15.1 functions of two or more...

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§15.1 FUNCTIONS OF TWO OR MORE VARIABLES §15.1 Functions of Two or More Variables After completing this section, students should be able to: Match equations of the form z = f (x, y) to graphs of surfaces and graphs of level curves. Describe the graphs of functions of three variables w = f (x, y, z) in terms of the level curves f (x, y, z) = k 124

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Page 1: 15.1 Functions of Two or More Variableslindagreen.web.unc.edu/files/2020/08/courseNotes_math233_2019F_… · §15.1 FUNCTIONS OF TWO OR MORE VARIABLES §15.1 Functions of Two or More

§15.1 FUNCTIONS OF TWO OR MORE VARIABLES

§15.1 Functions of Two or More Variables

After completing this section, students should be able to:

• Match equations of the form z = f (x, y) to graphs of surfaces and graphs of levelcurves.

• Describe the graphs of functions of three variables w = f (x, y, z) in terms of thelevel curves f (x, y, z) = k

124

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§15.1 FUNCTIONS OF TWO OR MORE VARIABLES

Example. Consider the function of two variables f (x, y) = pxy

1. What is its domain?

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§15.1 FUNCTIONS OF TWO OR MORE VARIABLES

For the function f (x, y) = pxy ...

2. What are its level curves?

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§15.1 FUNCTIONS OF TWO OR MORE VARIABLES

For the function f (x, y) = pxy ...

3. What does its graph look like?

END OF VIDEO

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§15.1 FUNCTIONS OF TWO OR MORE VARIABLES

Question. What do the lines in this contour map represent? Where should you go ifyou like steep hills? Mountain tops?

Example. What type of curve do you get when you intersect the graph of z = (x2 � y)2

with a horizontal plane?

Definition. A level curve of a surface z = f (x, y) is ...

Definition. A contour map of a surface z = f (x, y) is ...

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§15.1 FUNCTIONS OF TWO OR MORE VARIABLES

Example. Find the contour map for the surface z = (x2 � y)2.

A. B. C. D.Example. Find the 3-d graph of the surface z = (x2 � y)2.

1. 2. 3. 4.

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§15.1 FUNCTIONS OF TWO OR MORE VARIABLES

1. z = sinp

x2 + y2 2. z = 1x2+4y2 3. z = sin(x) sin(y)

4. z = x2y

2e�(x2+y

2) 5. z = x3 � 3xy

2 6. z = sin2(x) + y2

4

I. II. III.

IV. V. VI.

A. B. C.

D. E. F.

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§15.1 FUNCTIONS OF TWO OR MORE VARIABLES

Example. A contour map for a function f is shown. Use it to estimate the values off (�3, 3) and f (3,�2). What can you say about the shape of the graph?

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§15.1 FUNCTIONS OF TWO OR MORE VARIABLES

Functions of 3 or more variables

To visualize functions f (x, y, z) of three variables, it is handy to look at level surfaces.Example. f (x, y, z) = x

2 + y2 + z

2

(a) Guess what the level surfaces should look like.

(b) Graph a few level surfaces (e.g. x2 + y

2 + z2 = 10, x

2 � y2 + z

2 = 20, x2 � y

2 + z2 = 30)

on a 3-d plot.

Example. f (x, y, z) = x2 � y

2 + z2

1. Guess what the level surfaces should look like.

2. Graph a few level surfaces (e.g. x2 � y

2 + z2 = 0, x

2 � y2 + z

2 = 10, x2 � y

2 + z2 = 20)

on a 3-d plot.

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§15.2 LIMITS AND CONTINUITY

§15.2 Limits and Continuity

After completing this section, students should be able to:

1. Use a graph to build intuition for whether or not a limit of a function of twovariables exists.

2. Prove that a limit of a function of two variables does not exist by finding two pathsalong which the function approaches di↵erent values.

3. Prove that a limit of a function of two variables does exist by converting to polarcoordinates and using the squeeze theorem.

4. State the definition of continuity for functions of two variables in terms of limits.

5. Determine where a function is continuous.

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§15.2 LIMITS AND CONTINUITY

Recall LIMITS from Calculus 1:

Informally, limx!a

f (x) = L if the y-values f (x) get closer and closer to the same number L

when x approaches a from either the left of the right.

Does limx!0

f (x) exist for these functions?

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§15.2 LIMITS AND CONTINUITY

LIMITS for functions of two variables:

Informally, lim(x,y)!(a,b)

f (x, y) = L if the z-values f (x, y) ...

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§15.2 LIMITS AND CONTINUITY

For each function, decide if lim(x,y)!(0,0)

f (x, y) exists. The color is based on height: low is

blue and high is red.

