Math 1A: Homework 5 Solutions
July 23
1. Prove the following identities.
(a) sinh(x + y) = sinh(x) cosh(y) + sinh(y) cosh(x).
We have
=
=
=
=
=
sinh(x) cosh(y) + sinh(y) cosh(x)
x
y
y
x
e − e−x
e + e−y
e − e−y
e + e−x
+
2
2
2
2
x+y
x+y
x−y
−x+y
−x−y
x−y
e
−e
+ e−x+y − e−x−y
e
−e
−e
+e
+
4
4
x+y
−x−y
2e
− 2e
4
x+y
e
− e−x−y
2
sinh(x + y).
(b) tanh(x + y) =
tanh(x)+tanh(y)
.
1+tanh(x) tanh(y)
We have
tanh(x) + tanh(y)
1 + tanh(x) tanh(y)
=
sinh(x)
sinh(y)
+ cosh(y)
cosh(x)
sinh(x) sinh(y)
1 + cosh(x)
cosh(y)
=
sinh(x) cosh(y)+sinh(y) cosh(x)
cosh(x) cosh(y)
cosh(x) cosh(y)+sinh(x) sinh(y)
cosh(x) cosh(y)
sinh(x) cosh(y) + sinh(y) cosh(x)
cosh(x) cosh(y) + sinh(x) sinh(y)
sinh(x + y)
=
(∵ (a) above & 8(a) of Worksheet 4)
cosh(x + y)
= tanh(x + y).
=
(c) sinh(3x) = 4 sinh3 (x) + 3 sinh(x).
1
We have
sinh(3x) =
=
=
=
=
=
=
sinh(2x + x)
sinh(2x) cosh(x) + cosh(2x) sinh(x)
(∵ (a))
2
(2 sinh(x) cosh(x)) cosh(x) + (cosh (x) + sinh2 (x)) sinh(x) (∵ 8(b,c) Wksht4)
2 sinh(x) cosh2 (x) + cosh2 (x) sinh(x) + sinh3 (x)
3 sinh(x) cosh2 (x) + sinh3 (x)
3 sinh(x)(1 + sinh2 (x)) + sinh3 (x)
3 sinh(x) + 4 sinh3 (x).
(d) sinh(cosh−1 (x)) =
√
x2 − 1 for x ≥ 1.
Note that for x ≥ 1, cosh−1 (x) = ln(x +
√
x2 − 1) so we have
√
sinh(cosh−1 (x)) = sinh(ln(x + x2 − 1))
=
=
=
=
=
=
=
=
√
x2 −1))
√
2
− e−(ln(x+ x −1))
2
√
√
2
(x + x − 1) − (x + x2 − 1)−1
2
√
(x + x2 − 1)2 − 1
√
2(x + x2 − 1)
√
(x2 + 2x x2 − 1 + x2 − 1) − 1
√
2(x + x2 − 1)
√
x x2 − 1 + x2 − 1
√
(x + x2 − 1)
√
√
x x2 − 1 + ( x2 − 1)2
√
(x + x2 − 1)
√
√
x2 − 1(x + x2 − 1)
√
(x + x2 − 1)
√
x2 − 1.
e(ln(x+
2. Find the absolute maximum and minimum for each the following functions on the given
interval.
(a) f (x) = 2x3 − 3x2 − 12x + 1 on [−2, 3].
Observe that f is continuous on [−2, 3] since it is a polynomial so we can use the
closed interval method. For the critical points, we have:
• the end-points x = −2, 3.
• since f 0 (x) = 6x2 − 6x − 12, setting f 0 (x) = 0 gives x2 − x − 2 = 0 ⇒
(x + 1)(x − 2) = 0 ⇒ x = −1, 2. Both of these are inside the given interval.
2
• f is differentiable everywhere on (−2, 3). so no points come from the nondifferentiability.
Note that f (−2) = −3, f (−1) = 8, f (2) = −19 and f (3) = −8 so f attains an
absolute maximum of 8 at x = −1 and an absolute minimum of −19 at x = 2.
(b) f (x) =
sech(x)
1+tanh(x)
on [1, 5].
