Math 115 — Practice for Exam 3 · Fall 2016 3 4 more donuts 10 Winter 2012 3 6 bucket 12 Fall 2017 3 4 puddle 10 Winter 2015 2 7 10 Fall 2014 3 9 ice 8 Winter 2006 2 1 gift boxes

Math 115 — Practice for Exam 3

Generated December 12, 2018

Name: SOLUTIONS

Instructor: Section Number:

1. This exam has 16 questions. Note that the problems are not of equal difficulty, so you may want toskip over and return to a problem on which you are stuck.

2. Do not separate the pages of the exam. If any pages do become separated, write your name on themand point them out to your instructor when you hand in the exam.

3. Please read the instructions for each individual exercise carefully. One of the skills being tested onthis exam is your ability to interpret questions, so instructors will not answer questions about examproblems during the exam.

4. Show an appropriate amount of work (including appropriate explanation) for each exercise so that thegraders can see not only the answer but also how you obtained it. Include units in your answers whereappropriate.

5. You may use any calculator except a TI-92 (or other calculator with a full alphanumeric keypad).However, you must show work for any calculation which we have learned how to do in this course. Youare also allowed two sides of a 3′′ × 5′′ note card.

6. If you use graphs or tables to obtain an answer, be certain to include an explanation and sketch of thegraph, and to write out the entries of the table that you use.

7. You must use the methods learned in this course to solve all problems.

Semester Exam Problem Name Points Score

Fall 2007 3 5 Madam Whippy 10

Winter 2010 3 8 train+camera 12

Winter 2011 3 3 crankshaft hourglass 10

Fall 2016 3 4 more donuts 10

Winter 2012 3 6 bucket 12

Fall 2017 3 4 puddle 10

Winter 2015 2 7 10

Fall 2014 3 9 ice 8

Winter 2006 2 1 gift boxes 20

Fall 2011 3 5 social network 8

Fall 2008 3 1 Galleon 12

Winter 2002 3 4 naked mile 9

Fall 2013 2 1 swimming pool 7

Winter 2010 3 3 graduation party 12

Winter 2018 3 7 sea level 11

Fall 2016 3 5 10

Total 171

Recommended time (based on points): 195 minutes

5

5. (10 points) Madam Whippy’s ice-cream store has a vending machine that pumps out manillavanilla at a constant rate of 2 cm3 per second. If you are collecting the ice-cream in a cone ofmaximum radius 5 cm and height 10 cm, how fast is the radius of the surface of the ice-creamchanging when the height of ice cream in the cone is 6 cm? Assume that the ice cream is soft-serve and fills the cone with a flat surface.

[You may need: Volume of a cone of height h and radius r is given as V =1

3π r2 h.]

r

h

5

10

Using similar triangles,r

h=

5

10→ h = 2r

Thus V =2

3π r3 and

dV

dt= 2 πr2

dr

dt

and since we are givendV

dt= 2, we have

2 = 2 πr2dr

dt→

dr

dt=

1

πr2

which at h = 6 and r = 3 givesdr

dt=

1

9πcm/s.

University of Michigan Department of Mathematics Fall, 2007 Math 115 Exam 3 Problem 5 (Madam Whippy) Solution

Math 115 / Final (April 23, 2010) page 9

8. [12 points]A train is traveling eastward at a speed of 0.4 miles per minute along a long straight track,and a video camera is stationed 0.3 miles from the track, as shown in the figure. The camerastays in place, but it rotates to focus on the train as it moves.Suppose that t is the number of minutes that have passed since the train was directly northof the camera; after t minutes, the train has moved x miles to the east, and the camera hasrotated θ radians from its original position.

b−→

Train

b

Camera

x mi

0.3 mi

θ

a. [3 points] Write an equation that expresses the relationship between x and θ.

Solution:

tan(θ) =x

0.3, or θ = arctan

( x

0.3

)

b. [4 points] Suppose that seven minutes have passed since the train was directly north ofthe camera. How far has the train moved in this time, and how much has the camerarotated?

Solution: The train’s velocity is constant (i.e. dx

dt= 0.4), so we can use the formula

distance = velocity · time. Thus, the train has moved(

0.4 mi

min

)

(7min) = 2.8 miles.