A) lim(x,y)!(0,0)

x2 � y

2

x2 + y2 B) lim(x,y)!(0,0)

1 � x2

x2 + y2 + 1

C) lim(x,y)!(0,0)

x

x2 + y2 D) lim(x,y)!(0,0)

x3

x2 + y2

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§15.2 LIMITS AND CONTINUITY

Example. lim(x,y)!(0,0)

x2 � y

2

x2 + y2

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§15.2 LIMITS AND CONTINUITY

Example. lim(x,y)!(0,0)

1 � x2

x2 + y2 + 1

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§15.2 LIMITS AND CONTINUITY

Example. lim(x,y)!(0,0)

x

x2 + y2

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§15.2 LIMITS AND CONTINUITY

Example. lim(x,y)!(0,0)

x3

x2 + y2

END OF VIDEO

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§15.2 LIMITS AND CONTINUITY

Review. Informally, lim(x,y)!(a,b)

f (x, y) = L if the z-values f (x, y) ...

lim(x,y)!(0,0)

x2 � y

2

x2 + y2 lim(x,y)!(0,0)

x3

x2 + y2

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§15.2 LIMITS AND CONTINUITY

Does lim(x,y)!(0,0)

f (x, y) exist? The color is based on height: low is blue and high is red.

A) lim(x,y)!(0,0)

2xy

x2 + y2 B) lim(x,y)!(0,0)

xy2

x2 + y4

C) lim(x,y)!(0,0)

10 sin(xy)x + y + 20

D) lim(x,y)!(0,0)

3x2y � 3x

2 � 3y2

x2 + y2

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§15.2 LIMITS AND CONTINUITY

Example. Show that lim(x,y)!(0,0)

2xy

x2 + y2 does not exist.

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§15.2 LIMITS AND CONTINUITY

Example. Show that lim(x,y)!(0,0)

xy2

x2 + y4 does not exist.

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§15.2 LIMITS AND CONTINUITY

Example. Show that lim(x,y)!(0,0)

10 sin(xy)x + y + 20

does exist.

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§15.2 LIMITS AND CONTINUITY

Example. Show that lim(x,y)!(0,0)

3x2y � 3x

2 � 3y2

x2 + y2 does exist.

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§15.2 LIMITS AND CONTINUITY

Summary: for practical purposes, the best way to show that a limit does not exist is to:

For practical purposes, the best ways to show that a limit exists is to:

••

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§15.2 LIMITS AND CONTINUITY

Recall CONTINUITY from Calculus 1: A function f is continuous at the point x = a if:

1.

2.

3.

Recall from Calculus 1: common functions that are continuous include:

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§15.2 LIMITS AND CONTINUITY

CONTINUITY for functions of of two (or more) variables:

A function f (x, y) is continuous at the point (x, y) = (a, b) if:

1.

2.

3.

Common functions that are continuous include:

••••

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§15.2 LIMITS AND CONTINUITY

Example. Where is f (x, y) =

p4 � x2 +

p4 � y2

1 � x2 � y2 continuous?

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§15.3 PARTIAL DERIVATIVES

§15.3 Partial Derivatives

After completing this section, students should be able to:

• Compute partial derivatives.

• Use average rates of change to approximate partial derivatives.

• For functions of two variables, explain the geometric meaning of a partial deriva-tive as the slope of a tangent line to a curve in the intersection of a surface and aplane.

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§15.3 PARTIAL DERIVATIVES

Example. The wave heights h in the open sea depend on the speed ⌫ of the windand the length of time t that the wind has been blowing at that speed. So we writeh = f (⌫, t).

1. What is f (40, 20)?

2. If we fix duration at t = 20 hours andthink of g(⌫) = f (⌫, 20) as a functionof ⌫, what is the approximate valueof the derivative dg

d⌫

����⌫=40

?

3. If we fix wind speed at 40 knots, andthink of k(t) = f (40, t) as a function ofduration t, what is the approximatevalue of the derivative dk

dt

���t=20?

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§15.3 PARTIAL DERIVATIVES

Definition. For a function f (x, y) defined near (a, b), the partial derivatives of f at (a, b)are:

fx(a, b) = the derivative of f (x, b) with respect to x when x = a, and

fy(a, b) = the derivative of f (a, y) with respect to y when y = b.

In terms of the limit definition of derivatives, we have:

fx(a, b) = limh!0

fy(a, b) = limh!0

Geometrically, f (x, b) can be thought of as

So fx(a, b) = d

dxf (x, b)|x=a can be thought of as

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§15.3 PARTIAL DERIVATIVES

Note. To compute fx, we just take the derivative with x as our variable, holding allother variables constant. Similarly for the partial derivative with respect to any othervariable.

Example. f (x, y) = x

y. Find fx(1, 2) and fy(1, 2).

