Particle released from O with initial velocity A, and terminal velocity B. Resistance of motion is kv.

Equation of motion is

a = g - kv where a is acceleration.

Verify that v satisfies the equation d(ve^kt)/dx = ge^kt

No idea how to do it.

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- Sep 9th 2010, 08:21 PMLukybearMechanics
Particle released from O with initial velocity A, and terminal velocity B. Resistance of motion is kv.

Equation of motion is

a = g - kv where a is acceleration.

Verify that v satisfies the equation d(ve^kt)/dx = ge^kt

No idea how to do it. - Sep 10th 2010, 02:54 AMAckbeet
When you say "resistance of motion", is that a force? Because if so, your units in your equation of motion are incorrect. You'd be off by a factor of $\displaystyle m$. However, you can absorb the mass into the constant k. So let's assume the equation of motion you've given us is correct. Let me re-write it for you here:

$\displaystyle \displaystyle{\frac{dv}{dt}=g-kv,}$ or

$\displaystyle \displaystyle{\frac{dv}{dt}+kv=g.}$

Does this form suggest anything to you? Another way of thinking about this is to take the form you're supposed to prove, and use the product rule to expand out the LHS. Then compare that with the equation I just wrote down. - Sep 10th 2010, 05:59 AMLukybear
For the m factor, that just canceled by the equation since, its mkv and mg and ma. And sorry i still cannot figure out how to obtain expoential. If i integrate the function with respect to t, i would certainly get a exponential with further manipulation. However there is a constant of integration which would contain A, i.e. intial velocity A, so how would i get rid of that?

- Sep 10th 2010, 06:49 AMAckbeet
You need to think about this problem in terms of integrating factors. They're not asking you to integrate the DE, which means you shouldn't need an arbitrary constant of integration.

- Sep 11th 2010, 12:28 AMLukybear
Is it possible to just show me a solution for clarification?

- Sep 11th 2010, 02:37 AMAckbeet
No, that's not the way we work around here, generally. Let me do one more step that I suggested before. You're supposed to show that the velocity satisfies the equation

$\displaystyle \displaystyle{\frac{d}{dt}(v\,e^{kt})=g\,e^{kt}.}$

Just for fun, let's take the derivative on the LHS using the product rule:

$\displaystyle \displaystyle{e^{kt}\frac{dv}{dt}+k\,e^{kt}\,v=g\, e^{kt}.}$

What you have from the equation of motion is that

$\displaystyle \displaystyle{\frac{dv}{dt}+k\,v=g.}$

Compare these two equations. Does anything pop out at you? - Sep 11th 2010, 03:31 AMLukybear
So to answer the question, we are just multiplying given equation by e^kt. That is quite trivial. Thanks.

But could you hint me on the mechanical procedure to acquire d/dt (ve^kt) = ge^kt from

e^kt . dv/dt + ke^kt .v = ge^kt.

I understand that by observation, from your differentiation we are able to acheive this. But without prior knowledge, how are we able to do this? - Sep 11th 2010, 03:33 AMAckbeet
By using the integrating factor method.

- Sep 11th 2010, 04:23 AMLukybear
I understand that you do not provide solutions. But could you just show me one time how this method works? I am sure this problem is within my syllabus, but I do not compute this integrating factor method. Perhaps its lost in the tautology somewhere. If i could just see the process, mabey it would trigger my memory. Thanks.

- Sep 11th 2010, 04:57 AMHallsofIvy
I would suggest, rather than the integrating factor method, you rewrite the equation as

$\displaystyle \frac{dv}{g- kv}= dt$ and integrate both sides- if you really want to solve the equation. But the problem may well be within your syllabus and not require solving the equation at all. Just multiply use the product rule to differentiate $\displaystyle ve^{kt}$. - Sep 11th 2010, 08:29 AMAckbeet
Here's an example or two.

Reply to HallsofIvy:

That method is just fine for solving the DE, but that wasn't what was asked in the OP. Either method for solving is about as much work, probably.