math 116 — practice for exam 3 - university of michiganmconger/dhsp/116w18exam3... · winter 2015...

Math 116 — Practice for Exam 3

Generated April 13, 2018

Name: SOLUTIONS

Instructor: Section Number:

1. This exam has 12 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

Winter 2016 3 11 Hanoi Tower 5

Winter 2017 3 10 8

Winter 2004 2 6 Okefenokee 12

Winter 2011 2 2 microphone 14

Winter 2013 2 8 peanut 14

Fall 2009 3 4 ice sculpture 12

Winter 2015 2 5 8

Winter 2015 3 9 ladybugs1 10

Winter 2018 1 2 owl and dove 9

Fall 2012 1 6 tank 12

Winter 2012 1 9 lagoon 11

Winter 2017 3 1 tomatoes 6

Total 121

Recommended time (based on points): 122 minutes

Math 116 / Final (April 21, 2016) DO NOT WRITE YOUR NAME ON THIS PAGE page 12

11. [5 points] The Hanoi tower is made by rotating the region depicted below around the y-axis.The region is made up of infinitely many adjacent rectangles. The nth rectangle has width 1

and height an =1

n!(2n+ 1)where n = 0, 1, 2, 3, .... The rectangle touching the y-axis corre-

sponds to n = 0. Note that the y-axis is not to scale.

x1 2 3 4 5 6 7

y

a4

a3

a2

a1

a0

a5.. .

Compute the volume of the Hanoi Tower. Give an exact answer.

Solution: To compute the volume of the Hanoi Tower, we focus on each rectangle separately.The volume of the object made by the revolution of the nth rectangle is given by

[π(n+ 1)2 − πn2] · an = π(2n+ 1)

1

n!(2n+ 1)=

π

n!

The total volume is given by adding the volume of all those objects for n = 0, 1, 2, 3, ....

∞∑

n=0

π

n!= π ·

∞∑

n=0

1

n!= πe

University of Michigan Department of Mathematics Winter, 2016 Math 116 Exam 3 Problem 11 (Hanoi Tower) Solution

Math 116 / Final (April 24, 2017) do not write your name on this exam page 10

10. [8 points] The Taylor series centered at x = 0 for a function F (x) converges to F (x) for all xand is given below.

F (x) =

∞X

n=0

(−1)nx4n+1

(2n)!(4n+ 1)

a. [3 points] What is the value of F (101)(0)?Make sure your answer is exact. You do not need to simplify.

Answer: F(101)(0) =

b. [3 points] Find P9(x), the 9th degree Taylor polynomial that approximates F (x) nearx = 0.

c. [2 points] Use your Taylor polynomial from part b. to compute

limx→0

F (x)− x

2x5

University of Michigan Department of Mathematics Winter, 2017 Math 116 Exam 3 Problem 10 Solution

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University of Michigan Department of Mathematics Winter, 2004 Math 116 Exam 2 Problem 6 (Okefenokee) Solution

Math 116 / Exam 2 (March 2011) page 4

2. [14 points] A microphone at the point r = 0 detects sounds in a region enclosed by the cardioidr = 2 + 15

8cos θ. The microphone is placed in front of the stage at an auditorium to record a

musical band. Let d denote the smallest distance you must leave between the audience andthe microphone to avoid recording any noise from the public in attendance.

a. [5 points] Write an integral that computes the area of the shaded region A in terms ofθ1, θ2 and d.

Solution: The line x = −d in polar coordinates is r = −d

cos θ. Hence

A =

∫

θ2

θ1

1

2

(

−d

cos θ

)2

dθ −

∫

θ2

θ1

1

2

(

2 +15

8cos θ

)2

dθ

b. [4 points] Write a formula in terms of θ that computes the value of the slope of thetangent line to the cardioid.

Solution:

dy

dx=

y′

x′=

((2 + 15

8cos θ) sin θ)′

((2 + 15

8cos θ) cos θ)′

=(−15

8sin θ) sin θ + (2 + 15

8cos θ) cos θ

(−15

8sin θ) cos θ − (2 + 15

8cos θ) sin θ

c. [3 points] Find an exact expression for the values of 0 ≤ θ < 2π at which the cardioidhas a vertical tangent line. Full credit will not be given for decimal approximations.

Solution: x′ = (−15

8sin θ) cos θ − (2 + 15

8cos θ) sin θ = − sin θ(2 + 15

4cos θ) = 0.

sin θ = 0 then θ = 0, π.2 + 15

4cos θ = 0 then cos θ = −

8

15.

This yields θ = 0, π, θ1 = cos−1(

−8

15

)

, θ2 = 2π − cos−1(

−8

15

)

d. [2 points] Find the value of d. Show all your work.

