Math 1102: Calculus I (Math/Sci majors)
MWF 3pm, Fulton Hall 230
Homework 3 Solutions
Please write neatly, and show all work. Caution: An answer with no
work is wrong!
Do the following problems from the book: Chapter III, §6, p. 99
(4,8,14,22); Chapter IV, §1, p. 131 (2-12, evens), §2, p. 135 (1,3,5),
§3, p. 140 (5,6,7), §4, p. 145 (1,17,28); Chapter VIII, §1, p. 246 (21)
§6, p. 99: Find the derivatives of the following functions.
4.
d
1
(log x)5 = 5 (log x)4 ·
dx
x
8.
1
d
(log(ex + sin x)) = x
· (ex + cos x)
dx
e + sin x
14.
d
sin[(2x + 5)2 ] = cos(sin 5x) · 2(2x + 5) · 2
dx
22.
d
((sin x)(cos x)) = cos x · cos x + sin x · (− sin x) = cos2 x − sin2 x
dx
§1, p. 131: Find the following values:
2.
4.
√
2π
π
3
sin
= sin =
.
6
3
2
5π
π
π
1
sin π −
= sin
= sin = .
6
6
6
2
6.
2π
4π
π
1
cos π +
= cos
= − cos = − .
6
3
3
2
8.
√
5π
π
2
cos
= − cos = −
.
4
4
2
10.
tan
12.
π √
2π
= tan = 3.
6
3
π
π
π
= tan −
= − tan = −1.
tan 2π −
4
4
4
§2, p. 135:
1. Draw the graph of tan x for all values of x.
Solution. The function tan x is π-periodic:
tan(x + π) =
sin(x + π)
− sin x
sin x
=
=
.
cos(x + π)
− cos x
cos x
A picture of the graph of tan x for x between − π2 and π2 can be found on
p. 135, from which the general picture follows using the π-periodicity.
See Figure 1.
3. Draw the graph of cot x = 1/ tan x.
Solution. See Figure 1.
Draw the graphs of the following functions:
5a. y = sin x2 .
Solution. As x goes from 0 to 4π, x2 goes from 0 to 2π, and the point on
the unit circle that makes signed angle with the positive x-axis makes
6
4
2
-6
-4
2
-2
4
6
-2
-4
-6
Figure 1. The graphs of tan x (blue) and cot x (orange).
1.0
0.5
-15
-10
5
-5
10
15
-0.5
-1.0
Figure 2. The graphs of sin x2 (blue), sin x3 (orange),
and sin x4 (green).
a full cycle around the circle. Its y-coordinate goes from 0 to 1, back
to 0, to −1, and finally back to 0 again. The effect is stretch the graph
of sin x2 horizontally by a factor of 2. See Figure 2.
5b. y = sin x3 .
1.0
0.5
-15
-10
5
-5
10
15
-0.5
-1.0
Figure 3. The graphs of cos x2 (blue), cos x3 (orange),
and cos x4 (green).
Solution. See Figure 2.
5c. y = sin x4 .
Solution. See Figure 2.
5d. y = cos x2 .
Solution. See Figure 3.
5e. y = cos x3 .
Solution. See Figure 3.
5f. y = cos x4 .
Solution. See Figure 3.
§4, p. 140:
5. Find a formula for sin 3x in terms of sin x and cos x. Similarly for
cos 3x.
Solution. Using the addition formulae
sin(A + B) = sin A cos B + sin B cos A, and
cos(A + B) = cos A cos B − sin A sin B,
we have:
sin 3x = sin(x + 2x) = sin x cos 2x + sin 2x cos x
= sin x cos(x + x) + sin(x + x) cos x
= sin x(cos x cos x − sin x sin x) + (sin x cos x + sin x cos x) cos x
= sin x cos2 x − sin3 x + 2 sin x cos2 x.
