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Math 116 — Practice for Exam 3
Generated December 7, 2017
Name: SOLUTIONS
Instructor: Section Number:
- This exam has 11 questions. Note that the problems are not of equal difficulty, so you may want to skip over and return to a problem on which you are stuck.
- Do not separate the pages of the exam. If any pages do become separated, write your name on them and point them out to your instructor when you hand in the exam.
- Please read the instructions for each individual exercise carefully. One of the skills being tested on this exam is your ability to interpret questions, so instructors will not answer questions about exam problems during the exam.
- Show an appropriate amount of work (including appropriate explanation) for each exercise so that the graders can see not only the answer but also how you obtained it. Include units in your answers where appropriate.
- 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. You are also allowed two sides of a 3′′^ × 5 ′′^ note card.
- If you use graphs or tables to obtain an answer, be certain to include an explanation and sketch of the graph, and to write out the entries of the table that you use.
- You must use the methods learned in this course to solve all problems.
Semester Exam Problem Name Points Score Winter 2017 3 6 7 Fall 2016 3 7 Legendre 7 Fall 2015 3 4 8 Fall 2014 3 2 6 Winter 2012 2 9 wire 14 Winter 2014 2 5 telemarketing 7 Fall 2015 3 3 rumor 13 Winter 2016 2 8 hiking 13 Winter 2014 2 3 11 Fall 2013 3 6 6 Winter 2013 2 2 13 Total 105
Recommended time (based on points): 110 minutes
University of Michigan Department of Mathematics Winter, 2017 Math 116 Exam 3 Problem 6 Solution
Math 116 / Final (December 17, 2015) DO NOT WRITE YOUR NAME ON THIS PAGE page 5
- [8 points] Let f (x) = 3
1 + 2x^2. a. [5 points] Find the first 3 nonzero terms of the Taylor series for f centered at x = 0.
Solution: Using the Taylor series for (1 + y)^1 /^3 centered at y = 0,
√ (^3) 1 + y = 1 + 1 3 y +
3
3
y^2 +...
= 1 +
y 3
y^2 9
Substituting y = 2x^2 , √ (^3) 1 + 2x (^2) = 1 + 2 x 2 3
4 x^4 9
b. [3 points] For what values of x does the Taylor series converge? Solution: The binomial series for 3
1 + y converges when − 1 < y < 1. Substituting y = 2x^2 , this converges when 1 < 2 x^2 < 1, or −√^12 < x < √^12.
- [3 points] Determine the exact value of the infinite series
(−1)n+ (2n + 1)!
No decimal approximations are allowed. You do not need to show your work. Solution:
∑^ ∞
n=
(−1)n+ (2n + 1)!
∑^ ∞
n=
(−1)n(−1)^2 n+ (2n + 1)! = sin(−1).
University of Michigan Department of Mathematics Fall, 2015 Math 116 Exam 3 Problem 4 Solution
Math 116 / Final (December 12, 2014) page 4
- [6 points] Let f (x) = xe−x 2 . a. [4 points] Find the Taylor series of f (x) centered at x = 0. Be sure to include the first 3 nonzero terms and the general term.
Solution: We can use the Taylor series of ey^ to find the Taylor series for e−x 2 by substituting y = −x^2.
e−x 2 =
∑^ ∞
n=
(−x^2 )n n!
= 1 + (−x^2 ) + (−x^2 )^2 2!
(−x^2 )n n!
Therefore the Taylor series of xe−x 2 is
xe−x 2 =
∑^ ∞
n=
(−1)nx^2 n+ n! = x − x^3 + x^5 2!
(−1)nx^2 n+ n!
b. [2 points] Find f (15)(0).
Solution: We know that f^
(15)(0) 15! will appear as the coefficient of the degree 15 term of the Taylor series. Using part (a), we see that the degree 15 term has coefficient − 7!^1. Therefore f (15)(0) =
- [3 points] Determine the exact value of the infinite series
(−1)n 2 n n!
Solution: Notice that this is the Taylor series for ey^ applied to y = −2. Therefore, the series has exact value e−^2.
