Question

Question #a685f

Answer 1

The time at which the sphere is increasing at `0.04m^2s^-1`
is at `50/ pi` seconds.

The radius is `pi` metres.

Explanation:

I assume by the "plane" surface area you are referring to the flat surface of the hemisphere. The flat surface will clearly be a circle whose area is given by:

`A = pir^2=pi[r(t)]^2`

In the above we have made it explicit that `r` is time dependent.

We have that `(dr)/(dt)=0.02` from the question.

Integrate this with respect to `t` to get the explicit function of `r`:

`int(dr)/dtdt=int0.02dt=0.02t +C`

Assuming (as no other initial condition is stated) that the radius of the sphere is `0` when `t=0` then `C=0`.

So `r=0.02t`.

Using this information we can calculate the time derivative of `A`:

`(dA)/dt = d/dt{pi[r(t)]^2}=pid/(dt)(0.02t)^2`

`=0.0004pid/(dt)t^2=0.0008pit`

So, to work out the radius at which the circle's area is increasing at `0.04m^2 s^-1` we can set `(dA)/dt=0.04` and solve for `t`.

`-> (dA)/dt = 0.0008pit=0.04`

`-> t = 0.04/(0.0008pi)=50/pis`

So the time at which the sphere is increasing at `0.04m^2s^-1`
is at `50/ pi` seconds.

We know from above that `r=0.02t` so:

`r=0.02(50/pi)=pi m`

So the radius is `pi` metres.