So I have a sphere with radius $1\ cm$, and I'm pouring in water at $0.5\ cm^3/s$.
How would I find a the radius of the surface of the water at any given time? So in the end this would look like a quadratic equation relating time and radius, where the maxima would be where radius $= 1\ cm$. How would I go about using differentiation in solving for this equation?
Notice, let $r$ be the radius of the circular surface of water at any time $t$ then the volume of water poured (in a sphere of radius $R=1\ cm$ ) at the same time $t$ is given by using geometry $$\color{red}{V=\frac{\pi}{3}\left(2R^3-(2R^2+r^2)\sqrt{R^2-r^2}\right)}$$ Differentiating w.r.t. time $t$, we get $$\frac{dV}{dt}=\frac{\pi}{3}\left(\frac{(2R^2+r^2)r}{\sqrt{R^2-r^2}}-2r\sqrt{R^2-r^2}\right)\frac{dr}{dt}$$ $$\frac{dV}{dt}=\frac{\pi r^3}{\sqrt{R^2-r^2}}\frac{dr}{dt}$$$$\implies \frac{dr}{dt}=\frac{\sqrt{R^2-r^2}}{\pi r^3}\frac{dV}{dt}$$ Now, setting the values $R=1\ cm$ & constant rate of pouring water $\frac{dV}{dt}=0.5\ cm^3/s$, we get $$\frac{dr}{dt}=\frac{\sqrt{1-r^2}}{\pi r^3}(0.5)$$ $$\frac{2\pi r^3\ dr}{\sqrt{1-r^2}}=dt$$$$\implies \int \frac{2\pi r^3\ dr}{\sqrt{1-r^2}}=\int dt$$ $$\frac{-2\pi(2+r^2)\sqrt{1-r^2}}{3}=t+C$$ $\color{blue}{\text{Condition}}$: At initial time $t=0$, the radius of water surface $r=0$, hence, we get $$\frac{-2\pi(2+0)\sqrt{1-0}}{3}=0+C\implies C=\frac{-4\pi}{3}$$ Now, setting the value of $C$, we get $$\frac{-2\pi(2+r^2)\sqrt{1-r^2}}{3}=t-\frac{4\pi}{3}$$ $$\frac{2\pi(2+r^2)\sqrt{1-r^2}}{3}=\frac{4\pi}{3}-t$$ $$\color{red}{(1-r^2)(2+r^2)^2=\left(\frac{4\pi-3t}{2\pi}\right)^2}$$