1. ConcepTest Section 4.6 Question 1 A spherical snowball of radius r cm has surface area S cm 2. As the snowball gathers snow, its radius increases as in Figure 4.12. Approximately how fast, in cm 2 /min, is S increasing when the radius is 20 cm? (a) 4π · 3 2 (b) 4π · 20 2 (c) 4π · 46 2 (d) 8π · 3 · 2.5 (e) 8π · 20 · 2.5 (f) 8π · 46 · 1

2. ConcepTest Section 4.6 Answer 1 ANSWER (e). We want dS/dt when r = 20, that is, at t = 3. The surface area is given by S = 4πr 2, so differentiating gives To find dr/dt, we draw a tangent line to the curve at t = 3 and estimate its slope to be 2.5 cm/min. Thus COMMENT: Have students check the units of their answer.

3. ConcepTest Section 4.6 Question 2 A car is driving along a straight flat road when a plane flies overhead. Let x miles be the distance traveled by the car and y miles be the distance traveled by the plane at time t in hours since the plane was directly over the car. At time t, the distance D miles between the car and the plane is given by D 2 = x 2 + y 2 + 2 2. At one moment, the car has gone 3 miles and is moving at 60 mph, and the plane has gone 30 miles and is moving at 500 mph. To find the rate at which D is increasing at that time, you should: (a) Substitute x = 3, y = 30 into D 2 = x 2 + y 2 + 2 2. (b) Differentiate D 2 = x 2 + y 2 + 2 2 after substituting x = 3, y = 30. Then substitute dx/dt = 60, dy/dt = 500. (c) Differentiate D 2 = x 2 + y 2 + 2 2. Then substitute x = 3, y = 30, dx/dt = 60, dy/dt = 500. (d) None of the above.

4. ConcepTest Section 4.6 Answer 2 ANSWER (c). Note that substituting x = 3, y = 30 before differentiating, as in (b), gives zero instead of the correct answer. COMMENT: Have students calculate the answer, which 15,180 / = 502.385 mph.

5. ConcepTest Section 4.6 Question 3 Let P(t) be the population of California in year t. Then P’(2005) represents: (a)The growth rate (in people per year) of the population. (b)The growth rate (in percent per year) of the population. (c) The approximate number of people by which the population increased in 2005. (d) The approximate percent increase in the population in 2005. (e) The average yearly rate of change in the population since t = 0. (f) The average yearly percent rate of change in the population since t = 0.

6. ConcepTest Section 4.6 Answer 3 ANSWER (a). The derivative represents the rate of change of P(t) in people per year. Answer (c) is approximately correct. COMMENT: Using P, have students write expressions for each of the other rates.

7. ConcepTest Section 4.6 Question 4 Let g(t) be the average number of email messages, in millions per day, sent in a particular region in year t. What are the units and interpretations of the following quantities? (a) g(2005) (b) g ’(2005) (c) g -1 (20) (d) (g -1 )’(20)

8. ConcepTest Section 4.6 Answer 4 ANSWER (c). Differentiating x = 5 tanθ gives Since the light rotates at 2 revolutions per minute = 4π radians per minute, we know dθ/dt = 4π. Thus we can calculate dx/dt, the speed at which the spot is moving, for any angle θ. Differentiating any of the other relationships introduces dr/dt, whose values we cannot find as easily as we can find dθ/dt.

9. ConcepTest Section 4.6 Question 5 The foot of the ladder in Figure 4.13 moves away from the wall at a speed of 2 ft/min, causing the top of the ladder to slide down the wall without leaving it. Label each of the following statements as True or False and give a reason. (a) dx/dt and dy/dt have the same sign. (b)The top of the ladder ismoving faster and faster. (c)Keeping dx/dt constant,doubling x, y, and the length of the ladder, doubles dy/dt.

10. ConcepTest Section 4.6 Answer 5 ANSWER We know dx/dt = 2 ft/min. Let k be the (constant) length of the ladder in feet, so x 2 + y 2 = k 2. Differentiating with respect to t gives (a) False, since dx/dt > 0 and dy/dt < 0. Alternatively, this could be done without computation. Because the top of the ladder moves down, y decreases, so dy/dt is negative. We are given dx/dt = 2, which is positive, so dx/dt and dy/dt have different signs. (b) True, since as x increases, y decreases, so x/y increases. Thus the magnitude of dy/dt increases, which means that the top of the ladder is moving faster and faster. (c) False, since doubling x and y leaves the ratio x/y unchanged. Thus, dy/dt is unchanged. COMMENT: Have students carry out some of the calculation with specific values of x and y to confirm their answers.