University of Michigan Department of Mathematics Fall, 2014 Math 116 Exam 3 Problem 2 Solution
Math 116 / Exam 2 (March 19, 2012) page 11
c. [5 points] Find the mean length of wire between two consecutive flaws. Solution:
mean =
−∞
xf (x)dx =
0
c
x (1 + x)^3 dx = c lim b→∞
∫ (^) b
0
x (1 + x)^3 dx
u = x v′^ = (1 + x)−^3
u′^ = 1 v =
2(1 + x)^2
= c lim b→∞
−x 2(1 + x)^2
∣b 0 +
∫ (^) b
0
2(1 + x)^2
dx
= c lim b→∞
−x 2(1 + x)^2
2(1 + x)
∣∣b 0 =^
c 2
or
mean =
−∞
xf (x)dx =
0
c
x (1 + x)^3 dx = c lim b→∞
∫ (^) b
0
x (1 + x)^3 dx
u = 1 + x
= c lim b→∞
∫ (^) b+
1
u − 1 u^3
dx = c lim b→∞
∫ (^) b+
1
u−^2 − u−^3 dy
= c lim b→∞ −u−^1 + u−^2 2
∣b 1 +1 =^
c 2
d. [1 point] A second machine produces the same type of wire, but with a different probability density function (pdf). Which of the following graphs could be the graph of the pdf for the second machine? Circle all your answers. Solution: The graph on the left upper corner can’t be the density sincex is the distance between flaws, hence the probability density function can’t be positive for x < 0. The graph on the left lower corner can’t be the density since the area under the curve for x ≥ 0 is infinite (it has a positive horizontal asymptote). The graph on the right upper corner can’t be a density since it is negative on an interval.
x
y
x
y
x
y
University of Michigan Department of Mathematics Winter, 2012 Math 116 Exam 2 Problem 9 (wire) Solution
Math 116 / Exam 2 (March 24th, 2014) page 6
- [7 points] The Terrible Telemarketing corporation has realized that people often hang up on their telemarketing calls. After collecting data they found that the probability that someone will hang up the phone at time t seconds after the call begins is given by the probability density function p(t). The formula for p(t) is given below.
p(t) =
0 t < 0 te−ct 2 t ≥ 0 .
a. [5 points] Find the value of c so that p(t) is a probability density function. Solution: We must have
0 te
−ct^2 dt = 1. Let u = ct (^2) then du = 2ctdt. Thus we get the integral 1 2 c
0 e −udu = 1 2 c limN^ →∞^ −e −u|N 0 =^
1 2 c. Thus we have 1 =^
1 2 c so^ c^ =^
1
b. [2 points] What is the probability that someone will stay on the phone with a telemarketer for more than 4 seconds? Solution: The probability is
4 te
− 12 t^2 dt = e− (^8).
- [6 points] Consider the probability density function q(t) shown below.
q(t) =
0 t < 0 t 2 0 ≤^ t <^2 0 t ≥ 2
a. [4 points] What is the cumulative distribution function Q(t) of the density given by q(t)? Write your final answer in the answer blanks provided. Solution: Q(t) =
∫ (^) t −∞ q(s)^ ds^ =^
∫ (^) t ∫ (^) t^0 q(s)^ ds.^ If^ t <^ 0 then^ Q(t) = 0 and if 0^ ≤^ t <^ 2 then^ Q(t) = 0 q(s)^ ds^ =^
∫ (^) t 0
s 2 ds^ =^ t
(^2) /4. Then it follows that Q(t) = 1 for t ≥ 2.
If t < 0 then Q(t) = 0
If 0 ≤ t < 2 then Q(t) = t^2 / 4
If t ≥ 2 then Q(t) = 1
b. [2 points] What is the median of the distribution? Solution: The median is the number T such that Q(T ) = 12. Thus we want T 2 /4 = 12. Therefore T =
University of Michigan Department of Mathematics Winter, 2014 Math 116 Exam 2 Problem 5 (telemarketing) Solution
Math 116 / Exam 2 (March 21, 2016) DO NOT WRITE YOUR NAME ON THIS PAGE page 10
- [13 points] Brianne is hiking, and the temperature of the air in ◦C after she’s traveled x km is a solution to the differential equation
y′^ + y sin x = 0
a. [7 points] Find the general solution of the differential equation.
