Topic: Polar Equations
Shouldn't the graph of r Sin(t) = Sin (r Cos(t))
r=(Sin (r*Cos (t)))/(Sin (t) )
give me the same thing as y=f(x)=Sin (x) which is the Sine Curve?
y = r Sin(t)
x = r Cos(t) substitute into x,y equation
I tried entering as a polar equation "r(t)=", and then as a relation.
How does one graph the Sine function as a polar equation?\
r(t) - 1+Cos(t) Cardioid graphed just fine.
Why do you indicate the Polar Coordiantes as (θ, r) instead of (r, θ)?
Continuing on, I can graph the relation (x-3)^2 + (y-2)^2=4
A circle with center (3, 2) and radius 2
But after converting this to a Polar Equation 13/(2(3cos(t))+(2Sin(t))) I get a straight line?
Continuing on, I believe that the graph of the equation (4(sin(t))^2+9(cos(t))^2)^.5 should match the graph of the Parametric Equation: 2sin(t), 3cos(t) (I converted the Parametric to the Polar), but it doesn't.
I added my Graph file for both Parametrics and Polar.