An open can of oil is accidently 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 m, the thickness of the oil slick is decreasing at the rate of 4 mm/sec. How fast is it decreasing when the radius is 2 m?
As the volume of the open can is fixed (let us say $V$), let the radius of the circular disc in lake be $r$ and the thickness of the circular disc be $h$.
$$V=\pi r^2h$$
$$\frac{dr}{dt}=\frac{10}{100}$$
$$\frac{dV}{dt}=0=\pi(r^2\frac{dh}{dt}+2rh\frac{dr}{dt})………(1)$$
$$r^2\frac{dh}{dt}+2rh\frac{dr}{dt}=0$$
When $r=1,\frac{dh}{dt}=\frac{-4}{1000}$ and when $r=2,\frac{dh}{dt}=?$
$$(1)^2\times \frac{-4}{1000}+2(1)h\times \frac{10}{100}=0$$
$$h=\frac{2}{100}$$
When $r=2,$
$$(2)^2\frac{dh}{dt}+2(2)(\frac{2}{100})(\frac{10}{100})=0$$
$$\frac{dh}{dt}=-0.002 \mathrm{m/sec}$$
But the answer given in my book is $-0.0005$ m/sec.
Best Answer
The error in your method is that you found the height at the moment when $r=1$ and used it to find the change in height when $r=2$, however since the height is changing this doesn't work. It's easier to find a fixed value, i.e. the volume, as I've done below. We have
$$ h = \frac{V}{\pi r^2} \implies \frac{dh}{dt} = -\frac{2V}{\pi r^3} \frac{dr}{dt} $$
substitute in $r=1$ and $\frac{dh}{dt} = -0.004$
$$ -0.004 = -\frac{2V}{\pi (1)^3} 0.1 \implies \frac{2V}{\pi} = 0.04 $$
now use this volume when $r=2$
$$ \frac{dh}{dt} = -\frac{2V}{8 \pi } 0.1 = -\frac{2( 0.02\pi )}{8\pi}0.1 = -0.0005 $$