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4-71 QC: OSO/OVY T1: OSO October 25, 2018 17:24 SECTION 4.8 • • Related Rates 281
x and y are functions of time, t. By the Pythagorean Theorem, we have
2
2
[x(t)] + [y(t)] = 100.
Differentiating both sides of this equation with respect to time gives us
2
0 = d (100) = d { [x(t)] + [y(t)] 2 }
dt dt
′
′
= 2x(t)x (t) + 2y(t)y (t).
′
Solving for x (t), we obtain
y(t)
′
′
x (t) =− y (t).
x(t)
Since the height above ground of the top of the ladder at the point in question is 8
feet, we have that y = 8 and from the Pythagorean Theorem, we get
2
2
100 = x + 8 ,
so that x = 6. We now have that at the point in question,
y(t) 8 8
′
′
x (t) =− y (t) =− (−2) = .
x(t) 6 3
So, the bottom of the ladder is sliding away from the building at the rate
of 2.67 ft/sec.
EXAMPLE 8.3 Another Related Rates Problem

A car is traveling at 50 mph due south at a point 1 mile north of an intersection. A
y 50 2
police car is traveling at 40 mph due west at a point 1 mile east of the same
4
intersection. At that instant, the radar in the police car measures the rate at which
x
the distance between the two cars is changing. What does the radar gun register?
40 Solution First, we draw a sketch and denote the vertical distance of the first car
from the center of the intersection y and the horizontal distance of the police car x.
dx
(See Figure 4.95.) Notice that at the moment in question (call it t = t ), =−40,
0
dt
dy
FIGURE 4.95 since the police car is moving in the direction of the negative x-axis and =−50,
dt
Cars approaching an since the other car is moving in the direction of the negative y-axis. From the
intersection √
Pythagorean Theorem, the distance between the two cars is d = x + y . Since all
2
2
quantities are changing with time, we have
√
2 1∕2
2
d(t) = [x(t)] + [y(t)] = {[x(t)] + [y(t)] } .
2
2
Differentiating both sides with respect to t, we have by the chain rule that
′
′
′
2 −1∕2
2
d (t) = 1 {[x(t)] + [y(t)] } 2[x(t)x (t) + y(t)y (t)]
2
′
′
x(t)x (t) + y(t)y (t)
.
= √
[x(t)] + [y(t)] 2
2
′
′
1
Substituting in x(t ) = ,x (t ) =−40,y(t ) = 1 2 and y (t ) =−50, we have
0
0
0
0
4
1
1 (−40) + (−50)
−140
′
d (t ) = 4 √ 2 = √ ≈−62.6,
0
1 + 1 5
4 16
Copyright © McGraw-Hill Education from a stationary position.
so that the radar gun registers 62.6 mph. Note that this is a poor estimate of the
car’s actual speed. For this reason, police nearly always take radar measurements
In some problems, the variables are not related by a geometric formula, in which
case you will not need to follow the first two steps of our outline. In example 8.4, the
third step is complicated by the lack of a given value for one of the rates of change.
301