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Calc III: Determine how much work was done by an object in moving along an arc of the cycloid given by: Image attached

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I'm completely lost on this one, I don't even know what formula to use.

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Other answers:

|dw:1344456975045:dw| change everything into t's ... since parameter is given, it should be easy to find it.
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.
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
for a conservative field, |dw:1344457305326:dw| isn't this definition of conservative field?
|dw:1344457452297:dw| moreover F = -div V
\[\int\limits_C\vec F\cdot d\vec r=\int_a^b\vec F(\vec r(t))\cdot\vec r'(t)dt\]I don't see where you need \(\nabla f\) in this problem...
on nvm
so this is going to involve the fundamental theorem for line integrals
Also this is a friend of mines test result (I'm studying for finals) and I have no idea what she did.
\[\int\limits_C\nabla f\cdot d\vec r=f(\vec r(b))-f(\vec r(a))\]
she did what i suggested ... first find the curl of field (to make sure that it is conservative) though it isn't necessary.
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 ...
If finding the curl wasn't necessary, then where did she get these numbers from?
|dw:1344458052253:dw| use those values of t's ... and find the x coordinate and y coordinate. at t=0 and t=pi
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\[\vec r(\pi)=\langle3\pi,6\rangle\]\[\vec r(0)=\langle0,0\rangle\]
so\[W=\int\limits_C\vec F\cdot d\vec r=\int\limits_C\nabla f\cdot d\vec r=f(\vec r(\pi))-f(\vec r(0))\]\[\left(\frac13(3\pi)^3+\frac136^3\right)-0=9\pi^3+72\]
this way is super easy ... if you are preparing for finals ... try to work on both methods.
yes, always learn more than one way when possible !
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:\[W=\int\limits_C\vec F\cdot d\vec r=\int\limits_C\nabla f\cdot d\vec r=f(\vec r(b))-f(\vec r(a))\]
Sorry, my internet crashed. But yeah, that pretty much summed up all my questions. Thanks!

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