How to draw the equation $x = r\cos(\theta)$ and $y = r\sin(\theta)$

Question

Studying for calculus right now and in the notes it says a circle can have the equation x^2 + y^2 = r^2.

y = square root (r^2 - x^2)

or 

y = - (square root (r^2 - x^2))

It says that in parametric form:

x = rcos(theta)

and y = rsin(theta)

0<= theta <= 2pi

How did the notes get the parametric form, and how do I visualize / draw x = rcos(theta) and y = rsin(theta)?

Thanks in advance!

Answer 1

The pair of functions $x=r \cos{\theta}+a$ and $y=r \sin{\theta}+b$ trace out a circle of radius r about the point $(a,b)$, i.e. $(a,b)$ is the center of the circle. To see this, please refer to the definition of the $\sin{\theta}$ and $\cos{\theta}$ functions. The $(\cos{\theta},\sin{\theta})$ coordinates are the coordinates of the point on the unit circle that is also on the ray out of the origin that makes an angle $\theta$ with the positive x-axis. Thus $(\cos{\theta},\sin{\theta})$ traces out a unit circle for $\theta$ ranging between 0 and $2\pi$. Multiplying both coordinate functions by r stretches the circle out by a factor of r away from the origin. The parametric form can be written as \begin{equation} \{x(t),y(t)\}=\{ r \cos{\theta},r \sin{\theta}\} \end{equation} so that you can conclude that $x²+y²=r²$. To get a vector function that traces out a circle centered at the point $(a,b)$, you must simply translate the circle or radius r by a vector $(a,b)$.

Answer 2

We have that $r=\sqrt{x^2+y^2}, \theta=\arctan(\frac yx)$

We can see this in this shoddy diagram: