# Question #a685f

Jan 5, 2018

The time at which the sphere is increasing at $0.04 {m}^{2} {s}^{-} 1$
is at $\frac{50}{\pi}$ 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:

$A = \pi {r}^{2} = \pi {\left[r \left(t\right)\right]}^{2}$

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

We have that $\frac{\mathrm{dr}}{\mathrm{dt}} = 0.02$ from the question.

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

$\int \frac{\mathrm{dr}}{\mathrm{dt}} \mathrm{dt} = \int 0.02 \mathrm{dt} = 0.02 t + C$

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

So $r = 0.02 t$.

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

$\frac{\mathrm{dA}}{\mathrm{dt}} = \frac{d}{\mathrm{dt}} \left\{\pi {\left[r \left(t\right)\right]}^{2}\right\} = \pi \frac{d}{\mathrm{dt}} {\left(0.02 t\right)}^{2}$

$= 0.0004 \pi \frac{d}{\mathrm{dt}} {t}^{2} = 0.0008 \pi t$

So, to work out the radius at which the circle's area is increasing at $0.04 {m}^{2} {s}^{-} 1$ we can set $\frac{\mathrm{dA}}{\mathrm{dt}} = 0.04$ and solve for $t$.

$\to \frac{\mathrm{dA}}{\mathrm{dt}} = 0.0008 \pi t = 0.04$

$\to t = \frac{0.04}{0.0008 \pi} = \frac{50}{\pi} s$

So the time at which the sphere is increasing at $0.04 {m}^{2} {s}^{-} 1$
is at $\frac{50}{\pi}$ seconds.

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

$r = 0.02 \left(\frac{50}{\pi}\right) = \pi m$

So the radius is $\pi$ metres.