Question

The diameter and height of a right circular cylinder are found at a certain instant to be 10 centimeters and 20 centimeters, respectively. If the diameter is increasing at a rate of 1 centimeter per second, what change in height will keep the volume constant?

Answer 1

Explanation:

Here's the formula for the volume of a right circular cylinder:

`V=\pi r^2h`

Here's what we are given and what we need to find:

Given that d = 10 cm, h = 20 cm, dd/dt = 1 cm/sec

Need to find dh/dt when V is constant

Since our formula has a radius in it and not a diameter but the info given is a diameter, we can use the substitution that

`2r =d` so

`r=(1)/(2)d`

Now we can rewrite the formula in terms of diameter:

`V=\pi ((1)/(2)d)^2h` which simplifies down to

`V=(\pi )/(4)d^2h`

Now we will take the derivative of this equation with respect to time using the product rule. That derivative is

`(dV)/(dt)=(\pi )/(4)[d^2*(dh)/(dt)+2d(dd)/(dt)*h]`

Now we can fill in our values. Keep in mind that if the volume is constant, there is no change in the volume, so dV/dt = 0.

`0=(\pi )/(4)[(100(dh)/(dt)+2(10)(1)(20)]` and

`0=(\pi )/(4)(100(dh)/(dt)+400)`

Multiply both sides by pi/4 to get

`0=100(dh)/(dt)+400` and solve for dh/dt:

`-4=(dh)/(dt)`

Interpreted within the context of our problem, this means that the volume will be constant at those given values of diameter and height when the liquid in the cylinder is dropping at a rate of 4 cm/sec.