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Practice problems for Math 116 Exam 3, generated on December 7, 2017. The exam has 11 questions of varying difficulty, and the document provides instructions and solutions for each problem. The exam covers topics such as Taylor series, probability density functions, and mean length calculations. likely useful as study notes or exam preparation material for Math 116 students at the University of Michigan or other universities offering similar courses.
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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
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.
(−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
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) =
(−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
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).
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
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
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
t
y
t
y
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:
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
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