Using the fact that x = 2.8, we have θ = tan−1(2.8

0.3) ≈ 1.4641, so the camera has rotated

1.4641 radians in the clockwise direction.

c. [5 points] How fast is the camera rotating (in radians per minute) when t = 7?

Solution: We want to find dθ

dt. To do so, we will (implicitly) take the derivative of our

equation from part (a), with respect to t.

d

dttan(θ) =

d

dt

( x

0.3

)

, or1

cos2(θ)

dθ

dt=

1

0.3

dx

dt

We can then plug in dx

dt= 0.4 and θ = 1.4641:

1

cos2(1.4641)

dθ

dt=

0.4

0.3, so

dθ

dt=

0.4

0.3cos2(1.4641) ≈ 0.01513.

Thus, the camera is rotating at a speed of 0.01513 radians per minute (in the clockwisedirection) when t = 7.

University of Michigan Department of Mathematics Winter, 2010 Math 115 Exam 3 Problem 8 (train+camera) Solution


��

��

��

��

��

��

Crank shaft diagram (part a)

side view end view

r

Lh

h

θ

Hourglass diagram (part b)

H

R

3. [10 points] For each of the following determine the indi-cated quantity.

a. [4 points] In an internal combustion engine, pistonsare pushed up and down by a crank shaft similar tothe diagram shown to the right. As the shaft rotatesthe height of the piston, h, is related to the rotational

angle θ of the shaft by h = r cos θ +√

L2 − r2 sin2 θ,where r and L are constant lengths. If r = 10 cm,L = 15 cm, and h is decreasing at a rate of 5000 cm/swhen θ = 3π/4, how fast is θ changing then?

Solution: Using the chain rule, we know that

h′(t) =

(

−r sin θ − r2 sin θ cos θ√L2−r2 sin2 θ

)

· dθdt

. Thus if

θ = 3π/4, h′ = −5000, r = 10 and L = 15, we

have −5000 =

(

− 10√

2− 50

√225−50

)

· dθdt

≈ −3.29 · dθdt

. Thus dθdt

≈ 1500 radians/sec

b. [6 points] The lower chamber of an hourglass is shaped like a cone with height H in andbase radius R in, as shown in the figure to the right, above. Sand falls into this cone. Writean expression for the volume of the sand in the lower chamber when the height of the sandthere is h in (Hint: A cone with base radius r and height y has volume V = 1

3 π r2 y, and itmay be helpful to think of a difference between two conical volumes.). Then, if R = 0.9 in,H = 2.7 in, and sand is falling into the lower chamber at 2 in3/min, how fast is the heightof the sand in the lower chamber changing when h = 1 in?

Solution: The whole volume of the lower chamber is Vtot = 13 π R2 H. The volume of

the empty space above the sand is similarly Vemp = 13 π r2 (H − h), where r is the radius

at the height h. By comparing the similar triangles delimiting the full lower chamberand the empty top section, we see that r = R

H(H − h). Thus Vemp = 1

3 π R2

H2 (H − h)3.The volume of the lower, sand-filled region is therefore

V =1

3π

R2

H2

(

H3 − (H − h)3)

.

Then, differentiating, we have

dV

dt= π

R2

H2(H − h)2

dh

dt.

Thus, when dVdt

= 2, R = 0.9, H = 2.7, and h = 1,

2 = π1

32(1.7)2

dh

dt,

so that dhdt

= 18π(1.7)2

≈ 1.98 in/min.

University of Michigan Department of Mathematics Winter, 2011 Math 115 Exam 3 Problem 3 (crankshaft hourglass) Solution

Math 115 / Final (December 19, 2016) page 6

This problem continues the investigation of Xanthippe’s donuts.

4. [10 points] For your convenience, the graphs of p(t) and q(t) are reprinted below. Recall:

• The rate, in donuts per hour, at which Xanthippe makes donuts t hours after 7 am ismodeled by the function p(t).

• The rate, in donuts per hour, at which customers purchase donuts t hours after 7 am ismodeled by the function q(t).

• Assume that at 7 am, Xanthippe begins with no donuts in stock.

