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UAE-Math-Grade-12-Vol-1-SE-718383-ch4 GO01962-Smith-v1.cls November 6, 2018 2019 15:33 21:9
UAE_Math_Grade_12_Vol_1_SE_718383_ch4
October 22,
GO01962-Smith-v1.cls
Related Rates
Related Rates
CHAPTER 4 • •
4-70
280
• • Applications of Differentiation
Applications of Differentiation
CHAPTER 4
280
4.8
RELATED RATES
4.8 RELATED RATES 4-70
In this section, we present a group of problems known as related rates problems. The
In this section, we present a group of problems known as related rates problems. The
common thread in each problem is an equation relating two or more quantities that are
common thread in each problem is an equation relating two or more quantities that are
all changing with time. In each case, we will use the chain rule to find derivatives
all changing with time. In each case, we will use the chain rule to find derivatives of of
all terms in the equation (much as we did in section 2.8 with implicit differentiation).
all terms in the equation (much as we did in a previous section with implicit differen-
The differentiated equation allows us to determine how different derivatives (rates) are
tiation). The differentiated equation allows us to determine how different derivatives
(rates)
related. are related.
EXAMPLE 8.1
A Related Rates Problem
EXAMPLE 8.1 A Related Rates Problem
An oil tanker has an accident and oil pours out at the rate of 150 gallons per minute.
An oil tanker has an accident and oil pours out at the rate of 150 gallons per minute.
1 ′′ ′′
1
Suppose that the oil spreads onto the water in a circle at a thickness of of
Suppose that the oil spreads onto the water in a circle at a thickness 10 10 . (See
. (See
3 3
Figure 4.93.) Given that 1 ft equals 7.5 gl, determine the rate at which the
Figure 4.93.) Given that 1 ft equals 7.5 gl, determine the rate at which the
radius of the spill is increasing when the radius reaches 500 gallon.
radius of the spill is increasing when the radius reaches 500 gallon.
2
Solution Since the area of a circle of radius r is r , the volume of oil is given by
FIGURE 4.93 Solution Since the area of a circle of radius r is r 2, the volume of oil is given by
FIGURE 4.93
1
2
Oil spill V = (depth)(area) = 120 r ,
2,
1 r
Oil spill
V = (depth)(area) =
120
′′ =
since the depth is 1 1 ′′ 1 1 ft. Both volume and radius are functions of time, so
since the depth is 10 = 120 ft. Both volume and radius are functions of time, so
10 120
2
V(t) = [r(t)] .
2
V(t) = 120 [r(t)] .
120
Differentiating both sides of the equation with respect to t,weget
Differentiating both sides of the equation with respect to t,weget

′
′
V (t) = 2r(t)r (t).
′
′
V (t) = 120 2r(t)r (t).
120
3
The volume increases at a rate of 150 gl/min, or 150 = 20 ft /min. Substituting in
3
7.5
The volume increases at a rate of 150 gl/min, or 150 = 20 ft /min. Substituting in
′
V (t) = 20 and r = 500, we have 7.5
′
V (t) = 20 and r = 500, we have
20 = ′
2(500)r (t).
′
20 = 120 2(500)r (t).
120
′
Finally, solving for r (t), we find that the radius is increasing at the rate of
′
Finally, solving for r (t), we find that the radius is increasing at the rate of
2.4
≈ 0.76394 ft/min.
2.4 ≈ 0.76394 ft/min.

Although the details change from problem to problem, the general pattern of so-
Although the details change from problem to problem, the general pattern of so-
lution is the same for all related rates problems. Looking back, you should be able to
lution is the same for all related rates problems. Looking back, you should be able to
identify each of the following steps in example 8.1.
identify each of the following steps in example 8.1.
1. Make a simple sketch, if appropriate.
2. Set up an equation relating all of the
1. Make a simple sketch, if appropriate.relevant quantities.
2. Set up an equation relating all of the relevant quantities.respect to time (t).
3. Differentiate (implicitly) both sides of the equation with
3. Differentiate (implicitly) both sides of the equation with respect to time (t).
4. Substitute in values for all known quantities and derivatives.
4. Substitute in values for all known quantities and derivatives.
5. Solve for the remaining rate.
5. Solve for the remaining rate.
EXAMPLE 8.2 A Sliding Ladder
EXAMPLE 8.2 A Sliding Ladder
A 10-feet ladder leans against the side of a building. If the top of the ladder begins
to slide down the wall at the rate of 2 ft/sec, how fast is the bottom of the ladder
A 10-feet ladder leans against the side of a building. If the top of the ladder begins
sliding away from the wall when the top of the ladder is 8 ft off the ground?
to slide down the wall at the rate of 2 ft/sec, how fast is the bottom of the ladder
10 y
sliding away from the wall when the top of the ladder is 8 ft off the ground?
Solution First, we make a sketch of the problem, as seen in Figure 4.94. We have
10 y denoted the height of the top of the ladder as y and the distance from the wall to
x Solution First, we make a sketch of the problem, as seen in Figure 4.94. We have Copyright © McGraw-Hill Education Copyright © McGraw-Hill Education
the bottom of the ladder as x. Since the ladder is sliding down the wall at the rate of
denoted the height of the top of the ladder as y and the distance from the wall to
x
dy
FIGURE 4.94 the bottom of the ladder as x. Since the ladder is sliding down the wall at the rate of
=−2. (Note the minus sign here.) Observe that both
2 ft/sec, we must have that
Sliding ladder
FIGURE 4.94 2 ft/sec, we must have that dy dt =−2. (Note the minus sign here.) Observe that both
Sliding ladder dt
300
300
300 | Lesson 4-8 | Related Rates
300 | Lesson 4-8 | Related Rates
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