# Thread: Rate Problems!

1. ## Rate Problems!

A potter forms a piece of clay into a cylinder. As he rolls it, the length, L, of the cylinder increases and the radius, R, decreases. Assume that no clay is lost in the process. Suppose the length of the cylinder is increasing by 0.2 cm per second.

(a) What is the rate at which the volume is changing?
cm3/sec

(b) What is the rate at which the radius is changing when the radius is 3 cm and the length is 5 cm?

im not sure how the to relate the length and radius, help please!

2. Hello, mathlete!

A potter forms a piece of clay into a cylinder.
As he rolls it, the length, $\displaystyle L$, of the cylinder increases and the radius, $\displaystyle R$, decreases.
Assume that no clay is lost in the process.
Suppose the length of the cylinder is increasing by 0.2 cm/sec.

(a) What is the rate at which the volume is changing?
A trick question! . . . We are told that the volume is constant.

Therefore: .$\displaystyle \frac{dV}{dt} \;=\;0$

(b) What is the rate at which the radius is changing
when the radius is 3 cm and the length is 5 cm?
The volume of a cylinder is: .$\displaystyle V \;=\;\pi R^2L$

Since the volume is constant, we have: . $\displaystyle \pi R^2L \:=\:C$

Differentiate with respect to time (Product Rule):

. . $\displaystyle \pi R^2\!\cdot\!\frac{dL}{dt} + 2\pi RL\!\cdot\!\frac{dR}{dt} \;=\;0\quad\Rightarrow\quad \frac{dR}{dt} \;=\;-\frac{R}{2L}\!\cdot\!\frac{dL}{dt}$

We are given: .$\displaystyle R = 3,\;L = 5,\;\frac{dL}{dt} \:=\:0.2\text{ cm/sec}$

Therefore: . $\displaystyle \frac{dR}{dt} \;=\;-\frac{3}{2(5)}(0.2) \;=\;-0.06\text{ cm/sec}$