So I think I have found the solution to the following problem, but it feels like I just barely patched it together, if anyone could take a quick peek and make sure I wasn't making up my own math, I would be very appreciative!
So I started with the knowledge that the rate of change of mass is km(t).Quote:
When a liquid drop falls, it increases in mass. Mass at time $\displaystyle t$ is $\displaystyle m(t)$. Rate of growth of the mass of the liquid is $\displaystyle km(t)$ for positive constant $\displaystyle k$. When we apply the Laws of Motion, we get $\displaystyle (mv)' = gm(t)$, where $\displaystyle v$ is the velocity of the liquid drop, and $\displaystyle g$ is the acceleration due to gravity. The terminal velocity of the reaindrop is $\displaystyle \lim_{t \rightarrow \infty} v(t)$. Find an expression for the terminal velocity in terms of $\displaystyle g$ and $\displaystyle k$.
$\displaystyle \frac{dm}{dt} = km(t)$
$\displaystyle \frac{dm}{m(t)}=kdt$
$\displaystyle \ln|m(t)| = kt + c$
$\displaystyle m(t) = Ce^{kt}$
So now I used this result in conjunction with the information : $\displaystyle (m(t)v(t))'=gm(t)$.
$\displaystyle (m(t)v(t))' = gm(t)$
$\displaystyle v(t)m(t) = \int gm(t) dt = g \int m(t) dt$
Antiderivative of $\displaystyle m(t): \ \int m(t) dt = \int Ce^{kt} dt = \frac{C}{k}e^{kt} + Z$
$\displaystyle v(t)m(t) = \frac{gC}{k}e^{kt} + gZ$
$\displaystyle v(t) = \frac{\frac{gC}{k}e^{kt} + gZ}{Ce^{kt}}$
$\displaystyle v(t) = \frac{g}{k} + \frac{gZ}{Ce^{kt}}$
Now to find the terminal velocity, we take the limit: $\displaystyle \lim_{t \rightarrow \infty} v(t)$.
$\displaystyle Terminal \ Velocity = \lim_{t \rightarrow \infty} v(t) = \lim_{t \rightarrow \infty} (\frac{g}{k} + \frac{gZ}{Ce^{kt}}) = \frac{g}{k}$
Does this make good sense? Thanks!
