Parameterising a Line Segment to Evaluate an Integral

[bugatti79]

Homework Statement

Evaluate this integral along the line segment from (0,1) to (pi, -1) by parameterising this segment

Homework Equations

\int y sin x dx - cos x dy

The Attempt at a Solution

How would I parameterise this line segment?

 

[SammyS]

Homework Statement

Evaluate this integral along the line segment from (0,1) to (pi, -1) by parameterising this segment

Homework Equations

\int y sin x dx - cos x dy

The Attempt at a Solution

How would I parameterise this line segment?

Start by writing an equation for the line passing through those two points.

 

[bugatti79]

Start by writing an equation for the line passing through those two points.

y= \frac{-2x}{\pi} +1. How do I know whether to let x=t or not?

 

[tiny-tim]

hi bugatti79!

y= \frac{-2x}{\pi} +1. How do I know whether to let x=t or not?

you can let t = x, or t = y, or t = any function of x and y so long as it's strictly increasing (or decreasing) along the line

 

[SammyS]

y= \frac{-2x}{\pi} +1. How do I know whether to let x=t or not?

You may want t = 0 to correspond to x = -1 (the left end of the interval) and t = 1 to correspond to x = 1 (the right end of the interval).

That's a fairly common practice.

 

[bugatti79]

hi bugatti79!


you can let t = x, or t = y, or t = any function of x and y so long as it's strictly increasing (or decreasing) along the line

Very good, that's a good tip. Will keep that in mind.

You may want t = 0 to correspond to x = -1 (the left end of the interval) and t = 1 to correspond to x = 1 (the right end of the interval).

That's a fairly common practice.

I don't follow this to be honest. I am not sure what the line is. Is it an arc from (0,1) to (pi, -1)?

Since they don't tell you which path to use, you should ask yourself: will it depend on the path? And NO, it won't, since (ysin(x),-cos(x) is the gradient of F(x,y)=-ycos(x), and so the integral result is the difference between the values of F at the two points.

Does this mean the integral will evaluate to 0? I think that is only for closed simple curves and we are dealing with an open line hence it will be none 0..right?

I'm sorry, you said SEGMENT and I didn't realize. But my post will help you anyway, I hope.

What is the significant of whether it was a segment or not?

 

[SammyS]

Very good, that's a good tip. Will keep that in mind.

I don't follow this to be honest. I am not sure what the line is. Is it an arc from (0,1) to (pi, -1)?

Does this mean the integral will evaluate to 0? I think that is only for closed simple curves and we are dealing with an open line hence it will be none 0..right?

What is the significant of whether it was a segment or not?

The problem states LINE SEGMENT from (0,1) to (π, -1). That means the portion of the (straight) line passing through the points which is between (0,1) and (π, -1) including the endpoints.

The simplest parametrization is to let x(t) = t, and \displaystyle y(t)=\frac{-2x(t)}{\pi} +1=\frac{-2t}{\pi} +1\,, where t goes from 0 to π.

A parametrization I like is for t to go from 0 to 1. Then x(t) = π t, and \displaystyle y(t)=\frac{-2\pi t}{\pi} +1=-2t+1\,.

 

[bugatti79]

The problem states LINE SEGMENT from (0,1) to (π, -1). That means the portion of the (straight) line passing through the points which is between (0,1) and (π, -1) including the endpoints.

The simplest parametrization is to let x(t) = t, and \displaystyle y(t)=\frac{-2x(t)}{\pi} +1=\frac{-2t}{\pi} +1\,, where t goes from 0 to π.

A parametrization I like is for t to go from 0 to 1. Then x(t) = π t, and \displaystyle y(t)=\frac{-2\pi t}{\pi} +1=-2t+1\,.

Ok. I will work out both. Trying the first I get

\displaystyle \int y sin x dx - cos x dy = \int (\frac{-2 t}{\pi}+1) sin t dt+ \frac{2}{\pi} cos t dt=0...