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anonymous

  • one year ago

check my work

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  1. anonymous
    • one year ago
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    Find 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 \[\vec F=(2x-y+z)\hat i+(x+y-z^2)\hat j+(3x-2y+4z)\hat k\] z=0 so we have \[\vec F=(2x-y)\hat i+(x+y)\hat j+(3x-2y)\hat k\] Parametric equation of the ellipse, \[x=5\cos(\theta)\]\[y=4\sin(\theta)\] \[z=0\] \[\therefore dx=-5\sin(\theta)d\theta, dy=4\cos(\theta)d\theta, dz=0\] \[\therefore d\vec r=dx \hat i+dy \hat j +dz \hat k=-5\sin(\theta)d\theta \hat i+4\cos(\theta)d\theta \hat j\] \[\vec F.d \vec r=(2x-y)dx+(x+y)dy=2xdx-ydx+xdy+ydy\] \[\vec F. d \vec r=-25\sin(2\theta)d\theta+20\sin^2(\theta)d\theta+20\cos^2(\theta)d\theta+8\sin(2\theta)d\theta\]\[\vec F.d \vec r=(20-16\sin(2\theta))d\theta\] For an ellipse \[0\le \theta \le 2\pi\] therefore \[W=\int\limits_{C}(\vec F . d \vec r)=\int\limits_{0}^{2\pi}(20-16\sin(2\theta))d\theta\] \[W=[20\theta+8\cos(2\theta)]_{0}^{2\pi}=[(40\pi+8)-(0+8)]=40\pi+8-8=40\pi\]

  2. anonymous
    • one year ago
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    sorry should be \[\oint_{C} (\vec F . d \vec r)=\oint_{0}^{2\pi}(20-16\sin(2\theta)d\theta\] Since ellipse is a closed curve :)

  3. amistre64
    • one year ago
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    doesnt work require displacement? if you end up where you started ... my physics is rusty.

  4. anonymous
    • one year ago
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    Hmmm 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

  5. anonymous
    • one year ago
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    Also I think this field is non-conservative \[\vec \nabla \times \vec F \neq \vec 0\] 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

  6. amistre64
    • one year ago
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    2x 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.

  7. anonymous
    • one year ago
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    @amistre64 thx!!

  8. amistre64
    • one year ago
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    if you had any concern about it, where would your concern be focused on? your formulas appear fine, and your substitutions are good to me.

  9. amistre64
    • one year ago
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    the pun is nice to: 'check my work' and its a problem that involves finding the work done :)

  10. anonymous
    • one year ago
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    Oh, I didn't see that!! I guess that deserves 2 medals, 1 for maths and 1 for english!! :P

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