A spherical raindrop is falling with constant speed. As it falls it accumulates additional mass and its volume increases. The rate of change of the volume is proportional to its cross-sectional area (A=4πr^2), where r is its radius.
Formulate a differential equation model for the radius of the raindrop as a function of time. (Recall, V=(4/3)πr^3 for a sphere). Use k as your constant of proportionality.
dr/dt = ?
Does anyone know how to formulate a differential equation for radius and time? I am stuck on this question. Thank you
We are given that $\frac{dV}{dt}\propto4\pi r^2$, and that $V=\frac43\pi r^3$. The latter tells us that $\frac{dV}{dt}=\frac43\pi r^3\frac{dr}{dt}$, so $\frac{dV}{dt}\propto r^3\frac{dr}{dt}$.
So, we have $$r^3\frac{dr}{dt}\propto r^2$$$$\frac{dr}{dt}\propto\frac1r$$$$\frac{dr}{dt}=\frac kr$$