Given that a glass of water is filled to its fullest, $10\,cm$ in height, and that after three days the water level is at $9\,cm$ in height. Find when the glass will be empty.
The water is evaporating with a speed proportional to the surface area.
Own attempt, and where I'm stuck
$$\frac{dV}{dt}=-kS, \quad k > 0$$
$$dV = -kS\cdot dt$$
$$\qquad\qquad\qquad\qquad\qquad V = -kS\cdot t \quad\qquad\textit{From integrating}$$
Now, if the $S$, surface area, is constant, then I could easily calculate the change from $t=0$, and $t=5$ where $t$ is days after the initial scenario.
Here's where I am stuck. I have a suspicion, I would love to have it confirmed, and that is that the surface area does not matter; i.e. in some way, having a cylindrical glass with volume $V_\alpha$ and next to it a square glass with volume $V_\alpha$ means they will both be empty at $t = \beta$.
Otherwise, do you think I am to assume a shape, such as cylindrical or square by own choosing? (First class of calculus, if this helps in answering the question)
The red circled triangle has sides $R$ (the radius of the sphere), $r$ (the radius of the water surface) and $R-h$ (where $h$ is the water height). Pythagoras tells you that $R^2 = r^2 + (R-h)^2.$
Best Answer
If we take "evaporating with a speed proportional to the surface area" to mean that the rate of decrease in volume is proportional to the area of the upper surface of the water, then the shape of the glass can be cylindrical, spherical, conical, or just about any reasonable "cup" shape.
Consider the rate of evaporation, $\dfrac{dV}{dt}.$ We have $\dfrac{dV}{dt} = kA,$ where $A$ is the area of the upper surface of the water (not necessarily a constant) and $k$ is a constant.
Now consider how we might compute the volume of the glass up to height $h$ using the "disk" method: in effect, we assert that as depth of water rises (or falls) inside the glass, $\dfrac{dV}{dh} = A.$
From these two equations we can derive a value of $\dfrac{dh}{dt}.$ (One method is, use the Chain Rule to write $\dfrac{dV}{dh}\dfrac{dh}{dt} = \dfrac{dV}{dt},$ then solve for $\dfrac{dh}{dt}$ in that equation.) The resulting value of $\dfrac{dh}{dt}$ is constant, even if $A$ is a non-constant function of $h.$