∇T (x,y,z) =⟨Tx,Ty,Tz

⟩= T (x,y,z)⟨−2x,−6y,−18z⟩=−2T (x,y,z)⟨x,3y,9z⟩

∇T (1,−2,1) =−2T (1,−2,1)⟨1,−6,1⟩=−400e−22 ⟨1,−6,1⟩ .

• Useful technique: We recall that g(x) = eh(x) has the derivative

g′(x) =(

eh(x))′

= eh(x)h′(x) = g(x)h′(x).

(1) To get the desired rate of change, we need the unit vector u in the direction of the vector−→PQ = ⟨1,−2,1⟩,

u =

−→PQ

∥−→PQ∥=

⟨1,−2,1⟩∥⟨1,−2,1⟩∥

=⟨1,−2,1⟩√

6.

Thus by the Theorem 12.6.17, the rate of change of T at P(1,−2,1) in the direction of u is as follows:

DuT (1,−2,1) = ∇T (1,−2,1) ·u =−400e−22 ⟨1,−6,1⟩ · ⟨1,−2,1⟩√6

=−5600√6

.

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(2) The direction of the fastest increase is ∇T (1,−2,1) = −400e−22 ⟨1,−6,1⟩, exactly like the result statedin the Theorem 12.6.12 on the function of two variables.(3) Again, exactly like the result stated in the Theorem 12.6.12 on the function of two variables, the maximumrate of increase is the norm of ∇T (1,−2,1):

∥∇T (1,−2,1)∥=∥∥−400e−22 ⟨1,−6,1⟩

∥∥= 400e−22∥⟨1,−6,1⟩∥= 400√

38 e−22 ≈ 6.8782×10−7. □

We generalize the argument on the level curve in Remark 12.6.11 (4) to the function of three variables.Recall that for some constant c, the equation z = f (x,y) = c defines a level curve of the function z = f (x,y)with the height/level c on the xy–plane.For some constant k, the equation w = f (x,y,z) = k defines a level surface of the function w = f (x,y,z) withthe height/level k in the xyz–space. Suppose that u is any unit vector lying in the tangent plane to the levelsurface w = f (x,y,z) = k at a point (a,b,c) on the level surface. Then by the same argument developed inRemark 12.6.11, we have

0 = Du f (a,b,c) = ∇ f (a,b,c) ·u.

This occurs when ∇ f (a,b,c) is orthogonal to u. Since u is any unit vector on the tangent plane, it implies∇ f (a,b,c) should be the normal vector to the tangent plane at the point (a,b,c). Now that we have the normalvector to the tangent plane and the point (a,b,c) on the plane is given, we can find the equation of the tangentplane at the point (a,b,c):

0 = ∇ f (a,b,c) · ⟨x−a,y−b,z− c⟩=⟨

fx(a,b,c), fy(a,b,c), fz(a,b,c)⟩· ⟨x−a,y−b,z− c⟩

= fx(a,b,c)(x−a)+ fy(a,b,c)(y−b)+ fz(a,b,c)(z− c),i.e., 0 = fx(a,b,c)(x−a)+ fy(a,b,c)(y−b)+ fz(a,b,c)(z− c).

Recall from Section 10.5 Lines and Planes in Space that the equation of the plane containing a point (a,b,c)with normal vector ⟨α ,β ,γ⟩ (meaning this vector is normal/orthogonal/perpendicular to the plane) is

0 = ⟨α ,β ,γ⟩ · ⟨x−a,y−b,z− c⟩= α(x−a)+β (y−b)+ γ(z− c),i.e., 0 = α(x−a)+β (y−b)+ γ(z− c).

We summarize the argument in the following Theorem.

Theorem 12.6.19. Suppose that f (x,y,z) has continuous partial derivatives at the point (a,b,c) and also∇ f (a,b,c) = 0. Then ∇ f (a,b,c) is a normal vector to the tangent plane to the surface f (x,y,z) = k, at thepoint (a,b,c). Further, the equation of the tangent plane is

0 = fx(a,b,c)(x−a)+ fy(a,b,c)(y−b)+ fz(a,b,c)(z− c).

Definition 12.6.20. The normal line to the level surface at the point (a,b,c) on the surface is the linepassing through the point (a,b,c) and perpendicular to the tangent plane, i.e., parallel to the normal vec-tor ∇ f (a,b,c) =

⟨fx(a,b,c), fy(a,b,c), fz(a,b,c)

⟩.

We observe that the normal line to the level surface of w = f (x,y,z) passing through the point (a,b,c) has thefollowing parametric equations:

x = a+ fx(a,b,c)t, y = b+ fy(a,b,c)t, z = c+ fz(a,b,c)t. (12.6.1)

Example 12.6.21. Find the equation of the tangent plane and the normal line at the point (−2,1,−3) to theellipsoid

x2

4+ y2 +

z2

9= 3.

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ANSWER. The ellipsoid is the level surface (with k = 3) of the function

f (x,y,z) =x2

4+ y2 +

z2

9.

Therefore, we have

∇ f (x,y,z) =⟨

fx, fy, fz⟩=

⟨x2,2y,

2z9

⟩, and ∇ f (−2,1,−3) =

⟨−1,2,−2

3

⟩.

The Theorem 12.6.19 gives the equation of the tangent plane at (−2,1,−3) as

−1(x+2)+2(y−1)− 23(z+3) = 0, i.e., 3x−6y+2z+18 = 0.

By the equation (12.6.1) just above the example, parametric equations of the normal line are

x =−2− t, y = 1+2t, z =−3− 23

t.


Figure 12.5: Ellipsoid x2

4 + y2 + z2

9 = 3, Tangent Plane and Normal Line

As the last technique in this section, we use the argument developed for the tangent plane to the level surfacew = g(x,y,z) = k to find the tangent plane to the surface of z = f (x,y).The surface of z = f (x,y) can be viewed as the level surface of the new function w = g(x,y,z) = f (x,y)− zwith the height/level 0. By the Theorem 12.6.19, for the new function w = g(x,y,z), the equation of thetangent plane at the point (a,b, f (a,b)) on the level surface is obtained as follows:

0 = ∇g(a,b, f (a,b)) · ⟨x−a,y−b,z− f (a,b)⟩=⟨gx(a,b, f (a,b)),gy(a,b, f (a,b)),gz(a,b, f (a,b))

⟩· ⟨x−a,y−b,z− f (a,b)⟩

= gx(a,b, f (a,b))(x−a)+gy(a,b, f (a,b))(y−b)+gz(a,b, f (a,b))(z− f (a,b)).

But, since gx = fx, gy = fy and gz =−1, so the equation becomes

0 = fx(a,b)(x−a)+ fy(a,b)(y−b)− (z− f (a,b)),i.e., z = fx(a,b)(x−a)+ fy(a,b)(y−b)+ f (a,b).

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Example 12.6.22. Find an equation of the tangent plane to the surface of each function at (1,1,2).(1) z = f (x,y) =

√9−3x2 −2y2

(2) z = g(x,y) = 4−√

−x2 − y2 +8x+6y−8

ANSWER. We start with the gradient of f and g:

∇ f (x,y) =⟨

fx, fy⟩=−⟨3x,2y⟩

f (x,y), ∇ f (1,1) =− ⟨3,2⟩

f (1,1)=−⟨3,2⟩

2,

∇g(x,y) =⟨gx,gy

⟩=

⟨x−4,y−3⟩4−g(x,y)

, ∇g(1,1) =⟨−3,−2⟩

4−g(1,1)=−⟨3,2⟩

2.

• Useful technique: We recall that h(x) =√

k(x) has the derivative

h′(x) =(√

k(x))′

=k′(x)

2√

k(x)=

k′(x)2h(x)

.

(1) By the argument preceding this example, the equation of the tangent plane is

z = fx(1,1)(x−1)+ fy(1,1)(y−1)+ f (1,1) =−32(x−1)− (y−1)+2 =−3

2(x−3)− y.

(2) By the argument preceding this example, the equation of the tangent plane is

z = gx(1,1)(x−1)+gy(1,1)(y−1)+g(1,1) =−32(x−1)− (y−1)+2 =−3

2(x−3)− y,

which is exactly same as the one for (1) z = f (x,y) =√

9−3x2 −2y2 .How do we get this result? It’s because ∇ f (1,1) = −

⟨32 ,1⟩= ∇g(1,1). Thus, f and g share the tangent

plane z =−32 (x−3)− y, i.e., 3x+2y+2z = 9 at the point (1,1,2). See the figure 12.6. □

Figure 12.6: Upper Ball: Surface of w = g(x,y), Lower Ellipsoid: Surface of z = f (x,y)

Just as it is important to constantly think of ordinary derivatives as slopes of tangent lines and as instantaneousrate of change, it is crucial to keep in mind at all times the interpretations of gradients. Always think ofgradients as vector–valued functions whose values specify the direction of maximum increase of a functionand whose values provide normal vectors (to the level curves in two dimensions and to the level surfaces inthree dimensions).

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§12.7 Extrema of Functions of Several Variables.

□ 12.7.1 Definition and Second Derivatives Test.

Figure 12.7: Surface of z = f (x,y) = xe−x2/2−y3/3+y, Blue Dot: Local Maximum, Red Dot: Local Minimum

We start with the surface z = f (x,y) = xe−x2/2−y3/3+y. We observe that the graph has the peak at the bluedot (1,1,e1/6) and a valley at the red dot (−1,1,−e1/6). Such points are referred as local extrema, which aremathematically defined as follows.

Definition 12.7.1. We call f (a,b) a local maximum of f if there is an open disk R centered at (a,b), forwhich f (a,b)≥ f (x,y) for all (x,y) ∈ R. Similarly, f (a,b) is called a local minimum of f if there is an opendisk R centered at (a,b), for which f (a,b)≤ f (x,y) for all (x,y) ∈ R. In either case, f (a,b) is called a localextremum of f .

The idea here is the same as the local extrema for the functions of one variable, i.e., if f (a,b)≥ f (x,y) for all(x,y) “near” (a,b), we call f (a,b) a local maximum. If we look at the peak and valley carefully, we observeat both points, the tangent plane is horizontal to the xy–plane. It implies that at these points, the gradient of fshould be the zero vector. For this reason, we define the critical point of f (x,y) as follows. (We can think ofthis argument as the generalization of the one on the critical number for a function of one variable.)

Definition 12.7.2. The point (a,b) is a critical point of the function f (x,y) if the following conditions aresatisfied:

1. (a,b) is the domain of f , and2. either fx(a,b) = 0 and fy(a,b) = 0, or3. one or both of fx(x,y) and fy(x,y) do not exist at (a,b).

The second condition above corresponds to f ′(c) = 0 in Calculus I. The third condition above corresponds tothe one that f ′(c) is undefined in Calculus I.Recall from the Calculus I that the critical numbers are the candidates for local extrema:

1. Even if f (x) has a critical number x = c, f (x) may or may not have a local extremum at x = c.(Examples:f (x) = x2 has a critical number x = 0 and the local minimum at x = 0. g(x) = x3 has a critical number atx = 0 but it does not have a local extremum at x = 0.)

2. If f (x) has a local extremum at x = c, then x = c should be a critical number of f (x).

For the function f (x,y) of two variables, we can develop the exactly same argument.

Theorem 12.7.3. If f (x,y) is a local extremum at (a,b), then (a,b) must be a critical point of f (x,y).

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Example 12.7.4. Find all critical points of f (x,y) = xe−x2/2−y3/3+y and analyze each critical point graphi-cally.

ANSWER. We start with the first partial derivatives:

fx = (1− x2)e−x22−y3/3+y, fy = x(1− y2)e−x22−y3/3+y.

� Useful technique: g(x) = h(x)ek(x) has the derivative,

g′ =(

hek)′

= h′ek +hk′ek = ek (h′+hk′).

Since exponentials are always positive, we get(1) fx = (1− x2)e−x2/2−y3/3+y = 0 −→ 1− x2 = 0 −→ x =±1.(2) fy = x(1− y2)e−x2/2−y3/3+y = 0 −→ x(1− y2) = 0 −→ x = 0 or y =±1.But if x = 0, then fx = 0 and so there are no critical points with x = 0. That is, all critical points arecombinations of x =±1 and y =±1:

(1,1), (1,−1), (−1,1), (−1,−1).

From the figure 12.7, we observe that at (1,1), f (x,y) has the local maximum value f (1,1) = e1/6, while at(−1,1), f (x,y) has the local minimum value f (−1,1) =−e1/6.What can we say at (1,−1) and (−1,−1)? See the figure 12.8.(1) At (1,−1): Observe the intersection of the graph of f (x,y) and the plane x = 1. The intersection is a curvefrom the point A to the point B on the plane x = 1. The curve has the local minimum (red dot) at y = −1.That is, along the curve connecting A and B on the plane x = 1, the point (1,−1) gives the local minimum.But if we consider the intersection of the graph of f (x,y) and the plane y = −1, then the intersection is acurve from the point C to the point D on the plane y = −1. The curve has the local maximum (red dot) atx = 1. That is, along the curve connecting C and D on the plane y = −1, the point (1,−1) gives the localmaximum.Thus we can say that the local extrema at (1,−1) depends on the curves (one giving the local maximum at(1,−1) and one giving the local minimum at (1,−1)). In this case, we have the local maximum and the localminimum at (1,−1) simultaneously and the point (1,−1) is called the saddle point.(2) At (−1,−1): We have the exactly same feature as at (1,−1). That is, (−1,−1) is also a saddle point. □

Figure 12.8: Saddle Point S, Local Maxima A and B, Local Minima C and D

The formal definition of the saddle point is given as follows:

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Definition 12.7.5. The point P(a,b, f (a,b)) is a saddle point of z = f (x,y) if the following conditions aresatisfied:

1. (a,b) is a critical point of f ,2. every open disk centered at (a,b) contains

(1) points (x,y) in the domain of f for which f (x,y)< f (a,b) and also(2) points (x,y) in the domain of f for which f (x,y)> f (a,b).

How to find the saddle point? How to determine the local extrema? Above, we explained by using the graph.However, in general, it is not easy to find such important points by using the graph. Just as we developedand used the First Derivative Test and the Second Derivative Test on the function of a single variable inCalculus I, we introduce the Second Derivatives Test on the function of two variables as follows:

Theorem 12.7.6 (SECOND DERIVATIVES TEST). Suppose that f (x,y) has continuous second–order partialderivatives in some open disk containing the point (a,b) and that fx(a,b) = fy(a,b) = 0. Define the discrim-inant D for the point (a,b) by

D(a,b) = fxx(a,b) fyy(a,b)− [ fxy(a,b)]2 .

1. If D(a,b)> 0 and fxx(a,b)> 0, then f has a local minimum at (a,b).2. If D(a,b)> 0 and fxx(a,b)< 0, then f has a local maximum at (a,b).3. If D(a,b)< 0, then f has a saddle point at (a,b).4. If D(a,b) = 0, then no conclusion can be drawn.

