CHAPTER 14
Related Rates
14.1
14.2
14.3
14.4
The top of a 25-foot ladder, leaning against a vertical wall is slipping down the wall at the rate of 1 foot per second.
How fast is the bottom of the ladder slipping along the ground when the bottom of the ladder is 7 feet away from
the base of the wall?
Fig. 14-1
I Let y be the distance of the top of the ladder from the ground, and let x be the distance of the bottom of the
ladder from the base of the wall (Fig. 14-1). By the Pythagorean theorem, x
2 + y
2 = (25)
2
. Differentiating
with respect to time t, 2x • D,x + 2y • D,y = 0; so, x • D,x + y • D t y = 0. The given information tells us that
D,y = -1 foot per second. (Since the ladder is sliding down the wall, y is decreasing, and, therefore, its
derivative is negative.) When x = 7, substitution in x
2 + y
2 = (25)
2 yields y
2 = 576, y = 24. Substitution in x • D t x + y • D,y = 0 yields: 7 • D,x + 24 • (-1) = 0, D,x = ™ feet per second.
A cylindrical tank of radius 10 feet is being filled with wheat at the rate of 314 cubic feet per minute. How fast is
the depth of the wheat increasing? (The volume of a cylinder is nr
2 h, where r is its radius and h is its height.)
I Let V be the volume of wheat at time t, and let h be the depth of the wheat in the tank. Then V= ir(W)
2 h.
So, D,V = 1007T • D,h. But we are given that D t V= 314 cubic feet per minute. Hence, 314 = lOO-tr • D,h,
D,h = 314/(1007r). If we approximate TT by 3.14, then D,h = l. Thus, the depth of the wheat is increasing at
the rate of 1 cubic foot per minute.
A 5-foot girl is walking toward a 20-foot lamppost at the rate of 6 feet per second. How fast is the tip of her
shadow (cast by the lamp) moving?
Fig. 14-2
I Let x be the distance of the girl from the base of the post, and let y be the distance of the tip of her shadow
from the base of the post (Fig. 14-2). AABC is similar to ADEC. Hence, ABIDE = y/(y - x), f =
yl(y-x), 4 = y/(y-x), 4y-4x = y, 3y = 4x. Hence, 3 • D,y = 4 • D,x. But, we are told that D,x=-6
feet per second. (Since she is walking toward the base, x is decreasing, and D,x is negative.) So 3 • D,y =
4 • (-6), D t y = -8. Thus the tip of the shadow is moving at the rate of 8 feet per second toward the base of the
post.
Under the same conditions as in Problem 14.3, how fast is the length of the girl's shadow changing?
I Use the same notation as in Problem 14.3. Let ( be the length of her shadow. Then ( = y - x. Hence,
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Related Rates
14.1
14.2
14.3
14.4
The top of a 25-foot ladder, leaning against a vertical wall is slipping down the wall at the rate of 1 foot per second.
How fast is the bottom of the ladder slipping along the ground when the bottom of the ladder is 7 feet away from
the base of the wall?
Fig. 14-1
I Let y be the distance of the top of the ladder from the ground, and let x be the distance of the bottom of the
ladder from the base of the wall (Fig. 14-1). By the Pythagorean theorem, x
2 + y
2 = (25)
2
. Differentiating
with respect to time t, 2x • D,x + 2y • D,y = 0; so, x • D,x + y • D t y = 0. The given information tells us that
D,y = -1 foot per second. (Since the ladder is sliding down the wall, y is decreasing, and, therefore, its
derivative is negative.) When x = 7, substitution in x
2 + y
2 = (25)
2 yields y
2 = 576, y = 24. Substitution in x • D t x + y • D,y = 0 yields: 7 • D,x + 24 • (-1) = 0, D,x = ™ feet per second.
A cylindrical tank of radius 10 feet is being filled with wheat at the rate of 314 cubic feet per minute. How fast is
the depth of the wheat increasing? (The volume of a cylinder is nr
2 h, where r is its radius and h is its height.)
I Let V be the volume of wheat at time t, and let h be the depth of the wheat in the tank. Then V= ir(W)
2 h.
So, D,V = 1007T • D,h. But we are given that D t V= 314 cubic feet per minute. Hence, 314 = lOO-tr • D,h,
D,h = 314/(1007r). If we approximate TT by 3.14, then D,h = l. Thus, the depth of the wheat is increasing at
the rate of 1 cubic foot per minute.
A 5-foot girl is walking toward a 20-foot lamppost at the rate of 6 feet per second. How fast is the tip of her
shadow (cast by the lamp) moving?
Fig. 14-2
I Let x be the distance of the girl from the base of the post, and let y be the distance of the tip of her shadow
from the base of the post (Fig. 14-2). AABC is similar to ADEC. Hence, ABIDE = y/(y - x), f =
yl(y-x), 4 = y/(y-x), 4y-4x = y, 3y = 4x. Hence, 3 • D,y = 4 • D,x. But, we are told that D,x=-6
feet per second. (Since she is walking toward the base, x is decreasing, and D,x is negative.) So 3 • D,y =
4 • (-6), D t y = -8. Thus the tip of the shadow is moving at the rate of 8 feet per second toward the base of the
post.
Under the same conditions as in Problem 14.3, how fast is the length of the girl's shadow changing?
I Use the same notation as in Problem 14.3. Let ( be the length of her shadow. Then ( = y - x. Hence,
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