Question

A cylinder is being flattened so that its volume does not change. Find the rate of change of radius when r = 3 inches and h = 5 inches, if the height is decreasing at 0.6 in/sec. Hint: what is the rate of change of volume?

Answer 1

Answer:

The radius is increasing at a rate 0.18 inches per second.

Step-by-step explanation:

We are given the following in the question:

The volume of cylinder is constant.

$\dfrac{dV}{dt} = 0$

$\dfrac{dh}{dt} = -0.6\text{ inch per second}$

Instant radius= 3 inches

Instant height = 5 inches

Volume of cylinder =

$V = \pi r^2 h$

Rate of change of volume =

$\dfrac{dV}{dt} = \pi(2r\dfrac{dr}{dt}h + r^2\dfrac{dh}{dt})$

Putting all the values, we get,

$0 = \dfrac{22}{7}(2(3)\dfrac{dr}{dt}(5)+(3)^2(-0.6))\\\\30\dfrac{dr}{dt} = 9(0.6)\\\\\dfrac{dr}{dt} = \dfrac{9\times 0.6}{30} = 0.18$

Thus, the radius is increasing at a rate 0.18 inches per second.