I'm trying to solve this question:
Water is evaporating from an open container whose shape we have no information about. Let $v(t)$ represent the volume of the water at time $t$ and $h(t)$ the height. Also let $A(h)$ represent the area of the exposed water at height $h$. Assume: $$ \frac{\mathrm{d}v}{\mathrm{d}t}=-kA(h),\quad\text{ $k$ is positive}. $$ Prove that $$ \frac{\mathrm{d}h}{\mathrm{d}t}=-k. $$
I'm having a hard time with this problem because it requires manipulation of differentials which is conceptually hard to accept.
You have a solid whose cross-sectional area is given for each possible cut. Then, the volume of such a solid with height $h$ is
$$ V(h) = \int_0^h A(s)\text ds $$
Then, by chain rule,
$$\frac{\text dV}{\text dt} = \frac{\text dV}{\text dh}\frac{\text dh}{\text dt}$$
Using the integral above with the fundamental theorem of calculus, we have
$$ \frac{\text dV}{\text dh} = A(h) $$
while we are given that
$$ \frac{\text dV}{\text dt} = -kA(h) $$
Substituting these values into the chain rule equation gives
$$ -kA(h) = A(h) \frac{\text dh}{\text dt}$$
and solving this equation for $\frac{\text dh}{\text dt}$ gives the solution
$$ \frac{\text dh}{\text dt} = -k $$