Final Exam — Solutions
Laurence Field
Math 152, Section 31
March 12, 2012
Name: Solutions
Instructions: This exam has 8 questions for a total of 120 points.
The value of each part of each question is stated.
The time allowed is two hours.
Partial credit may be given for incorrect answers if your working is legible.
No books, notes or calculators are allowed.
Since there are no calculators, give your answers in exact form. No decimal approximation is needed.
Important: For question 8, answer EITHER part 8(a) OR part 8(b).
[Your score for question 8 will be the greater of your 8(a) score and your 8(b) score.]
Answer all parts of questions 1–7.
Question
Points
1
15
2
15
3
15
4
15
5
15
6
15
7
15
8
15
Total:
120
i
Score
1. Compute the following integrals.
√
Z
x
√
(a)
dx
x x+1
5
√
√
Solution: Let u = x x + 1 = x3/2 + 1, so that du = 23 x1/2 dx, so x dx =
Z
2
3
du. Then
√
Z
√
x
2
du
2
2
√
dx =
= log |u| + C = log(x x + 1) + C.
3
u
3
3
x x+1
Z
(b)
5
tan(3x) dx
Solution: Let u = cos(3x), so that du = −3 sin(3x) dx. Then
Z
Z
1
du
1
tan(3x) dx = −
= − logcos(3x) + C.
3
u
3
Z
π/2
sin5 x + cos x dx
(c)
5
−π/2
Solution: Note that sin5 x and cos x are odd and even functions of x respectively. Therefore
Z
π/2
Z
5
sin x + cos x dx = 2
−π/2
π/2
π/2
cos x dx = 2[sin x]0
= 2.
0
2. Consider the function f (x) = e1/x .
(a) Is f even, odd or neither?
1
Solution: Since
f (−x) = e−1/x = 1/f (x) 6= ±f (x),
f is neither even nor odd.
(b) Find all local extrema, points of inflection, and horizontal and vertical asymptotes of the graph of f .
Solution: As x → ±∞, 1/x → 0, so f (x) → 1, so there is a horizontal asymptote at y = 1.
The only gap in the domain is at 0. As x → 0+, 1/x → ∞, so f (x) → ∞, so x = 0 is a vertical
asymptote. As x → 0−, 1/x → −∞, so f (x) → 0.
For local extrema, we calculate
1
f 0 (x) = − 2 e1/x < 0,
x
so there are no local extrema as the function decreases on both branches (−∞, 0) and (0, ∞).
Now
2
1 + 2x 1/x
1
00
+ 3 e1/x =
e ,
f (x) =
x4
x
x4
1
8
so f 00 (x) vanishes at x = −1/2, is negative for x < −1/2 and positive otherwise. There is a
point of inflection at x = −1/2.
(c) Sketch the graph of f , showing all the features you found in (b).
4
Solution:
4
3
2
1
-4
-3
-2
-1
0
1
2
3
4
5
-1
(d) Write a formula for f −1 (y).
2
Solution: If y = f (x) = e1/x , then log y = 1/x, so x = 1/log y. Thus
f −1 (y) =
1
.
log y
3. (a) Give the definition of ax , where a > 0 and x is real but not necessarily rational. You should use the
exponential and/or natural logarithm functions in your answer.
3
Solution:
ax = exp(x log a) = ex log a .
(b) Let f be a continuous function on [a, b]. Let P be the partition
a = x0 < x1 < x2 < · · · < xn−1 < xn = b.
Let ∆xi be the length of the ith subinterval, that is, ∆xi = xi − xi−1 . Give the definition of Uf (P ),
the upper sum for f on the partition P .
Solution:
Uf (P ) = M1 ∆x1 + M2 ∆x2 + · · · + Mn ∆xn ,
2
4
where
Mi = max{f (x) : xi−1 ≤ x ≤ xi }
is the maximum value of f on the ith subinterval of P .
(c) Answer true or false to each statement. No justification is required.
(i) Let f be continuous on [a, b]. Then for all partitions P of [a, b] we have
Z b
f (x) dx ≤ Uf (P ).