How to understand the Theorem?Part I. In order to have D(a,b)> 0, we should have fxx(a,b) fyy(a,b)> 0. It occurs only when(1) fxx(a,b)> 0 and fyy(a,b)> 0 or (2) fxx(a,b)< 0 and fyy(a,b)< 0.� Case (1) fxx(a,b) > 0 and fyy(a,b) > 0: In this case, we observe that the surface z = f (x,y) is concaveupward in the plane y= b (from fxx(a,b)> 0) and also concave upward in the plane x= a (from fyy(a,b)> 0).Hence, the surface will look like an upward–opening paraboloid near the point (a,b) and thus f (x,y) has thelocal minimum at (a,b).� Case (2) fxx(a,b) < 0 and fyy(a,b) < 0: In this case, we observe that the surface z = f (x,y) is concavedownward in the plane y = b (from fxx(a,b) < 0) and also concave downward in the plane x = a (fromfyy(a,b) < 0). Hence, the surface will look like an downward–opening paraboloid near the point (a,b) andthus f (x,y) has the local maximum at (a,b).Part II. In order to have D(a,b)< 0, fxx(a,b) and fyy(a,b) should have the opposite signs (one positive andone negative). It implies that f (x,y) has the local minimum and the local maximum at (a,b) simultaneously.Hence, f (x,y) has the saddle point at (a,b).Be careful! Having fxx(a,b) > 0 and fyy(a,b) > 0 without having D(a,b) > 0 does not say that f (a,b) is alocal minimum.

Example 12.7.7. Locate and classify all critical points for f (x,y) = xye−x2/2−y2/2.

ANSWER. Part I. Critical Points.

fx = y(1− x2)e−x2/2−y2/2, fy = x

(1− y2)e−x2/2−y2/2.

Solving fx = 0, we have x = ±1 or y = 0. Solving fy = 0, we have x = 0 or y = ±1. So, both fx = 0 andfy = 0 are satisfied simultaneously at the points, (±1,±1) and (0,0), which are critical points.Part II. Classification of Critical Points.

fxx = (−3+ x2) f (x,y) = xy(−3+ x2)e−x2/2−y2/2,

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fyy = (−3+ y2) f (x,y) = xy(−3+ y2)e−x2/2−y2/2,

fxy =(1− x2)(1− y2)

xyf (x,y) = (1− x2)(1− y2)e−x2/2−y2/2,

D(x,y) = fxx(x,y) fyy(x,y)− f 2xy(x,y) =

[−(y2 +1

)x4 +

(−y4 +5y2 +2

)x2 −

(y2 −1

)2]

e−x2−y2.

By the Theorem 12.7.6, we deduce

fxx(0,0) = 0 = fyy(0,0), D(0,0) =−1 < 0 −→ saddle point at (0,0)

fxx(1,1) =−2e−1 = fyy(1,−1)< 0, D(1,1) = 4e−2 > 0 −→ local maximum at (1,1)

fxx(1,−1) = 2e−1 = fyy(1,−1)> 0, D(1,−1) = 4e−2 > 0 −→ local minimum at (1,−1)

fxx(−1,1) = 2e−1 = fyy(−1,1)> 0, D(−1,1) = 4e−2 > 0 −→ local minimum at (−1,1)

fxx(−1,−1) =−2e−1 = fyy(−1,−1)< 0, D(−1,−1) = 4e−2 > 0 −→ local maximum at (−1,−1)

That is, f (x,y) has the local maximum at (1,1) and (−1,−1) and the local minimum at (1,−1) and (−1,1)and the saddle point at (0,0). See the figure 12.9. □

Figure 12.9: Black Dot: Saddle Point at (0,0), Blue Dots at Peaks: Local Maximum at (1,1) and (−1,−1),Red Dots at Valleys: Local Minimum at (1,−1) and (−1,1)

Example 12.7.8. Locate and classify all critical points for f (x,y) = 2x2 − y3 −2xy.


fx = 2(2x− y) = 0, fy =−(3y2 +2x) = 0.

From the first equation, we get y = 2x. Putting this into the second equation, we get

3(2x)2 +2x = 0 −→ x = 0 or x =−16.

So we have two critical points (0,0) and(−1

6 ,−13

).

Part II. Classification of Critical Points.

fxx(x,y) = 4 > 0, fyy(x,y) =−6y, fxy(x,y) =−2,

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D(x,y) = fxx(x,y) fyy(x,y)− f 2xy(x,y) =−4(6y+1).

(1) At (0,0), we have D(0,0) =−4 < 0. By the Second Derivatives Test, f (x,y) has the saddle point at (0,0).(2) At

(−1

6 ,−13

), we have D

(−1

6 ,−13

)= 4 > 0 and fxx

(−1

6 ,−13

)= 4 > 0. Thus, by the Second Derivatives

Test, f (x,y) has this local minimum f (0,0) =− 154 at (0,0). See the figure 12.10. □

Figure 12.10: Black Dot: Saddle Point at (0,0,0), Red Dot: Local Minimum at (−1/6,−1/3,−1/54)

Example 12.7.9. Locate and classify all critical points for f (x,y) = x3 −2y2 −2y4 +3x2y.


fx(x,y) = 3x(x+2y) = 0, fy(x,y) =−4y−8y3 +3x2 = 0.

From the first equation, we have x = 0 or x =−2y. Putting these to the second equation, we get

0 =−4y−8y3 =−4y(2y2 +1), y = 0,

0 =−4y−8y3 +3(−2y)2 =−4y(1+2y2 −3y) =−4y(2y−1)(y−1), y =12

or y = 1.

Putting y = 1/2 and y = 1 back to x =−2y, we get x =−1 and x =−2, respectively. So we have the criticalpoints (0,0), (−1,1/2) and (−2,1).Part II. Classification of Critical Points.

fxx(x,y) = 6(x+ y), fyy(x,y) =−4(6y2 +1), fxy(x,y) = 6x,

D(x,y) = fxx(x,y) fyy(x,y)− f 2xy(x,y) =−36x2 −24(x+ y)(6y2 +1).

(1) At (−1,1/2): We have D(−1,1/2) = −6 < 0. By the Second Derivatives Test, f (x,y) has the saddlepoint at (−1,1/2).(2) At (−2,1): We have D(−2,1) = 24 > 0 and fxx(−2,1) =−6 < 0. By the Second Derivatives Test, f (x,y)has the local maximum at (−2,1).(3) At (0,0): We have D(0,0) = 0. So the Second Derivatives Test cannot give us any information. However,we can deduce some information from the function f (x,y) = x3−2y2−2y4+3x2y. When y = 0, the functionbecomes f (x,0) = x3, which has an inflection point at x = 0. It shows that there is no local extrema at thispoint (0,0). Hence, the critical point (0,0) has no significant graphical feature.As you can see from the figure 12.11, it is not easy to tell the local extrema and saddle point from the graph.So in this case, it’s good to use the analysis, i.e., the Second Derivatives Test. □

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Figure 12.11: Black Dot: Critical Point at (0,0), Red Dot: Saddle Point at (−1,1/2), Blue Dot: LocalMaximum at (−2,1)

□ 12.7.2 Linear Regression: Method of Least Squares.Skip. Please read the textbook.

□ 12.7.3 Steepest Ascent Algorithm: Estimating Local Extrema without Critical Points.Skip. Please read the textbook.

□ 12.7.4 Absolute Extrema.

Definition 12.7.10. We call f (a,b) the absolute maximum of f on the region R if f (a,b) ≥ f (x,y) for all(x,y) ∈ R. Similarly, f (a,b) is called the absolute minimum of f on the region R if f (a,b)≤ f (x,y) for all(x,y) ∈ R. In either case, f (a,b) is called an absolute extremum of f .

Recall from Calculus I that for a function of a single variable, if f is continuous on the closed interval [a,b],then f has the maximum and minimum value on [a,b] and the absolute extrema must occur at either criticalnumbers of f or at the endpoints of the interval [a,b]. We can extend this argument to the function of twovariables. We say that a region R ⊂ R2 is bounded if there is a disk that completely contains R.

Theorem 12.7.11 (EXTREME VALUE THEOREM). Suppose that f (x,y) is continuous on the closed andbounded region R ⊂ R2. Then f has both an absolute maximum and an absolute minimum on R. Further, anabsolute extremum may only occur at a critical point in R or at a point on the boundary of R.

If f (a,b) is an absolute extremum of f in R and (a,b) is an interior point of R, then (a,b) is also a localextremum of f , in which case, (a,b) must be a critical point. This implies that all of the absolute extremaof a function in region R occur either at critical points or on the boundary of the region. For this reason, inorder to find the absolute extrema on R , we look for (i) the local extrema inside R and (ii) the extrema on theboundary of R and compare all the extrema.We take the following steps:

Step 1. Critical Points: Find all critical points of f in the region R.Step 2. Extrema on Boundary of R: Find the maximum and minimum of f on the boundary of R.Step 3. Comparison: Compare the values of f at the critical points with the maximum and minimum values

of f on the boundary of R.

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Example 12.7.12. Find the absolute extrema of f (x,y) = 5+4x−2x2 +3y− y2 on the region R bounded bythe lines y = 2, y = x and y =−x.

ANSWER. Step 1. Critical Points.

fx = 4(1− x), fy = 3−2y,

which give x = 1 and y = 3/2. That is, we have only one critical point (1,3/2) which is inside the region Rwith the value f (1,3/2) = 9.25.Step 2. Extrema on Boundary of R.(1) On the boundary y= x on [0,2], the function becomes f (x,x) =−3x2+7x+5. Let g(x) = f (x,x). We findthe absolute extrema of g(x) on [0,2]. By Calculus I, g(x) has the absolute maximum g(7/6) = 9.08333 atx = 7/6 and the absolute minimum g(0) = 5 at x = 5. That is, only on the line y = x, f (x,y) has the absolutemaximum f (7/6,7/6) = 9.08333 and the absolute minimum f (0,0) = 5.(2) On the boundary y =−x on [−2,0], the function becomes f (x,−x) =−3x2 + x+5. Let h(x) = f (x,−x).We find the absolute extrema of h(x) on [−2,0]. By Calculus I, h(x) has the absolute maximum h(0) = 5 atx = 0. (Be careful! Even though h′(x) = 0 at x = 1/6, since 1/6 is not in the domain [−2,0], so it cannot bethe critical number.) h(x) has the absolute minimum h(−2) =−9 at x =−2. That is, only on the line y =−x,f (x,y) has the absolute maximum f (0,0) = 5 and the absolute minimum f (−2,2) =−9.(3) On the boundary y = 2 on [−2,2], the function becomes f (x,2) =−2x2 +4x+7. Let k(x) = f (x,2). Wefind the absolute extrema of k(x) on [−2,2]. By Calculus I, k(x) has the absolute maximum k(1) = 9 at x = 1and the absolute minimum k(−2) = −9 at x = −2. That is, only on the line y = 2, f (x,y) has the absolutemaximum f (1,2) = 9 and the absolute minimum f (−2,2) =−9.Step 3. Comparison. We collect all the values found above and compare them.

(1,3/2) (0,0) (7/6,7/6) (−2,2) (1,2)

f (x,y) 9.25 5 9.08333 −9 9

Based on the table, we conclude that on the region R, f (x,y) has the absolute maximum f (1,3/2) = 9.25 atthe critical point (1,3/2) inside the region R and the absolute minimum f (−2,2) =−9 at the boundary point(−2,2) on the boundary of R. See the figure 12.12. □

Figure 12.12: Left–Hand Side Figure: Region R with Black Dot (1,3/2) and Red Dot (−2,2), Right–HandSide Figure: Graph of f (x,y) on Region R with Absolute Maximum (Black Dot (1,3/2,9.25)) and AbsoluteMinimum (Red Dot (−2,2,−9)).

§12.8 Constrained Optimization and Lagrange Multipliers.Skip. Please read the textbook.

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Chapter 13

Multiple Integrals

§13.1 Double Integrals.

□ 13.1.1 Introduction.From Calculus I, we recall that the integral of f (x) over an interval I defined by I = { x ∈ R : a ≤ x ≤ b }is given by

ˆ b

af (x)dx or

ˆ x=b

x=af (x)dx or

ˆI

f (x)dx orˆ[a,b]

f (x)dx.

For a function f (x,y) of two variables x and y, we can consider a double integral of f (x,y). A double integralof f (x,y) over a region R ⊂ R2 is written by

¨R

f (x,y)dA.

□ 13.1.2 Double Integral over a Rectangle.

Theorem 13.1.1 (FUBINI’S THEOREM). Suppose that f (x,y) is integrable over the rectangle R defined by

R ={(x,y) ∈ R2 : a ≤ x ≤ b, c ≤ y ≤ d

}Then we can write the double integral of f (x,y) over R as either of the iterated integrals:

¨R

f (x,y)dA =

ˆ x=b

x=a

[ˆ y=d

y=cf (x,y)dy

]dx =

ˆ x=b

x=a

ˆ y=d

y=cf (x,y)dydx =

ˆ b

a

ˆ d

cf (x,y)dydx

or¨

Rf (x,y)dA =

ˆ y=d

y=c

[ˆ x=b

x=af (x,y)dx

]dy =

ˆ y=d

y=c

ˆ x=b

x=af (x,y)dxdy =

ˆ d

c

ˆ b

af (x,y)dxdy

Example 13.1.2. Evaluate˜

R(6x2 +4xy3)dA, where R = { (x,y) : 0 ≤ x ≤ 2, 1 ≤ y ≤ 4 }.

ANSWER. By Fubini’s Theorem 13.1.1, we have

¨R

f (x,y)dA =

ˆ x=2

x=0

[ˆ y=4

y=1(6x2 +4xy3)dy

]dx =

ˆ x=2

x=0

[6x2y+ xy4]y=4

y=1 dx

=

ˆ x=2

x=0(18x2 +255x)dx =

[6x3 +

2552

x2]x=2

x=0= 558.

or¨

Rf (x,y)dA =

ˆ y=4

y=1

[ˆ x=2

x=0(6x2 +4xy3)dx

]dy =

ˆ y=4

y=1

[2x3 +2x2y3]x=2

x=0 dy

=

ˆ y=4

y=1(16+8y3)dy =

[16y+2y4]y=4

y=1 = 558. □

Exercise 13.1.3. Evaluate˜

R(x+ y)dA, where R = { (x,y) : 1 ≤ x ≤ 2, 3 ≤ y ≤ 4 }.

83


□ 13.1.3 Geometrical Application.From Calculus I, we recall that the integral of y = f (x) = 5 over the interval

I = { x : 2 ≤ x ≤ 9 }

represents the area of the region under the curve y = f (x) = 5. We observe that the region is a rectangle withthe base 9−2 = 7 and the height 5. That is,

ˆ 9

25dx = 7×5 = 35,

which can be checked by computing the definite integral in the left–hand side.By the same argument, we observe that the integral of z = f (x,y) = 7 over a rectangle

R = { (x,y) : 2 ≤ x ≤ 5, 9 ≤ y ≤ 15 }

represents the volume of the box formed by the plane z = f (x,y) = 7 over R. We observe that the box has thelength 5−2 = 3, width 15−9 = 6 and height 7. That is,

¨R

7dA = 3×6×7 = 126,

which can be checked by computing the double integral in the left–hand side. We will discuss more applica-tions in later Sections.

□ 13.1.4 Double Integral over a General Region.From this topic, the order of integration is important.

