The set up for the question is as follows.
A circular oil slick is spreading on the sea. When its radius is $150 m$, the thickness of the slick is $0.01$ m and the radius is expanding at $0.2 m/min$.
I'm then asked to find the volume which I calculated to be $225\pi$$ m^2$ and the rate of change of the area of the slick when the radius is $150$m which I calculated to be $0.04\pi$ m$^2$/min which I don't even think I answered that correctly.
Lastly I'm asked "Assuming that, at each moment, the thickness of the slick is the same at each point, at what rate is the thickness decreasing when the radius is $150$ m?". I feel like this should be an easy question and I'm just missing something obvious.
I believe the solution is something along the lines of $\frac{dh}{dt}= \frac{dh}{dr} \frac{dr}{dt}$ but that's as much as I have.
I can't seem to work out $\frac{dh}{dr}$
Sorry for the bad formatting.
Thanks in advance
Best Answer
The volume of oil slick will be constant when it spreads on water, since nothing extra is being added. Now
$V= \pi r^2h$. assuming oil slick to be cylindrical
therefore , differentiating with respect to time-
$dV/dt=\pi ( 2rh.dr/dt+ r^2.dh/dt)$
since volume is not changing therefore $dV/dt=0$ and hence our expression reduce to-
$2h.dr/dt=-r.dh/dt$.
here $-dh/dt$ represents the rate at which height of slick or thickness of slick is decreasing. Now since question provides you the value of rate of increase of increase of radius of slick and thickness of slick when radius = $150m$, plug in them to get the answer as follows
$-dh/dt=(2h/r).dr/dt=2*(0.01m)*(0.2m/min)/(150m)=2.66*10^{-5}m/min$