# Derivatives for Polar Curves

We have thus far worked with polar curves, that is curves in the form $r = f(\theta)$. Recall the translation equations for coordinates in the $xy$-coordinate system to the polar coordinate system are as follows:

(1)Hence it follows that $x = f(\theta) \cos \theta$ and $y = f(\theta) \sin \theta$.

So now we have a set of parametric equations involving just $x$ and $\theta$. We can now use the same property for deriving derivatives for regular parametric curves, that is

(2)Hence we obtain:

(3)But recall that $f(\theta) = r$, and hence $f'(\theta) = \frac{dr}{d\theta}$, hence:

(4)## Example 1

**Determine the derivative of the polar curve $r = \sin \theta - 3$.**

We know that $r = \sin \theta - 3$, so it follows that $\frac{dr}{d\theta} = \cos \theta$. Hence we can plug these values in directly to our formula:

(5)## Example 2

**Determine the slope of the tangent of the polar curve $r = 2\cos \theta - 3$ at $\theta = \pi$.**

Once again, we know that $r = 2 \cos \theta - 2$ so $\frac{dr}{d\theta} = -2\sin \theta$. We can put these into our formula to obtain our derivative $\frac{dy}{dx}$:

(6)Now let's take $\frac{dy}{dx}$ and evaluate it $\theta = \pi$, we thus get:

(7)So in fact we get a vertical tangent at $\theta = \pi$, which is true as seen in this graph: