BROWN UNIVERSITYMATH 0350, HONORS CALCULUS

FINAL EXAMINSTRUCTOR: SAMUEL S. WATSON

Name:

Problem 1

Consider the regular tetrahedron with vertices at A = (0, 0, 0), B = (1, 0, 0),C = ( 12 ,

√3

2 , 0), and D = (12 ,√

36 ,√

63 ).

(a) Label the vertices on the figure shown.

(b) Find the angle between−→AD and the vector from A to the midpoint of BC.

You may express your answer in terms of radicals and inverse trig functions.

Solution

We label the figure as shown. Then

−−→AM =

−→AB +

12−→BC =

〈34

,

√3

4, 0

〉,

so−→AD · −−→AM =

〈12

,

√3

6,

√6

3

〉·〈

34

,

√3

4, 0

〉=

12

,

and AM =√

32 and AD = 1. Thus

θ = cos−1(

1/2(√

3/2)(1)

)= cos−1(1/

√3).

Final answer:

cos−1(

1√3

)

Problem 2

Find the value of t such that the points (4, 0, 2), (12, 3, 0), and (2− 2t,−t, 4) are collinear.

Solution

The points are collinear if the vector(12, 3, 0)− (4, 0, 2) = 〈8, 3,−2〉

from the first point to the second is parallel to the vector

〈−2− 2t,−t, 2〉

from the first point to the third. Writing this relation as

〈8, 3,−2〉 = λ〈−2− 2t,−t, 2〉

for some λ ∈ R, we see from the third component that λ = −1. Thus t = 3.

Final answer:

3

Problem 3

Suppose that C1 is a curve parametrized by r1(t) = 〈t2, t3〉 as t ranges from 0 to 1 and C2 is a curve parameterized byr2(t) = 〈t2, 34 t4〉 as t ranges from 0 to 1. Which is longer: C1 or C2?Note: this problem should be very light on calculation. Do not get into the weeds.

Solution

The length of C1 is ∫ 10

√(2t)2 + (3t2)2 dt,

while the length of C2 is ∫ 10

√(2t)2 + (3t3)2 dt.

Since 3t3 < 3t2, the first integral is larger. Therefore, C1 is the longer of the two curves.

Final answer:

C1

Problem 4

How many of the following numbers c have the property that there exists a direction u such that

(Du f )(1, 1) = c,

where f (x, y) = x2 + 3xy2?

8 − 3 7 −√

62 0 2 13 4 − 1 π

Explain your reasoning.

Solution

We have (Du f )(1, 1) = ∇ f · u = 〈2x + 3y2, 6xy〉 · u = 〈5, 6〉 · u =√

61 cos θ, where θ is the angle between 〈5, 6〉 andu. Thus the possible values of the directional derivative are the real numbers between −

√61 and

√61. Exactly 7 of

the given numbers fall in this range.

Final answer:

7

Problem 5

Suppose that P is a point selected uniformly at random from the set of points (x, y, z) satisfying x2 + y2 + z2 ≤ 1.Find the expected value of the distance from P to the z-axis. Express your answer in terms of π.

Solution

The pdf of P is some constant k over the solid unit sphere, and we know that∫∫∫

unit sphere k dV = 1, which implies

that k = 34π . Furthermore, the expected value of g(x, y, z) =√

x2 + y2 is

∫unit sphere

g(x, y, z)3

4πdV =

34π

∫ 2π0

∫ π0

∫ 10

√x2+y2︷ ︸︸ ︷

ρ sinφ ρ2 sinφdρ dφdθ

=3

4π

∫ 2π0

dθ∫ π

0sin2 φdφ

∫ 10ρ3 dρ

=

(3

4π

)(2π)

(π2

)(14

)=

3π16

Final answer:

3π16

Problem 6

Consider the solid R bounded by the cylinder x2 + y2 = 4, the surfacez = 6− x2 − 12 y2, and the plane x + z = −4. Find∫∫∫

R1 dV.

(Hint: use a suitable coordinate system, and remember our trick forevaluating

∫ 2π0 cos

2 θ dθ and∫ 2π

0 sin2 θ dθ.)

Solution

We have ∫∫∫R

1 dV =∫ 2π

0

∫ 20

∫ 6−r2 cos2 θ− 12 r2 sin2 θ−4−r cos θ

1r dz dr dθ

=∫ 2π

0

[5r2 − 1

4r4 cos2 θ− r

4

8sin2 θ+

13

r3 cos θ]2

0dθ

=∫ 2π

020− 4 cos2 θ− 2 sin2 θ+ 8

3cos θ dθ

= 40π− 4π− 2π = 34π .

Final answer:

34π

Problem 7

Consider the trapezoid D shown below. This trapezoid is the image of a rectangle under the transformation (x, y) =(uv, u), as shown.

T(u, v) = (uv, u)

R

12 2

1

3

u

v

y = 12

y = 2

y = xy = 13 x

D

x

y

(a) Based on the figure, along which side of R do you expect the Jacobian of this transformation to be largest?

(b) Calculate the Jacobian of T. Is the result actually bigger along the side you predicted in (a) than along the othersides?

(c) Use this coordinate transformation in the change of variables formula to calculate∫∫

D xy dx dy. Express youranswer as a mixed number.

Solution

(a) Whichever side maps to the top has the largest Jacobian, since those patches are biggest. Matching up vertices,we see that this is the right edge.

(b) The Jacobian is ∣∣∣∣det [v u1 0]∣∣∣∣ = | − u| = u,

which is indeed largest along the right edge.