0.5 1 1.5 2 2.5 3 3.5 4

20

40

60

80

100

y = p(t)

y = q(t)

t

y

a. [4 points] Estimate the total number of donuts produced by 10 am using a right-handRiemann sum with two equal subintervals. Be sure to write down all the terms in yoursum. Is your answer an underestimate or overestimate?

Solution: Each subinterval has width ∆t = 1.5. Therefore, a right-hand Riemann sumwith two equal subintervals is

∫3

0p(t) dt ≈ p(1.5) · 1.5 + p(3) · 1.5 = 60 · 1.5 + 40 · 1.5

Answer: donuts produced by 10 am ≈ 150

This is an (circle one) Overestimate Underestimate

b. [4 points] The number of donuts in stock t hours after 7 am is modeled by the functions(t). Estimate the t-values for all critical points of s(t) in the interval 0 < t < 4, andestimate all values of t in the interval 0 < t < 4 at which s(t) has a local extremum. Foreach answer blank write none if appropriate. You do not need to justify your answers.

Solution: We know s′(t) = p(t) − q(t). Since p(t) and q(t) are defined on 0 < t < 4,we only need to find where p(t)− q(t) = 0. In other words, where p(t) = q(t). From thegraph, we can see that s(t) goes from positive to negative at t = 1.2 and t = 3.1 andfrom negative to positive at t = 1.7.

Answer: Critical point(s) at t = 1.2, 1.7, 3.1

Local max(es) at t = 1.2, 3.1 Local min(s) at t = 1.7

c. [2 points] At what time is the number of donuts that Xanthippe has in stock the greatest?Round your answer to the nearest half hour. You do not need to justify your answer.

Answer: 10:00 am

University of Michigan Department of Mathematics Fall, 2016 Math 115 Exam 3 Problem 4 (more donuts) Solution


6. [12 points] A large bucket is left outside during a storm, and the bucket begins to fill withrain. The rain starts at midnight, at which point the bucket is empty. At 2am, the bucketsprings a leak and some water begins to drip out of it. The function r(t) is the rate at whichrain is falling into the bucket t hours after midnight, measured in in3/hr, while the functionℓ(t) is the rate at which water is leaking out of the bucket t hours after midnight, measured inin3/hr. These functions are graphed below.

1 2 3 4

10

20

30

t hours

in3/hr

r(t)

ℓ(t)

(2, 20)

(3, 5)

(3, 25)

(4, 30)

Be sure to include units in your answers to the following questions. No explanation isnecessary, but partial credit may be given for correct work. Assume the bucket is big enoughthat it never overflows during the storm.

a. [3 points] How much water was in the bucket at 3am?

Solution: The amount of water that has entered is the area under the solid curve from0 to 3, which is 32.5in3, and the amount that has exited is the area under the dashedcurve from 0 to 3, which is 12.5in3. Thus, there is 20in3 of water in the bucket at 3am.

b. [2 points] At what time was the amount of water in the bucket greatest?

Solution: This occurs when r(t) = ℓ(t) and r(t)− ℓ(t) is going from positive to negative,which is at t = 2.5, or 2:30am.

c. [3 points] What is the largest amount of water that was in the bucket between midnightand 4am?

Solution: This is the amount that was in the bucket at 2:30am, which is∫2.5

0(r(t) −

ℓ(t))dt = 25in3.

d. [2 points] At what time was the amount of water in the bucket increasing fastest?

Solution: This is when r(t)− ℓ(t) is largest, which is at 2am.

e. [2 points] Write an integral expressing the average rate at which rain fell into the bucketover the period from midnight to 4am. You do not need to evaluate your integral.

Solution:

1

4

∫4

0

r(t)dt

University of Michigan Department of Mathematics Winter, 2012 Math 115 Exam 3 Problem 6 (bucket) Solution


4. [10 points] Gabe the mouse is swimming alone in a very large puddle of water. He keeps track of hisswimming time by logging his velocity at various points in time. Gabe starts at a point on the edgeof the puddle and swims in a straight line with increasing speed. A table of Gabe’s velocity V (t), infeet per second, t seconds after he begins swimming is given below.

t 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6

V (t) 0 0.3 0.4 0.45 0.9 1.2 1.8 2.4 2.7 2.9 3 3.2 3.5

a. [3 points] Give a practical interpretation of the integral

∫5.5

1

V (t) dt in the context of the

problem. Be sure to include units.

