Name:
Sample Midterm 1
Midterm 1: September 22, 2014
1. Find the derivatives of the following:
√
(a) z = ln 3x − 2 Assuming the domain is appropriately restricted.
(b) y = x3 ex
(c) y = tan−1 x
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(d) y = log3 x2
(e) y = (x2 + 1)ln x
(f) y = 5 sinh2 x
(g) y = ln (coth x)
2. Evaluate the following:
Z
x4
(a)
dx
2x5 + π
Z
2
(b) (x + 3)ex +6x dx
Z
(c)
tan(x) dx
Z
1
dx
(d)
x2 − 6x + 13
√
Z 2/2
1
√
(e)
dx
1 − x2
0
Z
(f)
ex sinh ex dx
Z
2
(g)
x 2x dx
3. Find f −1 (x) and verify f −1 (f (x)) = x and f f −1 (x) = x
f (x) = 4x2 , x ≤ 0
4. Does f (z) = (z − 1)2 , where z ≥ 2 have an inverse? Why?
5. The population of bacteria grows at a rate proportional to the amount present.
(a) Write down the differential equation that describes how the population of bacteria changes over
time. Let y represent the number of bacteria cells , t represent time in minutes , and k be the
growth rate.
(b) Solve the differential equation from (a) using separation of variables.
(c) Initially there are 2 millions cells with a doubling time of 8 minutes. Find k.
(d) How many bacteria will there be after 30 minutes?
6. Solve the differential equations given y(1) = 0
xy 0 + (1 + x)y = e−x .
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7. A tank initially contains 50 gallons of brine, with 30 pounds of salt in solution. Water runs into the
tank at 3 gallons per minute and the well-stirred solutions run out at 2 gallons per minute. How long
will it be until there are 25 pounds of salt in the tank?
8. Use Euler’s Method with h = .25 to approximate the solution over the indicated interval
y 0 = x2 ,
y(0) = 0,
x ∈ [0, 1]
9. Find the solution to the differential equation
dy
= ay + b.
dt
10. Below is the slope field for the logistic growth model for a population, y .
dy
= ky (L − y)
dt
Where t represents time, L = 100, represents the carrying capacity, k = .2, represents the growth rate.
Sketch the solution on the slope field for y, given y(0) = 25.
What is
(a) lim y, if y(0) = L/2?
t→∞
(b) lim y, if y(0) = 0?
t→∞
(c) lim y, if y(0) = 2L?
t→∞
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The following tables will be included on the midterm:
Table 1: Trigonometric Functions and Their Derivatives
Dx cos
−1
Derivatives
−1
x= √
−1<x<1
1 − x2
Integrals
x
1
√
dx = sin−1
+C
a
a2 − x2
Z
1
x
1
dx = tan−1
+C
2 + x2
a
a
a Z
1
1
|x|
√
dx = sec−1
+C
a
a
x a2 + x2
Z
Identities
p
sin(cos x) = 1 − x2
p
cos(sin−1 x) = 1 − x2
p
sec(tan−1 x) = 1 + x2
(p
x2 − 1
if x ≥ 1
−1
p
tan(sec x) =
2
− x − 1 if x ≤ −1.
Table 2: Hyperbolic Functions and Their Derivatives
Derivatives
Inverse Derivatives
1
Dx sinh x = cosh x
Dx sinh−1 x = √
x2 + 1
1
Dx cosh x = sinh x
Dx cosh−1 x = √
x>1
2
x −1
1
Dx tanh x = sech2 x
Dx tanh−1 x =
−1 < x < 1
1 − x2
−1
Dx coth x = − csch2 x
Dx sech−1 x = √
0<x<1
x 1 − x2
Dx sech x = − sech x tanh x
Dx csch x = − csch x coth x
Note: The definitions of hyperbolic functions are not included on this chart (pg 374)
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−1