Ordinary DE Problem:

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Mar 7th 2010, 10:00 AM
Kasper

Ordinary DE Problem:

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!

Quote:

When a liquid drop falls, it increases in mass. Mass at time $t$ is $m(t)$ . Rate of growth of the mass of the liquid is $km(t)$ for positive constant $k$ . When we apply the Laws of Motion, we get $(mv)' = gm(t)$ , where $v$ is the velocity of the liquid drop, and $g$ is the acceleration due to gravity. The terminal velocity of the reaindrop is $\lim_{t \rightarrow \infty} v(t)$ . Find an expression for the terminal velocity in terms of $g$ and $k$ .

So I started with the knowledge that the rate of change of mass is km(t).

$\frac{dm}{dt} = km(t)$

$\frac{dm}{m(t)}=kdt$

$\ln|m(t)| = kt + c$

$m(t) = Ce^{kt}$

So now I used this result in conjunction with the information : $(m(t)v(t))'=gm(t)$ .

$(m(t)v(t))' = gm(t)$

$v(t)m(t) = \int gm(t) dt = g \int m(t) dt$

Antiderivative of $m(t): \ \int m(t) dt = \int Ce^{kt} dt = \frac{C}{k}e^{kt} + Z$

$v(t)m(t) = \frac{gC}{k}e^{kt} + gZ$

$v(t) = \frac{\frac{gC}{k}e^{kt} + gZ}{Ce^{kt}}$

$v(t) = \frac{g}{k} + \frac{gZ}{Ce^{kt}}$

Now to find the terminal velocity, we take the limit: $\lim_{t \rightarrow \infty} v(t)$ .

$Terminal \ Velocity = \lim_{t \rightarrow \infty} v(t) = \lim_{t \rightarrow \infty} (\frac{g}{k} + \frac{gZ}{Ce^{kt}}) = \frac{g}{k}$

Does this make good sense? Thanks!
Mar 7th 2010, 12:52 PM
CaptainBlack

Quote:

Originally Posted by Kasper

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).

$\frac{dm}{dt} = km(t)$

$\frac{dm}{m(t)}=kdt$

$\ln|m(t)| = kt + c$

$m(t) = Ce^{kt}$

So now I used this result in conjunction with the information : $(m(t)v(t))'=gm(t)$ .

$(m(t)v(t))' = gm(t)$

$v(t)m(t) = \int gm(t) dt = g \int m(t) dt$

Antiderivative of $m(t): \ \int m(t) dt = \int Ce^{kt} dt = \frac{C}{k}e^{kt} + Z$

$v(t)m(t) = \frac{gC}{k}e^{kt} + gZ$

$v(t) = \frac{\frac{gC}{k}e^{kt} + gZ}{Ce^{kt}}$

$v(t) = \frac{g}{k} + \frac{gZ}{Ce^{kt}}$

Now to find the terminal velocity, we take the limit: $\lim_{t \rightarrow \infty} v(t)$ .

$Terminal \ Velocity = \lim_{t \rightarrow \infty} v(t) = \lim_{t \rightarrow \infty} (\frac{g}{k} + \frac{gZ}{Ce^{kt}}) = \frac{g}{k}$

Does this make good sense? Thanks!

Without checking your algebra it looks OK (the fact that the terminal velocity does not depend on the constants of integration is encouraging)

The only comment I would make is that you made hard work of the first part, you should recognise the form of the solution of y'=ky straight off without having to go through seperation of variables.

CB
Mar 7th 2010, 06:10 PM
Kasper

Quote:

Originally Posted by CaptainBlack

Without checking your algebra it looks OK (the fact that the terminal velocity does not depend on the constants of integration is encouraging)

The only comment I would make is that you made hard work of the first part, you should recognise the form of the solution of y'=ky straight off without having to go through seperation of variables.

CB

Ah, I've been doing it all the time just to practice the method of seperation, but I think your right. Regardless, thanks for the confirmation!