1. Copyright © Cengage Learning. All rights reserved. 12 Further Applications of the Derivative

2. Copyright © Cengage Learning. All rights reserved. 12.5 Related Rates

3. 33 Recall that some basic facts about the rate of change of a quantity: Rate of Change of Q If Q is a quantity changing over time t, then the derivative dQ/dt is the rate at which Q changes over time. Quick Examples 1. If A is the area of an expanding circle, then dA/dt is the rate at which the area is increasing. 2. Words: The radius r of a sphere is currently 3 cm and increasing at a rate of 2 cm/s. Symbols: r = 3 cm and dr/dt = 2 cm/s.

4. 44 Example 1 – The Expanding Circle The radius of a circle is increasing at a rate of 10 cm/s. How fast is the area increasing at the instant when the radius has reached 5 cm? Solution: We have two related quantities: the radius of the circle, r, and its area, A. The first sentence of the problem tells us that r is increasing at a certain rate. When we see a sentence referring to speed or change, it is very helpful to rephrase the sentence using the phrase “the rate of change of.” Here, we can say The rate of change of r is 10 cm/s.

5. 55 Example 1 – Solution Because the rate of change is the derivative, we can rewrite this sentence as the equation Similarly, the second sentence of the problem asks how A is changing. We can rewrite that question: What is the rate of change of A when the radius is 5 cm? Using mathematical notation, the question is: What is when r = 5? cont’d

6. 66 Example 1 – Solution Thus, knowing one rate of change, dr/dt, we wish to find a related rate of change, dA/dt. To find exactly how these derivatives are related, we need the equation relating the variables, which is A = πr 2. To find the relationship between the derivatives, we take the derivative of both sides of this equation with respect to t. On the left we get dA/dt. On the right we need to remember that r is a function of t and use the chain rule. cont’d

7. 77 Example 1 – Solution We get Now we substitute the given values r = 5 and dr/dt = 10. This gives Thus, the area is increasing at the rate of 314 cm 2 /s when the radius is 5 cm. cont’d

8. 88 Related Rates Solving a Related Rates Problem A. The Problem 1. List the related, changing quantities. 2. Restate the problem in terms of rates of change. Rewrite the problem using mathematical notation for the changing quantities and their derivatives.

9. 99 Related Rates B. The Relationship 1. Draw a diagram, if appropriate, showing the changing quantities. 2. Find an equation or equations relating the changing quantities. 3. Take the derivative with respect to time of the equation(s) relating the quantities to get the derived equation(s), which relate the rates of change of the quantities.

10. 10 Related Rates C. The Solution 1. Substitute into the derived equation(s) the given values of the quantities and their derivatives. 2. Solve for the derivative required.

11. 11 Example 2 – The Falling Ladder Jane is at the top of a 5-foot ladder when it starts to slide down the wall at a rate of 3 feet per minute. Jack is standing on the ground behind her. How fast is the base of the ladder moving when it hits him if Jane is 4 feet from the ground at that instant? Solution: The first sentence talks about (the top of) the ladder sliding down the wall. Thus, one of the changing quantities is the height of the top of the ladder. The question asked refers to the motion of the base of the ladder, so another changing quantity is the distance of the base of the ladder from the wall.

12. 12 Example 2 – Solution Let’s record these variables and follow the outline above to obtain the solution. A. The Problem 1. The changing quantities are h = height of the top of the ladder b = distance of the base of the ladder from the wall 2. We rephrase the problem in words, using the phrase “rate of change”: The rate of change of the height of the top of the ladder is −3 feet per minute. cont’d

13. 13 Example 2 – Solution What is the rate of change of the distance of the base from the wall when the top of the ladder is 4 feet from the ground? We can now rewrite the problem mathematically: Find when h = 4. cont’d

14. 14 Example 2 – Solution B. The Relationship 1. Figure 45 shows the ladder and the variables h and b. Notice that we put in the figure the fixed length, 5, of the ladder, but any changing quantities, like h and b, we leave as variables. We shall not use any specific values for h or b until the very end. 2. From the figure, we can see that h and b are related by the Pythagorean theorem: h 2 + b 2 = 25. cont’d Figure 45

15. 15 Example 2 – Solution 3. Taking the derivative with respect to time of the equation gives us the derived equation: C. The Solution 1. We substitute the known values dh/dt = −3 and h = 4 into the derived equation: We would like to solve for db/dt, but first we need the value of b, which we can determine from the equation h 2 + b 2 = 25, using the value h = 4: 16 + b 2 = 25 cont’d

16. 16 Example 2 – Solution b 2 = 9 b = 3. Substituting into the derived equation, we get 2. Solving for db/dt gives Thus, the base of the ladder is sliding away from the wall at 4 ft/min when it hits Jack. cont’d