Question

A snowball, in the shape of a sphere, is melting so that the radius is decreasing at a uniform rate of 1cm/s. How fast is the volume decreasing when the radius equals 6cm?

Answer 1

We can write the first relationship (in `"cm/s"`) as:

`(dr)/(dt) = 1`

We know that the formula for the volume of a sphere is:

`V(r) = 4/3 pi(r(t))^3`

Also, the derivative of `V` with respect to `t`, `(dV)/(dt)`, implies that we also multiply by `(dr)/(dt)`, since `r = r(t)`, whose slope is `1`, and `(dV(r))/(dt) = (dV(r))/(dr(t))*(dr(t))/(dt) = (dV)/(dr)*(dr)/(dt)`.

Thus, if we differentiate this with respect to `t`, we can find a use for `(dr)/(dt)` through the Chain Rule.

`(dV)/(dt) = 4/3pi d/(dt)[(r(t))^3]`

`= 4/3pi d/(dr)[r^3] ((dr)/(dt))`

`= 4/3pi*3r^2((dr)/(dt))`

Finally, we can plug it in:

`= 4/3 pi (3r^2)(1) = color(green)(4 pir^2)`

Therefore, when `r = 6`, we get the rate of decrease in the volume while the radius is currently `"6 cm"`:

`(dV(6))/(dt) = V'(6) = 4 pi (36)`

`= color(blue)(144pi "cm"^3"/s")`

Answer 2

The volume is decreasing at a rate of `144picm^3.s^-1`

Explanation:

We are told that `(dr)/(dt)=-1"cm/s"`

We need to find `(dV)/(dt)` when `r=6"cm"`

We know:

`V=(4)/(3).pir^3`

So:

`(dV)/(dr)=(12)/(3).pir^2=4pir^2`

`(dV)/(dt)=(dV)/(dr).(dr)/(dt)`

`(dV)/(dt)=4pir^(2)xx(-1)=-4pir^2`

So when the radius `r=6"cm"` we can substitute this in :

`(dV)/(dt)=-4pi.6xx6=-144picm^3.s^-1`