math 50c final exam a name: answer key multiple choice. 1 ... · math 50c december 7, 2015 final...

Math 50C

December 7, 2015Final Exam A Name: Answer Key

D. Arnold

Multiple Choice. (45 points) For each of the following questions, determine the correctanswer then fill in the corresponding oval on your scantron.

(5pts) 1. Determine the slope of the tangent line to the surface f(x, y) = x2 + xy− 3y2 + 8 at thepoint (1, 1, 7) in the direction of the vector v = 〈2,−2〉.(a) 2

√2 (b) 2

√3 (c)4 4

√2

(d) −2√

5 (e) −5√

2

Solution: We’ll need the gradient of f(x, y) = x2 + xy − 3y2 + 8.

∇f(x, y) =

⟨∂f

∂x,∂f

∂

⟩= 〈2x+ y, x− 6y〉

Evaluate the gradient at the point (1, 1).

∇f(1, 1) = 〈3,−5〉

Next, we need a unit vector in the direction of v = 〈2,−2〉.

u =v

‖v‖

=〈2,−2〉‖〈2,−2〉‖

=〈2,−2〉√22 + (−2)2

=〈2,−2〉

2√

2

Finally, the slope of the tangent line in the direction of v is given by the directionalderivative.

Du(1, 1) = ∇f(1, 1) · u

= 〈3,−5〉 · 〈2,−2〉2√

2

=16

2√

2

=8

2√

2

= 4√

2

Math 50C/Final Exam A – of 15 – Name: Answer Key

(5pts) 2. Use the gradient to determine a vector normal to the surface z = xy at the point (1, 1, 1).

(a) 〈1, 1, 0〉 (b) 〈1,−1, 1〉 (c) 〈1, 1, 1〉(d)4 〈−1,−1, 1〉 (e) 〈−1, 1, 1〉Solution: First, we need a function such that z = xy is a level surface of that function.Consider:

f(x, y, z) = z − xyNote that the level surface f(x, y, z) = 0 is equivalent to the equation z = xy. Now, thegradient of f will be orthogonal to the level surface.

∇f(x, y, z) =

⟨∂f

∂x,∂f

∂y,∂f

∂z

⟩= 〈−y,−x, 1〉

Now, substitute the point (1, 1, 1).

∇f(1, 1, 1) = 〈−1,−1, 1〉Thus, 〈−1,−1, 1〉 is orthogonal to the surface z = xy at the point (1, 1, 1).

(5pts) 3. Which of the following points is a critical point of the function f(x, y) = 4x2−8x+4y+y2 + 8 ?

(a) (2, 1) (b)4 (1,−2) (c) (2,−1)

(d) (−1, 2) (e) (1, 2)

Solution: Set the first partials equal to zero and solve.

fx(x, y) = 8x− 8

0 = 8x− 8

x = 1

fy(x, y) = 4 + 2y

0 = 4 + 2y

y = −2

Hence, (1,−2) is a critical point.

(5pts) 4. The function f(x, y) = 3x2− 6x+ 2y2 + 4y has a critical point at (1,−1). What type ofextrema occurs at this critical point?

(a)4 Local minimum (b) Local maximum (c) Saddle point

(d) None of these

Solution: The first partials are:

fx(x, y) = 6x− 6 and fy(x, y) = 4y + 4

The second partials are:

fxx(x, y) = 6 and fyy(x, y) = 4 and fxy = 0

We can now use the second derivative test to classify the extrema at the critical oint(1,−1).

(x, y) fxx(x, y) fxx(x, y)fyy(x, y)− f 2xy(x, y) Classification

(1,−1) 6 (6)(4)− (0)2 = 24 Local minimum


(5pts) 5. Which of the following integrals determines the volume of the region in the first octantbounded by the planes y = 1 and x+ z = 2 ?

(a)

∫ 1

0

∫ 1

0

∫ 2−x

0

dz dy dx (b)

∫ 1

0

∫ 2

0

∫ 2−z

0

dz dx dy (c)

∫ 2

0

∫ 1

0

∫ x−2

0

dz dy dx

(d)4

∫ 2

0

∫ 1

0

∫ 2−z

0

dx dy dz (e) None of these

Solution: We first start with a sketch of the region in the first octant bounded by theplanes x+ z = 2 and y = 1.

