Question

A cylinder gets taller at a rate of 3 inches per second, but the radius shrinks at a rate of 1 inch per second. How fast is the volume of the cylinder changing when the height is 20 inches and the radius is 10 inches?

Answer 1

Volume of a cylinder is `V = pir^2h` where

`r` is the radius and `h` is the height of the cylinder,

We are given some rates of change:
`(dh)/(dt) = 3` `"in"`/`"s"` and `(dr)/(dt) = -1` `"in"`/`"s"`.

We seek `(dV)/dt` when `h=20` and `r = 10`.

Differentiate both sides of `V = pir^2h` with respect to `t`. We'll need the product rule on the right.

`d/dt(V) = d/dt(pir^2)h+pir^2 d/dt(h)`

`(dV)/dt = 2pir (dr)/dt h +pir^2(dh)/dt`

At the instant of interest,

`(dV)/dt = 2pi(10) (-1) (20) +pi(10)^2 (3) = -100pi` `"in"^3`/`"s"`

The volume is decreasing at a rate of `100pi` `"in"^3`/`"s"`

Answer 2

Volume is decreasing at a rate of `100pi \ i n^3s^(-1)`,
or approximately `314\ i n^3s^(-1)`

Explanation:

Let us setup the following variables:

# { (r, "Radius of cylinder at time t","(in)"), (h, "Height of cylinder at time t","(in)"), (V, "Volume of cylinder at time t", "(in"^3")"), (t, "time", "(sec)") :} #

Using the standard formula for volume of a cylinder:

`V = pir^2h`

Where both `r` and `h` are functions of `t`. Implicitly differentiating wrt `t`, we have via the product rule:

`d/dt V = (pir^2)(d/dt h) + (d/dt pir^2)(h)`

Then by the chain rule, we have:

`(dV)/dt = (pir^2)( (dh)/dt d/(dh) h) + ( (dr)/dt d/(dr) pir^2)(h)`
`\ \ \ \ \ \ = (pir^2)( (dh)/dt 1) + ( (dr)/dt 2pir)(h)`
`\ \ \ \ \ \ = pir^2 (dh)/dt + 2pirh (dr)/dt`

And we are given that

`(dh)/dt=3` and `(dr)/dt = -1`

So we have:

`(dV)/dt = 3pir^2 - 2pirh`

When `h=20` and `r=10` we have:

`[(dV)/dt]_(h=20, r=10) = 3pi(100) - 2pi(10)(20)`
`" " = 300pi - 400pi`
`" " = -100pi`

Hence, the volume is decreasing at a rate of `100 \ i n^3s^(-1)`