(c) We have ∫ 31

∫ 21/2

uv · u · u du dv =∫ 3

1v dv

∫ 21/2

u3 du = 151516

.

Problem 8

Find∫

C F · dr, where F = 〈y cos(xy), y3 + x cos(xy)〉 and C is the spiral curve

shown, which starts at (π/2, 1) and ends at the origin.

Solution

The function f (x, y) = sin(xy) + y4

4 has F as its gradient. Therefore, the fundamental theorem of vector calculusimplies that the desired integral is

f (0, 0)− f(π

2, 1)= 0−

(1 +

14

)= −5

4.

Final answer:

−54

Problem 9

(a) Consider the vector field F = 〈x2 − y, xy〉. Explain how to infer the signof∇ · F at the point (0.5, 0.5) based only on the vector plot of F shown. Thenconfirm your answer by calculating (∇ · F)(0.5, 0.5) exactly using the aboveexpression for F.

(b) Suppose now that F is a 3D vector field with the property that

F(x, y, 0) = 〈x2 − y, xy, 0〉

for all x and y. Explain how to infer the sign of the third component of∇×Fat the point (0.5, 0.5, 0) based only on the vector plot of F shown. Then con-firm your answer by calculating the third component of (∇× F)(0.5, 0.5, 0)exactly using the above expression for F.

1

1

Solution

(a) The arrow pointing away from (0.5, 0.5) is longer that the one pointing toward it. Therefore, the vector plotleads us to believe that the divergence at (0.5, 0.5) is positive.

We calculate ∇ · F = 2x + x = 3x = 32 > 0 at (0.5, 0.5), which comports with our graphical conclusion.

(b) A small paddle wheel situated at (0.5, 0.5, 0) with its axle in the z direction would rotate counterclockwiseviewed from above, because the vectors pushing on the top right part of the paddle wheel are longer than thevectors pushing on the bottom left. Therefore, we would expect the third component of∇× F to be positive at(0.5, 0.5, 0).

If we calculate the curl of F, we get

(∇× F) · k = ∂x(xy)− ∂y(x2 − y) = y + 1 =32

,

which is indeed positive.

Problem 10

Define the surface S to be the (whole) boundary of the solid cone r ≤ z ≤ 1. Find the flow outward through S of thevector field F = 〈y, x, z2〉.

Solution

We apply the divergence theorem to find that the given flow is equal to∫∫∫

R 2z dV, where R is the cone described.We can integrate over this cone in cylindrical coordinates to get∫ 2π

0

∫ 10

∫ 1r

2zr dz dr dθ =∫ 2π

0

∫ 10

r(1− r2)dr dθ =(

14

)(2π) =

π

2.

Final answer:

π2

Problem 11

Consider the surface S of points where x2 + y2 + z2 = 1 and x ≤√

32 . Suppose

that G = 〈2, 0, 0〉, which is equal to the curl of 〈0,−z, y〉.(a) Determine from the figure what the sign of the flow of G through S, fromthe inside to the outside, should be. Explain your reasoning.

(b) Find the flow of G through S (from the inside to the outside).

Solution

(a) The net flow of the region through the sphere is from the outside to the inside, since the contributions of all theoutward pointing arrows are canceled by corresponding inward-pointing errors on the opposite side of the sphere.Some of the contributions of inward-pointing vectors (in the back, where x is at its most negative) are not canceled.Therefore, the answer to (a) should be negative.

(b) Stokes’ theorem implies that the desired flow is equal to the line integral of 〈0,−z, y〉 along the boundary of thesurface. This boundary is the circle of radius

√1− (

√3

2 )2 = 12 centered at

(√3

2 , 0, 0)

and perpendicular to 〈1, 0, 0〉.The boundary orientation corresponding to the inside-to-outside surface orientation is clockwise as viewed from thepositive x-axis.

So we can use the parametrization(√

32 ,

12 cos t,−

12 sin t

)for this loop and calculate

∫ 2π0

〈0,

12

sin t,12

cos t〉·〈

0,−12

sin t,−12

cos t〉

dt = −π2

.

Alternative solution for (b): Stokes’ theorem allows us to write the desired flow as the flow (in the direction ofdecreasing x) of G through the disk of radius 12 centered along the x-axis at the point

(√3

2 , 0, 0)

. This flow is equalto −(2)(π/4) = −π/2. No integration!

Final answer:

−π2

BONUS (0 points)

The business end of a frother looks like the coil in the figure shown. Find aparameterized curve which looks as similar as possible to the one shown.

Note: the coil winds around a donut-shaped solid with the following prop-erty: the intersection of the solid with any level surface of the coordinatefunction θ is a disk of radius 0.2 which is centered at some point on the unitcircle in the xy-plane.

Solution

We define a curve r1(t) = 〈cos t, sin t, 0〉 which traces out the unit circle passing through the middle of the angularcross sections. We also define a curve r2 which winds around the sphere of radius 0.2 along a longitude line θ(t) = t.The point whose φ value is kt and whose θ value is given by t is 〈sin kt cos t, sin kt sin t, cos kt〉. We choose the valuek = 40 based on the number of times the coil winds around in a full revolution.

Altogether, we have

r1(t) + r2(t) = 〈cos t + 0.2 cos t sin 40t, sin t + 0.2 sin t sin 40t, cos 40t〉.

Final answer:

〈cos t + 0.2 cos t sin 40t, sin t + 0.2 sin t sin 40t, cos 40t〉

Additional space

Additional space

Additional space

fd@math0350_final-1: fd@math0350_final-2: fd@math0350_final-5: fd@math0350_final-6: