Calculus I, Fall 2012 - Solutions to Review Problems II
1. Find the following limits.
tan θ
a. limθ→0
.
θ
sin 2θ
b. limθ→0
.
sin 3θ
tan(θ + π/4) − 1
c. limθ→0
.
θ
1 − cos 2θ
d. limθ→0
.
θ
sin θ
, we have
a. From tan θ =
cos θ
sin θ
sin θ 1
tan θ
=
=
.
θ
θ cos θ
θ cos θ
From class, as θ → 0,
sin θ
→ 1.
θ
Further, as θ → 0,
cos θ → cos 0 = 1
(cos is a continuous function). Thus, as θ → 0,
tan θ
sin θ 1 1
=
→ 1 × = 1.
θ
θ
cos θ
1
That is,
tan θ
lim
= 1.
θ→0
θ
b. Note
sin 2θ 3θ 2
sin 2θ
=
.
(∗)
sin 3θ
2θ
sin 3θ 3
As θ → 0, certainly 2θ → 0 and 3θ → 0. Thus as θ → 0,
sin 2θ
sin 3θ
→ 1 and
→ 1,
2θ
3θ
3θ
1
and hence also
→ = 1. So the first two terms on the right side of (∗) converge
sin 3θ
1
to 1 as θ → 0. Therefore
sin 2θ
2
lim
= .
θ→0 sin 3θ
3
c. Let f (x) = tan x. Then
f (π/4 + θ) − f (π/4)
θ
tan(π/4 + θ) − tan(π/4)
= lim
θ→0
θ
tan(π/4 + θ) − 1
= lim
(since tan(π/4) = 1).
θ→0
θ
f 0 (π/4) = lim
θ→0
2
√
2
2
From class, tan x has
√ derivative sec x. So the limit is sec π/4. Now cos π/4 = 1/ 2
and thus sec π/4 = 2. We conclude
lim
θ→0
tan(π/4 + θ) − 1
= 2.
θ
Another approach is to use the trig. identity tan(φ + θ) =
tan(π/4 + θ) =
tan φ + tan θ
. This gives
1 − tan φ tan θ
tan(π/4) + tan θ
1 + tan θ
=
.
1 − tan(π/4) tan θ
1 − tan θ
A little algebra then shows
tan(π/4 + θ) − 1
2 tan θ
=
.
θ
θ(1 − tan θ)
tan θ
→ 1 as θ → 0. Further, tan θ → 0 as θ → 0 (tan is continuous).
By part a,
θ
Therefore, as θ → 0,
tan θ
1
1
2 tan θ
=2·
·
→2·1·
= 2,
θ(1 − tan θ)
θ
1 − tan θ
1−0
so the limit is 2.
cos θ − 1
cos 2θ − 1
d. From class,
→ 0 as θ → 0. Thus also
→ 0 as θ → 0. Therefore
θ
2θ
1 − cos 2θ
cos 2θ − 1 = −2
→ −2 × 0 = 0, as θ → 0.
θ
2θ
1 − cos 2θ
= 0.
That is, limθ→0
θ
2. Find all points where the tangent line to y = cos x + sin x is horizontal.
dy
dy
= − sin x + cos x. So the tangent line is horizontal (i.e.,
= 0) at all points x such
dx
dx
that cos x = sin x. This means that the point (cos x, sin x) on the unit circle corresponds
to an angle of π/4 or 5π/4 = π/4 + π. It follows that x = π/4 + nπ for n = 0, ±1, ±2, . . ..
3. Suppose f (1) = g(1) = 1 and f 0 (1) = −1, g 0 (1) = 2. Find the derivative at x = 1 of each
function.
a. f (x2 ).
b. f (g(x)).
c. f (x)2 g(f (x)).
d.
1
.
f (g(x))
a. The derivative at x is f 0 (x2 ) · 2x. So the derivative at 1 is 2f 0 (1) = −2.
3
b. By the chain rule, the derivative at 1 is
f 0 (g(1)) g 0 (1) = f 0 (1) g 0 (1)
= −1 · 2
= −2.
c. Using the product rule and the chain rule, the derivative at x is
2f (x)f 0 (x) g(f (x)) + f (x)2 g 0 (f (x))f 0 (x).
For x = 1, this is
2 · 1 · −1 · 1 + 12 · 2 · −1 = −4.
d. The derivative at x is −
1
f 0 (g(x)) g 0 (x). For x = 1, this is
f (g(x))2
1
− 2 · −1 · 2 = 2.
1
4. True or false.
a. The derivative of f (2x) is 2f 0 (x).
b. If f (x) = f (−x) then f 0 (x) = −f 0 (−x).
c. The derivative of f (1 + x) is f 0 (1 + x).
d. The derivative of f (1/x) is −1/f (x)2 .
a. False. The derivative of f (2x) is 2f 0 (2x).
d
(−x) = −f 0 (−x).
dx
d
c. True. The derivative of f (1 + x) is f 0 (1 + x) (1 + x) = f 0 (1 + x).
dx
d. False. The derivative of f (1/x) is f 0 (1/x) · −1/x2 = −f 0 (1/x)/x2 .
b. True. The derivative of f 0 (−x) is f 0 (−x)
√
d2 y
for y = sin x.
dx2
By the chain rule,
√ d √
dy
= cos x ( x)
dx
√ dx
cos x
d √
1
= √
(since ( x) = √ ).
dx
2 x
2 x
5. Find
Thus
√ d2 y
d dy d cos x
√
=
=
.
dx2
dx dx
dx 2 x
Applying the quotient rule, this is
√
√
√ d √
d
(cos x) 2 x − cos x (2 x)
dx
dx
. (∗)
4x
4
Since
d
(cos x) = − sin x, the chain rule gives
dx
√
√
d
sin x
(cos x) = − √ .
dx
2 x
Thus (∗) becomes
√
√
√
√
sin x √
cos x
cos x
√
√
√
−
2 x−
− sin x −
2 x
x
x
=
4x
√ 4x
√
− sin x cos x
√ .
