A particle subject to a force vector F(x,y) = (yi-xj) moves clockwise along the arc of a circle with radius 4 centered at the origin, that begins at (-4, 0) and ends at (0, 4) a) Find the work done by vector F.
all them question marks
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there? @dan815
k parametrize the curve w=int f.dr
r(t)=<x(t),y(t)> dr/dt=<x'(t),y'(t)> dr=<x'(t),y'(t)> * dt
f=<x,y>=<x(t),y(t) now f.dr is all in one variable and you can integrate over some time bound
in more general case what u are doing here is, u wanna see how much of your force is in the tangent component of your motion, because thats the component that matters wrt to our force
since word=f*d
you could also use green's theorem by completing the path along x axis
so (cost, sint) for the parametrization ?
you have to see your bounded area though, because greens theorem will give u integral all over bounded area curve
you could use greens theorem still though by subtracting an area, from the main integral
so would the integral be from pi to pi/2?
subtract the work done along x axis, lets work it both ways and compare the answers
curlf = d/dx(-x)-(d/dy)((y)= -2, :( not path independant so u gotta do long way
-2*k^ to be more accurate
\[-\int\limits_{\pi/2}^{\pi} (sinti-costj) (-sinti+costj)dt\]
ok that works
u can just say pi to pi.2 and get rid of negative sigh no problems
u will see that u have -(sin^2+cos^2) so
you ust have 1 |pi/2 to pi
which is pi/2
wait u set p something wrong
it shudnt be like that becase x=4costheta,y=4sintehta
you will get a factor of 16 that u are missing
so u get pi/2 * 16 = 8pi
oh yeah i missed the radius 4
happens when u look at megan fox too long
im getting 16pi by greens thm
or rose bottom
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