Figure 13.1: Region bounded by y = g1(x) and y = g2(x)

Theorem 13.1.4 (TYPE I. REGION BOUNDED BY y = g1(x) AND y = g2(x)). Suppose f (x,y) is continuouson the region R defined by

R = { (x,y) : a ≤ x ≤ b, g1(x)≤ y ≤ g2(x) }

for continuous functions g1 and g2, where g1(x)≤ g2(x) for all x ∈ [a,b]. Then

¨R

f (x,y)dA =

ˆ x=b

x=a

[ˆ y=g2(x)

y=g1(x)f (x,y)dy

]dx =

ˆ b

a

ˆ g2(x)

g1(x)f (x,y)dydx

See the figure 13.1. Be careful! When we change the order of the integration, the integral becomes different.

Number =¨

Rf (x,y)dA =

ˆ g2(x)

g1(x)

ˆ b

af (x,y)dxdy = Function of x

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Figure 13.2: Region bounded by y = x, y = 0, and x = 4


R(4ex2 − 5siny)dA, where R is the region bounded by graphs of y = x, y = 0,and x = 4.

ANSWER. See the figure 13.2. The region R is represented as follows:

R = { (x,y) : 0 ≤ x ≤ 4, 0 ≤ y ≤ x }.

By the Theorem 13.1.4, we deduce

¨R(4ex2

−5siny)dA =

ˆ 4

0

ˆ x

0(4ex2

−5siny)dydx =ˆ 4

0

[4ex2

y+5cosy]x

0dx

=

ˆ 4

0

(4ex2

x+5cosx−5)

dx = 2e16 +5sin4−22. □


R(3x2 +2y)dA, where R is the region bounded by graphs of y = 1− x2, y = 0.

Figure 13.3: Region bounded by x = h1(y) and x = h2(y)

Theorem 13.1.7 (TYPE II. REGION BOUNDED BY x = h1(y) AND x = h2(y)). Suppose f (x,y) is continuouson the region R defined by

R = { (x,y) : c ≤ y ≤ d, h1(y)≤ x ≤ h2(y) }

for continuous functions h1 and h2, where h1(y)≤ h2(y) for all y ∈ [c,d]. Then

¨R

f (x,y)dA =

ˆ y=d

y=c

[ˆ x=h2(y)

x=h1(y)f (x,y)dx

]dy =

ˆ d

c

ˆ h2(y)

h1(y)f (x,y)dxdy


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Figure 13.4: Region bounded by x = y2 and x = 2− y

Example 13.1.8. Let R be the region bounded by graphs of x = y2 and x = 2− y. Write˜

R f (x,y)dA as aniterated integral.

ANSWER. See the figure 13.4. The region R is represented as follows:

R = { (x,y) : −2 ≤ y ≤ 1, y2 ≤ x ≤ 2− y }.

By the Theorem 13.1.7, we deduce¨

Rf (x,y)dA =

ˆ 1

−2

ˆ 2−y

y2f (x,y)dxdy. □

Exercise 13.1.9. Find an integral equal to the volume of the solid bounded by the surfaces and evaluate theintegral:

z = 6− x− y, z = 0, x = 4− y2, x = 0.

□ 13.1.5 Changing Order of Integration.Sometimes we need to change the order of integration in order to evaluate a double integral.

Example 13.1.10. Evaluate the iterated integral´ 1

0

´ 1y ex2

dxdy.

Figure 13.5: Region R

ANSWER. See the figure 13.5. The region R in the double integral is interpreted as follows:

R = { (x,y) : 0 ≤ y ≤ 1, y ≤ x ≤ 1 }.

However, we can interpret the region R in the other way:

R = { (x,y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ x }.

Using this interpretation, the double integral becomesˆ 1

0

ˆ 1

yex2

dxdy =¨

Rex2

dA =

ˆ 1

0

ˆ x

0ex2

dydx =ˆ 1

0

[ex2

y]x

0dx =

ˆ 1

0ex2

xdx =12(e−1) . □

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Figure 13.6: Region R

Example 13.1.11. Change the order of integration:´ 1

0

´ √y0 f (x,y)dxdy.

ANSWER. See the figure 13.6. The region R in the double integral is interpreted as follows:

R = { (x,y) : 0 ≤ y ≤ 1, 0 ≤ x ≤√y }.

However, we can interpret the region R in the other way:

R = { (x,y) : 0 ≤ x ≤ 1, x2 ≤ y ≤ 1 }.

Using this interpretation, the double integral becomesˆ 1

0

ˆ √y

0f (x,y)dxdy =

¨R

f (x,y)dA =

ˆ 1

0

ˆ 1

x2f (x,y)dydx. □

Exercise 13.1.12. Change the order of integration:´ 2

1

´ lny0 f (x,y)dxdy.

□ 13.1.6 Linearity of Double Integral.We recall from Calculus Iˆ b

a(c f (x)+dg(x)) dx = c

ˆ b

af (x)dx+d

ˆ b

ag(x)dx,

ˆ b

af (x)dx =

ˆ p

af (x)dx+

ˆ b

pf (x)dx,

where a≤ p≤ b and c and d are any constants. Similarly, we have the following theorem for double integrals.

Figure 13.7: R = R1 ∪R2

Theorem 13.1.13 (LINEARITY OF DOUBLE INTEGRALS). Let f (x,y) and g(x,y) be integrable over theregion R ⊂ R2 and let c and d be any constant. Then, the following hold:

1.˜

R (c f (x,y)+dg(x,y)) dA = c˜

R f (x,y)dA+d˜

R g(x,y)dA2. If R = R1 ∪R2, where R1 and R2 are non–overlapping regions, then˜

R f (x,y)dA =˜

R1f (x,y)dA+

˜R2

f (x,y)dA


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§13.2 Area, Volume, and Center of Mass.

□ 13.2.1 Area.The integral

´ ba 1dx = b−a gives the length of the interval [a,b]. The double integral

˜R 1dA gives the area

of the region R.

Example 13.2.1. Find an area of the region R bounded by y = x2 and y = x+2.

. .x

.y

.−1 .2

.2

.4

.y = x2

.y = x+2

Figure 13.8: Region R bounded by y = x2 and y = x+2

ANSWER. The region R can be represented as follows:

R = { (x,y) : −1 ≤ x ≤ 2, x2 ≤ y ≤ x+2 }.

Hence, the area of R is obtained as follows:

A =

¨R

1dA =

ˆ 2

−1

ˆ x+2

x21dydx =

ˆ 2

−1(x+2− x2)dx =

92.

We note from Calculus I that the last integral represents the area of the region between two curves. That is,the area of R can be obtained as follows:

A =

ˆ 2

−1(upper curve− lower curve) dx =

ˆ 2

−1

(x+2− x2) dx =

92. □

Exercise 13.2.2. Use a double integral to compute the area of the region bounded by the curves y = 3x,y = 5−2x, and y = 0.

□ 13.2.2 Volume.We recall from Calculus I that the area under the curve y = f (x) over the interval [a,b] is obtained by theintegral

´ ba f (x)dx. By the same argument, we observe that the volume of the solid under the surface z =

f (x,y) over the region R is obtained by the double integral˜

R f (x,y)dA.

Example 13.2.3. Compute the volume of the solid bounded by x+2y+3z = 6 and three coordinate plates.

ANSWER 1. Step 1. Sketch and Setup Double Integral. The equation x+ 2y+ 3z = 6 represents a planepassing through (6,0,0), (0,3,0) and (0,0,2). The plane and the three coordinates plates, i.e., x = 0, y = 0and z = 0, forms a solid given in the figure 13.9. The solid is called a tetrahedron. We observe from thegiven equation

x+2y+3z = 6, i.e., z = 2− x3− 2y

3.

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. .x

.y

.6

.3

.O

.R

.x+2y = 6 with z = 0

Figure 13.9: Tetrahedron bounded by x+2y+3z = 6 and three coordinate plates and Region R

So the volume of the solid is obtained by the double integral

V =

¨R

(2− x

3− 2y

3

)dA,

where R is the region of the solid in the xy-plane.Step 2. Setup and Compute Iterated Integral. To express the double integral into the iterated integral, weneed to express the region R in the xy–plane in terms of x and y. When the plane x+ 2y+ 3z = 6 meets thexy–plane, the intersection becomes a line of which equation is x+2y = 6. So the region R in the xy–plane isthe region bounded by x+2y = 6, x = 0 and y = 0. That is,

R = { (x,y) : 0 ≤ x, 0 ≤ y, x+2y ≤ 6 }={(x,y) : 0 ≤ x ≤ 6, 0 ≤ y ≤ 6− x

2

}or R = { (x,y) : 0 ≤ x, 0 ≤ y, x+2y ≤ 6 }= { (x,y) : 0 ≤ y ≤ 3, 0 ≤ x ≤ 6−2y } .

Therefore, we deduce the following iterated integral from the double integral above:

V =

¨R

(2− x

3− 2y

3

)dA =

ˆ x=6

x=0

ˆ y= 6−x2

y=0

(2− x

3− 2y

3

)dydx =

ˆ x=6

x=0

(x2

12− x+3

)dx = 6

or V =

¨R

(2− x

3− 2y

3

)dA =

ˆ y=3

y=0

ˆ x=6−2y

x=0

(2− x

3− 2y

3

)dxdy =

ˆ y=3

y=0

(2y2

3−4y+6

)dy = 6 □

Remark 13.2.4. There are two more ways to solve the example above. In the solution above, we used theregion R in the xy–plane in the double integral. However, we CAN use the region P in the yz–plane or theregion Q in the xz–plane. Let us solve the example using the region P in the yz–plane and the region Q in thexz–plane.

ANSWER 2. USING REGION P IN yz–PLANE. When we use the region in the yz–plane, we should take thefunction of two variables y and z as the integrand of the double integral. That is,

x+2y+3z = 6, i.e., x = 6−2y−3z.


V =

¨P(6−2y−3z) dA,

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. .y

.z

.3

.2

.P

.2y+3z = 6 with x = 0

. .x

.z

.6

.2

.Q

.x+3z = 6 with y = 0

Figure 13.10: Regions P and Q

where P is the region of the solid in the yz–plane.By setting x = 0 in the given equation x+2y+3z = 6, we can obtain the line (boundary of the tetrahedron inthe yz–plane). So the region P in the yz–plane can be expressed as follows (See the figure 13.10.):

P = { (y,z) : 0 ≤ y, 0 ≤ z, 2y+3z ≤ 6 }={(y,z) : 0 ≤ y ≤ 3, 0 ≤ z ≤ 6−2y

3

}or P = { (y,z) : 0 ≤ y, 0 ≤ z, 2y+3z ≤ 6 }=

{(y,z) : 0 ≤ z ≤ 2, 0 ≤ y ≤ 6−3z

2

}.


V =

¨P(6−2y−3z) dA =

ˆ y=3

y=0

ˆ z= 6−2y3

z=0(6−2y−3z) dzdy =

ˆ y=3

y=0

(2y2

3−4y+6

)dy = 6

or V =

¨P(6−2y−3z) dA =

ˆ z=2

z=0

ˆ y= 6−3z2

y=0(6−2y−3z) dydz =

ˆ z=2

z=0

(9z2

4−9z+9

)dz = 6 □

ANSWER 3. USING REGION Q IN xz–PLANE. When we use the region in the xz–plane, we should take thefunction of two variables x and z as the integrand of the double integral. That is,

x+2y+3z = 6, i.e., y =6−3z− x

2.


V =

¨Q

6−3z− x2

dA,

where Q is the region of the solid in the xz–plane.By setting y = 0 in the given equation x+2y+3z = 6, we can obtain the line (boundary of the tetrahedron inthe xz–plane). So the region Q in the xz–plane can be expressed as follows (See the figure 13.10.):

Q = { (x,z) : 0 ≤ x, 0 ≤ z, x+3z ≤ 6 }={

(x,z) : 0 ≤ x ≤ 6, 0 ≤ z ≤ 6− x3

}or Q = { (x,z) : 0 ≤ x, 0 ≤ z, x+3z ≤ 6 }= { (x,z) : 0 ≤ z ≤ 2, 0 ≤ x ≤ 6−3z } .


V =

¨Q

6−3z− x2

dA =

ˆ x=6

x=0

ˆ z= 6−x3

z=0

6−3z− x2

dzdx =ˆ x=6

x=0

(3− x+

x2

12

)dx = 6

or V =

¨Q

6−3z− x2

dA =

ˆ z=2

z=0

ˆ x=6−3z

x=0

6−3z− x2

dxdz =ˆ z=2

z=0

(9z2

4−9z+9

)dz = 6 □

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As we can see, for the case of the volume of the tetrahedron, there are total 6 ways (3 ways depending on theregion and 2 ways for each region) to solve the problem.

Exercise 13.2.5. � Find the volume of the tetrahedron bounded by the plane 2x+ y+ z = 2 and the threecoordinates planes.� Find the volume of the solid lying in the first octant and bounded by the graphs of z = 4− x2, x+ y = 2,x = 0, y = 0 and z = 0.� Find the volume of the solid bounded by the graphs of z = 2, z = x2 +1, y = 0 and x+ y = 2.

□ 13.2.3 Moments and Center of Mass.We discuss a physical application of double integrals. Consider a thin, flat plate (a lamina) in the shape ofthe region R ⊂R2. From an engineering standpoint, it is often important to determine where you could placea support to balance the plate. We call this point the center of mass of the lamina. The center of mass of alamina is a point whose x– and y–component are called the moments with respect to the x–axis and y–axis,respectively. Let us introduce some definitions.

Definition 13.2.6. For a lamina with density function ρ(x,y) in the shape of the region R ⊂ R2,

1. the total mass m is obtained by

m =

¨R

ρ(x,y)dA

2. the moments, Mx and My, with respect to the x– and y–axis are, respectively,

Mx =

¨R

yρ(x,y)dA, My =

¨R

xρ(x,y)dA

3. the center of mass is the point (x, y), where

x =My

m=

˜R xρ(x,y)dA˜R ρ(x,y)dA

, y =Mx

m=

˜R yρ(x,y)dA˜R ρ(x,y)dA

Example 13.2.7. Find the mass and center of mass of a triangular lamina with vertices (0,0), (1,0), and(0,2) if the density function is ρ(x,y) = 1+3x+ y.

. .x

.y

.1

.2

.R

.2x+ y = 2

Figure 13.11: Triangular Lamina R

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ANSWER. The equation of the line connecting (1,0) and (0,2) is y = 2−2x. So the triangular lamina in thefigure 13.11 can be represented by the region R as follows:

R = { (x,y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ 2−2x }={

(x,y) : 0 ≤ y ≤ 2, 0 ≤ x ≤ 2− y2

}Let us use the first representation.(1) Total Mass m:

m =

¨R

ρ(x,y)dA =

ˆ x=1

x=0

ˆ y=2−2x

y=0(1+3x+ y) dydx = 4

ˆ x=1

x=0

(1− x2) dx =

83

(2) Center of Mass, (x, y):

x =1m

¨R

xρ(x,y)dA =38

ˆ x=1

x=0

ˆ y=2−2x

y=0x(1+3x+ y) dydx =

32

ˆ x=1

x=0

(x− x3) dx =

38

y =1m

¨R

yρ(x,y)dA =38

ˆ x=1

x=0

ˆ y=2−2x

y=0y(1+3x+ y) dydx =

14

ˆ x=1

x=0

(7−9x−3x2 +5x3) dx =

1116

Thus, the total mass is m = 83 and the center of mass is (x, y) =

(38 ,

1116

). □

Exercise 13.2.8. Find the mass and the center of mass of the lamina bounded by y = x3 and y = x2 with thedensity ρ(x,y) = 4.

