SOLUTIONS TO THE SECOND EXAM, MATH 114, FALL

2004

Multiple choice questions.

Question 1. Compute the partial derivative of the function

f(x, y, z) = e1−x cos(y) + z e−1/(1+y2)

with respect to x at the point (1, 0, π).

(a) -1

(b) −1/e

(c) 0

(d) π/e

(e) π

Answer 1. We compute

∂f

∂x= e1−x cos y(− cos y)

∂f

∂x|(1,0,π) = e1−1 cos 0(− cos 0) = e0(−1) = −1.

The correct answer is (a). 2

Question 2. The maximum value of (xy)6 on the ellipsex2

4+ y2 = 1 occurs at a point

(x, y) for which y2 is equal to

(a)√

2/3

(b) 1/2

(c) 2/3

(d) 5/11

(e) 10/11

1

Answer 2. We want to maximize f(x, y) = (xy)6 subject to the constraint g(x, y) =x2/4 + y2 − 1 = 0. The maximum occurs when

∇f = λ∇g,

g = 0.

We compute ∇f = 6x5y6ı + 6x6y5 and ∇g = (x/2)ı + 2y. So we must solve

6x5y6 =λx

26x6y5 = 2λy

x2

4+ y2 = 1

Solving the first two equations for λ we get λ = 12x4y6 and λ = 3x6y4, and so 12x4y6 = 3x6y4.Thus either x = 0, or y = 0, or xy 6= 0 and x2 = 4y2. Substituting the latter in the thirdequation we get 2y2 = 1. If either x or y is zero, then f = 0 and so the maximum of f occurswhen y2 = 1/2. The correct answer is (b). 2

Question 3. Evaluate the limit

lim(x,y)→(0,0)

xy

x2 + y2.

(a) −1

(b) 0

(c)1

2

(d) 1

(e) the limit does not exist

Answer 3. Along the line y = kx with k 6= 0 the limit equals to

limx→0

kx2

x2 + k2x2=

k

1 + k2.

Different k’s give rise to different limits and so the limit does not exist. The correct answeris (e). 2

Question 4. I see an object 3 miles to the East and 4 miles North. It appears to bemoving at 1 mile per minute in the Southwest direction. At how many miles per minute isit getting closer to me?

(a)1

5√

2

(b)1

5

(c)7

5√

2

(d)7

5

(e) 7√

2

Answer 4. I stand at (0, 0). The object starts at (3, 4) and travels in the direction of−ı− . Let −→r (t) denote the position of the object at time t. Then −→r (t) = −→r 0 +−→v 0t, where−→r 0 is the initial position vector and −→v 0 is the constant velocity vector of the object. So−→r 0 = 3ı + 4 and −→v 0 is a vector in the direction of −ı − with magnitude 1 (= the speedof the object). Thus

−→v 0 = − 1√2ı − 1√

2

−→r (t) =

(3 − t√

2

)ı +

(4 − t√

2

).

The distance between the object and me is |−→r (t)| and so the speed at which the object isgetting closer to me is ∣∣∣∣

d|−→r (t)|dt

|t=0

∣∣∣∣ .

We compute

|−→r (t)| =

√(3 − t√

2

)2

+

(4 − t√

2

)2

=

√t2 − 7

√2t + 25.

Differentiating with respect to t we get

d|−→r (t)|dt

=2t − 7

√2

2√

t2 − 7√

2t + 25,

and so ∣∣∣∣d|−→r (t)|

dt|t=0

∣∣∣∣ =

∣∣∣∣∣−7

√2

2 · 5

∣∣∣∣∣ =7

5√

2.

The correct answer is (c). 2

Question 5. The tangent plane to the graph of the function z = x2y + 1/(1 + y2) at thepoint (1, 1, 3/2) contains point (2, 2, t) for which value of t?

(a) 8 15

(b) 1 + 74

√2

(c) 4

(d) 5

(e) none of the above

Answer 5. The equation of the surface is f(x, y, z) = x2y + 1/(1 + y2) − z and so the equation ofthe tangent plane at P0 = (1, 1, 3/2) is

fx(P0) · (x − 1) + fy(P0) · (y − 1) −(

z − 3

2

)= 0.

Computing the partials we get fx = 2xy and fy = x2 − 2y/(1 + y2)2. Hence fx(P0) = 2 andfy(P0) = 1/2 and so the equation of the tangent plane is

2x +y

2− z = 1.

At (2, 2, t) this gives 4 + 1 − t = 1, i.e. t = 4. The correct answer is (c). 2

Question 6. A particle moves in a circle according to the equations ~r(t) = cos(t2)i + sin(t2)j.The magnitude of the normal component of the acceleration at time t is

(a) 0

(b) 1

(c) 6t

(d) t2

(e) 4t2

Answer 6. Recall that in the−→T−→N frame the acceleration decomposes as −→a = aT

−→T + aN

−→N . We

want to find the value of aN . We also have

aT =d|−→v |dt

aN =√

|−→a |2 − a2T ,

and so we have to compute |−→a |2 and aT .To determine −→a we differentiate −→r (t):

−→v = r′(t) = −2t sin(t2)ı + 2t cos(t2),

−→a = r′′(t) = (−2sin(t2) − 4t2 cos(t2))ı + (2 cos(t2) − 4t2 sin(t2)).

