lightburst Posted October 27, 2013 Share Posted October 27, 2013 You are given F(v) = -Cv^{2}, the force as a function of velocity, and you need to show that x - x_{0} = (m/C) ln(v_{0}/v) where m = mass and C is some constant. My question is why and how does definite integration work when you have 2 variables? I mean I was able to finish the problem using this as I've seen it be used in the textbook itself and also some math books, but I don't really understand or know how it works. In calculus classes, and also my book, definite integration is done using 1 variable. I mean we do multiple definite integrations but that's only 'on one side' and 'one at a time' and I've never really been taught definite integration on different variables on 'both sides' of the equation. I feel like because the right side is being evaluated from 0 to t, the other side should be evaluated from 0 to t also. Link to comment Share on other sites More sharing options...

studiot Posted October 27, 2013 Share Posted October 27, 2013 I feel like because the right side is being evaluated from 0 to t, the other side should be evaluated from 0 to t also. Yes this is sort of true, but presents a problem. The differential coefficient on the LHS is in terms of v (ie dv) and on the RHS in terms of t (ie dt) Yes you are integrating between the same two endpoints but one is in terms of t and one in terms of v and you need to match the differential coefficient and limits in your integral. Of course v and t are connected by an equation as are dv and dt Link to comment Share on other sites More sharing options...

lightburst Posted October 27, 2013 Author Share Posted October 27, 2013 I don't see why it wouldn't be possible to use t as a bound on the v side. I mean the bound and variables within the equation aren't the same thing, right? The bounds are just some constants, no? Also, they couldn't be the same end points because when you use, say the projectile motion/free fall equations, final velocity and time isn't the same and initial time = 0 and initial velocity isn't necessarily zero, no? Of course v and t are connected by an equation as are dv and dthmm I guess it makes better sense if like the bounds of the left hand side is a function of t, as in v(t). Is there perhaps a proof/theorem/property/particular topic I could look up? Link to comment Share on other sites More sharing options...

studiot Posted October 27, 2013 Share Posted October 27, 2013 (edited) In your problem you are given force as a function of velocity and asked about distance. Now velocity is a function of time, as is distance. So at any time, t, the particle or whatever has a specific velocity, v, and covered a specific distance, x. So t is the common variable running through the maths. If you know the time t you can calculate the velocity and therefore the foce, and the distance covered using the formula by substitution or change of variable. When you change the variable or substitute you also have to change the limits of integration so that if you are integrating say velocity between t = 0 and t=10 you need the velocity at t=0 and t=10 if you use a formula involving v not t. Edited October 27, 2013 by studiot Link to comment Share on other sites More sharing options...

gabrelov Posted November 19, 2013 Share Posted November 19, 2013 (edited) You are given F(v) = -Cv^{2}, the force as a function of velocity, and you need to show that x - x_{0} = (m/C) ln(v_{0}/v) where m = mass and C is some constant. My question is why and how does definite integration work when you have 2 variables? I mean I was able to finish the problem using this as I've seen it be used in the textbook itself and also some math books, but I don't really understand or know how it works. In calculus classes, and also my book, definite integration is done using 1 variable. I mean we do multiple definite integrations but that's only 'on one side' and 'one at a time' and I've never really been taught definite integration on different variables on 'both sides' of the equation. I feel like because the right side is being evaluated from 0 to t, the other side should be evaluated from 0 to t also. Nope thats not always the case, if we integrate one side with time the other doesn't necesarily with time. It maybe such state such as point 1 to 2 or what happened to the other side as we integrate it from time 1 to time 2. The above is a clear example as time passes the initial velocity becomes final velocity so you get from state 1 to state 2 of something after some time passes. In laymans term, example you have zero velocity at time zero, what would happen at velocity at time "t" offcourse in law of physics there should be a force to increase the velocity so there is acceleration. Edited November 19, 2013 by gabrelov Link to comment Share on other sites More sharing options...

swansont Posted November 19, 2013 Share Posted November 19, 2013 You can think of the speed integration limit as v(t), if that helps. Link to comment Share on other sites More sharing options...

daniton Posted November 21, 2013 Share Posted November 21, 2013 (edited) You are given F(v) = -Cv^{2}, the force as a function of velocity, and you need to show that x - x_{0} = (m/C) ln(v_{0}/v) where m = mass and C is some constant.My question is why and how does definite integration work when you have 2 variables?I mean I was able to finish the problem using this as I've seen it be used in the textbook itself and also some math books, but I don't really understand or know how it works. In calculus classes, and also my book, definite integration is done using 1 variable. I mean we do multiple definite integrations but that's only 'on one side' and 'one at a time' and I've never really been taught definite integration on different variables on 'both sides' of the equation.I feel like because the right side is being evaluated from 0 to t, the other side should be evaluated from 0 to t also.I think the case is solving the deferential equation: F(v)=MVdV/dx = CV^2 Edited November 21, 2013 by daniton Link to comment Share on other sites More sharing options...

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