Similarly,
cos 3x = cos(x + 2x) = cos x cos 2x + sin 2x sin x
= cos x cos(x + x) + sin(x + x) sin x
= cos x(cos x cos x − sin x sin x) + (sin x cos x + sin x cos x) sin x
= cos3 x − sin2 x cos x + 2 sin2 x cos x
6. Prove that sin(π/2 − x) = cos x, using only the addition formula for
the cosine.
Solution. We have
π
π
π
sin( − x) = sin cos(−x) + sin(−x) cos
2
2
2
= 1 · cos x + (− sin x) · 0 = cos x.
7. Prove the formulas
1
(cos(m − n)x − cos(m + n)x) ,
2
1
sin mx cos nx = (sin(m + n)x + sin(m − n)x) ,
2
1
cos mx cos nx = (cos(m + n)x + cos(m − n)x) .
2
sin mx sin nx =
Solution. The righthand sides of the above equations may be expanded using the addition formulae:
6
cos(m − n)x− cos(m + n)x = cos(mx − nx) − cos(mx + nx)
= cos mx cos nx + sin mx sin nx − (cos mx cos nx − sin mx sin nx)
= 2 sin mx sin nx.
Similarly, we have
sin(m + n)x+ sin(m − n)x = sin(mx + nx) + sin(mx − nx)
= sin mx cos nx + cos mx sin nx + sin mx cos nx − cos mx sin nx
= 2 sin mx cos nx,
and
cos(m + n)x+ cos(m − n)x = cos(mx + nx) + cos(mx − nx)
= cos mx cos nx − sin mx sin nx + cos mx cos nx + sin mx sin nx
= 2 cos mx cos nx.
The desired results follow after dividing the righthand sides above by 2.
§4, p. 145:
1. What is the derivative of cot x?
Solution. Since cot x = 1/ tan x, we may use the chain rule:
d
d
1
1
cos2 x 1
1
(cot x) =
(1/ tan x) = −
·
=
−
=− 2 .
2
2
2
2
dx
dx
tan x cos x
sin x cos x
sin x
1
17. Suppose θ is decreasing at the rate of 30
rad/sec. Find each of the
following:
17a. dy/dt, when θ = π/3 and x is constant, x = 12.
Solution. Since tan θ = xy , we may differentiate both sides:
Since
dθ
dt
dy
x − y dx
1
dθ
dt
dt
=
.
− 2 ·
2
cos θ dt
x
1
= − 30
, θ = π/3, x = 12, and dx
= 0, we have
dt
1
− 2
cos
π
4
12 dy
1
1 dy
dt
·− =
=
,
2
30
12
12 dt
so that
dy
12
= .
dt
15
√
17b. dz/dt, when θ = π/4 and y is constant, y = 10 2.
Solution. Since sin θ = yz , we have:
dy
z − y dz
dθ
dt
dt
cos θ ·
=
.
2
dt
z
√
2, the triangle is isosceles, so that x = y =
Since
θ
=
π/4
and
y
=
10
√
√
√
1
10 2, and z = 2 · 10 2 = 20. We have as well dθ
= − 30
, and dy
= 0,
dt
dt
so that
√
√
−10 2 dz
1
1
2 dz
dt
√ ·− =
=−
,
2
30
20
40 dt
2
and
2
dz
= .
dt
3
17c. dx/dt, when x = 1 if x and y are both changing but z is constant,
z = 2.
Solution. Since cos θ = xz , we have:
dx
z − x dz
dθ
dt
dt
− sin θ ·
=
.
dt √ z 2
√
Since x = 1 and z = 2, we have y = 3. Thus sin θ = y/z = 3/2. As
well, dz
= 0, so that
dt
√
dx
2
3
1
1 dx
−
· − = dt2 =
.
2
30
2
2 dt
Simplifying,
√
dx
3
=
.
dt
30
28. Two airplanes are flying in the same direction, and at a constant
altitude. At t = 0 airplane A is 1000ft vertically above airplane B. Airplane A travels at a constant speed of 600 ft/sec, and B at a constant
speed of 400 ft/sec. Find the rate of change of the angle of elevation θ
of A relative to B at time t = 10sec.
