Math 16B
1 (8 pts.)
Brian Osserman
Practice Exam 1
For each of the following pairs of statements, circle the correct statement.
(a) 2x /2y = 2x−y .
{z
}
|
2x /2y = 2x/y .
(b) (ln x)2 = 2 ln x.
ln(x2 ) = 2 ln x.
{z
}
|
correct
correct
(c) ln x + ln y = ln(x + y).
ln x + ln y = ln(xy).
|
{z
}
correct
(d) exy = (ex )y .
| {z }
exy = ex · ey .
correct
2 (8 pts.)
Find the equation for the tangent line at the point (2, 3) of the graph of the
function f (x) = 2x − e2−x .
Solution:
f 0 (x) = (ln 2)2x − e2−x · (−1) = (ln 2)2x + e2−x .
So
f 0 (2) = (ln 2)22 + e2−2 = 4 ln 2 + e0 = 4 ln 2 + 1.
This is the slope of the tangent line at (2, 3). So the equation is
y − 3 = (4 ln 2 + 1)(x − 2).
3 (20 pts.)
Calculate the derivatives of each of the following functions.
2
(a) e−x .
2
Solution: e−x · (−2x).
(b) x2 ln(5x).
Solution: 2x ln(5x) + x2 ·
1
5x
· 5.
(c) xex .
Solution: 1 · ex + x · ex .
(d) ln(x3 (x + 2)2 (x + 3)).
Solution:
d
d
ln(x3 (x + 2)2 (x + 3)) =
(ln(x3 ) + ln(x + 2)2 + ln(x + 3))
dx
dx
d
=
(3 ln(x) + 2 ln(x + 2) + ln(x + 3))
dx
3
2
1
=
+
+
.
x x+2 x+3
(e) (x + 1)5x .
Solution: (x + 1)5x ((ln(x + 1)) · 5 +
2
5x
x+1
· 1).
4 (15 pts.)
The number of a certain type of bacteria increases continuously at a rate
proportional to the number present. There are 100 present at a given time
and 800 present 3 hours later.
(a) Give a function P (t) to model the number of bacteria t hours after the
initial time.
Solution: P (t) = Cekt for some C, k. At t = 0, there are 100, so
100 = P (0) = Ce0 = C. So P (t) = 100ekt . At t = 3, there are 800, so
800 = P (3) = 100ek·3 . Dividing by 100, we get 8 = ek·3 . Taking ln, we
get ln 8 = k · 3, so k =
ln 8
,
3
and
P (t) = 100e(ln 8)/3·t .
Note: This simplifies further, since (ln 8)/3 = ln 81/3 = ln 2, so in fact
P (t) = 100e(ln 2)·t = 100 · 2t . However, you are not required to make
this simplification.
(b) How many will there be 10 hours after the initial time?
Solution: P (10) = 100 · 210 = 100 · 1024 = 102400. Note: You can
stop after you have substituted 10 for t in the equation for P (t). You
do not need to evaluate further.
(c) After how many hours are there 300 bacteria?
Solution: If P (t) = 100 · 2t = 300, then 2t = 3, and t = log2 3.
3
5 (12 pts.)
A watermelon is dropped (with no initial velocity) from a helicopter hovering at 1600 ft above the ground. Assume the acceleration due to gravity is
-32 ft/s2 .
(a) Give formulas for the velocity v and height h of the doomed watermelon, t seconds after it is dropped.
Solution: The acceleration a(t) is −32. But v 0 (t) = a(t) = −32, so
v(t) = −32t + C1 for some C1 . Since the initial velocity is 0, we have
0 = v(0) = −32(0) + C1 = C1 , so C1 = 0 and v(t) = −32t. Then
h0 (t) = v(t) = −32t, so h(t) = −16t2 + C2 for some C2 . We know
the initial height is 1600, so 1600 = h(0) = −16(0) + C2 = C2 , and
C2 = 1600. So h(t) = −16t2 + 1600.
(b) In how many seconds will the watermelon strike the ground?
Solution: Solve for h(t) = 0. We get 0 = h(t) = −16t2 + 1600, so
16t2 = 1600, and dividing through, t2 = 100, so t = 10 seconds (since
the solution has to be positive).
(c) What is the watermelon’s velocity as it strikes the ground?
Solution: It hits the ground at t = 10, and since v(t) = −32t, the
velocity is v(10) = −320 ft/s.
6 (7 pts.)
Find the function f (x) if f 0 (x) = 2e2x and f (0) = 3.
R
Solution: The (general) antiderivative of f is 2e2x dx. If u = 2x, then
du = 2dx, so
Z
eu du = eu + C = e2x + C.
f (x) =
Since f (0) = 3, we solve for C:
3 = f (0) = e2(0) + C = 1 + C,
so C = 2, and f (x) = e2x + 2.
4
7 (30 pts.)
Evaluate the following indefinite integrals.
(a)
R
x3 +3x2 +10
x+3
Solution:
Z
(b)
R
e−x
1−e−x
x3 + 3x2 + 10
dx =
x+3
R
(x3 + 3x2 ) + 10
dx
x+3
Z 3
Z
x + 3x2
10
=
dx +
dx
x+3
x+3
Z
Z
1
=
x2 dx + 10
dx
x+3
1 3
=
x dx + 10 ln |x + 3| + C.
3
Z
dx
Solution:
Z
(c)
dx
Z
1
e−x
dx
=
du (u = 1 − e−x , du = e−x dx)
1 − e−x
u
= ln |u| + C = ln |1 − e−x | + C.
x(x2 − 1)5 dx
Solution:
Z
Z
1
2
5
x(x − 1) dx =
u5 du (u = x2 − 1, du = 2xdx)
2
Z
1
1
1
=
u5 du = u6 + C = (x2 − 1)6 + C.
2
12
12
(d)
R
3
x2 ex dx
Solution:
Z
2 x3
xe
(e)
R
ln x
x
Z
1
eu du (u = x3 , du = 3x2 dx)
3
Z
1
1
1 3
=
eu du = eu + C = ex + C.
3
3
3
dx =
dx
5
Solution:
Z
ln x
dx =
x
=
6
Z
u du (u = ln x, du =
1
dx)
x
1 2
1
u + C = (ln x)2 + C.
2
2