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§15.3 PARTIAL DERIVATIVES

Notation. There are many notations for partial derivatives, including the following:

fx@ f

@x@z@x f1 D1 f Dx f

Note. Partial derivatives can also be taken for functions of three or more variables. Forexample, if f (x, y, z,w) is a function of 4 variables, then fz(3, 4, 2, 7) means:

END OF VIDEO

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§15.3 PARTIAL DERIVATIVES

Review. Which of the following is the limit definition of@ f

@x(1, 2)?

A. limh!0

f (1 + h, 2 + h) � f (1, 2)h

B. limh!0

f (1 + h, 2) � f (1, 2 + h)h

C. limh!0

f (1 + h, 2) � f (1, 2)h

D. limh!0

f (1, 2 + h) � f (1, 2)h

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§15.3 PARTIAL DERIVATIVES

Review. Based on the graph of z = f (x, y) shown

• is@ f

@x(1, 2) positive, negative, or 0?

• is@ f

@y(1, 2) positive, negative, or 0?

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§15.3 PARTIAL DERIVATIVES

Question. For f (x, y) = 3x2y + 5y

2 + cos(xy), find@ f

@x.

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§15.3 PARTIAL DERIVATIVES

Example. For f (x, y, z) = ex sin(y�z), find Dx f and Dz f .

Example. For z = f (x)g(y), find an expression for@z@x

.

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§15.3 PARTIAL DERIVATIVES

Example. For x2 � y

2 + z2 � 2z = 12, find

@z@y

.

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§15.3 PARTIAL DERIVATIVES

Notation. Second derivatives are written using any of the following notations:

Example. For f (x, y) = x2 + x

2y

2 � 2y2, calculate fxx, fxy, fyx, and fyy.

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§15.3 PARTIAL DERIVATIVES

Theorem. (Clairout’s Thm) Let f (x, y) be defined on a disk D containing (a, b). If fxy

and fyx are both defined and on D, then fxy(a, b) = fyx(a, b).

Example. Consider the function

f (x, y) =

8>><>>:xy

x2�y

2

x2+y2 if (x, y) , (0, 0)

0 if (x, y) = (0, 0)

It turns out that

• for all x D2 f (x, 0) =

• for all y D1 f (0, y) =

Therefore,

• D2,1 f (0, 0) =

• D1,2 f (0, 0) =

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§15.3 PARTIAL DERIVATIVES

Question. True or False: There is a function f with continuous second partial deriva-tives such that fx = x and fy = x.

Extra Example. The wave equation is given by

@2u

@t2 = a2@

2u

@x2

where u(x, t) represents displacement, t represents time, x represents the distance fromone end of the wave, and a is a constant.

Verify that the function u(x, t) = sin(x � at) satisfies the wave equation.

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§15.3 PARTIAL DERIVATIVES

In Calc 1, a function is di↵erentiable means that f0(x) exists.

Intuitively, when a function is is di↵erentiable, it has a tangent line that is a goodapproximation of the function.

Question. True or False (make a guess): A function f (x, y) of two variables is di↵eren-tiable means that fx and fy exists.

Question. True or False (make a guess): Intuitively, if a function of two variables isdi↵erentiable, then it has a tangent plance that is a good approximation of the function.

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§15.4 THE CHAIN RULE

§15.4 The Chain Rule

After completing the section, students should be able to:

• Compute partial derivatives using the Chain Rule, using formulas for functions,or using numerical information such as tables of values or contour graphs.

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§15.4 THE CHAIN RULE

Remember the Chain Rule for functions of 1 variable:

dy

dt=

dy

dx· dx

dt

Theorem. (Chain Rule, Case 1) Suppose z = f (x, y) is a di↵erentiable function of x andy and x = g(t) and y = h(t) are both di↵erentiable functions of t. Then

dz

dt=

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§15.4 THE CHAIN RULE

Example. w = x cos(4y2z), x = t

2, y = 1 � t, z = 1 + 2t. Find dw

dt.

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§15.4 THE CHAIN RULE

Theorem. (Chain Rule, Case 2) Suppose z = f (x, y) is a di↵erentiable function of x andy and x = g(s, t) and y = h(s, t) are both di↵erentiable functions of t. Then

@z@s=

@z@t=

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§15.4 THE CHAIN RULE

Example. z = er cos(✓), r = st,✓ =

ps2 + t2. Find @z@s and @z@t .

END OF VIDEO

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§15.4 THE CHAIN RULE

Review. The Chain Rule If u is a di↵erentiable function of variables x, y, z, and eachof x, y, and z are di↵erentable functions of variables r, t, then

@u@r=

@u@t=

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§15.4 THE CHAIN RULE

Example. w = lnp

x2 + y2 + z2, x = sin t, y = cos t, z = tan t. Finddw

dt.

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§15.4 THE CHAIN RULE

Example. u = xety, x = ↵2�, y = �2�, t = �2↵. Find

@u@↵

when ↵ = �1, � = 2, � = 1.