Observe that f is continuous on [1, 5] since 1+tanh(x) > 0 and sech(x) is bounded
so we can use the closed interval method. For the critical points, we have:
• the end-points x = 1, 5.
• we have
f 0 (x) =
=
=
=
=
=
(1 + tanh(x))(− sech(x) tanh(x)) − sech(x)(sech2 (x))
(1 + tanh(x))2
− sech(x) tanh(x) − sech(x) tanh2 (x) − sech3 (x)
(1 + tanh(x))2
− sech(x) tanh(x) − sech(x)(tanh2 (x) + sech2 (x))
(1 + tanh(x))2
− sech(x) tanh(x) − sech(x)(1)
(1 + tanh(x))2
− sech(x)(tanh(x) + 1)
(1 + tanh(x))2
− sech(x)
.
1 + tanh(x)
Since sech(x) 6= 0 for any x, we conclude that f has no stationary points.
• since 1 + tanh(x) > 0 on the given interval, f is differentiable everywhere on
(1, 5).
sech(5)
sech(1)
= e−1 and f (5) = 1+tanh(5)
= e−5 so f attains an
Note that f (1) = 1+tanh(1)
absolute maximum of e−1 at x = 1 and an absolute minimum of e−5 at x = 5.
Note: if we had simplified f (x), we could have obtained f (x) = e−x ; this would
have made our lives much easier but less interesting.
(c) f (x) = x +
1
x
on [0.5, 4].
Observe that f is continuous on [0.5, 4] since power functions are continuous
wherever defined so we can use the closed interval method. For the critical points,
we have:
• the end-points x = 0.5, 4.
• since f 0 (x) = 1− x12 , setting f 0 (x) = 0 gives 1−1/x2 = 0 ⇒ x2 = 1 ⇒ x = ±1.
Only x = 1 is inside the given interval.
• f is differentiable everywhere on (0.5, 4).
Note that f (0.5) = 2.5, f (1) = 2 and f (4) = 4.25 so f attains an absolute
maximum of 4.25 at x = 4 and an absolute minimum of 2 at x = 1.
3
(d) f (x) = 2 − |x| on [−4, 1].
Observe that f is continuous on [−4, 1] since |x| is continuous so we can use the
closed interval method. For the critical points, we have:
• the end-points x = −4, 1.
• when x < 0, we get f (x) = 2 − (−x) = 2 + x ⇒ f 0 (x) = 1 6= 0; when x > 0,
we get f (x) = 2 − (x) = 2 − x ⇒ f 0 (x) = −1 6= 0. Hence, f has no stationary
points.
• f is differentiable everywhere on (−4, 1) except at x = 0.
Note that f (−4) = −2, f (0) = 2 and f (2) = 1 so f attains an absolute maximum
of 2 at x = 0 and an absolute minimum of −2 at x = −4.
(e) f (x) = |x| + cos(x) on [−2π, 2π].
Observe that f is continuous on [−2π, 2π] since |x| and cos(x) are continuous so
we can use the closed interval method. For the critical points, we have:
• the end-points x = −2π, 2π.
• when x < 0, we get f (x) = −x + cos(x) ⇒ f 0 (x) = −1 − sin(x) = 0 ⇒
sin(x) = −1 ⇒ x = −π/2; when x > 0, we get f (x) = x + cos(x) ⇒ f 0 (x) =
1 − sin(x) = 0 ⇒ sin(x) = 1 ⇒ x = π/2.
• f is differentiable everywhere on (−2π, 2π) except at x = 0.
Note that f (±2π) = 1+2π, f (0) = 1 and f (±π/2) = π/2 so f attains an absolute
maximum of 1 + 2π at x = ±2π and an absolute minimum of 1 at x = 0.
√
(f) f (x) = x 9 − x2 on [−1, 3].
√
Observe that f is continuous on [−1, 3] since 9 − x2 is continuous wherever
defined so we can use the closed interval method. For the critical points, we have:
• the end-points x = −1, 3.
