Parameterization

[boneill3]

I have the line integral

\int y sin(x) i -cos(x) j

Over (0,1) (pi,-1) and I want to make the integral a function of a single value t.

Can someone please check my Parameterization

I'm using the Parameterization

r(t) = (1-t)(x0,y0) + t(x1,y1)

= (1-t)(0,1)+t(pi,-1)

= pi t 1-2t

so
x = pi t therefore x'(t) = pi

y = 1-2t therefore y'(t) = -2

regards

[jemidiah]

That parametrization is great, I didn't see any mistakes. Just to be clear, your integrand is a conservative vector field, so a parametrization doesn't matter if you're only interested in computing the integral.

[boneill3]

Because it is a conservative vector field I can use the potentail function.

I've calculated that as being


phi (x,y) = y-cos(x)

So if I plug in the values for

int F = phi (x1,y1)-phi (x0,y0)

That equals -1-cos(pi) - 1-cos(0)

= -2

I also had to compute it by another means so I tried my luck at the parameterization method because I definately need the practice.

[boneill3]

But I'm not sure if thats right because when I calculate by

\int f(t)x'(t)+g(t)y'(t) from 0 to 1

I get

(1-2t) sin(pi t)*(pi) + -cos(pi t)*(-2) from 0 to 1

= 0


So is

phi (x,y) = y-cos(x)

The right potential function ?

Not sure where I'm going wrong

[jemidiah]

Your potential function should be "-y cos(x)" not "y-cos(x)"; the negative sign seems to be throwing you off. With the proper function, the integral becomes

int F = phi (x1,y1)-phi (x0,y0) = +1cos(pi) + 1cos(0) = -1 + 1 = 0.

I didn't check the brute force path integral you gave because the results appear to match regardless .

[boneill3]

Your right.

I should have gone back to the start. I checked my paper work and I had it incorrect all along.
Thankyou