calculus help please
posted by stan on .
Find the points where the tangent line is horizontal or vertical. Find the area under one period of the curve.
x= r(deta  cos deta)
Y= r(1 sin deta)
deta= pie/3

%pi;/3 is not correct. There is a value of θ for the horizontal tangent, and another for the vertical tangent.
Given:
x= r(θ  cos θ)
Y= r(1 sin θ)
For the given parametric equations, r is a magnification factor, so for all intents and purposes, we can assume it to be 1.0. If it assumes any other value, the graph will be magnified or reduced accordingly.
To find dy/dx or dx/dy, we can apply the chain rule
dy/dx = (dy/dθ) / (dx/dθ)
and similarly
dx/dy = (dx/dθ) / (dy/dθ)
where dx/dθ = r(1+sinθ), and
dy/dθ = r(cosθ)
A vertical tangent occurs at a point where dy/dx=0, and a horizontal tangent occurs where dx/dy=0.
Note that at θ=π/2, both dx/dθ and dy/dθ are zero. Use l'hôpital's rule to evaluate.
You should get a complete cycle if you work with θ from 2π to 2π.
To help you visualize the function, I include two graphics plots:
http://img411.imageshack.us/img411/5971/1283050369a.png
http://img690.imageshack.us/img690/6584/1283050369b.png
Post your results for verifications if you wish.