• we have
√
f (x) = (1) 9 − x2 + x
0
(−2x)
√
2 9 − x2
√
x2
9 − x2 − √
9 − x2
(9 − x2 ) − x2
√
=
9 − x2
9 − 2x2
= √
.
9 − x2
=
Set f 0 (x) = 0 to get 9 − 2x2 = 0 ⇒ x = ± √32 . Only √32 belongs to the given
interval.
• f is differentiable everywhere on (−1, 3).
√
Note that f (−1) = − 8, f (3/sqrt2) = 92 and f (3) = 0 so f attains an absolute
√
maximum of 29 at x = √32 and an absolute minimum of − 8 at x = −1.
4
(g) f (x) = ln(x2 + x + 1) on [−1, 1].
Observe that x2 + x + 1 = (x + 1/2)2 + 3/4 > 0 for all x; in particular, this shows
that f is continuous on [−1, 1] so we can use the closed interval method. For the
critical points, we have:
• the end-points x = −1, 1.
• we have
f 0 (x) =
2x + 1
.
+x+1
x2
Set f 0 (x) = 0 to get 2x + 1 = 0 ⇒ x = −1/2. Note that this belongs to the
given interval.
• f is differentiable everywhere on (−1, 1).
Note that f (−1) = ln(1) = 0, f (−1/2) = ln(3/4) and f (1) = ln(3) so f attains
an absolute maximum of ln(3) at x = 1 and an absolute minimum of ln(3/4) at
x = −1/2.
(h) f (x) = x − 2 tan−1 (x) on [0, 4].
Observe that f is continuous on [0, 4] as tan−1 (x) is continuous everywhere so we
can use the closed interval method. For the critical points, we have:
• the end-points x = 0, 4.
• we have
f 0 (x) = 1 −
2
.
1 + x2
Set f 0 (x) = 0 to get 1 + x2 = 2 ⇒ x = ±1. Note that only x = 1 belongs to
the given interval.
• f is differentiable everywhere on (0, 4).
Note that f (0) = 0, f (1) = 1 − 2(π/4) = 1 − π/2 < 0 and f (4) = 4 − 2 tan−1 (4) >
4 − 2(π/2) = 4 − π > 0 so f attains an absolute maximum of (4 − 2 tan−1 (4))(≈
2.674) at x = 4 and an absolute minimum of (1 − π/2) at x = 1.
2
(i) f (x) = xe−x on [0, 4].
2
Observe that f is continuous on [0, 4] as e−x is continuous everywhere so we can
use the closed interval method. For the critical points, we have:
• the end-points x = 0, 4.
• we have
2
2
2
f 0 (x) = (1)e−x + (x)(−2xe−x ) = e−x (1 − 2x2 ).
2
2
2
2
Set f√0 (x) = 0 to get e−x (1 − 2x
√ ) = 0 ⇒ 1 − 2x = 0 ⇒ x = 1/2 ⇒ x =
±1/ 2. Note that only x = 1/ 2 belongs to the given interval.
• f is differentiable everywhere on (0, 4).
5
√
√
Note that f (0) = 0, f (1/ √
2) = e−1/2 / √
2 and f (4) = 4e−16 so f attains an
absolute maximum of e−1/2 / 2 at x = 1/ 2 and an absolute minimum of 0 at
x = 0.
(j) f (x) = 2 cos(x) − 5 sin(x) on [0, π/2].
Observe that f is continuous on [0, π/2] as both sin and cos are continuous everywhere so we can use the closed interval method. For the critical points, we
have:
• the end-points x = 0, 2π.
• we have
f 0 (x) = −2 sin(x) − 5 cos(x).
Set f 0 (x) = 0 to get 2 sin(x) = −5 cos(x) ⇒ tan(x) = −5/2. Note that
because tan(x) ≥ 0 on [0, π/2), there is no solution to tan(x) = −5/2 in
[0, π/2]. Hence, f has no stationary points in (0, π/2).
• f is differentiable everywhere on (0, 2π).
Note that f (0) = 2 and f (π/2) = −5 so f attains an absolute maximum of 2 at
x = 0 and an absolute minimum of −5 at x = π/2.
