Closed Line Integral Homework - Computing a Hypotenuse

In summary, I'm trying to find a parametrisation for this to work. The equation for a line is:((1- t)(a, b) + t(c, d))
  • #1
nayfie
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0

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:

[itex]((1 - t)(a, b) + t(c, d))[/itex]

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

[itex](a, b) + t(c, d)[/itex]

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 [itex](1 - t)[/itex] at the beginning comes from.

Thank you in advance!
 
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  • #2
hi nayfie! :smile:
nayfie said:
My lecturer told me that the equation for a line is:

[itex]((1 - t)(a, b) + t(c, d))[/itex]

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

[itex](a, b) + t(c, d)[/itex]

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:
 
  • #3
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.
 
  • #4
Thank you both for the replies, very helpful!
 

1. What is a closed line integral?

A closed line integral is a mathematical concept in which a function is integrated along a closed curve or path. It is used to calculate the total value of a vector field along a closed loop.

2. How is a closed line integral computed?

A closed line integral is computed by breaking the curve or path into smaller segments and approximating the value of the function at each point. These values are then added together to get the total value of the integral.

3. What is the purpose of computing a closed line integral?

The purpose of computing a closed line integral is to determine the total value of a vector field along a closed path. This can be useful in many applications such as calculating work done by a force, finding the net flux of a vector field, and determining the circulation of a fluid.

4. What is a hypotenuse?

A hypotenuse is the longest side of a right triangle, opposite the right angle. It is calculated using the Pythagorean theorem, which states that the square of the hypotenuse is equal to the sum of the squares of the other two sides.

5. How is a hypotenuse used in closed line integral homework?

In closed line integral homework, the hypotenuse is often used to represent the path or curve along which the integral is being computed. The length of the hypotenuse is important in determining the accuracy of the approximation method used to compute the integral.

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