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§15.4 THE CHAIN RULE

Example. A manufacturer modeled its yearly production P in millions of dollars as aCobb-Douglas function

P(L,K) = 1.47L0.65

K0.35

where L is the number of labor hours (in thousands) and K is the invested capital (inmillions of dollars). Suppose that when L is 30 and K is 8, the labor force is decreasingat a rate of 2000 labor hours per year and capital is increasing at a rate of $500,000 peryear. Find the rate of change of production per year. Note the units.

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§15.4 THE CHAIN RULE

Extra Example. Suppose f (x, y) = h(x2 � y2, 3xy

2 � 4), where h(u, v) is a di↵erentiablefunction.

1. Find equations for fx and fy in terms of hu and hv.

2. Use the contour graph for h to estimate fx and fy at the point (x, y) = (2, 1).

This is a contour graph for h not f .

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§15.5 DIRECTIONAL DERIVATIVES AND THE GRADIENT

§15.5 Directional Derivatives and the Gradient

After completing this section, students should be able to ...

• Find the directional derivative of a function of two or more variables, in thedirection of a vector.

• Find the gradient of a function of two or more variables.

• Explain how to calculate the directional derivative from the gradient.

• Find the direction of greatest increase for a function and the magnitude of thatincrease.

• Describe the relationship between the gradient vector and the level curves of afunction.

• Write down the equation for the tangent line for a level curve of a function of twovariables, and the equation of the tangent plane to the level surface of a functionof three variables, at a specified point.

• Write down the equation for a normal line to a level curve or level surface at aspecified point.

• Find the path of steepest descent on a surface.

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§15.5 DIRECTIONAL DERIVATIVES AND THE GRADIENT

Example. What is the approximate rate of change of the temperature function f (x, y)at Dubbo, New South Wales, in the direction of Sydney?

Definition. The directional derivative of f at (x0, y0) in the direction of the unit vector~u =< a, b > is given by:

D~u f (x0, y0) =

if this limit exists.

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§15.5 DIRECTIONAL DERIVATIVES AND THE GRADIENT

Equivalently,Definition. The directional derivative of f at (x0, y0) in the direction of the unit vector~u =< a, b > is given by:

Assuming f is di↵erentiable, what does the Chain Rule tell us about this derivative?

Theorem. For the unit vector ~u =< a, b >,

D~u f (x0, y0) = fx(x0, y0) + fy(x0, y0)

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§15.5 DIRECTIONAL DERIVATIVES AND THE GRADIENT

Note. For a vector ~v =< c, d > that is NOT a unit vector, the directional derivative inthe direction of ~v is defined as ...

Example. Find the directional derivative of f (x, y) = x cos(y) at (2, 0) in the direction of~v =< 3, 4 >

END OF VIDEO

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§15.5 DIRECTIONAL DERIVATIVES AND THE GRADIENT

Review. For a unit vector ~u =< a, b >, the directional derivative D~u f (x9, y0) is given bywhich of the following? (select all that apply)

A. limt!0

f (x0 + at, y0 + bt) � f (x0, y0)t

B.dg

dt

�����t=o

, where g(t) = f (x0 + at, y0 + bt)

C. a fx(x0, y0) + b fy(x0, y0)

D. The slope of the curve of intersection of the surface z = f (x, y) and the vertical planethat contains the vector < a, b >, if the curve is traversed in the direction of thevector < a, b >.

Question. True or False: It is possible for Du( f (x0, y0) to be positive for all unit vectors~u. (Assume f is di↵erentiable.)

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§15.5 DIRECTIONAL DERIVATIVES AND THE GRADIENT

Example. Find the directional derivative of f (x, y) =p

2x + 3y when (x, y) = (3, 1)

• in the direction ✓ = �⇡6

• in the direction of the vector ~w =< 5, 7 >

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§15.5 DIRECTIONAL DERIVATIVES AND THE GRADIENT

Definition. The gradient of the function f (x, y) at (x0, y0) is defined as:

r f (x0, y0) =

Note. For a unit vector ~u =< a, b >, D~u f (x0, y0) can be written in terms of the gradientas:

D~u f (x0, y0) =

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§15.5 DIRECTIONAL DERIVATIVES AND THE GRADIENT

Example. For f (x, y) = xe�y � y,

1. Find r f at the point (2, 0).

2. Find the directional derivative at (2, 0), in the direction of ~v =< 3,�4 >.

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§15.5 DIRECTIONAL DERIVATIVES AND THE GRADIENT

Question. For a di↵erentiable function f , at a point (x0, y0), in what direction is f

increasing most steeply? (i.e. for what unit vector ~u is D~u f maximal?) Answer interms of the gradient.

Theorem. The maximum value (most positive value) of D~u is and this maxi-mum occurs in the direction of .

The minimum value (most negative value) of D~u is and this minimum occursin the direction of .

