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anonymous
 one year ago
check my work
anonymous
 one year ago
check my work

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Find the work done in moving a particle once round the ellipse \[\frac{x^2}{25}+\frac{y^2}{16}=1\], z=0 under the field of force given by \[\vec F=(2xy+z)\hat i+(x+yz^2)\hat j+(3x2y+4z)\hat k\] z=0 so we have \[\vec F=(2xy)\hat i+(x+y)\hat j+(3x2y)\hat k\] Parametric equation of the ellipse, \[x=5\cos(\theta)\]\[y=4\sin(\theta)\] \[z=0\] \[\therefore dx=5\sin(\theta)d\theta, dy=4\cos(\theta)d\theta, dz=0\] \[\therefore d\vec r=dx \hat i+dy \hat j +dz \hat k=5\sin(\theta)d\theta \hat i+4\cos(\theta)d\theta \hat j\] \[\vec F.d \vec r=(2xy)dx+(x+y)dy=2xdxydx+xdy+ydy\] \[\vec F. d \vec r=25\sin(2\theta)d\theta+20\sin^2(\theta)d\theta+20\cos^2(\theta)d\theta+8\sin(2\theta)d\theta\]\[\vec F.d \vec r=(2016\sin(2\theta))d\theta\] For an ellipse \[0\le \theta \le 2\pi\] therefore \[W=\int\limits_{C}(\vec F . d \vec r)=\int\limits_{0}^{2\pi}(2016\sin(2\theta))d\theta\] \[W=[20\theta+8\cos(2\theta)]_{0}^{2\pi}=[(40\pi+8)(0+8)]=40\pi+88=40\pi\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0sorry should be \[\oint_{C} (\vec F . d \vec r)=\oint_{0}^{2\pi}(2016\sin(2\theta)d\theta\] Since ellipse is a closed curve :)

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1doesnt work require displacement? if you end up where you started ... my physics is rusty.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Hmmm that's true but since there's a field present it will affect the particle's motion as it moves around the ellipse, kinda like how a car moves around an elliptical road, friction present at every point will cause it to do some work. I think that is one way to think about it

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Also I think this field is nonconservative \[\vec \nabla \times \vec F \neq \vec 0\] So taking a particle to some point and bringing it back in the form of an elliptical path will create a difference in going to the point and return trip

amistre64
 one year ago
Best ResponseYou've already chosen the best response.12x dx 2 (5cos) (5sin) .... 25 (2 sin cos), which is simplified to a half angle ... it seems to me that all your substitutions are valid.

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1if you had any concern about it, where would your concern be focused on? your formulas appear fine, and your substitutions are good to me.

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1the pun is nice to: 'check my work' and its a problem that involves finding the work done :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh, I didn't see that!! I guess that deserves 2 medals, 1 for maths and 1 for english!! :P
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