Solution: The distance Gabe traveled, in feet, in between seconds 1 and 5.5 after he startedswimming.

b. [3 points] Estimate

∫5.5

1

V (t) dt by using a right-hand Riemann sum with 3 equal subdivisions.

Make sure to write down all terms in your sum.

Solution: If we divide the interval [1, 5.5] in three, we obtain ∆t =5.5− 1

3= 1.5. Then

Right(3) = (V (2.5) + V (4) + V (5.5))∆t = (1.2 + 2.7 + 3.2)(1.5) = (7.1)(1.5) = 10.65.

Answer=10.65 feet.

c. [1 point] Is your estimate from above an overestimate or an underestimate of the exact value of∫5.5

1

V (t) dt? Circle your answer.

Solution: overestimate underestimate not enough information

d. [3 points] Suppose Gabe wants to use a Riemann sum to calculate how far he traveled betweent = 1 and t = 5.5, accurate to within 0.15 feet. How many times would he have to measure hisvelocity in this interval in order to achieve this accuracy? Justify your answer.

Solution: Since V (t) is increasing in [1, 5.5] then |V (5.5) − V (1)|∆t ≤ 0.15, where ∆t is thepossible width of each interval in order for the estimate to be true. Hence ∆t ≤ 0.15

2.8. Then if

N is the number of times Gabe has to measure his velocity to attain its desired accuracy, then

N =5.5− 1

∆t≥

4.50.15

2.8

= 3(28) = 84

Answer= At least 84 times.

University of Michigan Department of Mathematics Fall, 2017 Math 115 Exam 3 Problem 4 (puddle) Solution

Math 115 / Exam 2 (March 24, 2015) page 8

7. [10 points] To aid in Elphaba’s escape, Walt has concocted a supplement that will make herstronger and more agile. The concentration of the supplement in Elphaba’s system, in mg/ml,t minutes after it is administered is given by the following formula:

T (t) =

{

at3 0 ≤ t ≤ 5

b(t− 6)2 + 10 5 < t ≤ 7

where a and b are constants.

a. [7 points] Given that T (t) is differentiable, find a and b. Give your answers in exact form.

Solution: Because the two pieces of the function are polynomials, the only point atwhich the function could fail to be differentiable is at t = 5. In order to be differentiableat t = 5, T (t) must be continuous there. So lim

t→5−T (t) = lim

t→5+T (t). Now,

limt→5−

T (t) = limt→5−

at3 = a(53) = 125a

andlimt→5+

T (t) = limt→5+

b(t− 6)2 + 10 = b(5− 6)2 + 10 = b+ 10.

So we must have 125a = b+ 10.

In order for T (t) to be differentiable at t = 5, the slope of the tangent line to the graphof y = at3 and t = 5 must be the same as the slope of the tangent line to the graph ofy = b(t− 6)2 +10 at t = 5. The slope of the tangent line to the graph of y = at3 at t = 5is 3at2 evaluated at t = 5, which is 75a. The slope of the tangent line to the graph ofy = b(t−6)2+10 at t = 5 is 2b(t−6) evaluated at t = 5, which is −2b. Hence 75a = −2b.

So in order for T (t) to be differentiable, we must have 125a = b + 10 and 75a = −2b.Solving these two equations simultaneously we find a = 4

65and b = −30

13.

Answer: a =

4

65 and b =−30

13

b. [3 points] Using the values of a and b you found in part (a), give a formula for the tangentline to the graph of y = T (t) at t = 5.

Solution: T (5) = a(53) = 4

65(125) = 100

13and T ′(5) = a(3(52)) = 4

65(75) = 60

13. So a

formula for the tangent line to the graph of y = T (t) at t = 5 is y = 100

13+ 60

13(t− 5).

Answer: y =

100

13+

60

13(t− 5) or

60

13t−

200

13

University of Michigan Department of Mathematics Winter, 2015 Math 115 Exam 2 Problem 7 Solution

Math 115 / Final (Dec 12, 2014) page 10

9. [8 points] Consider the family of functions given by

I(t) =At2

B + t2

where A and B are positive constants. Note that the first and second derivatives of I(t) are

I ′(t) =2ABt

(B + t2)2and I ′′(t) =

2AB(B − 3t2)

(B + t2)3.

a. [2 points] Find limt→∞

I(t). Your answer may include the constants A and/or B.