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§15.5 DIRECTIONAL DERIVATIVES AND THE GRADIENT

Example. For f (x, y) = xe�y � y at (2, 0), what is the maximum directional derivative

and what direction does it occur in?

Note. For f (x, y) = xe�y � y, consider the graph of r f on the x-y plane drawn below.

Since r f points in the direction of greatest increase, where do you expect the levelcurves to be?

-4.8 -4 -3.2 -2.4 -1.6 -0.8 0 0.8 1.6 2.4 3.2 4 4.8

Theorem. For a di↵erentiable function f of two variables, the level curves of f areto r f .

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§15.5 DIRECTIONAL DERIVATIVES AND THE GRADIENT

Question. Why is r f perpendicular to the level curves of f ?

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§15.5 DIRECTIONAL DERIVATIVES AND THE GRADIENT

Example. Use the gradient to find the equation of the tangent line to the level curvefor f (x, y) = x

2 � y2 at (2, 3).

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§15.5 DIRECTIONAL DERIVATIVES AND THE GRADIENT

Note. The tangent line for a level curve f (x, y) = k at the point (x0, y0) is given by theequation:

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§15.5 DIRECTIONAL DERIVATIVES AND THE GRADIENT

Everything we have done so far can also be done for functions of three or morevariables! For f (x, y, z), and unit vector ~u =< a, b, c >

• r f (x0, y0, z0) =

• D~u f (x0, y0, z0) =

• The direction of greatest increase at the point (x0, y0, z0) is , and thelargest value of the directional derivative is .

• The direction of greatest decrease at the point (x0, y0, z0) is , andthe most negative value of the directional derivative is .

• For any curve in the level surface of f that passes through the point (a, b, c),r f (a, b, c)is perpendicular to ...

• r f (x, y, z) is perpendicular to .

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§15.5 DIRECTIONAL DERIVATIVES AND THE GRADIENT

• The equation for a tangent plane to a level surface f (x, y, z) = k at the point (x0, y0, z0)is given by the equation:

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§15.5 DIRECTIONAL DERIVATIVES AND THE GRADIENT

Example. Find the equation of the tangent plane to the surface xy + yz + zx = 5 at thepoint (1, 2, 1).

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§15.5 DIRECTIONAL DERIVATIVES AND THE GRADIENT

Example. The hyperbolic paraboloid z = f (x, y) = 1+x2�3y

2 is shown below. A ball isreleased at the point (1, 2, 18) on the surface and it follows the path of steepest descentC.

(a) Find the equation of the projection of C onto the xy-plane.

(b) Find the equation of C on the surface.

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§15.6 TANGENT PLANES AND LINEAR APPROXIMATIONS

§15.6 Tangent Planes and Linear Approximations

After completing this section, students should be able to:

• Find the equation for the tangent plane of a surface z = f (x, y) at a point.

• Use the tangent plane to approximate a function.

• Compute the di↵erential and use it to estimate errors.

• Explain the relationship between the tangent plane, the linearization equation, andthe di↵erential.

• Compute the tangent plane from information about the function’s value and partialderivative values.

• Use linearization to interpolate between numerical values in a 2 x 2 talbe.

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§15.6 TANGENT PLANES AND LINEAR APPROXIMATIONS

Tangent line:

Tangent plane:

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§15.6 TANGENT PLANES AND LINEAR APPROXIMATIONS

Example. Find the tangent plane to the surface z = y2 � x

2 at the point (1, 2, 3).

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§15.6 TANGENT PLANES AND LINEAR APPROXIMATIONS

Find a general formula for the tangent plane of z = f (x, y) at the point (x0, y0, z0).

Notice that this formula is analogous to the formula for a tangent line for a function ofone variable.

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§15.6 TANGENT PLANES AND LINEAR APPROXIMATIONS

Usually, the tangent plane at a point (x0, y0, z0) is a good approximation of the surfacenear that point.

Example. Find the tangent plane for the surface z = f (x, y) when (x, y) = (1, 1), wheref (x, y) = 1 � xy cos(⇡y). Use it to approximate f (1.02, 0.97).

END OF VIDEO

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§15.6 TANGENT PLANES AND LINEAR APPROXIMATIONS

NEED TO CHANGE THIS UP – start with review of Calc 1 formulas like in 2019 Fallfiled in notes, change up rest so it is not so repetitive, make the four numbers problema recitation problem? and consider adding one about the change in a quantity giventhe change in inputs, making one change pos and one negExample. Suppose we have a surface z = f (x, y) and we know that f (1, 2) = 6, fx(1, 2) =�7 and fy(1, 2) = 4. What is the equation for the tangent plane to the surface at thepoint (1, 2, 6)?