Solution: Writing the equation as dy dx = −y sin x and separating the variables we get ∫ 1 y dy =
− sin xdx
ln |y| = cos x + C y = Aecos^ x
b. [2 points] If the temperature was 10 ◦C at the beginning of the hike, find T (x), the temperature of the air in ◦C after she’s traveled x km. Show your work. Solution: From (a), T (x) = Aecos^ x. Since T (0) = 10 we get A = (^10) e. Thus,
T (x) =
e
ecos^ x
c. [4 points] Brianne traveled 7 km on the hike. Using the information given in (b), find the coldest air temperature she encountered on the hike. Give an exact answer (i.e. no decimal approximations).
Solution: We want to find the minimum of the function T (x) = (^10) e ecos^ x^ over the interval [0, 7]. The critical points are 0, π, 2 π. Checking the outputs of T at those points and the endpoint 7 we find that the minimum is T (π) = (^10) e 2.
University of Michigan Department of Mathematics Winter, 2016 Math 116 Exam 2 Problem 8 (hiking) Solution
Math 116 / Exam 2 (March 24th, 2014) page 4
- [11 points] The graph of G(y) is shown below. Suppose that G′(y) = g(y). Consider the differential equation
dy dt = g(y).
− 3 − 2 − 1 1 2 3
− 2
2
y
G(y)
Note again that dy dt = g(y) and the given graph depicts G(y) not g(y).
a. [6 points] The differential equation has 3 equilibrium solutions. Find the 3 solutions and indicate whether they are stable or unstable by circling the correct answer. Equilibrium solution 1: − 2 Stable Unstable
Equilibrium solution 2: 0 Stable Unstable
Equilibrium solution 3: 2 Stable Unstable
b. [2 points] Circle the graph that could be the slope field of the above differential equation.
t
y
I
t
y
II
t
y
III
c. [3 points] Suppose y 1 (t), y 2 (t) and y 3 (t) are all solutions of the differential equation with different initial conditions as indicated below:
- y 1 (t) solves the differential equation with initial condition y(0) = −2.
- y 2 (t) solves the differential equation with initial condition y(0) = 1.5.
- y 3 (t) solves the differential equation with initial condition y(0) = − 2 .1.
Compute the following limits:
lim t→∞ y 1 (t) = − 2 lim t→∞ y 2 (t) = 0 lim t→∞ y 3 (t) = −∞ or DNE
University of Michigan Department of Mathematics Winter, 2014 Math 116 Exam 2 Problem 3 Solution
Math 116 / Exam 2 (March 20, 2013) page 4
- [13 points] a. [7 points] Consider the following differential equations:
A. y′^ = y(x − 2)^2 B. y′^ = y(x − 2) C. y′^ = −y(1 − y) D. y′^ = −y^2 (1 − y)
Each of the following slope fields belongs to one of the differential equations listed above. Indicate which differential equation on the given line. Find the equation of the equilibrium solutions and their stability. If a slope field has no equilibrium solutions, write none.
Differential equation: A
Equilibrium solutions and stability:
y = 0 unstable.
Differential equation: C
Equilibrium solutions and stability:
y = 1 unstable.
y = 0 stable.
University of Michigan Department of Mathematics Winter, 2013 Math 116 Exam 2 Problem 2 Solution
Math 116 / Exam 2 (March 20, 2013) page 5
b. [4 points] A bank account earns a p percent annual interest compounded continuously. Continuous payments are made out of the account at a rate of q thousands of dollars per year. Let B(t) be the amount of money (in thousands of dollars) in the account t years after the account was opened. Write the differential equation satisfied by B(t). Solution: dB dt
p 100
B − q.
c. [2 points] The slope field shown below corresponds to the differential equation satisfied by B(t) (for certain values of p and q). Sketch on the slope field below the solution to the differential equation that corresponds to an account opened with an initial deposit of 3 , 000 dollars.
Solution:
University of Michigan Department of Mathematics Winter, 2013 Math 116 Exam 2 Problem 2 Solution