Answer: limt→∞

I(t) = A

A researcher studying the ice cover over Lake Michigan throughout the winter proposes thatfor appropriate values of A and B, the function I(t) is a good approximation for the number ofthousands of square miles of Lake Michigan covered by ice t days after the start of December.For such values of A and B, a graph of y = I(t) for t ≥ 0 is shown below.

t

y

y = I(t)

Based on observations, the researcher chooses values of the parameters A and B so that thefollowing are true.

• y = 21 is a horizontal asymptote of the graph of y = I(t).

• I(t) is increasing the fastest when t = 25.

b. [6 points] Find the values of A and B for the researcher’s model.Remember to show your work carefully.

Solution: From part (a) above, we know the graph of y = I(t) has a horizontal asymp-tote at y = A. So A = 21.

I(t) is increasing fastest when I ′(t) is maximized. For any value of t at which I ′(t) ismaximized, t is a critical point of I ′(t), so I ′′(t) = 0 or I ′′(t) is undefined. The functionI ′′(t) is defined for all t, and I ′′(t) = 0 if and only if B − 3t2 = 0. So since I ′(t) ismaximized when t = 25, we have B − 3(25)2 = 0 so B = 3(25)2 = 1875. (Alternatively,B− 3t2 = 0 when t = ±

√

B/3. The positive solution is t t =√

B/3, so√

B/3 = 25 andB = 1875.)Thus, if A,B are chosen so that y = 21 is a horizontal asymptote of the graph, A = 21.If A,B are chosen so that I ′(t) is maximized at t = 25, 25 =

√

B/3. Thus, B = 1875.

Answer: A = 21 and B = 1875University of Michigan Department of Mathematics Fall, 2014 Math 115 Exam 3 Problem 9 (ice) Solution

2

1. (20 points) A small company called Maple, Inc. is designing a fancy gift box with a square base.The box must have a volume of 3000 cm3. The gift box has a lid which is to be made of a materialthat costs $1 per square centimeter. The material for the sides of the box costs $0.75 per cm2,and the material for the bottom is $0.80 per cm2.

(a) (10 pts.) What are the dimensions of the cheapest gift box the company can make?

Let x stand for the side of the square base/lid, h stand for the height of the box, and C standfor the cost of making one gift box.

We then have:

• x2h = 3000;

• (0.80 + 1)x2 + 4(0.75)xh = C;

Solving for h = 3000/x2 in the first equation, and substituting into the second, we obtainthe following cost equation:

C = 1.8x2 + +9000/x.

Solving for x in C ′(x) = (3.6x3− 9000)/x2 = 0, we obtain x = 25001/3

≃ 13.57 cm. So, forthis value of x, h = 3000/25002/3

≃ 16.29 cm.

SinceC ′′(x) = 3.6 + 18000/x3 > 0 for all x > 0,

we see that:

• C ′′(25001/3) > 0, and C has a local minimum at x = 25001/3,

• This local minimum is a global minimum since C(x) is concave up for x > 0.

So, the dimensions of the cheapest gift box are approximately

x ≃ 13.57 cm and h ≃ 16.29 cm,

where h is the height of the box, and x is the length of the side of the square base.

It turns out that Maple, Inc. also produces a cube-shaped wooden box to store jewelry. The costof producing q of these boxes is given by

C(q) = 8600 + 0.0001(q − 80)3(q + 90).

(b) (3 pts.) What is the marginal cost when 80 boxes are made? Show your work.

Marginal cost when 80 boxes are made: C ′(80)

Since,

• C ′(q) = 0.0003(q − 80)2(q + 90) + 0.0001(q − 80)3, and

• C ′(80) = 0;

then the marginal cost when 80 boxes are made is zero dollars per box.

(This problem continues on the next page.)

University of Michigan Department of Mathematics Winter, 2006 Math 115 Exam 2 Problem 1 (gift boxes) Solution

3

(This is a continuation of Problem 1).

(c) (3 pts.) The marginal cost of producing 95 of the cube-shaped jewelry boxes is about $13per box. Explain what this means in practical terms. (Your explanation should be under-standable to someone who does not know calculus or economics language).

