AP Calculus BC 4.4 Related Rates Overview

Review related-rates problems by defining changing quantities, differentiating their relationship and evaluating a stated instant.

[Maximum number: 3]

An ice sculpture melts in such a way that it can be modeled as a cone that maintains a conical shape as it decreases in size. The radius of the base of the cone is given by a twice-differentiable function r, where r(t) is measured in centimeters and t is measured in days. The table above gives selected values of $r^{\prime}(t)$ , the rate of change of the radius, over the time interval $0 \leq t \leq 12$ .

The height of the cone decreases at a rate of 2 centimeters per day. At time t=3 days, the radius is 100 centimeters and the height is 50 centimeters. Find the rate of change of the volume of the cone with respect to time, in cubic centimeters per day, at time t=3 days. (The volume V of a cone with radius r and height h is $V=\frac{1}{3} \pi r^{2} h$ .)

Write your responses to this question only on the designated pages in the separate Free Response booklet. Write your solution to each part in the space provided for that part.

Figure 1

Figure 2

$\frac{dV}{dt}=\frac{2}{3}\pi rh\frac{dr}{dt}+\frac{1}{3}\pi r^{2}\frac{dh}{dt}$

Product rule

1 point

Chain rule

1 point

$\left.\frac{dV}{dt}\right|_{t=3}=\frac{2}{3}\pi(100)(50)(-5)+\frac{1}{3}\pi(100)^{2}(-2)=-\frac{70,000\pi}{3}$

The rate of change of the volume of the sculpture at t=3 is $-\frac{70,000\pi}{3}$ cubic centimeters per day.

Answer

1 point

Scoring notes:

- The first 2 points could be earned in either order.
- A completely correct product rule with the correct differentials earns the product rule and chain rule points.
- The third point is earned for the correct substitution and answer $-\frac{70,000\pi}{3}$ .
- Units are not required or read in this part.

Total for part (d)

for question 4

9 points

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Covered syllabus points

4.4 Introduction to Related Rates question 1