□ 13.2.4 Moments of Inertia.The moment of inertia (also called the second moment) of a particle of mass m about an axis is defined tobe mr2, where r is the distance from the particle to the axis. We extend this concept to a lamina with densityfunction ρ(x,y) and occupying a region R by proceeding as we did for ordinary moments.

Definition 13.2.9. The moments of inertia, Ix and Iy, of the lamina about the x–axis and y–axis are, respec-tively,

Ix =

¨R

y2ρ(x,y)dA, Iy =

¨R

x2ρ(x,y)dA.

Example 13.2.10. Find the moments of inertia Ix and Iy for the lamina in the previous example: the triangularlamina with vertices (0,0), (1,0), and (0,2) and the density function is ρ(x,y) = 1+3x+ y.

ANSWER.

Iy =

¨R

x2ρ(x,y)dA =

ˆ x=1

x=0

ˆ y=2−2x

y=0x2 (1+3x+ y) dydx = 4

ˆ x=1

x=0

(x2 − x4) dx =

815

≈ 0.533

Ix =

¨R

y2ρ(x,y)dA =

ˆ x=1

x=0

ˆ y=2−2x

y=0y2 (1+3x+ y) dydx

=

ˆ x=1

x=0

(203−16x+8x2 +

16x3

3−4x4

)dx =

2815

≈ 1.8667

A comparison of the two moments of inertia shows that it is much more difficult to rotate the triangularlamina about the x–axis than about the y–axis. □

Exercise 13.2.11. Find the moments of inertial Ix and Iy for the lamina in the Exercise 13.2.8.

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§13.3 Double Integrals in Polar Coordinates.

□ 13.3.1 Polar Coordinates.We refresh our memory on the polar coordinates from the Section 9.4 Polar Coordinates.

Figure 13.12: Rectangular and Polar coordinates

Theorem 13.3.1 (CONNECTION BETWEEN POLAR AND CARTESIAN COORDINATES). Suppose a point Phas Cartesian coordinates (x,y) and polar coordinates (r,θ). Then we have

x = r cosθ , and y = r sinθ ⇐⇒ x2 + y2 = r2, and tanθ =yx.

□ 13.3.2 Double Integrals in Polar Coordinates.

Figure 13.13: Region R in polar coordinates

Theorem 13.3.2 (FUBINI’S THEOREM). Suppose that f (r,θ) is continuous on the region R given by

R = { (r,θ) : α ≤ θ ≤ β , g1(θ)≤ r ≤ g2(θ) } ,

where 0 ≤ g1(θ)≤ g2(θ) for all θ ∈ [α ,β ]. See the figure 13.13. Then,

¨R

f (r,θ)dA =

ˆ θ=β

θ=α

ˆ r=g2(θ)

r=g1(θ)f (r,θ)r dr dθ .

Example 13.3.3. Find the area of the region bounded by the curve r = 3+2sinθ .

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Figure 13.14: Region bounded by r = 3+2sinθ

ANSWER. For each fixed θ , r ranges from 0 to 3+2sinθ . To go all the way around the cardioid (graph ofr = 3+2sinθ : see the figure 13.14), exactly once, θ ranges from 0 to 2π . By Fubini’s Theorem 13.3.2, wehave

A =

¨R

dA =

ˆ 2π

0

ˆ 3+2sinθ

0r dr dθ =

ˆ 2π

0

[r2

2

]3+2sinθ

0dθ =

12

ˆ 2π

0(3+2sinθ)2 dθ = 11π. □

Exercise 13.3.4. Find the area of the region bounded by the curve r = 2−2cosθ .


R

√x2 + y2 dA, where R is the disk x2 + y2 ≤ 9.

Figure 13.15: Solid under z =√

x2 + y2 over x2 + y2 ≤ 9

ANSWER. See the figure 13.15. The disk R can be represented in polar coordinates as follows:

R = { (r,θ) : 0 ≤ r ≤ 3, 0 ≤ θ ≤ 2π } .

Using Fubini’s Theorem 13.3.2 with the representation, we have

¨R

√x2 + y2 dA =

ˆ θ=2π

θ=0

ˆ r=3

r=0

√r2 r dr dθ =

ˆ θ=2π

θ=0

ˆ r=3

r=0r2 dr dθ

=

ˆ θ=2π

θ=0

[r3

3

]r=3

r=0dθ =

ˆ θ=2π

θ=09dθ = 19π. □

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Figure 13.16: Two Cylinders


R e−x2−y2dA, where R is the disk x2 + y2 ≤ 4.

Example 13.3.7. Find the volume of the solid bounded by z = 7, z = 0, x2 + y2 = 25, and x2 + y2 = 4.

ANSWER 1 BASIC FORMULA. We observe that the solid is between two cylinders. See the figure 13.16.The large cylinder formed by x2 + y2 = 25, z = 0 and z = 7 has the volume

Vlarge = (Base Area) × (Height) = π(52)(7) = 175π.

The small cylinder formed by x2 + y2 = 4, z = 0 and z = 7 has the volume Vsmall = π(22)(7) = 28π . Thus,we get the volume of the solid,

V =Vlarge −Vsmall = 7π(52 −22)= 147π. □

ANSWER 2 RECTANGULAR COORDINATES. When we cut the solid horizontally, the cut sides have the ex-actly same shape with same area, which is the annulus R 1. The region R is expressed as

R ={(x,y) : 4 ≤ x2 + y2 ≤ 25

}={(x,y) : −2 ≤ x ≤ 2,

√4− x2 ≤ y ≤

√25− x2

}∪{(x,y) : −2 ≤ x ≤ 2, −

√25− x2 ≤ y ≤−

√4− x2

}= R1 ∪R2,

where R1 and R2 represent the first and second region, respectively. This separation implies¨R

f (x,y)dA =

¨R1

f (x,y)dA+

¨R2

f (x,y)dA.

From the figure 13.16, we observe that the height of the solid moves from z = 0 to z = 7. Hence, the volumeof the solid is obtained by

V =

¨R(7−0) dA =

¨R1

7dA+

¨R2

7dA =

ˆ x=2

x=−2

ˆ y=√

25−x2

y=√

4−x27dydx+

ˆ x=2

x=−2

ˆ y=−√

4−x2

y=−√

25−x27dydx

= 7ˆ x=2

x=−2

[√25− x2 −

√4− x2

]dx+7

ˆ x=2

x=−2

[−√

4− x2 +√

25− x2]

dx

= 14ˆ x=2

x=−2

[√25− x2 −

√4− x2

]dx,

which is not easy to compute. □1region between two circles having the same center and different radii

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ANSWER 3 POLAR COORDINATES. We use the polar coordinates and represent the region R as follows:

R ={(x,y) : 4 ≤ x2 + y2 ≤ 25

}= { (r,θ) : 0 ≤ θ ≤ 2π, 2 ≤ r ≤ 5 } .

Then, the volume of the solid is

V =

¨R

7dA =

ˆ θ=2π

θ=0

ˆ r=5

r=27r dr dθ

=

ˆ θ=2π

θ=0

[72

r2]r=5

r=2dθ =

ˆ θ=2π

θ=0

[72(52 −22)

]dθ =

72(52 −22)2π = 7π(52 −22) = 147π. □

As we can see, in this problem, the first method is the easiest one and the method using the rectangularcoordinates is the most difficult one in getting the volume.

Example 13.3.8. Find the volume of the solid below z = 8− x2 − y2 and above z = 3x2 +3y2.

Figure 13.17: Solid below z = 8− x2 − y2 and above z = 3x2 +3y2

ANSWER. See the figure 13.17. We first find the intersection of two surfaces, z = 8 − x2 − y2 and z =3x2 +3y2:

8− x2 − y2 = 3x2 +3y2 −→ 4x2 +4y2 = 8 −→ x2 + y2 = 2,

which is a circle centered at (0,0) with radius√

2 . It implies that a point (x,y,z) in the solid, precisely, (x,y),should be in the disk x2 + y2 ≤ 2. That is, as the base of the solid, we can take the region R:

R ={(x,y) : x2 + y2 ≤ 2

}.

On this region R, the surface of z = 8− x2 − y2 is higher than that of z = 3x2 +3y2. So, the volume V of thesolid is obtained by

V =

¨R

[8− x2 − y2 − (3x2 +3y2)

]dA =

¨R

[8−4(x2 + y2)

]dA.

When we use the following representation of R:

R ={(x,y) : x2 + y2 ≤ 2

}={(x,y) : −

√2 ≤ x ≤

√2 , −

√2− x2 ≤ y ≤

√2− x2

},

the volume V becomes

V =

¨R

[8−4(x2 + y2)

]dA =

ˆ x=√

2

x=−√

2

ˆ y=√

2−x2

y=−√

2−x2

[8−4(x2 + y2)

]dydx,

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which is not easy to compute. So we use the polar coordinates and express the region R as follows:

R ={(x,y) : x2 + y2 ≤ 2

}={(r,θ) : 0 ≤ r ≤

√2 , 0 ≤ θ ≤ 2π

}.

Using this representation, the volume V becomes

V =

¨R

[8−4(x2 + y2)

]dA =

ˆ θ=2π

θ=0

ˆ r=√

2

r=0

[8−4r2]r dr dθ

=

ˆ θ=2π

θ=0

ˆ r=√

2

r=0

[8r−4r3] dr dθ =

ˆ θ=2π

θ=0

[4r2 − r4]r=√

2r=0 dθ =

ˆ θ=2π

θ=04dθ = 8π. □

Example 13.3.9. Evaluate the iterated integral by converting to polar coordinates´ 2

0

´ √4−x2

−√

4−x2 e−x2−y2dydx.

ANSWER. The full expression of the integral is

ˆ 2

0

ˆ √4−x2

−√

4−x2e−x2−y2

dydx =ˆ x=2

x=0

ˆ y=√

4−x2

y=−√

4−x2e−x2−y2

dydx.

It implies that the region R of the integral is

R ={(x,y) : 0 ≤ x ≤ 2, −

√4− x2 ≤ y ≤

√4− x2

}.

If we sketch the region R in the xy–plane, we observe that R represents a quarter circle centered at the originwith radius 2 lying on the first quadrant of the xy–plane. When using the polar coordinates, R is representedas follows:

R ={(r,θ) : 0 ≤ r ≤ 2, −π

2≤ θ ≤ π

2

}.

Hence, the integral becomes in the polar coordinates

ˆ 2

0

ˆ √4−x2

−√

4−x2e−x2−y2

dydx =ˆ x=2

x=0

ˆ y=√

4−x2

y=−√

4−x2e−x2−y2

dydx

=

ˆ θ=π/2

θ=−π/2

ˆ r=2

r=0e−r2

r dr dθ =

ˆ θ=π/2

θ=−π/2

[−e−r2

2

]r=2

r=0

dθ

=

ˆ θ=π/2

θ=−π/2

1− e−4

2dθ =

π(1− e−4)

2. □

§13.4 Surface Area.Skip. Please read the textbook.

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§13.5 Triple Integrals.

□ 13.5.1 Formulas.

Box Q

Theorem 13.5.1 (FUBINI’S THEOREM). Suppose that f (x,y,z) is continuous on the box Q defined by

Q = { (x,y,z) : a ≤ x ≤ b, c ≤ y ≤ d, r ≤ z ≤ s } .

Then we can write the triple integral over Q as a triple iterated integral:˚

Qf (x,y,z)dV =

ˆ z=s

z=r

ˆ y=d

y=c

ˆ x=b

x=af (x,y,z)dxdydz =

ˆ z=s

z=r

ˆ x=b

x=a

ˆ y=d

y=cf (x,y,z)dydxdz

=

ˆ x=b

x=a

ˆ z=s

z=r

ˆ y=d

y=cf (x,y,z)dydzdx =

ˆ x=b

x=a

ˆ y=d

y=c

ˆ z=s

z=rf (x,y,z)dzdydx

=

ˆ y=d

y=c

ˆ z=s

z=r

ˆ x=b

x=af (x,y,z)dxdzdy =

ˆ y=d

y=c

ˆ x=b

x=a

ˆ z=s

z=rf (x,y,z)dzdxdy.

Moreover, if f (x,y,z) = g(x)h(y)k(z), then

˚Q

f (x,y,z)dV =

(ˆ x=b

x=ag(x)dx

)(ˆ y=d

y=ch(y)dy

)(ˆ z=s

z=rk(z)dz

).

Example 13.5.2. Evaluate the triple integral˝

Q 2xey sinzdV , where Q is the box defined by

Q = { (x,y,z) : 1 ≤ x ≤ 2, 0 ≤ y ≤ 1, 0 ≤ z ≤ π } .

ANSWER. By Fubini’s Theorem, we have˚

Q2xey sinzdV =

ˆ z=π

z=0

ˆ y=1

y=0

ˆ x=2

x=12xey sinzdxdydz

= 2(ˆ x=2

x=1xdx)(ˆ y=1

y=0ey dy

)(ˆ z=π

z=0sinzdz

)= 6(e−1). □

Exercise 13.5.3. Evaluate the triple integral˝

Q(x+ y+ z)dV , where Q is the box defined by

Q = { (x,y,z) : 0 ≤ x ≤ 1, −1 ≤ y ≤ 2, 0 ≤ z ≤ 3 } .

Exercise 13.5.4. Evaluate the triple integral˝

Q xyz2 dV , where Q is the box defined by

Q = { (x,y,z) : 0 ≤ x ≤ 1, 0 ≤ y ≤ 2, 0 ≤ z ≤ 3 } .

Solid Q bounded by Upper and Lower Surfaces

Theorem 13.5.5 (FIXED ONE VARIABLE). Suppose the solid Q can be written in the form

Q = { (x,y,z) : (x,y) ∈ R, g1(x,y)≤ z ≤ g2(x,y) } ,

where R is some region in the xy–plane and where g1(x,y)≤ g2(x,y) for all (x,y) in R. Then the triple integralover Q is expressed as follows:

˚Q

f (x,y,z)dV =

¨R

ˆ z=g2(x,y)

z=g1(x,y)f (x,y,z)dzdA.

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Figure 13.18: Tetrahedron bounded by 2x+ y+ z = 4 and three coordinates plates

Example 13.5.6. Evaluate˝

Q 6xydV , where Q is the tetrahedron bounded by the planes x = 0, y = 0, z = 0and 2x+ y+ z = 4.

ANSWER. See the figure 13.18. Let R be the region of Q on the xy–plane. Then the height of the tetrahedronturns to be z = 4− y−2x, i.e., when (x,y) of (x,y,z) ∈ Q moves over R, the z component moves from z = 0to z = 4− y−2x. So, the triple integral becomes

˚Q

6xydV =

¨R

ˆ z=4−y−2x

z=06xydzdA =

¨R

6xy(4− y−2x) dA,

which is computed by the method in Section 13.1 Double Integrals. Explicitly, the region R is expressed asfollows:

R = { (x,y) : 0 ≤ x ≤ 2, 0 ≤ y ≤ 4−2x } .

Using this expression, the triple integral becomes

˚Q

6xydV =

¨R

6xy(4− y−2x) dA =

ˆ x=2

x=0

ˆ y=4−2x

y=06xy(4− y−2x)dydx. □

Exercise 13.5.7. Evaluate˝

Q zdV , where Q is the tetrahedron bounded by the planes x = 0, y = 0, z = 0and x+ y+ z = 1.

