# Differential equation for a raindrop

I am learning differential equations using Single Variable Calculus by James Stewart. I was stuck by the 49th question in section 9.3, page 626. Here is the question

When a raindrop falls, it increases in size and so its mass at time t is a function of t, namely m(t). The rate of growth of the mass km(t) is for some positive constant k. When we apply New ton’s Law of Motion to the raindrop, we get (mv)'= gm, where v is the velocity of the raindrop (directed downward) and g is the acceleration due to gravity. The terminal velocity of the raindrop is lim(t->∞)v. Find an expression for the terminal velocity in terms of k and g.

And here is the provided solution:

I wonder why the equation in the first line of the solution: mv' + m'v = gm make sence?

I think mv' = gm, so does it means that m'v = 0?

In fact, Newton's law works in terms of $$\vec F=\frac {d\vec p}{dt}$$ where $\vec F$ is the resulting external force and $\vec p$ is momentum.
The expression for momentum writes $m\vec v$, where $m$ is the mass and $\vec v$ is the speed. In most cases the mass is constant, hence we simplify $\frac {d\vec p}{dt}=m\frac {d\vec v}{dt}$.
However, in your case mass is not constant, so we are forced to use the equation $$\vec F=m(t)\frac {d\vec v(t)}{dt}+\frac {dm(t)}{dt}\vec v.$$