Find the work

[mcwang719]

Find the work performed by F(x,y)=<y,-x> along the path C: x^2+y^2=1 from (0,1) to (1,0) going in the CLOCKWISE direction.

So I know W = \(\displaystyle \int {F \cdot dr}\)

I figured a couple of work problems so im not completely lost, but I'm having trouble finding the parameterization because that's what you need to figure out the problem right? r(t)=<?,?> If I had to guess, I would say <t^2,t^2>.

Also, the limits of integration would be from 0 to 1 right? And it says"going in the CLOCKWISE direction". Does that mean I have to do something special?

Any help would be greatly apprieciated!

 

[daon]

Find the work performed by F(x,y)=<y,-x> along the path C: x^2+y^2=1 from (0,1) to (1,0) going in the CLOCKWISE direction.

So i know W=\(\displaystyle \int {F \cdot dr}\) I figured a couple of work problems so im not completely lost, but i'm having trouble finding the parameterization because that's what you need to figure out the problem right? r(t)=<?,?> if i had to guess i would say <t^2,t^2>. also the limits of integration would be from 0 to 1 right? and it says"going in the CLOCKWISE direction" does that mean I have to do something special. any help would be greatly apprieciated!!!!!!!!!!!!

Since the interval is always positive: let x = t, then y = \(\displaystyle \sqrt{1 - t^2}\)
so r(t) = <t, \(\displaystyle \sqrt{1-t^2}\)>
dr = <1dt, \(\displaystyle \frac{1}{2\sqrt{1-t^2}} (-2t)\)dt>

And t goes from 0 to 1 (as you said) because x goes from 0 to 1 (x=t).

So, F.dr = <\(\displaystyle \sqrt{(1-t^2)}\), -t> ∙ <1dt, \(\displaystyle \frac{1}{2\sqrt{1-t^2}} (-2t)\)dt> = \(\displaystyle \sqrt{1-t^2}\)dt + \(\displaystyle \frac{t^2}{\sqrt{1-t^2}}\)dt = \(\displaystyle \frac{1}{\sqrt{1-t^2}}\)dt

So \(\displaystyle \L W= \int _0^1 \frac{dt}{\sqrt{1-t^2}}\).

Now please be cautious as its been over a year since I've attempted any Calc III material. Also, there are many ways to set the parameter here. For instance you can let x be cos(t). Either way you have a unit circle.

 

[mcwang719]

thanks daon!!! you've been really helpful!!!!