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Answer: Differential Equation: Initial Condition: University of Michigan Department of Mathematics. Fall, 2017 Math 116 Exam 3 Problem 5 (dunking booth) ...
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Semester Exam Problem Name Points Score
Fall 2016 2 2 5 Winter 2017 2 3 leaky boat 5 Fall 2017 3 5 dunking booth 5 Winter 2016 2 6 15 Fall 2017 3 1 foxes 6 Fall 2014 2 6 10 Fall 2015 2 5 10 Fall 2016 2 4 12 Winter 2018 3 12 6
Total 74
Recommended time (based on points): 72 minutes
Math 116 / Midterm (November 14, 2016) DO NOT WRITE YOUR NAME ON THIS PAGE page 3
h(w) = Aw^3 +
w
is a solution to the differential equation
w^2
dh dw
− 3 wh + B = 0
satisfying h(1) = 32. Show all your work, and write your final answers in the spaces provided.
Solution: A = 1 /^2
Hawk-I employees removed ballots from the ballot box continuously at a rate of 4 million ballots per hour. Those ballots were removed in proportion to the current ratio in the box. Hawk-I employees then instantly changed the the ballots voting for candidate B to vote for candidate A (leaving any votes for candidate A unchanged) before immediately returning the ballots to the box.
Assume that the ballot box always contains 100 million votes, and that the ballot box only contains votes for candidates A and B.
Write a differential equation that models a(t), the number of ballots voting for candidate A, in millions, in the ballot box t hours after Hawk-I began changing votes.
Solution:
da dt
a 25
University of Michigan Department of Mathematics Fall, 2016 Math 116 Exam 2 Problem 2 Solution
y = 4 − 4 x
x (cm)
y (cm)
Write, but do not evaluate, an expression involving one or more integrals that gives the mass, in grams, of the resulting solid.
Answer: Mass =
Answer: Differential Equation:
Initial Condition:
University of Michigan Department of Mathematics Fall, 2017 Math 116 Exam 3 Problem 5 (dunking booth) Solution
Math 116 / Exam 2 (March 21, 2016) DO NOT WRITE YOUR NAME ON THIS PAGE page 8
(^2) +A (^3) x solves the equation y′^ + 8y = 2xy is
b. [3 points] The function g is positive, decreasing and differentiable. The solution curves of the differential equation y′^ = e−xg(y) are
concave up concave down changing concavity
c. [3 points] Suppose that h(x) is an increasing differentiable function with h(0) = 0 and
xlim→∞ h(x) = 5. The value of the integral
0
(h(x))^4 h′(x) dx
diverges is 5^4 is 5^4 −
is 1 is 0
d. [3 points] Suppose a ≥ 1 is a constant, and the function h satisfies
x^1 /a^
≤ h(x) ≤
xa^
for
0 ≤ x ≤ 1. The integral
0
(h(x))^2 dx converges
always never sometimes
e. [3 points] The function f satisfies
x^3
≤ f (x) ≤
x
for x ≥ 1 and f (x) = g(x^2 ). The
integral
1
g(x) x
dx converges
always never sometimes
University of Michigan Department of Mathematics Winter, 2016 Math 116 Exam 2 Problem 6 Solution
Math 116 / Exam 2 (November 12, 2014) page 9
x
y
x
y
dy dx
= (y + 2)(y − 1)
dy dx
= (y − 2)(y + 1)
dy dx
= (y + 1)(y − 2)^2
dy dx
= (2 − x)(y + 1)
dy dx
= (x − 2)(y + 1)
dy dx
= (x − 1)(y − 2)
b. [6 points] Let r(θ) = k be a polar curve where k > 0 is a constant. Match the quantities on the left with their formulas (in terms of θ) on the right.
dy dθ
dx dθ
dy dx
(A.) k cos(θ)
(B.) − k cos(θ)
(C.) k sin(θ)
(D.) − k sin(θ)
(E.) tan(θ)
(F.) − tan(θ)
(G.) (^) tan(^1 θ)
(H.) − (^) tan(^1 θ)
University of Michigan Department of Mathematics Fall, 2014 Math 116 Exam 2 Problem 6 Solution
Math 116 / Exam 2 (November 18, 2015) DO NOT WRITE YOUR NAME ON THIS PAGE page 5
x
y
a. [4 points] On the slope field, carefully sketch a solution curve passing through the point (-1,-1).
Solution: See graph above.
b. [2 points] The slope field pictured above is the slope field for one of the following differ- ential equations. Which one? Circle your answer. You do not need to show your work.
dy dx
= cos x cos(2y)
dy dx
= sin x cos(2y)
dy dx
= cos x sin(2y)
dy dx
= sin x sin(2y)
c. [4 points] Find two equilibrium solutions to the differential equation you circled.
Solution: The equilibrium solutions of dy dx = cos x sin(2y) are the values of y such that sin(2y) = 0. Solving, we see that the equilibrium solutions are y = 0, ± π 2 , ±π,...
University of Michigan Department of Mathematics Fall, 2015 Math 116 Exam 2 Problem 5 Solution
Math 116 / Final (April 19, 2018) page 14
g′(−e) = 0 g(−e) = 2 g′′′(−e) = 30
P 3 (4) (−e) = 0 g(0) = ni^ P 3 (0) = 2 −^3 e^2 + 5e^3
i. y′^ = x^2 + y^2 D
ii. y′^ = y x B
iii. y′^ = − x y A
iv. y′^ = x(y^2 − 1) E
v. y′^ = x(1 − y^2 ) F
vi. y′^ = 3 x
(^2) + 2 y C
x
y (A)
x
y (D)
x
y (B)
x
y (E)
x
y (C)
x
y (F)
University of Michigan Department of Mathematics Winter, 2018 Math 116 Exam 3 Problem 12 Solution