My Math Forum parameterizing help

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 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)=. 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: USA Posts: 1,938 Thanks: 1006 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
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Quote:
 Originally Posted by fivestar The Curve goes from (1,0) to (2,pi/4).
Is the "curve" a straight line?

May 13th, 2017, 09:30 PM   #4
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Quote:
 Originally Posted by skipjack Is the "curve" a straight line?
It doesn't matter. The field is trivially shown to be conservative.

 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
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 Originally Posted by fivestar Im not sure what you mean. I just dont know how to make r(t). I know what to do from there.
have you covered Potential functions at all?

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: 3,160 Thanks: 866 "Conservative field" and "potential function" are physics terms. I prefer to use the mathematics terms "exact differential" and "anti-derivative", respectively.
May 15th, 2017, 12:23 PM   #9
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Quote:
 Originally Posted by romsek 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.
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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