□ 13.5.2 Applications: Volume, Mass and Center of Mass.

Volume

The volume V of a solid Q in the 3–dimensional space is obtained by the triple integral V =˝

Q dV .

Example 13.5.8. Find the volume of the tetrahedron Q bounded by x = 0, x = 2y, z = 0 and x+2y+ z = 2.

ANSWER. See the figure 13.19. We take the region R of Q on the xy–plane of Q as our base of the tetrahedronQ. The lower boundary of Q is the plane z = 0 and the upper boundary is the plane x+ 2y+ z = 2, that is,z = 2− x−2y. It implies

V =

˚Q

dV =

¨R

ˆ z=2−x−2y

z=0dzdA.

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Figure 13.19: Tetrahedron bounded by x = 0, x = 2y, z = 0 and x+2y+ z = 2

From Section 13.1 Double Integrals, the region R of Q on the xy–plane is expressed as follows:

R ={(x,y) : 0 ≤ x ≤ 1,

x2≤ y ≤ 1− x

2

}.

Therefore, we get

V =

˚Q

dV =

¨R

ˆ z=2−x−2y

z=0dzdA =

ˆ x=1

x=0

ˆ y=1−x/2

y=x/2

ˆ z=2−x−2y

z=0dzdydx =

13. □

Mass and Center of Mass

Theorem 13.5.9. Suppose a solid Q has the mass density ρ(x,y,z). Then

1. Total mass m is given by

m =

˚Q

ρ(x,y,z)dV.

2. It has the center of mass (x, y, z), where

x =Myz

m, y =

Mxz

m, z =

Mxy

m,

Myz =

˚Q

xρ(x,y,z)dV, Mxz =

˚Q

yρ(x,y,z)dV, Mxy =

˚Q

zρ(x,y,z)dV,

i.e., x =1˝

Q ρ(x,y,z)dV

˚Q

xρ(x,y,z)dV,

y =1˝

Q ρ(x,y,z)dV

˚Q

yρ(x,y,z)dV,

z =1˝

Q ρ(x,y,z)dV

˚Q

zρ(x,y,z)dV.

Example 13.5.10. Find the center of mass of the solid of constant mass density ρ bounded by the graphs ofz =

√x2 + y2 and z = 4.

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Figure 13.20: Solid bounded by z =√

x2 + y2 and z = 4

ANSWER. See the figure 13.20. Notice that the projection R of the solid onto the xy–plane is the disk ofradius 4 centered at the origin:

R ={(x,y) : −4 ≤ x ≤ 4, −

√16− x2 ≤ y ≤

√16− x2

}.

Further, for each (x,y) ∈ R, z ranges from the cone (z =√

x2 + y2 ) up to the plane z = 4. So, the total massm of the solid is given by

m =

˚Q

ρ(x,y,z)dV = ρ¨

R

ˆ z=4

z=√

x2+y2dzdA = ρ

ˆ x=4

x=−4

ˆ y=√

16−x2

y=−√

16−x2

ˆ z=4

z=√

x2+y2dzdydx =

64π3

,

which will be easily computed when we use the Cylindrical Coordinates (Section 13.6). We leave the centerof mass to the students. Practice and Exercise. □

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§13.6 Cylindrical Coordinates.

□ 13.6.1 Cylindrical Coordinates.The cylindrical coordinates can be viewed as the extension of the polar coordinates into the three–coordinatessystem. To be precise, in the cylindrical coordinate system, a point P(x,y,z) in three–dimensional spaceis represented by the ordered triple (r,θ ,z), where r and θ are polar coordinates of the projection of P ontothe xy–plane and z is the directed distance from the xy–plane to P. See the figure 13.21. To convert fromcylindrical to rectangular coordinates, we use the following relationship.

Figure 13.21: Cylindrical Coordinate System

Theorem 13.6.1 (RELATION BETWEEN RECTANGULAR AND CYLINDRICAL COORDINATES). A pointP(x,y,z) in the rectangular coordinates system corresponds to a point (r,θ ,z) in the cylindrical coordinatessystem by the following equations:

x = r cosθ , y = r sinθ , z = z.

So when we convert from rectangular to cylindrical coordinates, we use

r2 = x2 + y2, tanθ =yx, z = z.

Example 13.6.2. Write the given equation in cylindrical coordinates: (1) x2 + (y − 3)2 = 9 and (2) z =√x2 + y2 .

ANSWER. (1) In cylindrical coordinates, x = r cosθ and y = r sinθ . So the given equation becomes

(r cosθ)2 +(r sinθ −3)2 = 9, r2 cos2 θ + r2 sin2 θ −6r sinθ +9 = 9,

r2 −6r sinθ = 0, r = 6sinθ .

(2) In cylindrical coordinates, r2 = x2 + y2 and so the equation becomes

z =√

r2 = |r|. □

Example 13.6.3. Write the given equation in rectangular coordinates: z = r.

ANSWER. From the relation above, we have

z2 = r2 = x2 + y2, i.e., z2 = x2 + y2. □

Exercise 13.6.4. Write the given equation in cylindrical coordinates: (1) z = e−x2−y2and (2) x+2y+3z = 4.

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□ 13.6.2 Evaluating Triple Integrals with Cylindrical Coordinates.We can use cylindrical coordinates to simplify certain triple integrals. Suppose we can write the solid Q as

Q = { (r,θ ,z) : (r,θ) ∈ R, k1(r,θ)≤ z ≤ k2(r,θ) } , (13.6.1)

where k1(r,θ)≤ k2(r,θ) for all (r,θ) in the region R of the xy–plane defined by

R = { (r,θ) : α ≤ θ ≤ β , g1(θ)≤ r ≤ g2(θ) } , (13.6.2)

where 0 ≤ g1(θ)≤ g2(θ) for all θ ∈ [α ,β ]. Then we can write

˚Q

f (r,θ ,z)dV =

¨R

(ˆ z=k2(r,θ)

z=k1(r,θ)f (r,θ ,z)dz

)dA.

Using the polar coordinates, we express the integral in the full form as below.

Theorem 13.6.5 (REPRESENTATION OF TRIPLE INTEGRAL IN CYLINDRICAL COORDINATES). For Q andR given in equations (13.6.1) and (13.6.2) above, the triple integral

˝Q f (r,θ ,z)dV can be expressed as

follows:

˚Q

f (r,θ ,z)dV =

¨R

(ˆ k2(r,θ)

k1(r,θ)f (r,θ ,z)dz

)dA =

ˆ θ=β

θ=α

ˆ r=g2(θ)

r=g1(θ)

ˆ z=k2(r,θ)

z=k1(r,θ)f (r,θ ,z)r dzdr dθ .

Example 13.6.6. Set up the triple integral˝

Q f (x,y,z)dV in cylindrical coordinates, where Q is the solidabove the xy–plane and below z = 9− x2 − y2.

ANSWER. See the figure 13.22. In cylindrical coordinates, z = 9− x2 − y2 becomes z = 9− r2. We observethe surface of z = 9− x2 − y2 is the paraboloid and so Q has the circular base region. It implies

Q ={(r,θ ,z) : (r,θ) ∈ R, 0 ≤ z ≤ 9− r2 } , R = { (r,θ) : 0 ≤ θ ≤ 2π, 0 ≤ r ≤ 3 } .

Hence, the triple integral can be set up as follows:

˚Q

f (r,θ ,z)dV =

¨R

(ˆ 9−r2

0f (r,θ ,z)dz

)dA =

ˆ θ=2π

θ=0

ˆ r=3

r=0

ˆ z=9−r2

z=0f (r,θ ,z)r dzdr dθ . □

Figure 13.22: Solid between the xy–plane and z = 9− x2 − y2 and Solid between z =√

x2 + y2 and z = 2

Example 13.6.7. Evaluate´ x=2

x=−2

´ y=√

4−x2

y=−√

4−x2

´ 2z=√

x2+y2 (x2 + y2)3/2 dzdydx.

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ANSWER. See the figure 13.22. We find the solid Q first in the rectangular coordinates:

Q ={(x,y,z) : −2 ≤ x ≤ 2, −

√4− x2 ≤ y ≤

√4− x2 ,

√x2 + y2 ≤ z ≤ 2

}.

We introduce the region R on xy–plane as follows:

Q ={(x,y,z) : (x,y) ∈ R,

√x2 + y2 ≤ z ≤ 2

},

R ={(x,y) : −2 ≤ x ≤ 2, −

√4− x2 ≤ y ≤

√4− x2

}.

Now we express Q and R in cylindrical coordinates: (Observe that R is a disk x2 + y2 ≤ 4)

Q = { (r,θ ,z) : (r,θ) ∈ R, r ≤ z ≤ 2 } , R = { (r,θ) : 0 ≤ θ ≤ 2π, 0 ≤ r ≤ 2 } .

Using this representation, the integral becomesˆ x=2

x=−2

ˆ y=√

4−x2

y=−√

4−x2

ˆ 2

z=√

x2+y2(x2 + y2)3/2 dzdydx

=

ˆ θ=2π

θ=0

ˆ r=2

r=0

ˆ 2

z=r(r2)3/2r dzdr dθ =

ˆ θ=2π

θ=0

ˆ r=2

r=0

ˆ 2

z=rr4 dzdr dθ

=

ˆ θ=2π

θ=0

ˆ r=2

r=0

[r4z]2

z=r dr dθ =

ˆ θ=2π

θ=0

ˆ r=2

r=0r4(2− r)dr dθ

=

ˆ θ=2π

θ=0

[2r5

5− r6

6

]r=2

r=0dθ =

ˆ θ=2π

θ=0

3215

dθ =64π15

. □

Example 13.6.8. Set up and evaluate the triple integral˝

Q ze√

x2+y2 dV in the appropriate coordinate sys-tem, where Q is the solid inside x2 + y2 = 4, outside x2 + y2 = 1 and between z = 0 and z = 3.

Figure 13.23: Solid inside x2 + y2 = 4, outside x2 + y2 = 1 and between z = 0 and z = 3

ANSWER. See the figure 13.23. We use the cylindrical coordinate system. In cylindrical coordinates, x2 +y2 = 4 and x2 + y2 = 1 become r2 = 4, i.e., r = ±2, and r2 = 1, i.e., r = ±1, respectively. So the region Qinside x2 + y2 = 4, outside x2 + y2 = 1 and between z = 0 and z = 3 can be expressed as follows:

Q = { (r,θ ,z) : (r,θ) ∈ R, 0 ≤ z ≤ 3 } , R = { (r,θ) : 0 ≤ θ ≤ 2π, 1 ≤ r ≤ 2 } .

Hence, the triple integral can be set up as follows:˚Q

ze√

x2+y2dV =

¨R

(ˆ 3

0zer dz

)dA =

ˆ θ=2π

θ=0

ˆ r=2

r=1

ˆ z=3

z=0zerr dzdr dθ

=

(ˆ z=3

z=0zdz)(ˆ r=2

r=1err dr

)(ˆ θ=2π

θ=01dθ

)=

92

e2(2π) = 9πe2. □

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Q

√x2 + y2 dV , where Q is the region that lies inside the cylinder x2 + y2 = 16

and between the planes z =−5 and z = 4.


Q x2 dV , where Q is the solid that lies within the cylinder x2 + y2 = 1, abovethe plane z = 0, and below the cone z2 = 4x2 +4y2.

Exercise 13.6.11. Evaluate the iterated integral after changing coordinate systems:� ´ 3

−3

´ 0−√

9−x2

´ x2+z2

0

(x2 + z2) dydzdx.

� ´ 10

´ √1−x2

0

´ 41−x2−y2

(x2 + y2) dzdydx.

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§13.7 Spherical Coordinates.

□ 13.7.1 Spherical Coordinates.The spherical coordinate system simplifies the evaluation of triple integrals over regions bounded by spheresor cones.We start with a point P(x,y,z) in the rectangular coordinate system, i.e., the xyz–space. It is easy to put thepoint P(x,y,z) in the rectangular coordinate system, which is shown in Figure 1. For the point P(x,y,z), wecan deduce three quantities

1. distance between O and P, denoted by ρ ,2. angle from the positive z–axis to the line OP, denoted by ϕ ,3. angle from the positive x–axis to the line OQ, denoted by θ .

See the figure 13.24. These three quantities forms the spherical coordinate system. The combinationof these three quantities, (ρ,ϕ ,θ), corresponds to the point P(x,y,z) and so we may say (ρ ,ϕ ,θ) is therepresentation of the point P(x,y,z) in the spherical coordinate system.

Figure 13.24: Spherical Coordinate System

Since ρ represents the distance, it should satisfy ρ ≥ 0. Since ϕ represents the angle from the positive z–axisto the line OP, it should satisfy 0 ≤ ϕ ≤ π . We take 0 ≤ θ ≤ 2π as the range of the θ .Let us find the specific relationship between the rectangular coordinates and the spherical coordinates.

Conversion from Spherical to Rectangular Coordinates. Suppose a point P is given in the sphericalcoordinates, P(ρ,ϕ ,θ). Let us find its representation in the rectangular coordinates P(x,y,z).We observe that the angle between OP and OQ is π

2 −ϕ . It implies that the angle between OP and PQ is ϕ .(Why?) So, from the triangle OPQ, we get

∥OQ∥= ∥OP∥sinϕ , and ∥PQ∥= ∥OP∥cosϕ ,

where ∥OP∥, ∥OQ∥ and ∥PQ∥ mean the lengths of the line segment OP, OQ and PQ, respectively.It is easy to see z = ∥PQ∥, i.e.,

z = ∥PQ∥= ∥OP∥cosϕ , z = ∥OP∥cosϕ .

Again from the triangle having the angle θ on the xy–plane, we observe

x = ∥OQ∥cosθ , and y = ∥OQ∥sinθ .

When we put ∥OQ∥= ∥OP∥sinϕ into the equations, we get

x = ∥OQ∥cosθ = ∥OP∥sinϕ cosθ , and y = ∥OQ∥sinθ = ∥OP∥sinϕ sinθ .

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Since the length of OP is ∥OP∥= ρ , hence, we finally deduce

x = ∥OP∥sinϕ cosθ = ρ sinϕ cosθ ,y = ∥OP∥sinϕ sinθ = ρ sinϕ sinθ ,z = ∥OP∥cosϕ = ρ cosϕ .

That is, in the rectangular coordinates, we have P(ρ sinϕ cosθ ,ρ sinϕ sinθ ,ρ cosϕ).

Conversion from Rectangular to Spherical Coordinates. Suppose a point P is given in the rectangularcoordinates, P(x,y,z). Let us find its representation in the spherical coordinates P(ρ,ϕ ,θ).Since ρ represents the length of OP, we get

ρ = ∥OP∥=√

x2 + y2 + z2 , i.e., ρ =√

x2 + y2 + z2 .

Since z is equal to the length of PQ, we get

z = ∥PQ∥= ∥OP∥cosϕ =√

x2 + y2 + z2 cosϕ , i.e., cosϕ =z√

x2 + y2 + z2,

i.e., 0 ≤ ϕ = cos−1

(z√

x2 + y2 + z2

)≤ π.

From the triangle having the angle θ on the xy–plane, we observe

x = ∥OQ|cosθ =√

x2 + y2 cosθ , cosθ =x√

x2 + y2, 0 ≤ θ = cos−1

(x√

x2 + y2

)≤ 2π.