Lf (P ) ≤
2
a
Solution: True (by definition of the integral).
(ii) If a function f is concave down, then it is decreasing.
2
Solution: False, for example: f (x) = −x2 .
(iii) Let f be continuous on (a, b) and let c ∈ (a, b). If f 0 (x) → +∞ as x → c, then the graph of f
has a vertical tangent at c.
2
Solution: True (by definition of a vertical tangent).
(iv) If f is differentiable and one-to-one on (a, b), then the inverse function f −1 is differentiable
everywhere in its domain.
2
Solution: True (by a theorem from class).
4. Consider the region R bounded by the curves y = x3 and y = x2 .
(a) Find the area A of the region R.
4
Solution: The curves intersect where x3 = x2 , i.e. x = 0, 1. The area is
Z
A=
1
(x2 − x3 ) dx =
0
1 3 1 4
x − x
3
4
1
=
0
1
.
12
(b) Find the coordinates (x, y) of the centroid of R.
Solution: We have
1
Z
x (x2 − x3 ) dx =
xA =
0
and
Z
yA =
0
1
6
1
20
1 4
1
(x − x6 ) dx =
.
2
35
Thus x = 3/5 and y = 12/35.
(c) Using Pappus’ theorem or otherwise, find the volume obtained by revolving R around the line
x = −2.
3
5
Solution: Since the region is entirely to the right of the line, and the distance from the centroid
to the line is 3/5 − (−2) = 13/5, the volume is
V = 2π
13 1
13π
·
=
.
5 12
30
5. In this question, consider a hemispherical bowl of radius 3 inches.
(a) Find a formula for V (h), the volume (in cubic inches) of the water in the bowl when the bowl is
filled to a depth of h inches.
8
Solution: The bowl is obtained by revolving the semicircle x2 + (y − 3)2 = 9, y ≤ 3 around the
y-axis. Using the disk method,
Z
h
V (h) =
π(9 − (y − 3)2 ) dy =
0
π 2
π
[9y − y 3 ]h0 = [9h2 − h3 ].
3
3
(b) The bowl has a small hole in the bottom and is leaking water at a rate of 1 cubic inch per minute.
How quickly is the depth of the water in the bowl decreasing when the depth is 1 inch?
7
Solution: By the chain rule,
dV
dV dh
dh
=
= π(6h − h2 ) .
dt
dh dt
dt
At the relevant point, −1 = dV /dt = π(6 − 1) dh/dt and so dh/dt = −1/5π, so the depth is
decreasing at 1/5π inches per minute.
6. (a) A sum of $1000 is kept in a savings account which earns continuously compounded interest at the
rate of 6% per year.
(i) What is the balance of the account after 1 year?
3
Solution: In the notation from class,
A(t) = A(0) ert ,
so the balance in dollars is
A(1) = 1000 e0.06 .
(ii) How many years from the initial deposit will it take for the balance to reach $3000?
Solution: We want to find t satisfying
3000 = A(t) = 1000 e0.06t ,
so t = (log 3)/0.06 years is how long it will take.
Z
(b) Consider the function f (x) =
2
x
1
dt.
1 + t2
4
4
(i) Show that f is one-to-one. What is the range of f −1 ?
4
Solution: By the first fundamental theorem of calculus,
f 0 (x) =
1
> 0,
1 + x2
so f is increasing, hence one-to-one. The range of f −1 is the domain of f , namely R.
(ii) Calculate (f −1 )0 (0).
4
Solution: Since f (2) = 0, f −1 (0) = 2, so
1
= 1 + 22 = 5.
f 0 (2)
(f −1 )0 (0) =
7. (a) Consider the function f whose graph is depicted below.
y
y = f (x)
1
3
Z
8
5
6
x
x
f (t) dt, showing all local extrema and points of inflection.
Sketch the graph of the function g(x) =
0
Solution:
y
local maximum
y = g(x)
1
local minimum
point of inflection
3
5
6
x
(b) The position of a particle P moving on a circle is described by an angle θ(t) which depends on
time t. Suppose that θ00 (t) = − sin(t/a), where a > 0 is a constant. Suppose also that θ0 (0) = a and
θ(0) = 0.
