Closed Line Integral Homework - Computing a Hypotenuse

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Homework Help Overview

The original poster is attempting to compute a closed line integral over a triangular region, specifically focusing on finding a parametrization for the hypotenuse after calculating two sides. There is a discussion regarding different equations for a line and their implications for the line integral's value.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster compares two equations for a line and questions why one parametrization yields a negative value while the other is positive. Participants provide insights into the meaning of the equations and their geometric interpretations.

Discussion Status

The discussion is ongoing, with participants offering clarifications about the parametrizations and their respective behaviors. The original poster expresses gratitude for the insights received, indicating a productive exchange of ideas.

Contextual Notes

The original poster notes that the problem may not fit traditional homework formats, and there is an emphasis on understanding the parametrization's derivation and its impact on the integral's evaluation.

nayfie
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Homework Statement



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Homework Equations



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The Attempt at a Solution



This isn't really a proper homework question so I'll just write my problem here:

I'm trying to compute a closed line integral over a triangular region. I have calculated two of the sides, but am now left with the hypotenuse. I'm trying to find a parametrisation for this to work.

My lecturer told me that the equation for a line is:

((1 - t)(a, b) + t(c, d))

Yet I had always known the equation of a line to be:

(a, b) + t(c, d)

I have tried computing this line integral using both parametrisations. The first one works, but the second one doesn't. (They both give the same value but the first one is negative and the second one is positive. For the line integral to give what I believe is the correct answer, the line integral for this section needs to be negative).

Can somebody explain why the first equation works but the second one doesn't? Also can you please give some insight into how the first one works? I don't understand where the (1 - t) at the beginning comes from.

Thank you in advance!
 
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hi nayfie! :smile:
nayfie said:
My lecturer told me that the equation for a line is:

((1 - t)(a, b) + t(c, d))

Yet I had always known the equation of a line to be:

(a, b) + t(c, d)

the first is for the line from (a,b) to (c,d)

the second is for the line from (a,b) and parallel to the line from (0,0) to (c,d) :wink:
 
Specifically, with (1- t)(a, b)+ t(c, d) when t=0, that is (a, b) and when t= 1 that is (c, d) so that is the line through points (a, b) and (c, d). With (a, b)+t(c, d) when t= 0, that is (a, b) and when t= 1 it is (a+c, b+ d). The "slope" is (b+d- (b))/(a+b- (a))= d/b, the same as the first line so they are parallel as tiny-tim said, but, in general, does not contain the point (c, d). (It will contain (c, d) if and only if we can solve a+ tc= c, b+ dt= d for the same value of t. That is, we must have (c-a)/c= (d- b)/d.)

The form you give is, as you say, a line but, in general, not the specific line you want.
 
Thank you both for the replies, very helpful!
 

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