The cost of producing 96 boxes is about $13 more than the cost of producing 95 boxes.

(d) (4 pts.) Let R and P denote, respectively, the revenue and the profit of Maple, Inc. fromselling q of the cube-shaped jewelry boxes. Fill in the blank and circle the right choice in theparagraph below, as indicated.

If the profit P is maximized when 95 jewelry boxes are sold, then

R′(95) = 13 dollars per box (fill in the blank), and P ′′(95) must be

POSITIVE / NEGATIVE / ZERO (circle the appropriate choice).

University of Michigan Department of Mathematics Winter, 2006 Math 115 Exam 2 Problem 1 (gift boxes) Solution


5. [8 points] The owners of a social networking site are concerned about their profit margins,so they develop the graphs for marginal revenue and marginal cost in terms of the number ofusers on the site. Use the graphs below to answer the following questions. The dashed graphis marginal revenue and the solid line is marginal cost.

0 200 400 600 800 1000

0

2

4

6

8

Users in thousands

Dol

lars

per

thou

sand

user

s

a. [3 points] At which approximate number(s) of users is marginal cost equal to marginalrevenue?

Solution: At around 450, 575, and 825 thousand users, the two curves intersect, somarginal cost is equal to marginal revenue at those three points.

b. [5 points] Which of your answers from part (a) maximizes profit? Clearly justify youranswer.

Solution: In order to maximize profit, we need to decide which of the points abovecould be local maxima. Since profit is revenue minus cost, we can find the sign of thederivative of profit by seeing whereMR−MC is positive and negative. SinceMR > MC

for when the number of users is less than 450,000 and between 580,000 and 800,000 users,and MR < MC everywhere else, 450 thousand users and 800 thousand users are localmaxima of the profit function. In addition, since the area between the marginal cost andmarginal revenue curves is greater between 450 and 580 (and counted negatively) thanthe area between 580 and 800, at 450 thousand users profit will be maximized.

University of Michigan Department of Mathematics Fall, 2011 Math 115 Exam 3 Problem 5 (social network) Solution


1. [12 points] In the 17th century, a ship’s navigator would estimate the distance the ship hastraveled using readings of the ship’s velocity, v(t), in knots (nautical miles per hour). Supposethat between noon and 3:00 pm a certain galleon is traveling with the wind and against theocean current, and that its velocity is given as the difference between the wind velocity w(t)and the velocity of the ocean current c(t), so that v(t) = w(t) − c(t), where t is in hours sincenoon. Consider the wind and ocean velocities for various times between noon and 3:00 p.m.,given by the graphs below:

c(t)

w(t)

t1 t2 t3 t4 t5 t6 t7

a. [1 point] Using integral notation write an expression giving the distance the ship traveledfrom noon to 3:00 pm. Give units.

Solution: d =

∫3

0

v(t)dt, with the distance in nautical miles.

b. [1 point] Using integral notation write an expression giving the average velocity of theship between noon and 3:00 pm. Give units.

Solution: vav =1

3

∫3

0

v(t)dt, with the distance in nautical miles/hour, or knots.

c. [2 points] For what intervals was the ship’s velocity positive?

Solution: The ship’s velocity is positive when w(t) > c(t), which happens on the inter-vals (t1, t3) and (t5, t7).

d. [2 points] For what t values was the ship not moving towards its destination?

Solution: Since this happens when the ship’s velocity is zero or negative, the t valuesare [t0, t1], [t3, t5] and t7.

e. [2 points] For what intervals was the ship’s velocity increasing?

Solution: The ship’s velocity is increasing when the acceleration is positive, and sincea(t) = v′(t) = w′(t) − c′(t), in order for a(t) > 0 we need w′(t) > c′(t), i.e. that theslope of the tangent line to w(t) is greater than the slope of the tangent line to c(t). Thishappens on the intervals (t0, t2) and (t4, t6).

f. [4 points] Please circle each integral which is positive and underline each integral whichis negative.

