anonymous
  • anonymous
Calc III: Determine how much work was done by an object in moving along an arc of the cycloid given by: Image attached
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
1 Attachment
anonymous
  • anonymous
I'm completely lost on this one, I don't even know what formula to use.
experimentX
  • experimentX
|dw:1344456883598:dw|

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experimentX
  • experimentX
|dw:1344456975045:dw| change everything into t's ... since parameter is given, it should be easy to find it.
experimentX
  • experimentX
the other method, ... check if the field is conservative or not, if the field is conservative, the difference in the potential should give you the work done.
anonymous
  • anonymous
Well, the field is conservative, and the parameter is 0 to pi, but what I'm not understanding is what to do with the F*dr
experimentX
  • experimentX
for a conservative field, |dw:1344457305326:dw| isn't this definition of conservative field?
experimentX
  • experimentX
|dw:1344457452297:dw| moreover F = -div V
anonymous
  • anonymous
\[\int\limits_C\vec F\cdot d\vec r=\int_a^b\vec F(\vec r(t))\cdot\vec r'(t)dt\]I don't see where you need \(\nabla f\) in this problem...
anonymous
  • anonymous
on nvm
anonymous
  • anonymous
so this is going to involve the fundamental theorem for line integrals
anonymous
  • anonymous
Also this is a friend of mines test result (I'm studying for finals) and I have no idea what she did.
anonymous
  • anonymous
\[\int\limits_C\nabla f\cdot d\vec r=f(\vec r(b))-f(\vec r(a))\]
experimentX
  • experimentX
she did what i suggested ... first find the curl of field (to make sure that it is conservative) though it isn't necessary.
experimentX
  • experimentX
if the field is conservative, and you know the potential ... the line integral of Field along the path (or so called work done in physics) = change in potential ...
anonymous
  • anonymous
If finding the curl wasn't necessary, then where did she get these numbers from?
experimentX
  • experimentX
|dw:1344458019983:dw|
experimentX
  • experimentX
|dw:1344458032365:dw|
experimentX
  • experimentX
|dw:1344458052253:dw| use those values of t's ... and find the x coordinate and y coordinate. at t=0 and t=pi
anonymous
  • anonymous
Work is defined as\[W=\int\limits_C\vec F\cdot d\vec r=\int\limits_C\nabla f\cdot d\vec r=f(\vec r(b))-f(\vec r(a))\]the potential function for the conservative vector field is\[f=\frac13x^3+\frac13y^3\]and the line is\[\vec r(t)=\langle3(t-\sin t),3(1-\cos t),\rangle\]we have\[\vec r(\pi)=\langle3\pi,6\rangle\]\[\vec r(0)=\langle0,0\rangle\]
anonymous
  • anonymous
so\[W=\int\limits_C\vec F\cdot d\vec r=\int\limits_C\nabla f\cdot d\vec r=f(\vec r(\pi))-f(\vec r(0))\]\[\left(\frac13(3\pi)^3+\frac136^3\right)-0=9\pi^3+72\]
anonymous
  • anonymous
questions?
experimentX
  • experimentX
this way is super easy ... if you are preparing for finals ... try to work on both methods.
anonymous
  • anonymous
yes, always learn more than one way when possible !
anonymous
  • anonymous
btw when I said "work is defined as" I only meant the first part of the expression\[W=\int\limits_C\vec F\cdot d\vec r\]the next part only follows if the force \(\vec F\) is conservative:\[W=\int\limits_C\vec F\cdot d\vec r=\int\limits_C\nabla f\cdot d\vec r=f(\vec r(b))-f(\vec r(a))\]
anonymous
  • anonymous
Sorry, my internet crashed. But yeah, that pretty much summed up all my questions. Thanks!
experimentX
  • experimentX
yw

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