- anonymous

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

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- anonymous

##### 1 Attachment

- anonymous

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

- experimentX

|dw:1344456883598:dw|

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## More answers

- experimentX

|dw:1344456975045:dw|
change everything into t's ... since parameter is given, it should be easy to find it.

- 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

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

for a conservative field, |dw:1344457305326:dw|
isn't this definition of conservative field?

- experimentX

|dw:1344457452297:dw|
moreover F = -div V

- 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

on nvm

- anonymous

so this is going to involve the fundamental theorem for line integrals

- anonymous

Also this is a friend of mines test result (I'm studying for finals) and I have no idea what she did.

##### 1 Attachment

- anonymous

\[\int\limits_C\nabla f\cdot d\vec r=f(\vec r(b))-f(\vec r(a))\]

- 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

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

If finding the curl wasn't necessary, then where did she get these numbers from?

##### 1 Attachment

- experimentX

|dw:1344458019983:dw|

- experimentX

|dw:1344458032365:dw|

- 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

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

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

questions?

- experimentX

this way is super easy ... if you are preparing for finals ... try to work on both methods.

- anonymous

yes, always learn more than one way when possible !

- 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

Sorry, my internet crashed.
But yeah, that pretty much summed up all my questions. Thanks!

- experimentX

yw

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