Now, fix a value of z between 0 and 2, then draw an arrow of integration in the y-direction. Note that we enter at y = 0 and leave at y = 1. Finally, draw an arrow inthe x-direction, noting we enter the region at x = 0 and leave the region at x = 2 − z.Hence the volume of the region is given by the integral:∫ 2

0

∫ 1

0

∫ 2−z

0

dx dy dz


(5pts) 6. Which of the following integrals determines the volume of the region bounded above bythe sphere x2 + y2 + z2 = 4 and below by the plane z = 0 ?

(a)

∫ 2π

0

∫ 2

0

∫ √4−r20

dz dr dθ (b)4

∫ 2π

0

∫ 2

0

∫ √4−r20

r dz dr dθ(c)

∫ 2π

0

∫ 4

0

∫ √4−r20

r dz dr dθ

(d)

∫ 2π

0

∫ 2

0

∫ √r2−40

r dz dr dθ(e) None of these

Solution: We start with a sketch of the region.

We use cylindrical coordinates. Fix θ, starting with θ = 0 and ending with θ = 2π.Next, r-values go from r = 0 to r = 2. Finally, an arrow of integration in the z-directionindicates that the z-values go from z = 0 to z =

√4− r2. Hence, the volume of the

region is given by the following integral. Recall that dV = r dz dr dθ.∫ 2π

0

∫ 2

0

∫ √4−r20

r dz dr dθ


(5pts) 7. The volume of a sphere of radius 3 is determined by which of the following integrals?

(a)4

∫ 2π

0

∫ π

0

∫ 3

0

ρ2 sinφ dρ dφ dθ (b)

∫ 2π

0

∫ π

−π

∫ 3

0

ρ sin2 φ dρ dφ dθ

(c)

∫ 2π

0

∫ π

0

∫ 3

0

ρ cosφ dρ dφ dθ (d)

∫ 2π

0

∫ π

0

∫ 3

0

ρ2 cosφ dρ dφ dθ

(e) None of these

Solution: We start with a sketch.

We use spherical coordinates. Fix a value of θ, starting with θ = 0 and ending withθ = 2π. Next, fix a value of φ, starting with φ = 0 and ending with φ = π. Finally,note that the final arrow of integration starts at ρ = 0 and ends at ρ = 3. Thus,the volume is given by the following integral. Recall that with spherical coordinates,dV = ρ2 sinφ dρ dφ dθ. ∫ 2π

0

∫ π

0

∫ 3

0

ρ2 sinφ dρ dφ dθ


(5pts) 8. Which of the following is the region of integration for the following integral ?∫ 1

0

∫ 2−y

y

f(x, y)dxdy

(a)

0 1 20

1

2

x

y(b)

0 1 20

1

2

x

y

(c)4

0 1 20

1

2

x

y(d)

0 1 20

1

2

x

y

Solution: Note that working with the following region, if we fix a value of y between 0and 1, the run an arrow of integration in the x-direction, the arrow enters the region atx = y and leaves the region at x = 2− y.

0 1 20

1

2

x

y

(y, y) (2− y, y)

Hence, this is the region of integration for the integral∫ 1

0

∫ 2−y

y

f(x, y) dx dy


(5pts) 9. Which of the following vectors fields is conservative?

(a) 〈x+ y, y − x〉 (b)4 〈y2, 2xy〉 (c) 〈y2, x2〉(d) 〈ex cos y, ex sin y〉 (e) 〈x, xy〉Solution: Take the field F(x, y) = 〈y2, 2xy〉 and note that:

∂

∂x2xy = 2y

∂

∂yy2 = 2y

Hence, F is conservative.

5 ptsEssay Questions. (70 points) Do each of the following questions on a separate sheet ofpaper. When finished, arrange your solutions in order, then staple this page atop yourwork.

(10pts) 1. Consider the surface f(x, y) = x2 + y2 − 9. Find the equation of the line tangent to thesurface at the point P (1,−1,−7) in the direction of the vector v = 〈−5, 12〉.Solution: Take the gradient of f(x, y) = x2 + y2 − 9.

∇f(x, y) =

⟨∂f

∂x,∂f

∂y

⟩= 〈2x, 2y〉

Evaluate at the point (1,−1).

∇f(1,−1) = 〈2,−2〉

Now we need a unit vector in the direction of v = 〈−5, 12〉.

u =v

‖v‖

=〈−5, 12〉√25 + 144

=〈−5, 12〉

13

Now we can find the slope of the tangent line in the direction of v.