That is,

0 ≤ ρ =√

x2 + y2 + z2 ,

0 ≤ ϕ = cos−1

(z√

x2 + y2 + z2

)≤ π,

0 ≤ θ = cos−1

(x√

x2 + y2

)= sin−1

(y√

x2 + y2

)≤ 2π.

We summarize as follows:

Theorem 13.7.1 (RELATIONSHIP BETWEEN RECTANGULAR AND SPHERICAL COORDINATES). A pointP(x,y,z) in the rectangular coordinate system corresponds to a point (ρ ,ϕ ,θ) in the spherical coordinatesystem by the following equations:

x = ρ sinϕ cosθ , y = ρ sinϕ sinθ , z = ρ cosϕ .

It is worthwhile to memorize the equation,

ρ =√

x2 + y2 + z2 , or ρ2 = x2 + y2 + z2.

Example 13.7.2. Find rectangular coordinates for the point described by (ρ,ϕ ,θ) = (8,π/4,π/3) in spheri-cal coordinates.

ANSWER. By the Theorem 13.7.1 above, we get

x = 8sinπ4

cosπ3= 2

√2 , y = 8sin

π4

sinπ3= 2

√6 , z = 8cos

π4= 4

√2 .

Hence, we get (x,y,z) = (2√

2 ,2√

6 ,4√

2 ). See the figure 13.25. □

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Figure 13.25: Spherical Coordinate System

Exercise 13.7.3. The point is given (0,2√

3 ,−2) in rectangular coordinates. Find spherical coordinates forthis point.

Example 13.7.4. Rewrite the equation of the cone z2 = x2 + y2 in spherical coordinates.

ANSWER. Putting x = ρ sinϕ cosθ , y = ρ sinϕ sinθ , and z = ρ cosϕ into the equation, we get

(ρ cosϕ)2 = (ρ sinϕ cosθ)2 +(ρ sinϕ sinθ)2 ,

simply, 0 = ρ2 (cos2 ϕ − sin2 ϕ)= ρ2 (cosϕ − sinϕ)(cosϕ + sinϕ)

−→ ρ = 0 or cosϕ − sinϕ = 0 or cosϕ + sinϕ = 0

−→ ρ = 0 or ϕ =π4

or ϕ =3π4. □

□ 13.7.2 Triple Integrals in Spherical Coordinates.We can use spherical coordinates to simplify a triple integral.

Theorem 13.7.5 (REPRESENTATION OF TRIPLE INTEGRAL IN SPHERICAL COORDINATES). The tripleintegral

˝Q f (r,θ ,z)dV can be expressed as follows:˚

Qf (x,y,z)dV =

˚Q

f (ρ sinϕ cosθ ,ρ sinϕ sinθ ,ρ cosϕ)ρ2 sinϕ dρ dϕ dθ .

Note that dV = ρ2 sinϕ dρ dϕ dθ .

Example 13.7.6. Evaluate˝

Q e(x2+y2+z2)3/2

dV , where Q is the unit ball:

Q ={(x,y,z) : x2 + y2 + z2 ≤ 1

}.

ANSWER. In spherical coordinates, Q has the representation

Q ={(ρ,ϕ ,θ) : ρ2 ≤ 1

}= { (ρ ,ϕ ,θ) : 0 ≤ ρ ≤ 1, 0 ≤ ϕ ≤ π, 0 ≤ θ ≤ 2π } .

Hence, the triple integral can be set up and computed as follows:˚

Qe(x

2+y2+z2)3/2dV =

˚Q

eρ3dV =

ˆ θ=2π

θ=0

ˆ ϕ=π

ϕ=0

ˆ ρ=1

ρ=0eρ3

ρ2 sinϕ dρ dϕ dθ =4π(e−1)

3. □

Remark 13.7.7. It would have been extremely awkward to evaluate the integral in the Example 13.7.6 aboveusing rectangular coordinates rather than spherical coordinates. In rectangular coordinates the iterated inte-gral would have been

˚Q

e(x2+y2+z2)3/2

dV =

ˆ x=1

x=−1

ˆ y=√

1−x2

y=−√

1−x2

ˆ z=√

1−x2−y2

z=−√

1−x2−y2e(x

2+y2+z2)3/2dzdydx.

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Example 13.7.8. Use spherical coordinates to find the volume of the solid that lies above the cone z =√x2 + y2 and below the sphere x2 + y2 + z2 = z.

Figure 13.26: Solid between cone z =√

x2 + y2 and sphere x2 + y2 + z2 = z

ANSWER. See the figure 13.26. We express the equations of cone and sphere in the spherical coordinatesystem. (1) Cone z =

√x2 + y2 becomes ϕ = π

4 . (Why?) (2) Sphere x2 + y2 + z2 = z becomes ρ = cosϕ .(Why?) So, the solid Q has the representation:

Q ={(ρ,ϕ ,θ) : 0 ≤ θ ≤ 2π, 0 ≤ ϕ ≤ π

4, 0 ≤ ρ ≤ cosϕ

}.

Using this representation, the volume of the solid Q is obtained as follows:

V =

˚Q

dV =

ˆ θ=2π

θ=0

ˆ ϕ=π/4

ϕ=0

ˆ ρ=cosϕ

ρ=0ρ2 sinϕ dρ dϕ dθ =

π8. □

Example 13.7.9. Find the volume lying inside the sphere x2 + y2 + z2 = 2z and inside the cone z2 = x2 + y2.

ANSWER. We express the equations of cone and sphere in the spherical coordinate system. (1) Cone z =√x2 + y2 becomes ϕ = cosϕ . (Why?) (2) Sphere x2 + y2 + z2 = 2z becomes ρ = 2cosϕ . (Why?) So, the

solid Q has the representation:

Q ={(ρ ,ϕ ,θ) : 0 ≤ θ ≤ 2π, 0 ≤ ϕ ≤ π

4, 0 ≤ ρ ≤ 2cosϕ

}.

Using this representation, the volume of the solid Q is obtained as follows:

V =

˚Q

dV =

ˆ θ=2π

θ=0

ˆ ϕ=π/4

ϕ=0

ˆ ρ=2cosϕ

ρ=0ρ2 sinϕ dρ dϕ dθ = π. □

Example 13.7.10. Evaluate the triple iterated integral´ 2−2

´ √4−x2

0

´√4−x2−y2

0(x2 + y2 + z2) dzdydx.

ANSWER. See the figure 13.27. Let us find the expression of the integral solid Q from the given tripleintegral:

Q ={(x,y,z) : −2 ≤ x ≤ 2, 0 ≤ y ≤

√4− x2 , 0 ≤ z ≤

√4− x2 − y2

}.

Can you imagine the shape of the solid Q? When we ignore the z–component of the points in Q, we observea upper–half disk x2 + y2 ≤ 4. Explicitly,

Q ={(x,y,z) : (x,y) ∈ R, 0 ≤ z ≤

√4− x2 − y2

}, R =

{(x,y) : −2 ≤ x ≤ 2, 0 ≤ y ≤

√4− x2

} of 126


Figure 13.27: Solid bounded by z =√

4− x2 − y2 and y = 0 and z = 0

and R is the upper–half disk x2 +y2 ≤ 4. That is, the projection of Q onto the xy–plane is the upper–half diskx2 + y2 ≤ 4.From the expression of Q, we also observe that the solid is under the curve z=

√4− x2 − y2 , which is a upper

sphere of the ball centered at the origin with the radius 4, because the equation implies z2 = 4− x2 − y2, i.e.,x2 + y2 + z2 = 4.Now, we use the spherical coordinates and express the solid and the integral:

Q ={(ρ ,ϕ ,θ) : 0 ≤ θ ≤ π, 0 ≤ ϕ ≤ π

2, 0 ≤ ρ ≤ 2

}. (Why?)

Hence, we get

ˆ 2

−2

ˆ √4−x2

0

ˆ √4−x2−y2

0

(x2 + y2 + z2) dzdydx =

˚Q

(x2 + y2 + z2) dV =

˚Q

ρ2 dV

=

ˆ θ=π

θ=0

ˆ ϕ=π/2

ϕ=0

ˆ ρ=2

ρ=0ρ2 (ρ2 sinϕ

)dρ dϕ dθ =

32π5

.□

§13.8 Change of Variables in Multiple Integrals.Skip. Please read the textbook.

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Chapter 14

Vector Calculus

§14.1 Vector Fields.Skip. Please read the textbook.

§14.2 Line Integrals.

□ 14.2.1 Line Integral.We start with the evaluation theorem on how to compute the line integral.

Theorem 14.2.1 (EVALUATION THEOREM).

1. Function of Two Variables. Suppose that f (x,y) is continuous in a region D containing the curve Cand that C is described parametrically by (x(t),y(t)) for a ≤ t ≤ b, where x(t) and y(t) have continuousfirst derivatives. Then,

ˆC

f (x,y)ds =ˆ t=b

t=af (x(t),y(t))

√[x′(t)]2 +[y′(t)]2 dt.

2. Function of Three Variables. Suppose that f (x,y,z) is continuous in a region D containing the curveC and that C is described parametrically by (x(t),y(t),z(t)) for a ≤ t ≤ b, where x(t), y(t) and z(t) havecontinuous first derivatives. Then,

ˆC

f (x,y,z)ds =ˆ t=b

t=af (x(t),y(t),z(t))

√[x′(t)]2 +[y′(t)]2 +[z′(t)]2 dt.

Example 14.2.2. Find the mass of a spring in the shape of the helix defined parametrically by x = 2cos t,y = t, z = 2sin t, for 0 ≤ t ≤ 6π with density ρ(x,y,z) = 2y.

ANSWER. Parametrically, the density becomes ρ(x,y,z) = 2y = 2t. The arc length element ds is given by

ds =√

[x′(t)]2 +[y′(t)]2 +[z′(t)]2 dt =√

5 dt.

Hence, the mass of the spring is obtained as follows:

m =

ˆC

ρ(x,y,z)ds =ˆ t=6π

t=02t√

5 dt = 36π2√

5 . □

Example 14.2.3. Evaluate the line integral´

C(2x2 −3yz)ds, where C is the curve defined parametrically byx = cos t, y = sin t, z = cos t, with 0 ≤ t ≤ 2π .

ANSWER. Parametrically, the integrand becomes f (x,y,z) = 2x2−3yz= 2cos2 t−3sin t cos t. The arc lengthelement ds is given by

ds =√[x′(t)]2 +[y′(t)]2 +[z′(t)]2 dt =

√1+ sin2 t dt.

Hence, the line integral becomesˆ

C(2x2 −3yz)ds =

ˆ t=2π

t=0

(2cos2 t −3sin t cos t

)√1+ sin2 t dt ≈ 6.9922. □

111


Theorem 14.2.4 (SEPARATION). Suppose that f (x,y,z) is a continuous function in some region D containingthe oriented curve C. Then, if C is piecewise–smooth, with C =C1 ∪C2 ∪ ·· ·∪Cn, where C1, C2, . . . , Cn areall smooth and where the terminal point of Ci is the same as the initial point of Ci+1, for i = 1,2, . . . ,n− 1,we have

1.´−C f (x,y,z)ds =

Ć f (x,y,z)ds.

2.´

C f (x,y,z)ds =´

C1f (x,y,z)ds+

Ć2

f (x,y,z)ds+ · · ·+´

Cnf (x,y,z)ds.

Theorem 14.2.5 (LENGTH OF CURVE). For any piecewise–smooth curve C (in two or three dimensions),Ć 1ds gives the arc length of the curve C.

Theorem 14.2.6 (EVALUATION THEOREM). Suppose that f (x,y,z) is continuous in a region D containingthe curve C and that C is described parametrically by (x(t),y(t),z(t)) for a ≤ t ≤ b, where x(t), y(t) and z(t)have continuous first derivatives. Then,

ˆC

f (x,y,z)dx =ˆ t=b

t=af (x(t),y(t),z(t))x′(t)dt

ˆC

f (x,y,z)dy =ˆ t=b

t=af (x(t),y(t),z(t))y′(t)dt

ˆC

f (x,y,z)dz =ˆ t=b

t=af (x(t),y(t),z(t))z′(t)dt

Example 14.2.7. Compute the line integral´

C(4xz+2y)dx, where C is the line segment (1) from (2,1,0) to(4,0,2) and (2) from (4,0,2) to (2,1,0).

ANSWER. We solve only (1): First, we parameterize the line segment from (2,1,0) to (4,0,2) as follows:

x = 2+(4−2)t = 2+2t, y = 1+(0−1)t, z = 0+(2−0)t = 2t, 0 ≤ t ≤ 1.

Then the integrand becomes

4xz+2y = 4(2+2t)(2t)+2(1− t) = 16t2 +14t +2.

The element dx becomesdx = x′(t)dt = 2dt.

Hence, the line integral is obtained as follows:

ˆC(4xz+2y)dx =

ˆ t=1

t=0

(16t2 +14t +2

)2dt =

863. □

Remark 14.2.8 (VERY IMPORTANT NOTATION). For the convenience, we usually writeˆ

Cf (x,y,z)dx+

ˆC

g(x,y,z)dy+ˆ

Ch(x,y,z)dz =

ˆC

f (x,y,z)dx+g(x,y,z)dy+h(x,y,z)dz.

□ 14.2.2 Application: Work.

Theorem 14.2.9 (WORK DONE). The work done W by the force field F(x,y,z) in moving a particle alongthe curve C is obtained as follows:

W =

ˆC

F(x,y,z) · dr,

where dr = ⟨dx, dy, dz⟩ is the element of the parametric curve C.

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Example 14.2.10. Compute the work done by the force field F(x,y,z) = ⟨4y, 2xz, 3y⟩ acting on an object asit moves along the helix defined parametrically by x = 2cos t, y = 2sin t and z = 3t, from the point (2,0,0)to the point (−2,0,3π).

ANSWER. By the formula, the work done is given by

W =

ˆC

F(x,y,z) · dr =ˆ

C4ydx+2xzdy+3ydz.

We observe the point (2,0,0) occurs when t = 0 and the point (−2,0,3π) occurs when t = π . Hence withdx =−2sin t dt, dy = 2cos t dt and dz = 3dt, the line integral becomes

W =

ˆC

4ydx+2xzdy+3ydz

=

ˆ t=π

t=04(2sin t)(−2sin t)dt +2(2cos t)(3t)(2cos t)dt +3(2sin t)(3)dt

=

ˆ t=π

t=0[4(2sin t)(−2sin t)+2(2cos t)(3t)(2cos t)+3(2sin t)(3)] dt

=

ˆ t=π

t=0

[−16sin2 t +24t cos2 t +18sin t

]dt = 36−8π +6π2. □

Example 14.2.11. Compute the work done by the force field F(x,y) = ⟨y,−x⟩ acting on an object as it movesalong the parabola y = x2 −1 from (1,0) to (−2,3).

ANSWER. By the formula, the work done is given by

W =

ˆC

F(x,y) · dr =ˆ

Cydx− xdy.

We use x = t and y = t2 − 1 as parametric equations for the curve, with t ranging from t = 1 to t = −2. Inthis case, dx = dt and dy = 2t dt. Hence the work is

W =

ˆC

ydx− xdy =ˆ t=−2

t=1

[(t2 −1)1− t(2t)

]dt = 6. □

§14.3 Independence of Path and Conservative Vector Fields.Skip. Please read the textbook.

