Derivatives for Polar Curves

# 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)
\begin{align} \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} \end{align}

Hence we obtain:

(3)
\begin{align} \frac{dy}{dx} = \frac{f'(\theta)\sin\theta + f(\theta)\cos \theta}{f'(\theta) \cos \theta - f(\theta)\sin \theta} \end{align}

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

(4)
\begin{align} \frac{dy}{dx} = \frac{f'(\theta)\sin\theta + f(\theta)\cos \theta}{f'(\theta) \cos \theta - f(\theta)\sin \theta} \frac{dy}{dx} = \frac{\frac{dr}{d\theta}\sin\theta + r\cos \theta}{\frac{dr}{d\theta} \cos \theta - r\sin \theta} \end{align}

## 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)
\begin{align} \frac{dy}{dx} = \frac{\frac{dr}{d\theta}\sin\theta + r\cos \theta}{\frac{dr}{d\theta} \cos \theta - r\sin \theta} \frac{dy}{dx} = \frac{\cos \theta \cdot \sin \theta + (\sin \theta - 3)\cos \theta}{\cos \theta \cdot \cos \theta - (\sin \theta - 3) \sin \theta} \end{align}

## 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)
\begin{align} \frac{dy}{dx} = \frac{\frac{dr}{d\theta}\sin\theta + r\cos \theta}{\frac{dr}{d\theta} \cos \theta - r\sin \theta} \frac{dy}{dx} = \frac{(-2\sin\theta) \cdot \sin\theta + (2\cos \theta - 3)\cos \theta}{(-2\sin \theta) \cdot \cos \theta - (2 \cos \theta - 3) \sin \theta} \end{align}

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

(7)
\begin{align} \frac{(-2(0)) \cdot 0 + (-2 - 3)(-1)}{(-2(0)) \cdot (-1) - (2(-1) - 3) (0)} \\ = \frac{5}{0} \end{align}

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