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Calculus1 18 Online
OpenStudy (anonymous):

how do you find dy/dx of r=1+cos(theta) as a function of theta?

OpenStudy (dan815):

derivative of cos = - sin

OpenStudy (anonymous):

That's just the derivative of 1+cos(theta) right? So it would be -sin(theta)

OpenStudy (dan815):

are u sure it wants dy/dx or dr/dtheta?

OpenStudy (anonymous):

I didnt think it was possible to find dy/dx, but thats what the worksheet says

OpenStudy (anonymous):

Yeah, of course it is (possible) :)

OpenStudy (anonymous):

It's probably just a typo, aka using variables interchangeably. That happens a lot.

OpenStudy (dan815):

i think it means derivative as a function of theta so ya u can just say its -sin (theta)

OpenStudy (anonymous):

yep

OpenStudy (anonymous):

But where's the adventure in that :D \[\huge x=r\cos(\theta)\]\[\huge y=r\sin(\theta)\]

OpenStudy (dan815):

wherse the fun in that |dw:1364950377114:dw|

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