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§15.6 TANGENT PLANES AND LINEAR APPROXIMATIONS

Example. Find the tangent plane for the surface z = sin(x)+ y2 at (x, y) = (0, 2) and use

it to approximate sin(0.2) + (1.9)2

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§15.6 TANGENT PLANES AND LINEAR APPROXIMATIONS

This method of approximating a function’s value with the height of the tangent planecan be written in terms of a linearization equation or in the language of di↵erentials.

Definition. The linearization of f (x, y) is written as:

Recall: In Calc 1, for a function y = f (x), the di↵erential was defined as d f = f0(x)dx.

Definition. The di↵erential for the function y = f (x, y) is:

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§15.6 TANGENT PLANES AND LINEAR APPROXIMATIONS

� f represents the actual change in a function f (x, y) � f (a, b). d f represents the corre-sponding change in the tangent plane between (a, b) and (x, y).

Di↵erentials are useful for estimating errors and making approximations, because thedi↵erential is linear in all its variables, and linear functions are easier to calculate.

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§15.6 TANGENT PLANES AND LINEAR APPROXIMATIONS

Example. Use di↵erentials to estimate the amount of metal in a closed cylindrical canthat is 10 cm high and 4 cm in diameter if the metal in the top and bottom is 0.1 cmthick and the metal in the sides is 0.05 cm thick.

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§15.6 TANGENT PLANES AND LINEAR APPROXIMATIONS

Example. Four positive numbers, each less than 50, are rounded to the first decimalplace and multiplied together.

1. Use di↵erentials to estimate the size of the maximum possible error in the computedproduct that might result from the rounding.

2. If the four numbers round to 20, 10, 30, and 5, what is the estimated maximumrelative error? percent error?

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§15.6 TANGENT PLANES AND LINEAR APPROXIMATIONS

When the tangent plane approximation works and when it doesn’t.

Note. Anytime fx and fy exist at a point (a, b), it is possible to write down an equationfor a tangent plane for z = f (x, y) at the point where (x, y) = (a, b).

• Usually ...

• Sometimes ...

Definition. (Informal definition) A function is f is called di↵erentiable at (a, b) if ...

Theorem. If fx and fy exist in near (a, b) and are at (a, b), then f isdi↵erentiable at (a, b).

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§15.6 TANGENT PLANES AND LINEAR APPROXIMATIONS

Example. Here is an example of a function that is not di↵erentiable at (x, y) = (0, 0).

f (x, y) =

8>><>>:

xy

x2+y2 if (x, y) , (0, 0)

0 if (x, y) = (0, 0)

Question. In what sense is the tangent plane not a good approximation to the function?

Question. Is this function continuous at (0, 0)?

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§15.6 TANGENT PLANES AND LINEAR APPROXIMATIONS

Question. What was the relationship between di↵erentiability and continuity in Cal-culus 1?

Question. What is the relationship between di↵erentiability and continuity for func-tions of several variables?

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§15.6 TANGENT PLANES AND LINEAR APPROXIMATIONS

Recap: The tangent plane, the linearization, and the di↵erential are related as follows:

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§15.6 TANGENT PLANES AND LINEAR APPROXIMATIONS

Note. So far we have considered tangent planes for functions defined explicitly by anequation of the form z = f (x, y).

How can we define tangent planes for functions that are defined implicitly by anequation of the form F(x, y, z) = 0? (Hint: remember level surfaces and the gradient.)

Note. The tangent plane for the surface F(x, y, z) = 0 is given by the equation:

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§15.6 TANGENT PLANES AND LINEAR APPROXIMATIONS

Example. Find the equation for the tangent plane to the surface xyz2 = 6 at the point

(3, 2, 1).

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§15.6 TANGENT PLANES AND LINEAR APPROXIMATIONS

Note. Any surface that is defined explicitly by the equation z = f (x, y) can also bedefined implicitly by the equation F(x, y, z) = 0, where F is related to f as follows:

We now have two ways to find a tangent plane for z = f (x, y):

1. Use the formula from the beginning of this class for the surface z = f (x, y)

2. Use the formula for the tangent plane to the level surface F(x, y, z) = 0, forF(x, y, z) =...

Verify that both methods give the same equation.

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§15.7 MAXIMUMS AND MINIMUMS

§15.7 Maximums and Minimums

After completing this section, students should be able to:

• Define maximum and minimum values and points.

• Define critical points.

• Explain the relationship between critical points and extreme points.

• Locate the critical points for a function.

• Use the Second Derivatives Test to find local maximum and minimum points andsaddle points for a function.

• Give conditions for when a function is guaranteed to have an absolute maximumand minimum value.

• Find absolute maximum and minimum values for a continuous function on aclosed bounded region.

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§15.7 MAXIMUMS AND MINIMUMS

Recall: For a function f (x) of one variable, f has a local maximum at x = a if ...

and f has a local minimum at x = a if ...

Definition. A function of two variables f (x, y) has a local maximum at (a, b) if ...

and f (x, y) has a local minimum at (a, b) if ...