Solution: d = −x(θ1) = −(2 + 15

8cos θ1) cos θ1 =

8

15

University of Michigan Department of Mathematics Winter, 2011 Math 116 Exam 2 Problem 2 (microphone) Solution

Math 116 / Exam 2 (March 20, 2013) page 12

8. [14 points]

a. [6 points] Find a definite integral that computes the shaded area outside the circle r = 5

2

and inside the curve given by r = 2 + sin 2θ in the graph below.

Solution:

Area =1

2

∫ 5π

12

π

12

(

(2 + sin(2θ))2 −

(

5

2

)2)

dθ.

Here we found the limits of integration by solving for where 5/2 = 2+ sin 2θ for θ in thefirst quadrant.

b. [4 points] Find the polar coordinates (r, θ) of the points where the line y =√

3 x intersectsthe graph of r = 2 + b sin 2θ. Here the constant 0 < b < 2. Your answers may include b.

P1=

P2=

Solution: We want tan θ =√

3, so θ =π

3,4π

3. In the first case, we have

P1 = (r, θ) =

(

2 + b sin

(

2π

3

)

,π

3

)

=

(

2 +

√

3

2b,π

3

)

.

P2 = (r, θ) =

(

2 + b sin

(

8π

3

)

,4π

3

)

=

(

2 +

√

3

2b,4π

3

)

.

University of Michigan Department of Mathematics Winter, 2013 Math 116 Exam 2 Problem 8 (peanut) Solution


c. [4 points]

i) Find the equation in polar coordinates of the line x = 0.

Solution: θ =π

2.

ii) Find the equation in polar coordinates of the line y = 4.

Solution: y = r sin θ = 4, so r =4

sin θ.

University of Michigan Department of Mathematics Winter, 2013 Math 116 Exam 2 Problem 8 (peanut) Solution

Math 116 / Final (December 17, 2009) page 6

4. [12 points] During a party, the host discovers that he has been robbed of his favorite goldhallway clock, and immediately calls the police. The last time the host noticed the clock wasstill hanging, it was 6:00 p.m. When the police arrive, Officer Tom notices a decorative icesculpture. When he first arrives at 9:00 p.m., the ice sculpture’s temperature is 27oF. Afterquestioning others at the party, Officer Tom again takes the temperature of the ice sculptureat 10:00 p.m., and finds it to be 29oF. The temperature of the room has remained a constant68oF all day.

a. [2 points] Let S be the temperature of the sculpture, measured in degrees Fahrenheit.Assuming the sculpture obeys Newton’s Law of Heating and Cooling, write a differentialfor dS

dt, where t is the number of hours since 9:00 p.m. Your answer may contain an

unknown constant, k.

Solution:dS

dt= k(S − 68) (or some appropriate variation thereof)

b. [7 points] Using separation of variables, and the information provided about the sculpture,solve the differential equation to find S(t), where t is the number of hours since 9:00 p.m.Your answer should contain no unknown constants.

Solution:dS

S−68 = kdt, which gives ln |S − 68| = kt + C, or S = Aekt + 68. When t = 0,

S = 27, which gives us 27 = A + 68, so A = −41, leaving S = −41ekt + 68. We use theother condition to solve for k. At t = 1, S = 29, so 29 = −41ek(1) + 68. We solve for k:41ek = 39, or ek = 39

41 , which leaves us with k = ln(

3941

)

≈ −0.05001.

S = −41e−0.05001t + 68

c. [3 points] Company FunIce provided the sculpture, and it was delivered by their employeeBill. For each ice sculpture they produce, the company guarantees the temperature willbe exactly 18oF upon delivery. If the sculpture was 18oF at delivery, and Bill left directlyafterwards, should he be considered as a possible suspect of the robbery? Briefly justifyyour answer.

Solution: We solve for when the temperature of the sculpture was S = 18. We solvefor t: 18 = −41e−0.05001t + 68, or 50

41 = e−0.05001t. This gives us t ≈ −3.96823 hours. Thesculpture was delivered nearly four hours before 9:00 p.m., putting the delivery near 5:00p.m. Since Bill left immediately after delivery, he should not be considered as a suspect,since the clock was still there at 6:00 p.m.

University of Michigan Department of Mathematics Fall, 2009 Math 116 Exam 3 Problem 4 (ice sculpture) Solution


5. [8 points] The graph of a slope field corresponding to a differential equation is shown below.

t

y

-2 -1 1 2

-2

-1

1

2

-2 -1 1 2

-2

-1

1

2

-2 -1 1 2

-2

-1

1

2

-2 -1 1 2

-2

-1

1

2

-2 -1 1 2

-2

-1

1

2

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-2

-1

1

2

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-2

-1

1

2

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1

2

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-2

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1

2

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2

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-2

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1

2

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1

2

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-2

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2

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-2

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1

2

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2

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2

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-2

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2

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-2

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2

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2

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-2

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2

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2

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2

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-2

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2

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2

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2

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2

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2

For all parts of this problem, no work is required and very little partial credit will be given.

a. [2 points] On the slope field, sketch a solution curve passing through the point (0, 0).

