APPM 1345
Exam 3 Solutions
Spring 2014
1. (10 points) Find the 100th derivative of the following functions.
1
t
(a) f (t) = 7
(b) h(x) = ln(13x) + ln
13x
Solution:
(a) f (t) = 7t , f 0 (t) = 7t (ln 7), f 00 (t) = 7t (ln 7)2 , f 000 (t) = 7t (ln 7)3 , . . . , f (100) (t) = 7t (ln 7)100
(b) First simplify the function.
h(x) = ln(13x) + ln
1
13x
= (ln 13 + ln x) + (ln 1 − ln 13 − ln x) = ln 1 = 0
h(100) (x) = 0
2. (16 points) For each of the following curves, find an equation of the tangent line at the given value.
(a) y = x ln x, x = e
(b) y = (tan x)ln x , x =
π
4
Solution:
(a) The y-coordinate at x = e is y(e) = e ln e = e. The tangent slope is
y = x ln x
1
y 0 = x · + ln x = 1 + ln x
x
y 0 (e) = 1 + ln e = 2
The equation of the tangent line is y = e + 2(x − e) = 2x − e.
(b) The y-coordinate is y(π/4) = (tan(π/4))ln(π/4) = 1ln(π/4) = 1. The tangent slope can be found
using logarithmic differentiation:
y = (tan x)ln x
ln y = ln(tan x)ln x = (ln x) ln(tan x)
1 dy
sec2 x
1
·
= ln x ·
+ ln(tan x) ·
y dx
tan x
x
2
dy
sec x ln(tan x)
ln x
= (tan x)
ln x ·
+
dx
tan x
x
π dy π
2
4
= (1) ln
· + ln(1) ·
= 2 ln
dx x=π/4
4
1
π
4
The equation of the tangent line is y = 1 + 2 ln
π π
x−
4
4
3. (22 points) Find the values of x where the following curves have horizontal tangents.
(a) y = x2 ln (2/x)
(b) y =
log6 x
−6
x
(c) y = (sin x)esin x , 0 ≤ x ≤ 2π
Solution:
(a)
y = x2 ln(2/x)
−2
0
2 x
+ 2x ln(2/x) = −x + 2x ln(2/x)
y =x ·
2 x2
Solve y 0 = 0.
0 = −x + 2x ln(2/x)
0 = x(−1 + 2 ln(2/x))
x = 0 is not a solution because x is not in the domain of the function. Here is the only solution:
0 = −1 + 2 ln(2/x)
1
= ln(2/x)
2
2
e1/2 =
x
2
x = 2e−1/2 = √
e
(b)
log6 x
ln x
−6=
−6
x
x ln 6
1 x · x1 − ln x
·
y0 =
ln 6
x2
1 − ln x
= 2
x (ln 6)
y=
Solve y 0 = 0.
0 = 1 − ln x
x= e
(c)
y = (sin x)esin x
y 0 = sin x(cos x)esin x + (cos x)esin x
Solve y 0 = 0.
0 = (cos x)esin x (sin x + 1)
Since esin x > 0, either cos x = 0 or sin x = −1 in [0, 2π].
π 3π
,
2 2
x=
4. (10 points) Let y =
ex
.
+1
ex
(a) Show that y is an increasing function and therefore one-to-one.
(b) Find the inverse function of y.
Solution:
(a) Show that y 0 > 0.
y=
ex
(ex + 1)ex − ex · ex
e2x + ex − e2x
ex
0
⇒
y
=
=
=
ex + 1
(ex + 1)2
(ex + 1)2
(ex + 1)2
Since y 0 > 0 for all x, y is an increasing function and one-to-one.
(b) Solve for x, then swap x and y.
ex
ex + 1
x
e y + y = ex
y=
ex y − ex = −y
−y
y
=
ex =
y−1
1−y
y
x = ln
1−y
x
y −1 = ln
1−x
5. (10 points) Evaluate the following limits.
−1
(a) lim tan
u→0
u
(e )
(b) lim cos
t→1+
−1
(ln t)
−1
(c) lim sin
x→∞
Solution:
(a) lim tan−1 (eu ) = tan−1 lim eu = tan−1 (1) = π/4
u→0
u→0
−1
−1
(b) lim cos (ln t) = cos
lim ln t = cos−1 (0) = π/2
t→1+
t→1+
1 − x2
1 − x2
1
π
−1
−1
−1
(c) lim sin
= sin
lim
= sin
−
= −
2
2
x→∞ 1 + 2x
x→∞
1 + 2x
2
6
1 − x2
1 + 2x2
6. (20 points) Evaluate the following integrals.
Z
Z
e3x
(a)
tan(3x) dx
(b)
dx
1 + e6x
Z
(c)
1
eπ
sin(ln x)
dx
x
Solution:
(a) Let u = cos(3x), du = −3 sin(3x) dx ⇒ −du/3 = sin(3x) dx.
Z
Z
Z
sin(3x)
du
1
1
1
tan(3x) dx =
dx = −
= − ln |u| + C = − ln|cos(3x)| + C
cos(3x)
3
u
3
3
(b) Let u = e3x , du = 3e3x dx ⇒ du/3 = e3x dx.
Z
Z
1
1
1
e3x
du
3x
dx
=
=
arctan
u
+
C
=
arctan
e
+C
1 + e6x
3
1 + u2
3
3
(c) Let u = ln x, du = dx/x. Change limits to u(1) = 0, u (eπ ) = π.
Z eπ
Z π
π
sin(ln x)
dx =
sin u du = − cos u 0 = − cos π + cos 0 = 1 + 1 = 2
x
1
0
7. (12 points) Alvin mistakenly left his smartphone outside overnight in the Alaskan cold. The phone had a
temperature of −30◦ C when he brought it inside in the morning. Alvin wants to check his messages but
the phone won’t work properly until it warms up to 0◦ C. If the phone reaches a temperature of −25◦ C
after 2 minutes in Alvin’s 20◦ C house, how long will it take before Alvin can use his phone?
(Use the approximations ln 2 ≈ 0.7, ln 3 ≈ 1.1, and ln 5 ≈ 1.6 to calculate your answer.)
Solution: We are given T0 = −30, TS = 20, T (2) = −25. We wish to find t when T = 0. First find k.
T − TS = (T0 − TS )ekt
−25 − 20 = (−30 − 20)ek(2)
−45 = −50e2k
9
= e2k
10
2k = ln(9/10)
1
k = ln(9/10)
2
Now solve for t when T = 0.
0 − 20 = (−30 − 20)ekt
2
= ekt
5
kt = ln(2/5)
ln(2/5)
2 ln(2/5)
2(ln 2 − ln 5)
2(ln 2 − ln 5)
t=
=
=
=
k
ln(9/10)
ln 9 − ln 10
2 ln 3 − ln 2 − ln 5
2(0.7 − 1.6)
2(−0.9)
≈
=
= 18 min
2(1.1) − 0.7 − 1.6
2.2 − 2.3
The actual value of t is about 17.4 min.