May 13th, 2017, 05:15 PM  #1 
Member Joined: Aug 2016 From: illinois Posts: 46 Thanks: 0  parameterizing help
I have a line integral problem. The function is F(X,Y)=<tany,xsec^2y>. The Curve goes from (1,0) to (2,pi/4). I don't know how to find r(t).
Last edited by skipjack; May 13th, 2017 at 08:40 PM. 
May 13th, 2017, 08:30 PM  #2 
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,497 Thanks: 755 
If the curve is left unspecified, the first thing you should suspect is that your field is conservative and that only the endpoints matter. See whether you can figure out the potential function whose gradient is your field and then just take the difference of the potential at the endpoints. Last edited by skipjack; May 13th, 2017 at 08:42 PM. 
May 13th, 2017, 09:13 PM  #3 
Global Moderator Joined: Dec 2006 Posts: 18,053 Thanks: 1395  
May 13th, 2017, 09:30 PM  #4 
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,497 Thanks: 755  
May 14th, 2017, 05:32 PM  #5 
Member Joined: Aug 2016 From: illinois Posts: 46 Thanks: 0 
I'm not sure what you mean. I just don't know how to make r(t). I know what to do from there.
Last edited by skipjack; May 15th, 2017 at 04:14 AM. 
May 14th, 2017, 06:06 PM  #6  
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,497 Thanks: 755  Quote:
Conservative fields are the gradients of potential functions. The line integral of a conservative field depends only on the values at the end points of the curve. So it's exact shape is irrelevant. Either you use this fact, or they give you more info on what the curve looks like because w/o more info you can't parameterize the thing. If you like go ahead and parameterize it as a straight line as skipjack suggested. You'll get the same answer as any other curve. $r(t) = \left \{1+t,~\dfrac \pi 4 t \right\},~t \in [0,1]$  
May 14th, 2017, 06:17 PM  #7 
Member Joined: Aug 2016 From: illinois Posts: 46 Thanks: 0 
Yea, we covered how to find potential functions from conservative fields but Okay thank you I think I was just overthinking it.
Last edited by skipjack; May 15th, 2017 at 04:16 AM. 
May 15th, 2017, 02:48 AM  #8 
Math Team Joined: Jan 2015 From: Alabama Posts: 2,731 Thanks: 707 
"Conservative field" and "potential function" are physics terms. I prefer to use the mathematics terms "exact differential" and "antiderivative", respectively.

May 15th, 2017, 12:23 PM  #9 
Global Moderator Joined: Dec 2006 Posts: 18,053 Thanks: 1395  I suggested that method because of the "parameterizing help" title. The "potential" method involves using a function that isn't defined for all values of y, so the route isn't entirely arbitrary.


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