Solution. Since A has a constant speed of 600 ft/sec, its distance
from its starting point is given by 600t ft after t seconds. Similarly,
B’s distinct from its starting point is 400t. Thus we have the picture:
We see that
1000
5
tan θ =
= .
200t
t
Differentiating, we have
dθ
5
sec2 θ ·
= − 2,
dt
t
so that
5
dθ
=− 2 2 .
dt
t sec θ
At t = 10, tan θ = 1/2. Since 1 + tan2 θ = sec2 θ, we have sec2 θ =
1 + 41 = 54 . Plugging into the formula for dθ
, we have
dt
5
1
dθ =− 2 5 =− .
dt
25
10 ·
t=10
4
Chapter VIII, §1, p. 246:
21a. Show that the derivatives of the hyperbolic sine and hyperbolic
cosine functions satisfy
d
(cosh t) = sinh t, and
dt
d
(sinh t) = cosh t.
dt
Solution. Since
d
(et )
dt
= et , we use the chain rule:
d
d et + e−t
1 d t
d −t
(cosh t) =
=
(e ) + (e ) =
dt
dt
2
2 dt
dt
Similarly,
d
d et − e−t
1 d t
d −t
(sinh t) =
=
(e ) − (e ) =
dt
dt
2
2 dt
dt
21b. Show that for all t we have cosh2 t − sinh2 t = 1.
1 t −t
(e −e ) = sinh t.
2
1 t −t
(e +e ) = cosh t.
2
Solution. We compute:
2
2
cosh t − sinh t =
et + e−t
2
2
−
et − e−t
2
2
1 2t
e + 2et e−t + e−2t − (e2t − 2et e−t + e−2t )
4
1
= (e2t + 2 + e−2t − e2t + 2 − e−2t )
4
1
= (4) = 1.
4
=
Problems A, B, and C concern the function f (x), defined by

 x2 sin x1 if x 6= 0
f (x) =

0
if x = 0.
When you finish these problems, you will have found out that a differentiable function may have a derivative that is not continuous!
Problem A. Show that f is continuous at x = 0.
Solution A. In order to show that f is continuous at x = 0, we need
to see that
lim f (x) = f (0).
x→0
We have seen in class (using the squeeze property) that
1
lim x sin
x→0
x
exists and is equal to 0. Since as well limx→0 x = 0, the product rule
for limits applies, and we conclude
1
1
1
2
lim x sin
= lim x · x sin
= lim x · lim x sin
= 0.
x→0
x→0
x→0
x→0
x
x
x
Problem B. Show that f is differentiable at x = 0, and compute f 0 (0).
Solution B. We compute the difference quotient for f at 0:
f (0 + h) − f (0)
f (h)
=
.
h
h
For h non-zero, this is equal to
h2 sin h1
1
= h sin
.
h
h
As we’ve seen, the limit of the above expression as h goes to 0 exists,
and is equal to 0. Thus
f (0 + h) − f (0)
1
0
f (0) = lim
= lim h sin
= 0,
h→0
h→0
h
h
and f is differentiable at 0.
Problem C. Show that the derivative of f is not continuous at x = 0.
(Hint: Use the chain rule to find an expression for f 0 (x) when x 6= 0.
Compare lim f 0 (x) and f 0 (0).)
x→0
Solution C. For x 6= 0, we may compute the derivate f 0 (x) using the
chain and product rules:
d
1
1
1
−1
2
2
x sin
= 2x · sin
+ x · cos
· 2
dx
x
x
x
x
1
1
= 2x sin
− cos
.
x
x
1
1
Claim: As x goes to 0, the function L(x) = 2x sin
−
cos
does
x
x
not have a limit: If it did, then L(x) − 2x sin x1 would have a limit
as x goes to 0, since the limit
of a difference is the difference of limits.
But this is equal to cos x1 , which has no limit as x goes to 0. Thus
lim f 0 (x) does not exist, and the function f 0 is not continuous.
x→0