Solution: ∫t3

t1

v(t)dt

∫t7

t5

v(t)dt

∫t7

t0

w(t)dt

∫t5

t3

c(t)dt

University of Michigan Department of Mathematics Fall, 2008 Math 115 Exam 3 Problem 1 (Galleon) Solution

Ì

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University of Michigan Department of Mathematics Winter, 2002 Math 115 Exam 3 Problem 4 (naked mile) Solution

Math 115 / Exam 2 (November 12, 2013) page 2

1. [7 points] Liam wants to build a rectangular swimming pool behind his new house. The poolwill have an area of 1600 square feet. He will have 8-foot wide decks on two sides of the pooland 10-foot wide decks on the other two sides of the pool (see the diagram below).

10 10

8

8

ℓ

w

a. [4 points] Let ℓ and w be the length and width (in feet) of the pool area including thedecks as shown in the diagram. Write a formula for ℓ in terms of w.

Solution: The area of the pool needs to be 1600 sq ft, so

(ℓ− 2(10))(w − 2(8)) = 1600

Solving this for ℓ gives

ℓ =1600

w − 16+ 20

b. [3 points] Write a formula for the function A(w) which gives the total area (in squarefeet) of the pool and the decks in terms of only the width w. Your formula should notinclude the variable ℓ. (This is the function Liam would minimize in order to find theminimum area that his pool and deck will take up in his yard. You do not need to do theoptimization in this case.)

Solution: The pool and decks together make a rectangle of length ℓ and width w. Thearea A of the rectangle is A = ℓw. Substituting the formula from part (a) gives

A(w) =

(

1600

w − 16+ 20

)

w

University of Michigan Department of Mathematics Fall, 2013 Math 115 Exam 2 Problem 1 (swimming pool) Solution


3. [12 points]Scott is having a graduation party, and his mom wants to order individual cakes for the guests.Each cake is a right circular cylinder with radius R centimeters, height H centimeters, andvolume 250 cubic centimeters. In addition,

• there is a fixed cost of $3 per cake;

• the entire side of the cake will have maize icing with blue candy “M”s, which costs$0.02 per square centimeter; and

• the entire top of the cake will have blue icing, which costs $0.01 per square centimeter.

Recall that the volume of a right circular cylinder with radius R and height H is V = πR2H.

a. [4 points] Find a formula for the cost C of one cake, in terms of its radius R.

Solution: C(R) = 3 + 0.02(area of the side) + 0.01(area of the top)

= 3 + 0.02(2πRH) + 0.01(πR2)

Solving for H in terms of R in the volume formula, gives H = 250/πR2, so

C(R) = 3 + 0.02(2πR250

πR2) + 0.01πR2

= = 3 +10

R+ 0.01πR2.

b. [8 points] What radius and height should Scott’s mom choose for the cakes if she wishesto minimize her costs? What is the minimum price for one cake? (To get credit, you mustfully justify your answer using algebraic work.)

Solution: We need to find critical points to find the minimum cost, using the firstderivative:

C ′(R) =−10

R2+ 0.02πR.

A reasonable domain is R > 0, and C ′ is only undefined when R = 0, which is outsidethe domain. We set C ′ = 0 to find any other critical points.

−10

R2+ 0.02πR = 0

−10 + 0.02πR3 = 0

R3 =10

0.02π

R =

(

10

0.02π

)1/3

≈ 5.4193

Since C ′′(R) = 20

R3 +0.02π, we have C ′′(5.4193) > 0, so R = 5.4193 is a local minimum. Itmust also be a global minimum, since C is continuous on the domain and there is only onecritical point. Then H = 250

π5.41932 ≈ 2.7096, and C = 3 + 10

5.4193 + 0.01π(5.4193)2 ≈ 5.77.

radius = 5.4193cm height = 2.7096cm cost = $5.77

University of Michigan Department of Mathematics Winter, 2010 Math 115 Exam 3 Problem 3 (graduation party) Solution


7. [11 points] A group of meteorologists observe that the sea level is rising by observing a pieceof a rock in the sea.

Only the tip of the rock is visible,and as the sea water rises, less andless of the rock is above water.Let h and r be the height and ra-dius (in inches), respectively, of thepart of the rock that is above thesea. The volume of the rock (incubic inches) is then given by theformula

V =π

2(1 + r

2)h.

h

r

a. [8 points] The meteorologists notice that, as the level of the sea is rising, the radius andvolume of the rock are changing. A year after they started taking the measurements,the radius and height of the rock are 5 and 46 inches, respectively. They notice that atthat time, the radius is decreasing at a rate of 0.05 inches per year, which makes thevolume change at a rate of 80 cubic inches per year. At what rate is the height of the rockchanging at that time? Be sure to include units.