Du(1,−1) = ∇f(1,−1) · u

= 〈2,−2〉 · 〈−5, 12〉13

= −34

13


Now, if we walk one unit in the direction of v = 〈−5, 12〉, the unit vector in thatdirection, u = 〈−5/13, 12/13〉, tells us that our x-value changes by −5/13 of a unit andour y-value changes by 12/13 of a unit. Because we know the slope if −34/13 in thisdirection, a vector in the direction of the tangent line is

w =

⟨− 5

13,12

13,−34

13

⟩Now, a sketch will help us focus.

P (1,−1,−7) X(x, y, z)

w =⟨− 5

13, 1213,−34

13

⟩

Hence, the equation of the line is:

−−→PX = tw

〈x− 1, y + 1, z + 7〉 = t

⟨− 5

13,12

13,−34

13

⟩

Comparing components, the parametric equations of the tangent line are:

x = 1− 5

13t

y = −1 +12

13t

z = −7− 34

13t

(10pts) 2. Use the gradient to find the equation of the plane tangent to the surface z = x2+3y2−12at the point P (−1, 1,−8).

Solution: First, we need a function which has a level surface z = x2+3y2−12. Consider:

f(x, y, z) = x2 + 3y2 − z

The level surface f(x, y, z) = 12 is equivalent to the given surface z = x2 + 3y2 − 12 .Now, the gradient is orthogonal to the level surface.

∇f(x, y, z) = 〈2x, 6y,−1〉

And evaluating at the point (−1, 1,−8):

∇f(−1, 1,−8) = 〈−2, 6,−1〉

A sketch will help us focus.


P (−1, 1,−8)

X(x, y, z)

〈−2, 6,−1〉

Select an arbitrary point X(x, y, z) on the tangent plane. Then, the equation of thetangent plane is:

−−→PX · 〈−2, 6,−1〉 = 0

〈x+ 1, y − 1, z + 8〉 · 〈−2, 6,−1〉 = 0

−2x− 2 + 6y − 6− z − 8 = 0

Hence, the equation of the tangent plane is 2x− 6y + z + 16 = 0.

(10pts) 3. Evaluate the integral ∫ 4

0

∫ 1

0

∫ 2

2y

2 cosx2√z

dx dy dz

Include a sketch of the region of integration.

Solution: The difficulty is the fact we can’t integrate in this order, so we’ll have to changethe order of integration. To do that, we first have to sketch the region of integration.Note the order of events in the integral:∫ 4

0

∫ 1

0

∫ 2

2y

2 cosx2√z

dx dy dz

1. Fix a z-value between 0 and 4.

2. Fix a y-value between 0 and 1.

This leads to the following sketch.


Now, the x-values go from x = 2y to x = 2. This leads to the following sketch.

- / 'x.4

Now, let’s change the order of integration, as shown in the following image.


nX,

\= ,Lttao

Fix a value of z between 0 and 4. Then draw an arrow of integration in the y-direction,starting at x = 0 and ending at x = 2. Draw a final arrow of integration in the x-direction, starting at y = 0 and ending at y = x/2. This gives us the following orderchange to the integral. ∫ 4

0

∫ 2

0

∫ x/2

0

2 cosx2√z

dy dx dz

Now we can integrate with respect to y. Finally!∫ 4

0

∫ 2

0

∫ x/2

0

2 cosx2√z

dy dx dz =

∫ 4

0

∫ 2

0

[2 cosx2√

zy

]x/20

dx dz

=

∫ 4

0

∫ 2

0

[2 cosx2√

z

x

2

]dx dz

=

∫ 4

0

∫ 2

0

x cosx2√z

dx dz

Note that Dx(1/2) sinx2 = x cosx2.

=

∫ 4

0

[ 12

sinx2√z

]20

dz

=

∫ 4

0

[1

2√z

sinx2]20

dz

=1

2sin 4

∫ 4

0

z−1/2 dz

=1

2sin 4

[2z1/2

]40

= 2 sin 4


(10pts) 4. Evaluate the line integral ∫C

(x+ 2y − 3z) ds,

where C is the line segment joining the origin to the point (1,−2, 2) by parametrizingthe path in terms of arc length.

Solution: Start with a sketch to help us focus. Let X(x, y, z) be an arbitrary point onthe segment and let s represent the distance already travelled.