§14.4 Green’s Theorem.Skip. Please read the textbook.

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§14.5 Curl and Divergence.In this section, we briefly discuss the curl and divergence, which are generalizations of the notion of deriva-tive applied to vector fields.

□ 14.5.1 Curl.

Definition 14.5.1. The curl of the vector field F(x,y,z) = ⟨F1(x,y,z),F2(x,y,z),F3(x,y,z)⟩ is the vector fielddefined by

CurlF =

(∂F3

∂y− ∂F2

∂ z

)i+(

∂F1

∂ z− ∂F3

∂x

)j+(

∂F2

∂x− ∂F1

∂y

)k,

defined at all points at which all the indicated partial derivatives exist.

We memorize the equation CurlF = ∇×F:

∇×F =

∣∣∣∣∣∣∣∣∣i j k∂∂x

∂∂y

∂∂ z

F1 F2 F3

∣∣∣∣∣∣∣∣∣=(

∂F3

∂y− ∂F2

∂ z

)i+(

∂F1

∂ z− ∂F3

∂x

)j+(

∂F2

∂x− ∂F1

∂y

)k = CurlF.

Example 14.5.2. Compute CurlF for (1) F(x,y,z) = x2i−3xyj+xk and (2) F(x,y,z) =⟨xy2,3y2z2,2x− zy3⟩.

ANSWER. From the definition, we need the following derivatives:

1. Differentiate F1 with respect to only y and z: ∂F1∂y and ∂F1

∂ z

2. Differentiate F2 with respect to only x and z: ∂F2∂x and ∂F2

∂ z

3. Differentiate F3 with respect to only x and y: ∂F3∂x and ∂F3

∂y

(1) We compute those partial derivatives mentioned above:

∂F1

∂y=

∂∂y

(x2)= 0,

∂F1

∂ z=

∂∂ z

(x2)= 0,

∂F2

∂x=

∂∂x

(−3xy) =−3y,∂F2

∂ z=

∂∂ z

(−3xy) = 0,

∂F3

∂x=

∂∂x

(x) = 1,∂F3

∂y=

∂∂y

(x) = 0.

Putting them into the definition, we have

CurlF =

(∂F3

∂y− ∂F2

∂ z

)i+(

∂F1

∂ z− ∂F3

∂x

)j+(

∂F2

∂x− ∂F1

∂y

)k

= (0−0) i+(0−1) j+(−3y−0)k =−j−3k.

(2) We compute those partial derivatives mentioned above:

∂F1

∂y=

∂∂y

(xy2)= 2xy,

∂F1

∂ z=

∂∂ z

(xy2)= 0,

∂F2

∂x=

∂∂x

(3y2z2)= 0,

∂F2

∂ z=

∂∂ z

(3y2z2)= 6y2z,

∂F3

∂x=

∂∂x

(2x− zy3)= 2,

∂F3

∂y=

∂∂y

(2x− zy3)=−3zy2.

Putting them into the definition, we have

CurlF =

(∂F3

∂y− ∂F2

∂ z

)i+(

∂F1

∂ z− ∂F3

∂x

)j+(

∂F2

∂x− ∂F1

∂y

)k

=(−3zy2 −6y2z

)i+(0−2) j+(0−2xy)k =−9y2zi−2j−2xyk. □

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Example 14.5.3. Compute the curl for (1) F(x,y,z) = xi+ yj and (2) G(x,y,z) = yi− xj.

ANSWER. A simple computation shows

CurlF = ⟨0,0,0⟩= 0, CurlG = ⟨0,0,−2⟩=−2k. □

Definition 14.5.4. If the vector field F satisfies ∇×F = 0, i.e., CurlF = 0, at a point, then we say that thevector field F is irrotational at that point. (That is, the fluid does not tend to rotate near the point.)

□ 14.5.2 Divergence.Another important measure on the behavior of a fluid is the divergence.

Definition 14.5.5. The divergence of the vector field F(x,y,z) = ⟨F1(x,y,z),F2(x,y,z),F3(x,y,z)⟩ is the scalarfunction defined by

DivF(x,y,z) =∂F1

∂x+

∂F2

∂y+

∂F3

∂ z,

defined at all points at which all the indicated partial derivatives exist.

We memorize the equation DivF = ∇ ·F:

∇ ·F =

⟨∂∂x

,∂∂y

,∂∂ z

⟩· ⟨F1,F2,F3⟩=

∂F1

∂x+

∂F2

∂y+

∂F3

∂ z= DivF.

Example 14.5.6. Compute the divergence for the given vector field:

1. F(x,y,z) = xezi+ yz2j+(x+ y)k.2. F(x,y,z) =

⟨y2,x2ez,cos(xy)

⟩.

ANSWER. By the definition, we have

1. DivF =∂F1

∂x+

∂F2

∂y+

∂F3

∂ z=

∂∂x

(xez)+∂∂y

(yz2)+ ∂

∂ z(x+ y) = ez + z2 +0 = ez + z2.

2. DivF =∂F1

∂x+

∂F2

∂y+

∂F3

∂ z=

∂∂x

(y2)+ ∂

∂y

(x2ez)+ ∂

∂ z(cos(xy)) = 0+0+0 = 0. □

Example 14.5.7. Compute the divergence for F(x,y,z) = xi+ yj and G(x,y,z) = yi− xj.

ANSWER. By the definition, we have

DivF = 2, DivG = 0. □

Definition 14.5.8. If the vector field F satisfies ∇ ·F = 0, i.e., DivF = 0, throughout some region D, then wesay that the vector field F is source–free or incompressible.

Example 14.5.9. If f (x,y,z) is a scalar function and F(x,y,z) is a vector field, determine whether eachoperation is a scalar function, a vector field, or undefined: (1) ∇× (∇F), (2) ∇× (∇ ·F), (3) ∇ · (∇ f ).

We can now summarize a number of equivalent properties for three–dimensional vector fields.

Theorem 14.5.10 (CONSERVATIVE VECTOR FIELDS). Suppose F(x,y,z) = ⟨F1(x,y,z),F2(x,y,z),F3(x,y,z)⟩is a vector field whose components F1, F2 and F3 have continuous first partial derivatives throughout all ofR3. Then the followings are equivalent1:

1. F(x,y,z) is conservative.

1equal in amount or value; like, equal, same; being essentially equal to something

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2.´

C F · dr is independent of path.3.´

C F · dr = 0 for every piecewise–smooth closed curve C.4. ∇×F = 0.5. F(x,y,z) is a gradient field (i.e., F = ∇ f for some potential function f ).

§14.6 Surface Integrals.Skip. Please read the textbook.

§14.7 The Divergence Theorem.Skip. Please read the textbook.

§14.8 Stokes’ Theorem.Skip. Please read the textbook.

§14.9 Applications of Vector Calculus.Skip. Please read the textbook.

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Chapter 15

Appendix: Distances between Various Objects

§15.1 Two–Dimensional Space.

□ 15.1.1 Two Points P(x0,y0) and Q(x1,y1).

.

.distance d

.P(x0,y0)

.Q(x1,y1)

Using the vector−→PQ, the distance d between P and Q is the magnitude of the vector

−→PQ:

d = ∥−→PQ∥= ∥⟨x1 − x0,y1 − y0⟩∥=√(x1 − x0)2 +(y1 − y0)2

Theorem 15.1.1. The distance d between two points P(x0,y0) and Q(x1,y1) is obtained by

d =√

(x1 − x0)2 +(y1 − y0)2 .

□ 15.1.2 Line L : y = mx+b and Point P(x0,y0) not lying on L.

.

.Line L : y = mx+b.P(x0,y0)

.Q(x1,y1).90◦

.distance d

The line L : y = mx+b can be parameterized by

L : x = t, y = mt +b,

which is parallel to the vector v = ⟨1,m⟩. Two vectors−→PQ = ⟨x1 − x0,y1 − y0⟩ and v = ⟨1,m⟩ should be

perpendicular, i.e.,x1 − x0 +(y1 − y0)m = 0. (15.1.1)

Since Q(x1,y1) lies on the line L, soy1 = mx1 +b. (15.1.2)

Solving two equations (15.1.1) and (15.1.2) for x1 and y1, we have

x1 =x0 +(y0 −b)m

1+m2 , y1 =b+(x0 +my0)m

1+m2 .

Thus, the distance d between P and Q (which is the distance between P and the line L) is the magnitude ofthe vector

−→PQ and it is computed by

d = ∥−→PQ∥= ∥⟨x1 − x0,y1 − y0⟩∥=√(x1 − x0)2 +(y1 − y0)2 =

|mx0 +b− y0|√1+m2

.

117


Theorem 15.1.2. The distance d between the line L : y = mx+ b and the point P(x0,y0) not lying on L isobtained by

d =|mx0 +b− y0|√

1+m2.

Example 15.1.3. Find the distance between the line y = 2x+3 and the point P(0,5).

ANSWER. By the Theorem 15.1.2, the distance d is computed by

d =|mx0 +b− y0|√

1+m2=

|2(0)+3−5|√1+22

=2√5. □

□ 15.1.3 Parallel Lines L : y = mx+b and M : y = mx+ c.

.

.Line L : y = mx+b .Line M : y = mx+ c

.P(x0,y0)

.Q(x1,y1)

.90◦

.90◦

.distance d

Applying the Theorem (15.1.2) to the point P(x0,y0) and the line M : y = mx+ c, we deduce the distance,

d =|mx0 + c− y0|√

1+m2=

|c−b|√1+m2

,

where y0 = mx0 +b is used.

Theorem 15.1.4. The distance d between parallel lines L : y = mx+b and M : y = mx+ c is obtained by

d =|c−b|√1+m2

.

Example 15.1.5. Find the distance between two parallel lines L : y = 2x+3 and M : y = 2x+7.

ANSWER. By the Theorem 15.1.4, the distance d is computed by

d =|c−b|√1+m2

=|7−3|√1+22

=4√5. □

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§15.2 Three–Dimensional Space.

□ 15.2.1 Two Points P(x0,y0,z0) and Q(x1,y1,z1).

.

.distance d

.P(x0,y0,z0)

.Q(x1,y1,z1)

Using the vector−→PQ, the distance d between P and Q is the magnitude of the vector

−→PQ:

d = ∥−→PQ∥= ∥⟨x1 − x0,y1 − y0,z1 − z0⟩∥=√

(x1 − x0)2 +(y1 − y0)2 +(z1 − z0)2

Theorem 15.2.1. The distance d between two points P(x0,y0,z0) and Q(x1,y1,z1) is obtained by

d =√(x1 − x0)2 +(y1 − y0)2 +(z1 − z0)2 .

□ 15.2.2 Line L and Point Q(x0,y0,z0) not lying on L:L : x = x1 +at, y = y1 +bt, z = z1 + ct.

.

.Line L : x = x1 +at, y = y1 +bt, z = z1 + ct.Q(x0,y0,z0)

.P(x1,y1,z1)

.R(x2,y2,z2).90◦

.distance d

We choose two points P(x1,y1,z1) and R(x2,y2,z2) lying on the line L. Then by the formula on the crossproduct, we have

∥−→PQ×−→PR∥= ∥−→PQ∥∥−→PR∥sinθ = ∥−→PR∥(distance d), d =

∥−→PQ×−→PR∥

∥−→PR∥. (15.2.1)

The vector v = ⟨a,b,c⟩ is parallel to the line L and points P and R are on the line L. Thus,−→PR is parallel to

the vector v = ⟨a,b,c⟩ and so−→PR = sv,

for some scalar s. It implies that the formula (15.2.1) on the distance d can be simplified by

d =∥−→PQ×−→

PR∥∥−→PR∥

=∥s(−→

PQ×v)∥

∥sv∥=

∥−→PQ×v∥∥v∥

. (15.2.2)

From the formula (15.2.2), we observe that to compute the distance, we need to choose only one point P onthe line L and use the vector v parallel to the line L.

Theorem 15.2.2. The distance d between the line L and the point Q(x0,y0,z0) not lying on L : x= x1+at, y=y1 +bt, z = z1 + ct is obtained by

d =∥−→PQ×v∥

∥v∥,

where v = ⟨a,b,c⟩ and P is any point on the line L.

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Example 15.2.3. Find the distance between the line L and the point Q(3,−1,7) not lying on L:

L : x = 2−2t, y = 2+ t, z =−1+6t.

ANSWER. The vector v = ⟨−2,1,6⟩ is parallel to the line L. When we choose a point P(2,2,−1) from theline L by setting t = 0, we obtain the vector

−→PQ = ⟨1,−3,8⟩. By the Theorem (15.2.2), the distance between

the line L and the point Q is obtained by

d =∥−→PQ×v∥

∥v∥=

∥⟨1,−3,8⟩×⟨−2,1,6⟩∥∥⟨−2,1,6⟩∥

=∥⟨−26,−22,−5⟩∥

∥⟨−2,1,6⟩∥=

√1185√

41. □

□ 15.2.3 Parallel Lines L and M:L : x = x0 +at,y = y0 +bt,z = z0 + ct and M : x = x1 +a′s,y = y1 +b′s,z = z1 + c′s.

.

.Line L : x = x0 +at,y = y0 +bt,z = z0 + ct

.Line M : x = x1 +a′s,y = y1 +b′s,z = z1 + c′s

.P(x0,y0,z0)

.Q(x1,y1,z1).90◦

.90◦.distance d

We fix one of the lines and choose a point from the other line. Note that there exists a particular scalar k suchthat v′ = ⟨a′,b′,c′⟩= k ⟨a,b,c⟩= kv by the condition that L and M are parallel.(1) Let us fix the line L and choose a point Q(x1,y1,z1) from the line M. We choose a point P(x0,y0,z0) fromthe line L and obtain the vector

−→PQ. By the Theorem 15.2.2, the distance between the line L and the point Q

is obtained by

d =∥−→PQ×v∥

∥v∥,

which is exactly same as the Theorem 15.2.2.(2) Let us fix the line M and choose a point P(x0,y0,z0) from the line L. We choose a point Q(x1,y1,z1) fromthe line M and obtain the vector

−→PQ. By the Theorem 15.2.2, the distance between the line M and the point

P is obtained by

d =∥−→PQ×v′∥

∥v′∥=

∥k(−→

PQ×v)∥

∥kv∥=

∥−→PQ×v∥∥v∥

,

which is exactly same as the Theorem 15.2.2.For this reason, we can use either the vector parallel to the line L or the vector parallel to the line M in theTheorem 15.2.2. If we choose P from the line L, then we should choose Q from the other line M and viceversa.

Theorem 15.2.4. The distance d between parallel lines L : x = x0 + at,y = y0 + bt,z = z0 + ct and M : x =x1 +a′s,y = y1 +b′s,z = z1 + c′s is obtained by

d =∥−→PQ×v∥

∥v∥, or d =

∥−→PQ×v′∥∥v′∥

,

where v = ⟨a,b,c⟩. If we choose P from the line L, then we should choose Q from the other line M and viceversa.

Example 15.2.5 (MIDTERM EXAM PROBLEM). Find the distance between two lines L and M:

L : x = 1+2t, y= 3, z =−1−4t,M : x = 2− s, y = 2, z = 3+2s.

