How fast does a circular disc of spilled oil spread? 
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.
 A: 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 $r=2$
$$ \frac{dh}{dt} = -\frac{2V}{8 \pi } 0.1 = -\frac{2( 0.02\pi )}{8\pi}0.1 = -0.0005 $$
A: The problem is that you took the thickness at r=1m and considered it the thickness at r=2m. 
You need to calculate the new thickness at r=2m from the volume. 
A: After I fixed a mistake in my computations, I got $-0.0005$m/sec.
Since $r'(t)=\frac1{10}$ and $r(0)=1$ we get $r(t)=\frac1{10}t+1$. Using $V=\pi r^2(t)h(t)$ we get 
$$
h(t)=\frac{V}{\pi r^2(t)}\text{ and }h'(t)=-\frac{2Vr'(t)}{\pi r^3(t)}=-\frac{V}{5\pi \left(\frac1{10}t+1\right)^3}.
$$
Since
$$
-\frac4{1000}=h'(0)=-\frac{V}{5\pi \left(\frac1{10}\cdot 0+1\right)^3}=-\frac{V}{5\pi}
$$
we get $V=\frac{2\pi}{100}$ and therefore 
$$
h'(t)=-\frac{4}{1000\left(\frac1{10}t+1\right)^3}.
$$
Now consider $2=r(t)$ iff $t=10$ and we get
$$
h'(10)=-\frac{4}{1000\left(\frac1{10}\cdot 10+1\right)^3}=-\frac{4}{8000}=-0.0005.
$$
A: You have that $h$ changes so $h$ at $r=1$ is not the same as $h$ at $r=2$.
You have that at $r=1$ that $$V=\pi\frac{2}{100}=\frac{\pi}{50}$$
And since $V$ doesn't change for  $r=2$ $$\frac{\pi}{50}=2^2h\pi=4h\pi\\\frac{1}{200}=h$$
Now plugging in your formula
$$2^2\frac{dh}{dt}+2(2)\left(\frac1{200}\right)\frac{10}{100}=0$$
From this follows that
$$\frac{dh}{dt}=-0.0005\frac{m}{\sec}$$
