Question

An open can of oil is accidentally dropped into a lake; assume the oil spreads over the surface as a circular disc of uniform thickness whose radius increases steadily at the rate of $10$ cm$/$sec. At the moment when the radius is $1$ meter, the thickness of the oil slick is decreasing at the rate of $4$mm$/$sec, how fast it is decreasing when the radius is $2$ meters.

Answer 1

As the volume of the open can is fixed (lets say $V$), let the radius of the circular disc in lakh be $r$ and the thickness of the circular disc be $h$.

So we have:

$V=\pi r^{2}h$

$\Rightarrow \frac{dV}{dt}=2rh\frac{dr}{dt}+r^2\frac{dh}{dt}=0$

It is given that (converting everything into meters)

$\frac{dr}{dt}=0.1$, at $r=1$, $\frac{dh}{dt}=-0.004$

Substituting into the main equation, we get:

$0.2h=0.004\Rightarrow h=0.02$

Now, we have $r=2$ and assume that $h=0.02$, so from the equation we get:

$0.008+4\frac{dh}{dt}=0\Rightarrow \frac{dh}{dt}=-0.002$

Thus, when the radius is $2$ meters, the thickness is decreasing at $2$ mm/sec.