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§15.7 MAXIMUMS AND MINIMUMS

Recall: For a function f (x) of one variable, a critical number is ...

Recall: For a function f (x) of one variable, max and min points are related to criticalnumbers.

• If f (x) has a local max or min at x = a, then a is / is not / may or may not be acritical number.

• If a is a critical number, then f (x) does / does not /may or may not have a localmax or min at a.

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§15.7 MAXIMUMS AND MINIMUMS

Definition. For a function f (x, y) of two variables, the point (a, b) is a critical point if

Max and min points are related to critical points:

• If f (x, y) has a local max or min at (a, b), then (a, b) is / is not /may or may not bea critical point.

• If (a, b) is a critical point, then f (x, y) does / does not / may or may not have alocal max or min at (a, b).

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§15.7 MAXIMUMS AND MINIMUMS

Recall: The 2nd derivative test for functions of one variable:

Suppose f00(x) is continuous in an open interval around a. Suppose f

0(a) = 0.

If f00(a) > 0, then f has a local at a.

If f00(a) < 0, then f has a local at a.

Definition. The discriminant of f at (a, b) is given by

D =

Theorem. (Second Derivatives Test) Suppose that the second partial derivatives of f

exist and are continuous on a disk around (a, b), and suppose that fx(a, b) = fy(a, b) = 0.Then

(a) If D > 0 and fxx(a, b) > 0, then f has a at (a, b).

(b) If D > 0 and fxx(a, b) < 0, then f has a at (a, b).

(c) If D < 0 then f has a at (a, b).

(d) If D = 0, then .

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§15.7 MAXIMUMS AND MINIMUMS

Example. Find the local maxes and mins and saddles for

f (x, y) = x2 + y

2 + x2y + 4

.

END OF VIDEOS

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§15.7 MAXIMUMS AND MINIMUMS

Question. What is the di↵erence between a local maximum point and an absolutemaximum point?

Question. A critical point for f (x, y) is a point (x0, y0) such that (select all that apply)

A. Either fx(x0, y0) or fy(x0, y0) DNE, or else either one or the other is 0.

B. Either fx(x0, y0) or fy(x0, y0) DNE, or else both of them are 0.

C. The tangent plane to z = f (x, y) doesn’t exist at (x0, y0), or else it is horizontal.

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§15.7 MAXIMUMS AND MINIMUMS

Question. If (x0, y0) is a critical point for the function f (x, y), then:

A. f must have a local max or min at (x0, y0)

B. f cannot have a local max or min at (x0, y0)

C. f may or may not have a local max or min at (x0, y0).

Question. If f has a local maximum value at (x0, y0), then:

A. f must have a critical point at (x0, y0)

B. f cannot have a critical point at (x0, y0)

C. f may or may not have a critical point at (x0, y0).

Question. If a DIFFERENTIABLE function f has a maximum or minimum value at(x0, y0), then (select all that are true)

A. fx(x0, y0) = 0 and fy(x0, y0) = 0 .

B. D~u f (x0, y0) = 0 for all unit vectors ~u.

C. The tangent plane to f at (x0, y0) is horizontal.

D. r f (x0, y0) = 0

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§15.7 MAXIMUMS AND MINIMUMS

Question. Suppose f (x, y) has continuous first and second partial derivatives and(x0, y0) is a critical point. If fxx(x0, y0) > 0, then

A. f has a local max at (x0, y0)

B. f has a local min at (x0, y0)

C. f has a saddle at (x0, y0)

D. There is not enough information to determine the behavior of f at (x0, y0)

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§15.7 MAXIMUMS AND MINIMUMS

Theorem. (Second Derivatives Test) Suppose that the second partial derivatives of f

exist and are continuous on a disk around (a, b), and suppose that fx(a, b) = fy(a, b) = 0.Then

(a) If D > 0 and fxx(a, b) > 0, then f has a at (a, b).

(b) If D > 0 and fxx(a, b) < 0, then f has a at (a, b).

(c) If D < 0 then f has a at (a, b).

(d) If D = 0, then .

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§15.7 MAXIMUMS AND MINIMUMS

Example. Find the local maximum points, local minimum points, and saddle pointsfor f (x, y) = x

3 � 3x + 3xy2.

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§15.7 MAXIMUMS AND MINIMUMS

Example. Find the point on the plane 2x � y + z = 3 that is closest to the point (0, 1, 1).

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§15.7 MAXIMUMS AND MINIMUMS

Note. Recall, in Calc 1:

• If f is continuous and [a, b] is a closed interval, then f (x) (circle one) must /may ormay not achieve an absolute maximum and minimum value on [a, b].

• The abs. maximum value of f on [a, b] occurs (circle one) at one of the end pointsa or b / at a critical point / either at a critical point or at one of the end points a or b.

