MATH 1914 Homework 6 Solutions
Section 2.8 Problems
1. The area of a triangle with sides of lengths a and b and contained angle θ is
1
A = ab sin(θ)
2
(a) If a = 4 cm, b = 9 cm, and θ increases at a rate of 0.5 rad/min, how fast is the area
increasing when θ = π/3.
Solution:
dA
1
dθ
= abcos(θ)
dt
2
dt
1
= (4)(9) cos(π/3)(0.5)
2
18
cm2 /rad
=
4
(b) If a = 4 cm, b increases at a rate of 1.5 cm/min, and θ increases at a rate of 0.5 rad/min,
how fast is the area increasing when b = 9 cm and θ = π/3.
Solution:
1
db 1
dθ
dA
= a sin(θ) + ab cos(θ)
dt
2
dt 2
dt
1
1
= (4) sin(π/3)(1.5) + (4)(9) cos(π/3)(0.5)
2√
2
3 3+9
cm2 /rad
=
2
(c) If a increases at a rate of 3.5 cm/min, b increases at a rate of 1.5 cm/min, and θ increases
at a rate of 0.5 rad/min, how fast is the area increasing when a = 4 cm, b = 9 cm and
θ = π/3
Solution:
dA
1
da 1
db 1
dθ
= b sin(θ) + a sin(θ) + ab cos(θ)
dt
2
dt 2
dt 2
dt
1
1
1
= (9) sin(π/3)(3.5) + (4) sin(π/3)(1.5) + (4)(9) cos(π/3)(0.5)
2
2!
2
√
√
√
63 3 3 3
=
+
+ 92 cm2 /rad
8
2
1
2. At noon, car A is 150 km west of car B. Car A is driving east at 25 km/h and car
B is driving north at 35 km/h. How fast is the distance between the cars changing at 4:30
pm?
Solution:
y
z
150 - x
z 2 = (150 − x)2 + y 2
dx
dy
dz
= −2(150 − x) + 2y
2z
dt
dt
dt
4:30 is 4.5 hours after noon, so car A has traveled (4.5)(25) = 112.5 miles and car B has
traveled (4.5)(35) = 157.5 miles. Using the equation z 2 = (150 − x)2 + y 2 we get z ≈ 161.9
miles. Plugging everything in we get
2z
dz
dt
dz
dt
dz
dt
dz
dt
dx
dy
+ 2y
dt
dt
dy
dx
−2(150 − x) dt + 2y dt
=
2z
−2(150 − 112.5)(25) + 2(157.5)(35)
=
2(161.9)
= −2(150 − x)
≈ 28.258km/h
3. A boat is pulled into a dock by a rope attached to the bow of the boat and passing
through a pulley on the dock that is 1 n higher than the bow of the boat. If the rope is
pulled in at a rate of 0.75 m/s, how fast is the boat approaching the dodck when it is 12 m
from the dock?
2
y (rope)
z
x
Solution:
We have
dx
2
2
2
want dt when x = 12m. We know x + (1) = y , giving:
2x
dy
dt
= −0.75 m/s and we
dy
dx
= 2y
dt
dt
dx
y dy
=
dt
x dt
and at x = 12m
122 + 12 = y 2 =⇒ y =
√
145
Plugging everything in we get:
dx
=
dt
√
√
145
−3 145
(0.75) =
m/s
12
48
4. A swimming pool is 20 ft wide, 40 ft long, 3 ft deep at the shallow end, and 9 ft deep at
its deepest point. A cross-section is shown in the figure below. If the pool is being filled at
a rate of 0.65 ft3 /min, how fast is the water level rising when the depth of the deepest point
is 5 ft?
Solution:
3
The figure is drawn without the top 3 feet. V = 12 (b + 12)h(20) = 10(b + 2)h and from
similar triangles,
x
6
y
8
= and =
h
6
h
3
so b = x + 12 + y = h + 12 + 11h
. Therefore,
3
110h2
11h
h = 240h +
V = 10 24 +
3
3
dh
0.8
3
and os 0.8 = dV
= 240 + 220h
. When h = 5, dh
= 240+5(220/3)
= 2275
ft/min. 5. Gravel is
dt
3
dt
dt
3
being dumped from a conveyer belt at a rate of 20 ft /min and its coarseness is such that if
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 increasing when the pile is 10 ft high?
