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.
To find the points where the tangent line is horizontal or vertical, we need to find the derivatives of x and y with respect to deta, set them equal to zero, and solve for deta.
Given:
x = r(deta - cos(deta))
y = r(1 - sin(deta))
We will now find the derivatives of x and y with respect to deta:
dx/deta = r(1 + sin(deta))
dy/deta = r*cos(deta)
To find where the tangent line is horizontal, we need to set dx/deta equal to zero and solve for deta:
r(1 + sin(deta)) = 0
1 + sin(deta) = 0
sin(deta) = -1
Since sin(deta) is equal to -1 at deta = -pi/2 and deta = 3pi/2, we have two possible points where the tangent line is horizontal.
To find where the tangent line is vertical, we need to set dy/deta equal to zero and solve for deta:
r*cos(deta) = 0
cos(deta) = 0
Since cos(deta) is equal to 0 at deta = pi/2 and deta = 3pi/2, we have two possible points where the tangent line is vertical.
Next, to find the area under one period of the curve, we need to integrate the function y with respect to x. However, since the given parametric equations are in terms of deta, we need to express the area in terms of deta.
To find the limits of integration for one period, we need to find the values of deta where the curve completes one full cycle. In this case, we are given deta = pi/3, which represents the endpoint of one period.
So, the limits of integration for one period will be from deta = 0 to deta = pi/3.
The area under the curve can be found by integrating the y-coordinate function with respect to deta from the lower limit to the upper limit:
Area = ∫[0, π/3] r(1 - sin(deta)) deta
Integrating this expression will give us the area under one period of the curve.