The problem says — Water is running into a conical reservoir, $10$ cm deep and $5$ cm in radius at the rate of $1.5$ c.c. per minute. 1) At what rate is the water level rising when the water is $4$ cm deep? 2) At what rate is the area of the water surface increasing when the water is $6$ cm deep? 3) At what rate is the wetted surface of the reservoir increasing when the water is $8$ cm deep?
Now, I didn’t have a problem with parts 1) and 2). For the 3) part however, I worked out the wetted surface to be $W=\frac{\sqrt{5} h^2 \pi}{4}$
Then I differentiated it with respect to time — $\frac{dW}{dt} = \frac{\sqrt{5}\pi h}{2} \frac{dh}{dt}$
$\frac{dh}{dt}$ was obtained in part 1) to be $\frac{6}{\pi h^2}$. I substituted this to finally get $\frac{dW}{dt} = \frac{3\sqrt{5}}{\pi h}.$
But this answer is wrong according to my text. Where have I gone wrong?
[Note: The text gives $\frac{dW}{dt}=\frac{\sqrt{5}\pi h}{2}$]
Hint. The wetted surface should be given by $$S=πr\sqrt{64+r^2},$$ where $r$ represents the radius of the water surface at time $t.$ When we note that $h/2=r,$ we obtain $$S=\frac{πh^2\sqrt 5}{4},$$ so that $$S'=\frac{π\sqrt 5}{2}hh'.$$ Now since the volume of water is $V=πr^2h/3,$ we get that $$3/2=V'=\frac{π}{6}h^2h',$$ which gives $h'=6/πh^2.$ Substituting in the expression for $S'$ gives $$S'=\frac{3\sqrt 5}{h}.$$