Solution: See graph above.

b. [2 points] If you approximated the value of y(1) using Euler’s method starting at thepoint (0, 0), would your approximated value be an overestimate or underestimate? Circleyour answer.

Solution:

OVERESTIMATE UNDERESTIMATE

c. [4 points] Which of the following differential equations could correspond to the slope fieldabove? Circle all that apply.

Solution:

dydt

= (t− 1)(y − 1.1)(y + 1.8)2

dydt

= (t− 1)(y + 1.1)(y − 1.9)2

dydt

= (t− 1)(0.9− y)(y + 1.8)2

dydt

= (t− 1)(y − 0.9)(y + 1.9)4



dydt

= (1− t)(y − 1.1)(y + 1.9)4


Math 116 / Final (April 23, 2015) page 9

9. [10 points] Vic is planning to put ladybugs in his garden to eat harmful pests. The ladybugexpert at the gardening store claims that the number of ladybugs in his garden can be modeledby the differential equation

dL

dt=

L

20−

L2

100

where L is the number of ladybugs, in hundreds, in Vic’s garden, t days after they are intro-duced.

a. [4 points] Find the equilibrium solutions to this differential equation and indicate theirstability.

Solution: To find the equilibrium solutions we set the right hand side of the equationabove equal to 0. The resulting equation simplifies to L(5 − L) = 0. There is then anunstable equilibrium at L = 0 and a stable equilibrium at L = 5.

b. [2 points] If Vic starts his garden with 50 ladybugs, what will the long term populationof ladybugs in his garden be according to the differential equation above?

Solution: The long term population is 500 ladybugs, as the solution to the differentialequation above with initial condition L(0) = .5 will tend to the stable equilbrium atL = 5 as t → ∞.

The long term population is 500 ladybugs

c. [4 points] For what value of b is the function L(t) = 5ebt(

4 + ebt)

−1a solution to this

differential equation.

Solution: We can compute thatdL

dt= 5bebt

(

4 + ebt

)

−1

− 5e2bt(

4 + ebt

)

−2

=20bebt

(1 + ebt)2

andL

20−

L2

100=

5ebt(

4 + ebt)

−1

20−

25e2bt(

4 + ebt)

−2

100=

ebt

(1 + ebt)2. These two expres-

sions will be equal provided that b =1

20.

b = 1/20

University of Michigan Department of Mathematics Winter, 2015 Math 116 Exam 3 Problem 9 (ladybugs1) Solution

Math 116 / Exam 1 (February 5, 2018) page 4

2. [9 points]

When Alejandra and Brontel were childrenthey spent summer mornings chasing birds inflight. One memorable day they encounteredan owl. The following graph shows the veloci-ties a(t) of Alejandra (solid) and b(t) of Bron-tel (dashed), measured in meters per second,t seconds after the owl took off. The area ofeach region is given.

a. [1 point] How far (in meters) do Alejandraand Brontel chase the owl?

Solution: Summing the areas un-der either curve gives a total dis-tance of 10 m.

1 2 3 4 5

1

2

3

4

5

a(t)

b(t)

2.1

4.6

5.4

2.5

t

v

b. [5 points] Suppose the owl ascends to a height of h meters according to h(t) =√

t wheret is seconds since it went airborne. Let L(h) be the number of meters Alejandra is aheadof Brontel when the owl is h meters above ground. Write an expression for L(h) involvingintegrals and compute L′(2).

Solution: The owl is h meters above the ground at time t = h2. Thus,

L(h) =

∫h2

0

a(t)− b(t) dt.

We compute L′(h) using the second fundamental theorem of calculus.

L′(h) = 2ha(h2)− 2hb(h2).

So we haveL′(2) = 2 · 2a(4)− 2 · 2b(4) = 4 · 3− 4 · 1 = 8,

where we get the values of a(2) and b(2) from the graph.

c. [3 points] The next bird to pass is a dove. This time Alejandra runs twice as fast andBrontel runs three times as fast as they did when chasing the owl. How much faster (inm/s) is Brontel than Alejandra on average in the first 5 seconds?

Solution: The integral that represents this average is

1

5

∫5

0

3b(t)− 2a(t) dt =3

5

∫5

0

b(t) dt−2

5

∫5

0

a(t) dt.