We are given dV
= 20 ft3 /min. We know V = 13 πr2 h and that the diameter, d, satisfies that
dt
4
d = h, so for radius, r, 2r = h, so r = h2 . We want
dh
dt
when h = 10. Thus
1
V = πr2 h
3
2
1
h
V = π
h
3
2
h3 dh
V = π
12 dt
dV
3h2 π dh
=
dt
12 dt
h2 π dh
dV
=
dt
4 dt
Plugging in the given values we get:
20 =
100π dh
dh
4
=⇒
=
ft/min
4 dt
dt
5π
6. Two carts, A and B are connected by a rope 39 ft long that passes over a pulley P
(see the below figure). The point Q is on the floor 12 ft directly beneath P and between the
carts. Cart A is being pulled away from Q at a speed of 2 ft/s. How fast is cart B moving
toward Q at the instant when cart A is 5 ft from Q?
5
Solution:
6
We have
dx
dt
= −2 ft/s and want dy
when x = −5. By the Pythagorean Theorem,
dt
p
√
x2 + 122 + y 2 + 122 = 39
So by differentiating:
dx
2y
dy
2x
√
+ p
=0
2 x2 + 122 dt
2 y 2 + 122 dt
p
−x y 2 + 122 dx
dy
= √
dt
y x2 + 122 dt
When x = −5 we need to find y.
p
p
39 = (−5)2 + 122 + y 2 + 122
p
39 = 13 + y 2 + 122
p
26 = y 2 + 122
√
So y = 532. So plugging everything in we get:
dy
−5(26)
−10
=√
(−2) = √
ft/s
dt
532(13)
133
7
Section 2.9 Problems
7. Find the differential dy and evaluate dy for the given values of x and dx:
y=
x−1
, x = 2, dx = 0.03
x+1
Solution:
(x + 1) − (x − 1)
dx
(x + 1)2
2
=
dx
(x + 1)2
dy =
At x = 2 and dx= 0.03:
2
dy = (0.03)
9
8. Use linear approximation (or differentials) to estimate the number: (2.999)5 .
Solution:
Let f (x) = x5 at a = 3. Then f 0 (x) = 5x4 and f (3) = 243, and f 0 (3) = 405. Then L(x) =
243+405(x−3) so x5 ≈ 243+405(x−3) near 3. So (2.999)5 ≈ 243+405(2.999−3) ≈ 242.595.
9. On p. 431, of Physics: Calculus, 2d ed., by Eugene p
Hecht (Pacific Grove, CA: Brooks/Cole,
2000), in the course of deriving the formula T = 2π L/g for the period of a pendulum of
length L, the author obtains the equation aT = −g sin θ for the tangential acceleration of
the bob of the pendulum. He then says, ”for small angles, the value θ in radians is very
nearly the value of sin θ; they differ by less than 2% out to about 20o .
(a) Verify the the linear approximation at 0 for the sine function:
sin x ≈ x
Solution:
Let f (x) = sin(x), f 0 (x) = cos(x). So f (0) = 0 and f 0 (0) = 1. Thus f (x) ≈ f (0) + f 0 (0)(x −
0) = 0 + 1(x − 0) = x.
10. Suppose that the only information we have about a function f is that f (1) = 5 and the
graph of its derivative is as shown below.
(a) Use a linear approximation to estimate f (0.9) and f (1.1)
Solution:
The graph show f 0 (1) = 2, so
L(x) = f (1) + f 0 (1)(x − 1)
= 5 + 2(x − 1)
= 2x + 3
8
Thus f (0.9) ≈ L(0.9) = 4.8 and f (1.1) ≈ L(1.1) = 5.2.
(b) Are your estimates in part (a) too large or too small? Explain.
Solution:
(b) From the graph we see that f 0 (x) is positive and decreasing. This means that the slopes
of the tangent lines are positive but the tangents are becoming less steep. So the tangent
lines lie above above the curve. Thus the estimates in part 9a) are too large.
9