Physics Help Forum Resistive Force Calculus Help

 Kinematics and Dynamics Kinematics and Dynamics Physics Help Forum

 Dec 10th 2013, 08:33 AM #1 Junior Member   Join Date: Dec 2013 Posts: 9 Resistive Force Calculus Help I have the differentialequation dv/dt = g - kv^n and need to solve it with the initial condition that v(o) = v.inital. I tried seperating the variables so i get dv = g-kv^n dt and then 1/(g-kv^n)dv = dt. Then I wanted to integrate both sides but I am not sure how or if this is even correct. Please help!
 Dec 10th 2013, 04:37 PM #2 Physics Team     Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,310 I don't believe there is a single, closed-form solution for this integral that works for all values of n. If you know that n = 0, or n=1, or n=2 it's pretty straight-forward, but not for higher values of n.
Dec 10th 2013, 07:12 PM   #3

Join Date: Apr 2008
Location: On the dance floor, baby!
Posts: 2,451
 Originally Posted by Jokerpoker I have the differentialequation dv/dt = g - kv^n and need to solve it with the initial condition that v(o) = v.inital. I tried seperating the variables so i get dv = g-kv^n dt and then 1/(g-kv^n)dv = dt. Then I wanted to integrate both sides but I am not sure how or if this is even correct. Please help!
The solution for general n is a hypergeometric function. Could you tell us where this problem came from? Maybe that will help.

-Dan
__________________
Do not meddle in the affairs of dragons for you are crunchy and taste good with ketchup.

See the forum rules here.

 Dec 11th 2013, 12:50 AM #4 Junior Member   Join Date: Dec 2013 Posts: 9 It is just from my physics text book, where it is related to newtons second law, and the mass cancels out on both sides. So how would you solve it (if possible where n is just n, and if not then for example n = 2)?
Dec 11th 2013, 05:46 AM   #5

Join Date: Apr 2008
Location: On the dance floor, baby!
Posts: 2,451
 Originally Posted by Jokerpoker It is just from my physics text book, where it is related to newtons second law, and the mass cancels out on both sides. So how would you solve it (if possible where n is just n, and if not then for example n = 2)?

Let v = sqrt(g/k) * u. Then dv = sqrt(g/k) du. The integral becomes:

So

Which you can finish from here. (The solution for v can be put into the form of an inverse hyperbolic tangent for a "cleaner" solution, but I think the log format is easier to grasp.)

-Dan
__________________
Do not meddle in the affairs of dragons for you are crunchy and taste good with ketchup.

See the forum rules here.

 Dec 11th 2013, 07:44 AM #6 Junior Member   Join Date: Dec 2013 Posts: 9 Im not really sure how you arrive at v = sqrt(g/k)*u? Im assuming you are using u to mean v.initial? - How can you just replace v with u? - Thanks by the way, this is really helpful.
Dec 11th 2013, 09:10 AM   #7

Join Date: Apr 2008
Location: On the dance floor, baby!
Posts: 2,451
 Originally Posted by Jokerpoker Im not really sure how you arrive at v = sqrt(g/k)*u? Im assuming you are using u to mean v.initial? - How can you just replace v with u? - Thanks by the way, this is really helpful.
It's not a speed it's a change of variables. As far as v = sqrt(g/k) u, this substitution gives the integrand as 1/(1 - u^2), which is easier to integrate than when the constants were in with it.

-Dan
__________________
Do not meddle in the affairs of dragons for you are crunchy and taste good with ketchup.

See the forum rules here.

 Dec 11th 2013, 09:51 AM #8 Junior Member   Join Date: Dec 2013 Posts: 9 Okay then Im not sure how you arrive at the substitue u. If I was to substitute something with u, I would say u = what ever I want to substitute?
Dec 11th 2013, 05:02 PM   #9

Join Date: Apr 2008
Location: On the dance floor, baby!
Posts: 2,451
 Originally Posted by Jokerpoker Okay then Im not sure how you arrive at the substitue u. If I was to substitute something with u, I would say u = what ever I want to substitute?
Okay then, let u = sqrt{k/g} v. I'm not sure why this is a sticking point. Would you be more comfortable with x = sqrt{k/g} v? Or is it that you are having trouble with why I'm choosing that substitution?

-Dan
__________________
Do not meddle in the affairs of dragons for you are crunchy and taste good with ketchup.

See the forum rules here.

 Dec 11th 2013, 11:56 PM #10 Junior Member   Join Date: Dec 2013 Posts: 9 Yeah, Im definitely having a hard time seeing why/how you chose that substitution?

 Tags calculus, force, resistive

 Thread Tools Display Modes Linear Mode

 Similar Physics Forum Discussions Thread Thread Starter Forum Replies Last Post jjorg Kinematics and Dynamics 9 Dec 25th 2013 03:09 PM furor celtica Kinematics and Dynamics 1 Oct 28th 2011 08:40 AM purakanui Special and General Relativity 1 Jan 16th 2011 06:24 PM pramod Kinematics and Dynamics 1 May 1st 2009 03:01 AM C.E Advanced Mechanics 0 Apr 28th 2009 02:20 PM