=
−
4x
4x x
dy .
6. If x + y = 3, find
dx x=4
√
√
Differentiate both sides of x + y = 3 with respect to x. This gives
√
√
1
1 dy
√ + √
= 0,
2 x 2 y dx
or
√
y
dy
= −√ .
dx
x
√
√
√
√
For x = 4, x = 2 and x + y = 3 then implies y = 1. Thus
dy 1
=− .
dx x=4
2
1
mv 2 where m is the object’s mass and v its
2
velocity. If the object is accelerating at a rate of 10 10 m/s2 , at what rate is the kinetic
energy changing when the object’s velocity is 30 m/s?
7. The kinetic energy of an object is K =
dK
when v = 30 (meters per second). We’re told the object accelerates
dt
dv
at the constant rate of 10 (meters per second squared), that is,
= 10. By the chain
dt
rule,
We want to find
dK
dK dv
=
dt
dv dt
dK
= 10
.
dv
Now
d 1
dK
=
mv 2
dv
dv 2
= mv.
Hence
dK
= 10 mv and the rate of change of kinetic energy is 300m when v = 30.
dt
5
8. A 10 ft. ladder leaning against a vertical wall starts to slip. If the bottom of the ladder
is sliding away from the wall at a rate of 1 ft./sec., find the rate at which the top of the
ladder is falling along the wall. Write c for the speed of light (c = 983, 571, 056 ft./sec.).
How high in terms of c is the top of the ladder from the ground when it is falling at a
rate of 2c ft./sec? Something is wrong here! What is going on?
Write x for the distance from the bottom of the ladder to the wall and y for the height
of the top of the ladder along the wall as in the picture.
T
T
T
T
T
T
T
T
y
T 10
T
T
T
T
TT
x
dx
dy
= 1 and want to find
. We have
dt
dt
2
2
2
x + y = 10 (Pythagoras’ Theorem). Differentiating with respect to t gives
Now x and y are functions of time t. We’re given
2x
dx
dy
+ 2y
= 0.
dt
dt
Thus
dy
x dx
=−
dt
y dt
x
=− .
y
From x2 + y 2 = 102 , we see x =
p
100 − y 2 . Hence
p
dy
100 − y 2
=−
.
dt
y
p
So 2c =
(∗)
100 − y 2
when the ladder is falling at twice the speed of light. Squaring both
y
sides,
4c2 =
100 − y 2
y2
That is, 4c2 y 2 = 100 − y 2 , or 4c2 y 2 + y 2 = 100. Solving for y, we find
y=√
10
.
4c2 + 1
6
This is nonsense—the ladder does not go faster than the speed p
of light! Look closely at
p
100 − y 2
equation (∗). As y → 0+ , 100 − y 2 → 10. So for y very small,
is essentially
y
p
10
100 − y 2
. Thus
becomes arbitrarily large (i.e., goes to infinity) as y → 0+ . Does
y
y
this mean the ladder crashes into the ground at infinite speed? No! At some point during
its fall, the ladder loses contact with the wall. From that point on, our analysis doesn’t
apply.
Comment. What if the ladder is constrained to stay in contact with the wall, say its
top is fixed to a smooth groove in the wall, and the bottom moves at constant speed?
Then does the ladder smash into the ground at infinite speed? This scenario requires
that forces approaching infinity be applied to the ladder. It is not realistic.
9. An ice cube is melting so that its volume is shrinking at a rate of 0.1 in.3 /min. At what
rate is the surface area of the cube changing when its side is 0.2 in?
Write x for the sidelength of the cube, V for its volume and S for its surface area. We
1
dV
=−
(the minus sign is because the volume
have V = x3 and S = 6x2 . We’re given
dt
10
dS
1
is shrinking) and want to find
when x = . Now
dt
5
dV
dV dx
=
dt
dx dt
dx
= 3x2
dt
(from V = x3 ).
This means that
−
1
dx
1
dx
= 3x2
, or −
=
.
10
dt
30x2
dt
Hence
dS
dS dx
=
dt
dx dt
dx
= 12x
dt
= 12x −
=−
Thus when x =
(from S = 6x2 ).
1 30x2
2
.
5x
1
, the rate of change of surface area is −2 (square inches per minute).
5
10. A person h ft. tall walks away from a street lamp that is 20 ft. high. If the length of
the person’s shadow is increasing at a rate of 1 ft./sec., how fast (in terms of h) is the
person walking?
Write x for the person’s distance from the street lamp and s for the length of the person’s
ds
dx
= 1 and want to find
. Look closely at the diagram.
shadow. We have
dt
dt
7
H
HH
H
20 − h
20
HH
H
HH
H
HH
H
HH
H
HH
θ HH
x
h
h
x
H
HH
θ H
H
s
Note
cot θ =
(cot =
HH
HH
x
s
and cot θ =
20 − h
h
1
adj.
=
). Thus
opp.
tan
x
s
20 − h =
and x =
s.
20 − h
h
h
The person’s height h is a constant. Hence
dx
20 − h ds
=
dt
h
dt
20 − h
ds
=
(since
= 1)
h
dt
(in feet per second).