3. Two sides of a triangle have lengths 12 m and 15 m. The angle between them is
increasing at a rate of 2◦ /min. How fast is the length of the third side changing when
the angle between the sides of fixed length is 60◦ .
Let θ be the angle (in radians) between the two sides and let x be the length (in metres)
π
π
= 2 × 180
= 90
rad/min.
of the third side. We are given that dθ
dt
By the law of cosines, we have
x2 = 122 + 152 − 2(12)(15) cos(θ) = 369 − 360 cos(θ).
√
Plug in θ = π/3 to get x2 = 369 − 360(1/2) = 369 − 180 = 189 ⇒ x = 189. By
differentiating both sides of the relationship above with respect to t, we then have
2x
dx
dθ
= 360 sin(θ)
dt
√ dt
dx
360( 3/2)(π/90)
√
=
dt
2 189
π
= √ m/min.
3 7
4. The angle of elevation of the sun is decreasing at a rate of 0.25 rad/hr. How fast is the
shadow cast by a 400-ft-tall building increasing when the angle of elevation of the sun
is π/6?
6
Let θ be the angle of elevation (in radians) of the sun and let x be the length (in feet)
= −0.25 rad/hr. We then have
of the shadow. We are given that dθ
dt
400
= tan(θ) ⇒ x = 400 cot(θ).
x
Differentiate both sides of the relationship above with respect to t to get
dx
dθ
= −400 csc2 (θ)
dt
dt
= −400 csc2 (π/6)(−0.25)
= 100(2)2 = 400 ft/hr.
5. The volume of an ice cube is decreasing at a rate of 2 cm3 /min. How fast is the surface
area decreasing when the length of an edge is 5 cm?
Let V, S and L be the volume, surface area and length of the cube. We then have
V = L3 ⇒ L = V 1/3 ; S = 6L2 ⇒ S = 6V 2/3 .
When L = 5, we have V = 53 = 125; we are also given that
both sides of the relationship above with respect to t to get
dV
dt
= −2. Differentiate
dS
dV
= 6(2/3)V −1/3
dt
dt
−1/3
= (4)(125)
(−2) = −8/5 = −1.6 cm2 /min.
6. The minute hand on a watch is 8 mm long and the hour hand is 4 mm long. How fast
is the distance between the tips of the hands changing at one o’clock?
Let θ be the angle (in radians) between the two hands and let x be the distance (in
mm) between the tips. By the law of cosines, we have
x2 = 82 + 42 − 2(8)(4) cos(θ) = 80 − 64 cos(θ).
At one o’clock, the hour
hand by θ = 2π/12 = π/6 rad.
q hand is√ahead ofpthe minute
p
√
√
This shows that x = 80 − 64( 3/2) = 80 − 32 3 = 4 5 − 2 3.
Since the hour hand covers an angle of 2π in 12 hours, its angular speed is 2π/12 = π/6
rad/hr. The minute hand covers an angle of 2π in 1 hour, its angular speed is 2π rad/hr.
We therefore have dθ
= π/6 − 2π = −11π
.
dt
6
Differentiate both sides of the relationship above with respect to t to get
2x
dx
dθ
= 64 sin(θ)
dt
dt
dx
64(1/2)(−11π/6)
p
=
√
dt
2(4 5 − 2 3)
22π
= − p
√ ≈ −18.59 mm/hr.
3 5−2 3
7
7. Gravel is being dumped from a conveyor belt at a rate of 30 ft3 /min, and its coarseness
is such that it forms a pile in the shape of a cone whose base diameter and height are
always equal. How fast is the height of the pile rising when the pile is 10 ft high?
Let h, D and V be the height, diameter and volume respectively of the conical pile.
Since we are given that h = D, we have
1
π
V = π(D/2)2 h = h3
3
12
We are given that dV
= 30 and that h = 10. Differentiate both sides of the relationship
dt
above with respect to t to get
dV
3π 2 dh
=
h
dt
12 dt
4 dV
dh
=
dt
πh2 dt
4
6
=
(30) =
ft/min.
2
π(10)
5π
8