I've been attempting to solve this question for the last hour but have had no success. The question asks for the rate at which lateral surface area changes in respect to time for a right circular cylinder.
$\frac{dV}{dt}, \frac{dh}{dt}$, and the radius are all given, yet the numbers don't seem to work out from what I've seen.
Since I don't know latex I'll just link the pictures of my work and the question itself. I've tried it multiple ways but I never get the right answer. However, depending on what given I use I get two different answers whose average is the correct answer. The teacher said that this is a tricky question with an unexpected solution so I'm really anxious to find out how its done.
Thank you in advanced.
Work and Question
WE have $$s=2\pi rh \implies \frac {ds}{dt} = 2\pi (r \frac {dh}{dt} + h \frac {dr}{dt})=2\pi (12+ h \frac {dr}{dt}) $$
$$ v=\pi r^2h \implies \frac {dv}{dt}=\pi (2r\frac {dr}{dt}h+r^2 \frac {dh}{dt})$$
$$ \frac {dv}{dt} = \pi (12\frac {dr}{dt}h+ 72)=96\pi$$
$$ \implies \frac {dr}{dt}h =2$$
Thus $$\frac {ds}{dt} = 2\pi (12 + 2) = 28\pi $$