P (0, 0, 0) X(x, y, z) Q(1,−2, 2)

s

First, we need a unit vector in the direction of motion.

u =

−→PQ

‖−→PQ‖

=〈1,−2, 2〉

3Now, the equation of the segment is:

−−→PX = su

〈x, y, z〉 = s 〈1/3,−2/3, 2/3〉Comparing components, the equation of the segment is

x =1

3s

y = −2

3s

z =2

3s

where the arc length will vary from s = 0 to s = 3. Hence, the line integral becomes:∫C(x+ 2y − 3z) ds =

∫ 3

0

[1

3s+ 2

(−2

3s

)− 3

(2

3s

)]ds

=

∫ 3

0

[1

3s− 4

3s− 2s

]ds

=

∫ 3

0

[−3s] ds

Now we can integrate with respect to s.

= −3

2s2∣∣∣∣30

= −27

2


(10pts) 5. Evaluate ∫C

yx dx+ xz dy + xy dz,

where C is the path defined by r(t) = 〈t, t2, t3〉, for 0 ≤ t ≤ 2.

Solution: Because r(t) = 〈t, t2, t3〉, we have:

x = t, dx = dt

y = t2, dy = 2t dt

z = t3, dz = 3t2 dt

Now, we are integrating from t = 0 to t = 2.∫C

yx dx+ xz dy + xy dz =

∫ 2

0

(t2)(t)(dt) + (t)(t3)(2t dt) + (t)(t2)(3t2 dt)

=

∫ 2

0

(t3 + 2t5 + 3t5) dt

=

∫ 2

0

(t3 + 5t5

)dt

=1

4t4 +

5

6t6∣∣∣∣20

=1

4(16) +

5

6(64)

=172

3

(10pts) 6. Show that the vector field F(x, y) = (tan y)i + (x sec2 y)j is conservative, then evaluate∫CF ·dr where C is any curve going from the point P (1, 0) to the point Q(2, π/4) in the

xy-plane.

Solution: Note that:

∂

∂x(x sec2 y) = sec2 y

∂

∂y(tan y) = sec2 y

Hence, the vector field F(x, y) = (tan y)i + (x sec2 y)j is conservative. Therefore, thereexists an f such that:

F = ∇f⟨tan y, x sec2 y

⟩=

⟨∂f

∂x,∂f

∂y

⟩


Comparing the first components:

∂f

∂x= tan y

Integrate with respect to x.

f(x, y) = x tan y + φ(y)

Next, compare the second components.

∂f

∂y= x sec2 y

Substitute the current f .

∂

∂y(x tan y + φ(y) = x sec2 y

Differentiate with respect to y.

x sec2 y + φ′(y) = x sec2 y

Thus,

φ′(y) = 0

φ(y) = K,

where K is an arbitrary constant. Substitute the function φ(y) = K back into the latestf .

f(x, y) = x tan y +K

Now, if C is any curve connecting the point P (1, 0) to the point Q(2, π/4), we can usethe Fundamental Theorem of Line Integrals to evaluate the integral.∫

CF · dr =

∫C∇f · dr

= f(2, π/4)− f(1, 0)

=(

2 tanπ

4+K

)− (1 tan 0 +K)

= 2


(10pts) 7. Use Green’s Theorem to evaluate the integral∫C

x2y2 dx+ 4xy3 dy

where C is the boundary of the triangle with vertices (0, 0), (1, 3), and (0, 3).

Solution: Using Green’s Theorem,∫C

x2y2 dx+ 4xy3 dy =

∫∫D

(∂

∂x(4xy3)− ∂

∂y(x2y2)

)dA

=

∫∫D

(4y3 − 2x2y

)dA

Now, an image to help us focus.

x

y

3

0 1

(x, 3)

(x, 3x)

Fix x between x = 0 and x = 1, then draw an arrow of integration in the y-direction.The arrow enters the region at y = 3x and leaves the region at y = 3. Hence, our integralbecomes: ∫∫

D

(4y3 − 2x2y

)dA =

∫ 1

0

∫ 3

3x

(4y3 − 2x2y

)dy dx

Integrate with respect to y.

=

∫ 1

0

[y4 − x2y2

]33xdx

=

∫ 1

0

[(81− 9x2)− (81x4 − 9x4)

]dx

=

∫ 1

0

(81− 9x2 − 72x4) dx

= 81x− 3x3 − 72

5x5∣∣∣∣10

= 81− 3− 72

5

=318

5

math 50c final exam a name: answer key multiple choice. 1 ... · math 50c december 7, 2015 final...

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