(i) Find a formula for the angle θ(t) at time t.
5
5
P
θ(t)
Solution: Integrating,
θ0 (t) = a cos(t/a) + C
for some constant C. Since θ0 (0) = a, C = 0. Integrating again,
θ(t) = a2 sin(t/a) + D
for some constant D. Since θ(0) = 0, D = 0. Therefore θ(t) = a2 sin(t/a).
(ii) For which values of a does the particle P eventually visit every point on the circle?
2
Solution: For this to happen, we need the range of θ to cover every angle on the circle,
that is, a√full range of 2π radians. Since the range of θ is [−a2 , a2 ], this happens iff a2 ≥ π,
i.e., a ≥ π.
8. Important: For question 8, answer EITHER part 8(a) OR part 8(b).
[Your score for question 8 will be the greater of your 8(a) score and your 8(b) score.]
EITHER:
(a) Suppose f is a function with domain (− π2 , π2 ). Suppose also that, for all x,
f 00 (x) + f (x) = 0.
(i) Show that
d 0
f (x) cos x + f (x) sin x = 0.
dx
4
Solution:
d 0
f (x) cos x + f (x) sin x = f 00 (x) cos x − f 0 (x) sin x + f 0 (x) sin x + f (x) cos(x)
dx
= [f 00 (x) + f (x)] cos x = 0.
(ii) Hence prove that
5
d f (x) sec x = C sec2 x
dx
for some constant C.
Solution: Integrating the result in (i),
f 0 (x) cos x + f (x) sin x = C
6
(∗)
for some constant C. Multiplying by sec2 x,
f 0 (x) sec x + f (x) sec x tan x = C sec2 x.
d
f (x) sec x .
Now observe that the left-hand side is dx
(iii) By taking indefinite integrals of both sides of (∗), prove that
6
f (x) = C sin x + D cos x
for some constant D.
Solution: Integrating (∗),
f (x) sec x = C tan x + D
for some constant D. Multiplying both sides by cos x, we get the result.
OR:
(b) Let f be a nonnegative continuous function and a, b nonnegative constants. Suppose that, for all
x ≥ 0,
Z x
f (x) ≤ a + b
f (t) dt.
(†)
0
Z
(i) Let g(x) =
x
f (t) dt. Show that g 0 (x) ≤ a + b g(x) for all x ≥ 0.
2
0
Solution: By the first fundamental theorem of calculus, for all x ≥ 0
g 0 (x) = f (x) ≤ a + b g(x)
where the inequality follows from (†).
(ii) Show that
3
d
g(t) e−bt = g 0 (t) − b g(t) e−bt .
dt
Solution:
d
g(t) e−bt = g 0 (t) e−bt + g(t) (−be−bt ) = g 0 (t) − b g(t) e−bt .
dt
(iii) Using the results of (i) and (ii), show that for all t ≥ 0.
d g(t) e−bt ≤ a e−bt .
dt
Solution: By (ii) and then (i),
d g(t) e−bt = g 0 (t) − b g(t) e−bt ≤ (a + b g(t) − b g(t)) e−bt = a e−bt .
dt
7
3
(∗∗)
(iv) By integrating both sides of (∗∗) from 0 to x, show that for all x ≥ 0,
g(x) e−bx ≤
4
a −bx
−e
+1 .
b
Solution: Since the integral is order-preserving, by integrating (∗∗) from 0 to x ≥ 0 we get
Z x
a
g(x) e−bx − g(0) ≤
a e−bt dt = −e−bx + 1 .
b
0
Now observe that g(0) = 0 by definition of g, and we get the result.
(v) Deduce from (iv) and (†) that, for all x ≥ 0,
3
f (x) ≤ a ebx .
Solution: For any x ≥ 0, by (†),
f (x) ≤ a + bg(x) ≤ a + bebx
a −bx
−e
+1 ,
b
where the second inequality follows from (iv). The right-hand side simplifies as a−a+a ebx =
a ebx , whence the result follows.
8