Each of these integrals is equal to 10 as we see from the graph. Hence

1

5

∫5

0

3b(t)− 2a(t) dt =10

5= 2.

Answer: 2 m/s

University of Michigan Department of Mathematics Winter, 2018 Math 116 Exam 1 Problem 2 (owl and dove) Solution

Math 116 / Exam 1 (October 10 , 2012) page 10

6. [12 points] A tank whose base is at ground level has lateral walls in the form of a trapezoid3 meters wide at the bottom, 4 meters wide at the top, and 3 meters high, and has a lengthof 7 meters, as shown in the figures below. The tank contains water up to a level of 1 meter.The density of water is 1000 kg per m3.

a. [8 points] Write an expression that approximates the work done in lifting a horizontalslice of water with thickness ∆h that is at a height of h meters above the ground to thetop of the tank. Use g = 9.8 m/s2 for the acceleration due to gravity.

Solution: Workslice ≈ 1000(13h+ 3)(7)g(3− h)∆h ≈ 68600(1

3h+ 3)(3− h)∆h

b. [4 points] Write an expression for the work required to pump all the water in the tank tothe top of the tank. You do not need to evaluate the expression. Include units.

Solution: Workwater =

∫

1

0

68600

(

1

3h+ 3

)

(3− h)dh Joules.

University of Michigan Department of Mathematics Fall, 2012 Math 116 Exam 1 Problem 6 (tank) Solution

Math 116 / Exam 1 (February 6, 2012) page 10

9. [11 points] In the following problems show all your work to receive full credit.

a. [7 points] The population of an invasive aquatic plant in a circular lagoon has densitygiven by δ(r) = 20(1−e−r

2

) kg/m2, where r is the distance in meters from its center. Thelagoon has radius R meters. Find the exact amount of plants living at the lake.

Solution: Since density is a function of radius, we will slice the biomass into rings withthickness ∆r. The mass of one ring is approximately

δ(r) · 2πr∆r.

So now we add up all the slices, and we get total mass

∫

R

0

20(

1− e−r

2)

· 2πr dr = 20π

∫

R

0

2r(

1− e−r

2)

dr

= 20π

∫

R2

0

(

1− e−u

)

du

= 20π(

u+ e−u

)∣

∣

R2

0

= 20π(R2 + e−R

2

− 1).

b. [4 points] Let

F (x) =

∫

x

0

√

e2t − 1dt.

Find the exact value of the length of the curve on 0 ≤ x ≤ 1.

Solution: The formula for length of an arc is

L =

∫

1

0

√

1 + (F ′(x))2 dx.

By the second FTC, F ′(x) =√e2x − 1. Therefore the length of the arc is

L =

∫

1

0

√

1 + (√

e2x − 1)2 dx =

∫

1

0

√

1 + (e2x − 1) dx

=

∫

1

0

exdx

= e− 1.

University of Michigan Department of Mathematics Winter, 2012 Math 116 Exam 1 Problem 9 (lagoon) Solution

Math 116 / Final (April 24, 2017) do not write your name on this exam page 2

1. [6 points] Suppose that the yield of the tomato plants in a particular Michigan garden is afunction of the amount of water that the plants receive (from rainfall and irrigation).Let T (w) be the seasonal yield (in pounds) of the tomato plants in a season when the plantsreceive w gallons of water every week. A portion of the graph of T 0(w) (the derivative ofT (w)) is shown below.Note that T 0(w) is linear for 50 ≤ w ≤ 70. Let A be the area of the region between thew–axis and the graph of T 0(w) for 0 ≤ w ≤ 30, and let B be the area of the region betweenthe w–axis and the graph of T 0(w) for 30 ≤ w ≤ 50,

10 20 30 40 50 60 70

−3

−2

−1

1

2

3

4

A

B

w

T 0(w)

a. [2 points] If the tomato plants yield 150 pounds of tomatoes when the plants receive 70gallons of water every week, how many pounds of tomatoes would the plants yield in aseason when they receive 30 gallons of water each week? (Your answer may involve theconstants A and B.)

Answer:

b. [2 points] In order to maximize the yield of the tomato plants, how many gallons ofwater should the plants receive each week? (Round to the nearest 5 gallons.)

Answer:

c. [2 points] Consider the integral

Z30

10

T0(w) dw.

Rank the following four estimates of the value of this integral in order from least togreatest by writing them in the correct order on the answer blanks below:

LEFT(10) RIGHT(10) TRAP(10) MID(10)

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University of Michigan Department of Mathematics Winter, 2017 Math 116 Exam 3 Problem 1 (tomatoes) Solution

math 116 — practice for exam 3 - university of michiganmconger/dhsp/116w18exam3... · winter 2015...

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