Solution:

V =π

2(1 + r

2)h

dV

dt=

π

2

(

2rdr

dt

)

h+π

2(1 + r

2)dh

dt

dV

dt= πrh

dr

dt+

π

2(1 + r

2)dh

dt

−80 = π(5)(46)(−0.05) +π

2(1 + (5)2)

dh

dt

dh

dt=

−80 + 11.5π

13π≈ −1.07

Answer: The height of the part of the rock that is above the sea is (circle one)

Increasing Decreasing Not Enough Information

at a rate of 1.07 inches per year.

University of Michigan Department of Mathematics Winter, 2018 Math 115 Exam 3 Problem 7 (sea level) Solution


b. [3 points] Meteorologists discover that the sea level increases as a function of the averagetemperature registered during the year. Let R(T ) be the sea level (in centimeters) if theaverage temperature next year is T degrees Celsius (◦C). Circle the one statement belowthat is best supported by the equation

R′(16) = 0.5.

(I) The sea level will rise by about 0.5 centimeters if the average temperature next yearis 16 degrees Celsius.

(II)The sea level will rise by approximately 0.1 centimeters if the average temperaturenext year rises from 15.8◦C to 16◦C next year.

(III) The average temperature next year needs to increases by about 0.5◦C in order for thesea level to rise to a level of 16 centimeters next year.

(IV) The sea level will rise by about 0.2 centimeters if the average temperature next yearrises from 16◦C to 16.2◦C next year.

(V) The sea level rises by 0.5 centimeters for every additional degree Celsius the averagetemperature rises above 16◦C.

University of Michigan Department of Mathematics Winter, 2018 Math 115 Exam 3 Problem 7 (sea level) Solution


5. [10 points] The table below gives several values of a function q(u) and its first and secondderivatives. Assume that all of q(u), q′(u), and q′′(u) are defined and continuous for all realnumbers u.

u 0 1 2 3 4 5 6

q(u) 30 23 19 20 24 25 24

q′(u) 0 −6 −2 1 3 1 −2

q′′(u) −9 5 4 3 2 −5 0

Compute each of the following. Do not give approximations. If it is not possible to find thevalue exactly, write not possible.

a. [2 points] Compute

∫

2

5

q′′(t) dt.

Solution:∫

2

5q′′(t) dt = q′(2)− q′(5) = −2− 1 = −3.

Answer:

∫

2

5

q′′(t) dt = −3

b. [2 points] Compute

∫

5

1

(−2q′′(u) + 2u) du.

Solution:∫

5

1(−2q′′(u)+2u) du = (−2q′(5)+52)−(−2q′(1)+12) = (−2+25)−(12+1) =

10.

Answer:

∫

5

1

(−2q′′(u) + 2u) du = 10

c. [2 points] Suppose that q(u) is an even function. Compute

∫

5

−5

q(u) du.

Solution:∫

5

−5q(u) du = 2

∫

5

0q(u) du. This cannot be computed exactly.

Answer:

∫

5

−5

q(u) du = not possible

d. [2 points] Suppose that q(u) is an even function. Compute

∫

5

−5

(

q′(u) + 7)

du.

Solution:∫

5

−5(q′(u) + 7) du = (q(5) + 7 · 5)− (q(−5) + 7 · (−5)). Since q(5) = q(−5), we

have∫

5

−5(q′(u) + 7) du = q(5)− q(−5) + 7 · 10 = 70.

Answer:

∫

5

−5

(

q′(u) + 7)

du = 70

e. [2 points] Compute the average value of −5q′(u) on the interval [1, 4].

Solution: Average value = 1

4−1

∫

4

1−5q′(u) du = 1

3[−5q(4)− (−5q(1))] = 5

3[q(1)− q(4)] =

−5

3.

Answer:

−5

3

University of Michigan Department of Mathematics Fall, 2016 Math 115 Exam 3 Problem 5 Solution

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