Using this we compute |−→v | = 2t and aT = (2t)′ = 2. Similarly we have |−→a |2 = 4 + 16t2 and so

aN =√

4 + 16t2 − 4 = 4t.

The correct answer is (e). 2

Question 7. Which of the quantities is nearest to the value of

exp

(0.003

1.001

)cos(0.002) ?

(a) 1

(b) 1.001

(c) 1.002

(d) 1.003

(e) 1.000006

Answer 7. To approximate the value we will linearize the function f(x, y, z) = ex

y cos(z) at(0, 1, 0). For the partials of f we compute

fx =1

yex/y cos(z), fx|(0,1,0) = 1

fy = xex/y cos(z), fy|(0,1,0) = 0

fz = −ex/y sin(z), fz|(0,1,0) = 0

Since f(0, 1, 0) = 1 we get

L(x, y, z) = 1 + 1 · (x − 0) + 0 · (y − 1) + 0 · (z − 0) = 1 + x

for the linearization. Therefore

f(.003, 1.001, .002) ∼ L(.03, 1.001, .002) = 1 + .003 = 1.003.

The correct answer is (d). 2

Free response questions.

Free Response Question 1. A cardboard rocket takes off horizontally from the edge of a threefoot high table, accelerating in the horizontal direction at twice the acceleration of gravity (as wellas being subject to gravitational acceleration) until it hits the floor. How far from the table doesit get?

Free Response Answer 1. Let g denote the acceleration of gravity. The vertical accelerationof the rocket is the gravitational one, i.e. y ′′(t) = −g. The horizontal acceleration is given tobe twice the gravitational one, i.e. x′′(t) = 2g. Hence the total acceleration is −→a (t) = 2gı − g.Integrating with respect to time we get the velocity:

−→v (t) = −→v 0 + 2gtı − gt,

with −→v 0 being the initial velocity. Since the rocket takes off from rest, we have −→v 0 =−→0 . Integrating

again with respect to time we get the position vector

−→r (t) = −→r 0 + gt2 ı− gt2

2

with −→r 0 being the initial position. Since the table is 3 feet high we have −→r 0 = 3 and so

−→r (t) = gt2ı +

(3 − gt2

2

).

The rocket hits the ground when y(t) = 3− gt2

2 = 0, i.e. when t =√

6/g. When the rocket hits the

ground the horizontal distance travelled is x(√

6/g) = g(√

6/g)2 = 6. 2

Free Response Question 2. Find all absolute minima of the function y4 +xy3 on the trianglex ≥ 0, x + y ≤ 1, x − y ≤ 1.

Free Response Answer 2. The critical points of f inside the region occur when fx = fy = 0.We compute

fx = y3 fy = 4y3 + 3xy2.

Setting fx = 0 gives y = 0. If y = 0, then fy = 0. So the critical points inside the region are allpoints (x, 0) with 0 < x < 1.

On the boundary x = 0, −1 ≤ y ≤ 1 we have f(0, y) = y4. The minimum of y4 on [−1, 1] occursat y = 0. So on the x = 0, −1 ≤ y ≤ 1 the minimum of f occurs at (0, 0).

On the boundary x = 1 − y, 0 ≤ y ≤ 1 we have f(1 − y, y) = y4 + (1 − y)y3 = y3. Thus theminimum of f on x = 1 − y, 0 ≤ y ≤ 1 occurs at (1, 0).

On the boundary x = 1 + y, 0 ≤ y ≤ 1 we have f(1 + y, y) = y4 + (1 + y)y3 = 2y4 + y3. Wecompute

d(2y4 + y3)

dy= 8y3 + 3y2.

The critical points are y = 0 and y = −3/8. The value of f(1 + y, y) at −3/8 is negative, and thevalues f(1 + y, y) at the end points y = 0 and y = 1 is non-negative. Thus the minimum of f(x, y)on x = 1 + y, 0 ≤ y ≤ 1 occurs at (5/8,−3/8) and is equal to f(−3/8, 5/8) = −27/2048.

Consequently, the minimum of f on the whole region will occur at one of the points (x, 0),0 ≤ x ≤ 1 or at the point (−3/8, 5/8). Since f(x, 0) = 0 for all 0 ≤ x ≤ 1, it follows that the globalminimum of f occurs at (−3/8, 5/8) and is equal to −27/2048. 2

Free Response Question 3. The functions R = Al

r2and V = πlr2 determine the resistance and

volume of a wire in terms of its length, l, and cross-sectional radius, r (where A is a constant ofthe material). If we consider these equations as specifying l and r as functions of R and V , thenwhat is ∂l/∂R?

Free Response Answer 3. Thinking of l and r as functions of R and V we compute the partial

derivative of the identities V = πlr2 and R = Al

r2with respect to R to get

0 = π∂l

∂Rr2 + 2πlr

∂r

∂R

1 = A∂l

∂R

1

r2− 2A

∂r

∂R

l

r3

Multiplying the first equation by A and the second by πr4 and adding up the results we get

πr4 = (Aπr2 + Aπr2)∂l

∂R,

or∂l

∂R=

r2

2A=

√V

2√

πRA.

2