Question. How do we generalize the idea of a closed interval to R2?

On which of these would you expect a continuous function to always have a max anda min?

Definition. (Informal definition) A closed set is one that ...

Definition. A bounded set is one that ...

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§15.7 MAXIMUMS AND MINIMUMS

Theorem. (Extreme Value Theorem) If f (x, y) is on a ,set D, then f attains its absolute max and abs min value.

The absolute max and min occur either at or on .

Note. To find the abs max and min of f on D

1.

2.

3.

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§15.7 MAXIMUMS AND MINIMUMS

Example. Find the abs max and min of f (x, y) = x2 � 2xy + 2y on the triangular region

with vertices (0, 2), (3, 0), (3, 2).

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§15.7 MAXIMUMS AND MINIMUMS

Example. Maximize f (x, y) = 2x3 + y

4 on the disk D = {(x, y)|x2 + y2 1}.

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§15.7 MAXIMUMS AND MINIMUMS

Example. Find the absolute maximum and absolute minimum of the function f (x, y) =3x

2 � 2y2 � 4y on the region bounded by the curves y = x

2 and y = 3.

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§15.8 LAGRANGE MULTIPLIERS

§15.8 Lagrange Multipliers

After completing this section, students should be able to:

• explain what type of problem Lagrange Multipliers can be used for

• give a qualitative explanation of why the maximum value of a function f (x, y)subject to a constraint condition g(x, y) = k occurs where r f is parallel to rg.

• use Lagrange multipliers to solve constrained optimization problems

• use Lagrange multipliers to maximize a function on the boundary of a region

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§15.8 LAGRANGE MULTIPLIERS

Example. Find the rectangle of largest area that can be inscribed in the ellipsex

2

4+

y2

9= 1

(with sides parallel to the x and y axes)

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§15.8 LAGRANGE MULTIPLIERS

Consider the graph of f (x, y) = 4xy and the graph ofx

2

4+

y2

9= 1 in 3-dimensions (on

the left).

-5 -4 -3 -2 -1 0 1 2 3 4 5

-3

-2

-1

1

2

3

Consider the graph of the level curves of f (x, y) = 4xy and the graph ofx

2

4+

y2

9= 1 in

2-dimensions (on the right).

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§15.8 LAGRANGE MULTIPLIERS

Theorem. (Lagrange multipliers) To find the max and min values of f (x, y) subjectto the constraint g(x, y) = k (assuming these extreme values exist and rg , 0 on theconstraint curve g(x, y) = k), we need to:

1. find all values (x0, y0) such that r f is parallel to rg, i.e. where r f = �rg for some� 2 R.

2. compare the size of f (x0, y0) on all these candidate points.

Note. In practice, we set up the equations:

and solve this system of equations for x, y, and �.

Note. For this to work, g must be ”smooth” (have continuous gx and gy) Otherwise wemust also check ”corners” and ”cusps” where gx or gy don’t exist or have discontinu-ities.

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§15.8 LAGRANGE MULTIPLIERS

Note. Method of Lagrange multipliers also works to maximize or minimize functionsf (x, y, z) of 3 variables, subject to a constraint surface g(x, y, z) = k.

We setfx(x, y, z) = �gx(x, y, z)

fy(x, y, z) = �gy(x, y, z)

fz(x, y, z) = �gz(x, y, z)

g(x, y, z) = k

and solve this system of equations for x, y, z, and �.

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§15.8 LAGRANGE MULTIPLIERS

Example. Find the rectangle of largest area that can be inscribed in the ellipsex

2

4+

y2

9= 1,

using Lagrange multipliers.

END OF VIDEO

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§15.8 LAGRANGE MULTIPLIERS

Example. The graph below depicts the level curves of the function f (x, y) in thin blueand black curves, and a path g(x, y) = 5 with a thick red curve. Where would youexpect to find the maximum value of f (x, y) subject to the constraint that g(x, y) = 5?Why?

The idea behind Lagrange multipliers is that the maximum of z = f (x, y) subject to theconstraint g(x, y) = k (for some constant k) has to occur where r f is torg.

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§15.8 LAGRANGE MULTIPLIERS

Example. Find the point on the plane 2x � y + z = 3 that is closest to the point (0, 1, 1).

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§15.8 LAGRANGE MULTIPLIERS

Example. Maximize f (x, y) = 2x3 + y

4 on the disk D = {(x, y)|x2 + y2 1}.

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§15.8 LAGRANGE MULTIPLIERS

Extra Example. A lidless cardboard box is to be made with a volume of 4 cubic meters.Find the dimensions of the box that requires the least amount of cardboard.

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§15.8 LAGRANGE MULTIPLIERS

Extra Example. Find the absolute maximum and minimum values of f (x, y) = x2+ y

2�2y + 1 on the semicircular region R = {(x, y)|x2 + y

2 4, y � 0}

237