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I was wondering how you can make a polar equation parametric, and I just don't get it. My book says that for $r = f(\theta)$, $x = f(t) \cos t$ and $y = f(t) \sin t$, but this makes absolutely no sense to me.

If you need an example problem, $r = \frac{4}{2 - \cos \theta}$ would work very well. Thanks in advance.

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In your example, $f(\theta) = \dfrac{4}{2 - \cos \theta}$. Therefore, your parametric equations would be $$\begin{align*} x(t) &= f(t) \cos t = \frac{4}{2 - \cos t} \cos t = \frac{4 \cos t}{2 - \cos t}, \\ y(t) &= f(t) \sin t = \frac{4}{2 - \cos t} \sin t = \frac{4 \sin t}{2 - \cos t}. \end{align*}$$ That is it.

The reason why this works is because a polar curve can be interpreted as a special kind of parametrization, where the parameter $t$ is the angle $\theta$ as measured counterclockwise from the positive $x$-axis; and the radius $r$ is broken into its horizontal ($x$) and vertical ($y$) components via the relationship $x = r \cos \theta$, $y = r \sin \theta$. Since $r$ is a function of angle $\theta$, then $x$ and $y$ become a function of $\theta$ and in turn the parameter $t = \theta$.

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In two dimensions you have:

$$x=r\cos(\theta)$$ $$y=r\sin(\theta)$$

So, taking the square of both equations and summing them you get:

$$r^2=x^2+y^2$$

$$r = \sqrt{x^2+y^2}$$

Now we'll get the relation for $\theta$. Dividing the second equation over the first one:

$$\frac{y}{x}=\frac{\sin (\theta)}{\cos (\theta)}$$

$$\frac{y}{x}=\tan(\theta)$$

$$\theta=\arctan \frac{y}{x}$$

There you have the inverse change of variables for polar coordinates.

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