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ANSWER. We observe that the lines L and M are parallel to the vector v = ⟨1,0,−2⟩. We choose pointsP(1,3,−1) and Q(2,2,3) from each line L and M, respectively, and form the vector

−→PQ = ⟨1,−1,4⟩ and use

the Theorem 15.2.4:

d =∥−→PQ×v∥

∥v∥=

∥⟨1,−1,4⟩×⟨1,0,−2⟩∥∥⟨1,0,−2⟩∥

=∥⟨2,6,1⟩∥∥⟨1,0,−2⟩∥

=

√41√5. □

□ 15.2.4 Skew Lines L and M (Version 1):L : x = xL +aLt, y = yL +bLt, z = zL + cLt andM : x = xM +aMs, y = yM +bMs, z = zM + cMs.

.

.Line L : x = xL +aLt,y = yL +bLt,z = zL + cLt

.Line M : x = xM +aMs,y = yM +bMs,z = zM + cMs

.P(x0,y0,z0)

.Q(x1,y1,z1)

.90◦

.90◦

.distance d

Let PL(xL,yL,zL) and QM(xM,yM,zM) be the points on L and M given in the parametric equations of L and Mand let vL = ⟨aL,bL,cL⟩ and vM = ⟨aM,bM,cM⟩.Let P(x0,y0,z0) and Q(x1,y1,z1) be points on L and M, respectively, such that the line passing through bothP and Q is perpendicular to both lines L and M. Then the length of the line segment of PQ or the magnitudeof the vector

−→PQ becomes the distance between skew lines L and M.

We observe(1) The vector

−→PQ is perpendicular to both lines L and M, i.e., perpendicular to both vectors vL = ⟨aL,bL,cL⟩

and vM = ⟨aM,bM,cM⟩. It implies that−→PQ is parallel to the cross product vL ×vM, i.e.,

−→PQ = k (vL ×vM) ,

where k is a scalar.(2) If we introduce a unit vector, we observe

−→PQ =± vL ×vM

∥vL ×vM∥(distance d between P and Q) =±d

vL ×vM

∥vL ×vM∥. (15.2.3)

Suppose P(x0,y0,z0) and Q(x1,y1,z1) are obtained when t = t0 and s = s0, i.e.,

x0 = xL +aLt0, y0 = yL +bLt0, z0 = zL + cLt0,x1 = xM +aMs0, y1 = yM +bMs0, z1 = zM + cMs0.

Then, the vector−→PQ is obtained by

−→PQ = ⟨x1 − x0,y1 − y0,z1 − z0⟩

= ⟨xM +aMs0 − xL −aLt0, yM +bMs0 − yL −bLt0, zM + cMs0 − zL − cLt0⟩= ⟨xM − xL, yM − yL, zM − zL⟩+ s0 ⟨aM,bM,cM⟩− t0 ⟨aL,bL,zL⟩

=−−−→PLQM + s0vM − t0vL.

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Combining with the result (15.2.3), we deduce

±dvL ×vM

∥vL ×vM∥=−→PQ =

−−−→PLQM + s0vM − t0vL −→ ±d

vL ×vM

∥vL ×vM∥=−−−→PLQM + s0vM − t0vL.

Applying the dot product with vL ×vM on both sides, we get

±dvL ×vM

∥vL ×vM∥· (vL ×vM) =

−−−→PLQM · (vL ×vM)+ s0vM · (vL ×vM)− t0vL · (vL ×vM)

±d∥vL ×vM∥2

∥vL ×vM∥=−−−→PLQM · (vL ×vM) , ±d∥vL ×vM∥=−−−→

PLQM · (vL ×vM) .

Therefore, we deduce the formula on the distance d between skew lines L and M:

d =|−−−→PLQM · (vL ×vM) |

∥vL ×vM∥

Theorem 15.2.6. The skew lines L and M given by

L : x = xL +aLt, y = yL +bLt, z = zL + cLt, and M : x = xM +aMs, y = yM +bMs, z = zM + cMs

has the distance

d =|−−−→PLQM · (vL ×vM) |

∥vL ×vM∥where vL = ⟨aL,bL,cL⟩, vM = ⟨aM,bM,cM⟩ and PL(xL,yL,zL) and QM(xM,yM,zM).

Example 15.2.7. Find the distance between skew lines L and M:

L : x = 3t, y = 9− t, z = 2+ t,M : x =−6−3s, y =−5+2s, z = 10+4s.

ANSWER. The lines L and M are skew. (Check this out by yourself!)We observe that the lines L and M are parallel to the vectors vL = ⟨3,−1,1⟩ and vM = ⟨−3,2,4⟩, respectively.The lines L and M also pass through the points PL(0,9,2) and QM(−6,−5,10), respectively, and so we havethe vector

−−−→PLQM = ⟨−6,−14,8⟩.

Now, the Theorem 15.2.6 implies

d =|−−−→PLQM · (vL ×vM) |

∥vL ×vM∥=

| ⟨−6,−14,8⟩ · (⟨3,−1,1⟩×⟨−3,2,4⟩) |∥⟨3,−1,1⟩×⟨−3,2,4⟩∥

=1

∥⟨−6,−15,3⟩∥

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣−6 −14 8

3 −1 1

−3 2 4

∣∣∣∣∣∣∣∣determinant

∣∣∣∣∣∣∣∣absolute

=270

3√

30= 3

√30 . □

□ 15.2.5 Plane Π : ax+by+ cz = d and Point P(x0,y0,z0) not lying on Π.

.

.P(x0,y0,z0)

.Q(x1,y1,z1)

.90◦

.distance D .Plane Π : ax+by+ cz = d

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The vector−→PQ is parallel to the normal vector n = ⟨a,b,c⟩ of the plane Π. So there exists a unique1 scalar s0

such that−→PQ = s0n −→ x1 = x0 +as0, y1 = y0 +bs0, z1 = z0 + cs0.

Since Q(x1,y1,z1) is on the plane Π, we have

d = ax1 +by1 + cz1 = a(x0 +as0)+b(y0 +bs0)+ c(z0 + cs0)

= ax0 +by0 + cz0 + s0(a2 +b2 + c2)= ax0 +by0 + cz0 + s0∥n∥2

s0 =d − (ax0 +by0 + cz0)

∥n∥2 .

The distance D between the plane Π and the point P(x0,y0,z0) is the magnitude of the vector−→PQ and so we

have

D = ∥−→PQ∥= ∥s0n∥= |s0|∥n∥=∣∣∣∣d − (ax0 +by0 + cz0)

∥n∥2

∣∣∣∣∥n∥= |d − (ax0 +by0 + cz0) |∥n∥

.

Theorem 15.2.8. The distance D between the plane Π : ax+by+ cz = d and the point P(x0,y0,z0) not lyingon Π is obtained by

D =|ax0 +by0 + cz0 −d|

∥n∥=

|ax0 +by0 + cz0 −d|√a2 +b2 + c2

.

Example 15.2.9. Find the distance between the plane Π : 2x− 3y+ z = 6 and the point P(1,0,0) not lyingon Π.

ANSWER. The plane Π has the normal vector n = ⟨2,−3,1⟩ and the Theorem 15.2.8 implies

D =|ax0 +by0 + cz0 −d|

∥n∥=

|2(1)−3(0)+1(0)−6|√14

=4√14

. □

□ 15.2.6 Plane Π : ax+by+cz = d and Line L : x = x0+a′t,y = y0+b′t,z = z0+c′t not intersecting withΠ.We choose the point P(x0,y0,z0) from the line L and use the Theorem 15.2.8.

Theorem 15.2.10. The distance D between the plane Π : ax + by+ cz = d and line L : x = x0 + a′t,y =y0 +b′t,z = z0 + c′t not intersecting with Π is obtained by

D =|ax0 +by0 + cz0 −d|

∥n∥=

|ax0 +by0 + cz0 −d|√a2 +b2 + c2

.

Example 15.2.11. Find the distance between the plane Π : 2x − 3y− 3z = 6 and the line L : x = 3t,y =9+ t,z = 2+ t not intersecting with Π.

ANSWER. The plane Π has the normal vector n = ⟨2,−3,−3⟩ and the line L passes through the pointP(0,9,2). The Theorem 15.2.10 implies

D =|ax0 +by0 + cz0 −d|

∥n∥=

|2(0)−3(9)−3(2)−6|√22

=

√39√22

. □

1radically distinctive and without equal; the single one of its kind

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□ 15.2.7 Parallel Planes Π0 : ax+by+ cz = d and Π1 : a′x+b′y+ c′z = e.

.

.P(x0,y0,z0)

.Q(x1,y1,z1)

.90◦

.90◦

.distance D

.Plane Π0 : ax+by+ cz = d

.Plane Π1 : a′x+b′y+ c′z = e

We fix one of the planes and choose a point from the other plane. Note that there exits a particular scalar ksuch that n′ = ⟨a′,b′,c′⟩= k ⟨a,b,c⟩= kn by the condition that Π0 and Π1 are parallel. It implies

Π1 : a′x+b′y+ c′z = e ⇐⇒ Π1 : kax+ kby+ kcz = e ⇐⇒ Π1 : ax+by+ cz =ek.

(1) Let us fix the plane Π0 having the normal vector n = ⟨a,b,c⟩ and choose a point Q(x1,y1,z1) from theplane Π1. Then the Theorem 15.2.8 implies

D =|ax1 +by1 + cz1 −d|

∥n∥=

|e/k−d|∥n∥

=|e/k−d|√

a2 +b2 + c2.

(2) Let us fix the plane Π1 having the normal vector n′ = ⟨a′,b′,c′⟩= kn and choose a point P(x0,y0,z0) fromthe plane Π0. Then the Theorem 15.2.8 implies

D =|a′x0 +b′y0 + c′z0 − e|

∥n′∥=

|kax0 + kby0 + kcz0 − e|k∥n∥

=|ax0 +by0 + cz0 − e/k|

∥n∥=

|d − e/k|∥n∥

.

As we can notice, two results in (1) and (2) are exactly same. Thus, we conclude

D =|d − e/k|

∥n∥=

|d − e/k|√a2 +b2 + c2

.

Theorem 15.2.12. The distance D between two parallel planes Π0 : ax+by+cz= d and Π1 : a′x+b′y+c′z=e is obtained by

D =|d − e/k|

∥n∥=

|d − e/k|√a2 +b2 + c2

.

If a′ = a, b′ = b and c′ = c, then k = 1. The Theorem 15.2.12 gives a corollary.

Corollary 15.2.13. The distance D between two planes having the same normal vectors, i.e., Π0 : ax+by+cz = d and Π1 : ax+by+ cz = e, is obtained by

D =|d − e|∥n∥

=|d − e|√

a2 +b2 + c2.

Example 15.2.14. Find the distance between parallel planes Π0 : 2x−3y+ z = 6 and Π1 : 4x−6y+2z = 8.

ANSWER. The planes Π0 and Π1 have parallel normal vectors n0 = ⟨2,−3,1⟩ and n1 = ⟨4,−6,2⟩ withn1 = 2n0. The Theorem 15.2.12 implies

D =|d − e/k|√

a2 +b2 + c2=

|6−8/2|√22 +(−3)2 +12

=2√14

. □

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□ 15.2.8 Appendix: Skew Lines L and M (Version 2):L : x = xL +aLt, y = yL +bLt, z = zL + cLt andM : x = xM +aMs, y = yM +bMs, z = zM + cMs.

.

.Line L : x = xL +aLt,y = yL +bLt,z = zL + cLt

.Line M : x = xM +aMs,y = yM +bMs,z = zM + cMs

.P(x0,y0,z0)

.Q(x1,y1,z1)

.90◦

.90◦

.distance d

Let us develop an argument using the formula on the distance between a plane and a point not lying on theplane.Let PL(xL,yL,zL) and QM(xM,yM,zM) be the points on L and M given in the parametric equations of L and Mand let vL = ⟨aL,bL,cL⟩ and vM = ⟨aM,bM,cM⟩.Let P(x0,y0,z0) and Q(x1,y1,z1) be points on L and M, respectively, such that the line passing through bothP and Q is perpendicular to both lines L and M, i.e., perpendicular to vL and vM. Then we can find a plane Πcontaining the line L and having the vector

−→PQ as its normal vector. Once we find the equation of the plane

Π, we use the Theorem 15.2.8 on the distance between the plane Π and any point on the line M.Since

−→PQ is perpendicular to both vL and vM, so

−→PQ is parallel to the cross product vL ×vM, i.e.,

−→PQ = s(vL ×vM) ,

for some scalar s. It implies that the plane Π containing the line L and having−→PQ as its normal vector is

expressed by the equation,Π : (vL ×vM) · ⟨x− xL,y− yL,z− zL⟩= 0,

where the point PL(xL,yL,zL) is chosen arbitrary from the line L.When we arbitrarily choose the point QM(xM,yM,zM) on the line M, the Theorem 15.2.8 implies the distancebetween Π and the point QM,

D =|(vL ×vM) · ⟨xM − xL,yM − yL,zM − zL⟩ |

∥vL ×vM∥=

|−−−→PLQM · (vL ×vM) |∥vL ×vM∥

,

which is exactly same as the Theorem 15.2.6.Let us rewrite the result above. If vL ×vM = ⟨a,b,c⟩, then the equation of the plane Π becomes

Π : a(x− xL)+b(y− yL)+ c(z− zL) = 0 −→ Π : ax+by+ cz = d,

where d = axL +byL + czL. The result above becomes

D =|−−−→PLQM · (vL ×vM) |

∥vL ×vM∥=

|axM +byM + czM −d|√a2 +b2 + c2

.

Theorem 15.2.15. The skew lines L and M given by

L : x = xL +aLt, y = yL +bLt, z = zL + cLt, and M : x = xM +aMs, y = yM +bMs, z = zM + cMs

has the distance

d =|−−−→PLQM · (vL ×vM) |

∥vL ×vM∥where vL = ⟨aL,bL,cL⟩, vM = ⟨aM,bM,cM⟩ and PL(xL,yL,zL) and QM(xM,yM,zM).

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Personally, I do not recommend this argument on finding the distance between skew lines. Please try to usethe Theorem 15.2.6 from the Subsection 15.2.4 Distance between Skew Lines (Version 1).

Example 15.2.16. Find the distance between skew lines L and M:

L : x = 3t, y = 9− t, z = 2+ t,M : x =−6−3s, y =−5+2s, z = 10+4s.

ANSWER. The lines L and M are skew. (Check this out by yourself!)We observe that the lines L and M are parallel to the vectors vL = ⟨3,−1,1⟩ and vM = ⟨−3,2,4⟩, respectively.The lines L and M also pass through the points PL(0,9,2) and QM(−6,−5,10), respectively, and so we havethe vector

−−−→PLQM = ⟨−6,−14,8⟩.

The plane Π containing the line L and having vL ×vM = ⟨−6,−15,3⟩ as its normal vector is obtained by

Π : −6(x−0)−15(y−9)+3(z−2) = 0 −→ Π : 2x+5y− z = 43.

Applying the Theorem 15.2.15 with the point QM(−6,−5,10) on the line M, we have the distance

D =|axM +byM + czM −d|√

a2 +b2 + c2 ∥=

|2(−6)+5(−5)−10−43|√22 +52 +(−1)2

=90√30

= 3√

30 ,

which is exactly same result as in the Example 15.2.7 on page 122. □

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