Question

find dy/dx of
r(theta) = 4cos(theta)
at theta = pi/3 and pi/2

Answer 1

r(∅)=4 cos∅
r^' (∅)=4 d/dx cos∅
r^' (∅)=-4 sin∅
r^' (π/3)=-4 sin(π/3)=-4∙[√3)/2]=-2∙√3)
r^' (π/2)=-4 sin(π/2)=-4∙1=-4

Answer 2

isn't that just [dtheta/dx\] and not dy/dx?
i am thinking it's (f'theta*sin theta+ fcos theta)/(f'theta*cos theta - f sin theta)

Answer 3

I was assuming that r(theta)=y and (theta)=x therefore dy/dx=dr/dtheta? Which is basically what I did in my calculations...

Answer 4

okay is see thanks!

Answer 5

You're welcome!