Section 10.4 Calculus with Polar Curves

In one sense, we don’t have much to do in this section. Since we can think of polar functions as parameterized Cartesian functions, we can rely on the results from Section 10.3 Calculus with Parametric Curves for much of this! In fact, that’s exactly what we’ll do.

When we consider a polar function \(r=f(\theta)\text{,}\) we’ll also think of it as a parametric one:

\begin{align*} x \amp = f(\theta)\cos(\theta)\\ y \amp = f(\theta)\sin(\theta) \end{align*}

Slopes and arc lengths will follow pretty directly from our results for parametric curves, but we’ll need to think about the areas bounded by polar curves a bit more.

Subsection Slopes of Tangent Lines

Activity 10.4.1 Slopes on a Polar Flower.

Consider the polar flower defined by

\begin{equation*} r=2\sin(3\theta) \end{equation*}

for \(0\leq \theta \leq \pi\text{.}\) For this activity, refer to the graphing utility below.

(a)

How many points on this polar flower do you think have horizontal tangent lines?

(b)

How many points on this polar flower do you think have vertical tangent lines?

(c)

Make a conjecture about the slope of the tangent line when \(\theta = \dfrac{5\pi}{6}\) by looking at the graph only.

(d)

Find an expression for \(\dydx\text{.}\)

Solution.

\begin{align*} \dydx \amp = \frac{\frac{d}{d\theta}\left(2\sin(3\theta)\sin(\theta)\right)}{\frac{d}{d\theta}\left(2\sin(3\theta)\cos(\theta)\right)}\\ \amp = \frac{2\sin(3\theta)\cos(\theta) + 6\cos(3\theta)\sin(\theta)}{6\cos(3\theta)\cos(\theta)-2\sin(3\theta)\sin(\theta)}\\ \amp = \frac{\sin(3\theta)\cos(\theta) + 3\cos(3\theta)\sin(\theta)}{3\cos(3\theta)\cos(\theta)-\sin(3\theta)\sin(\theta)} \end{align*}

(e)

Pick one of the values of \(\theta\) where you thought this curve might have a horizontal or vertical tangent line. Evaluate \(\dydx\) at that value, and check to see if you were correct.

(f)

Evaluate \(\dydx\) when \(\theta=\dfrac{5\pi}{6}\text{.}\) How close was your guess at the slope?

Solution.

\begin{align*} \dydx\bigg|_{\theta = \frac{5\pi}{6}} \amp = \frac{\sin\left(\frac{15\pi}{6}\right)\cos\left(\frac{5\pi}{6}\right) + 3\cos\left(\frac{15\pi}{6}\right)\sin\left(\frac{5\pi}{6}\right)}{3\cos\left(\frac{15\pi}{6}\right)\cos\left(\frac{5\pi}{6}\right)-\sin\left(\frac{15\pi}{6}\right)\sin\left(\frac{5\pi}{6}\right)}\\ \amp= \frac{-\frac{\sqrt{3}}{2}}{-\frac{1}{2}} \\ \amp = \sqrt{3} \end{align*}

Subsection Arc Lengths

We could, similarly, dive right into our Parametric Arc Length formula, but let’s pause for a second. It might get a bit messy working with each specific \(\frac{dx}{d\theta}\) and \(\frac{dy}{d\theta}\text{,}\) so we’ll start with this in a more general sense.

Consider some polar function named \(r\text{.}\)

We’ll use the product rule to find the derivatives needed for the arc length formula:

\begin{align*} \frac{dx}{d\theta} \amp = \frac{d}{d\theta}\left(r\cos(\theta)\right)\\ \amp = r'\cos(\theta)-r\sin(\theta)\\ \frac{dy}{d\theta} \amp = \frac{d}{d\theta}\left(r\sin(\theta)\right)\\ \amp = r'\sin(\theta)+r\cos(\theta) \end{align*}

Let’s now find use these in our arc length formula!

\begin{align*} \ell \amp = \int_{\theta=\alpha}^{\theta=\beta} \sqrt{\left( r'\cos(\theta)-r\sin(\theta) \right)^2 + \left( r'\sin(\theta)+r\cos(\theta) \right)^2}\;d\theta\\ \amp = \int_{\theta=\alpha}^{\theta=\beta} \sqrt{ (r'\cos(\theta))^2 - 2rr'\sin(\theta)\cos(\theta) + (r\sin(\theta))^2 + (r'\sin(\theta))^2 + 2rr'\sin(\theta)\cos(\theta) + (r\cos(\theta))^2 }\;d\theta\\ \amp = \int_{\theta=\alpha}^{\theta=\beta} \sqrt{ r^2(\sin^2(\theta)+\cos^2(\theta)) + (r')^2(\sin^2(\theta)+\cos^2(\theta)) }\;d\theta\\ \amp = \int_{\theta=\alpha}^{\theta=\beta} \sqrt{r^2 + (r')^2}\;d\theta \end{align*}

Definition 10.4.2 Polar Arc Length.

If \(r(\theta)\) is a continuous function on \([\alpha, \beta]\) and differentiable on \((\alpha,\beta)\text{,}\) then the length of the polar curve \(r(\theta)\) between \(\theta=\alpha\) ,and \(\theta=\beta\) is:

\begin{equation*} \ell = \int_{\theta=\alpha}^{\theta=\beta}\sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2}\;d\theta\text{.} \end{equation*}

Example 10.4.3 Petal Length.

For the polar graphs that we’ve been calling “Polar Flowers,” we can call each loop that the polar curve makes (between the points where it crosses the pole) petals.

We’ll consider, again, the polar flower \(r=2\sin(3\theta)\text{.}\)

(a)

Find the interval of \(\theta\) that defines the first petal of the polar flower.

Hint.

We’re really thinking about where \(2\sin(3\theta) = 0\text{.}\)

(b)

Set up an integral for the length of the curve tracing out that first petal.

Solution.

\begin{align*} \ell \amp = \int_{\theta=0}^{\theta = \pi/3} \sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2}\;d\theta \\ \amp = \int_{\theta=0}^{\theta = \pi/3} \sqrt{(2\sin(3\theta))^2 + (6\cos(3\theta))^2}\;d\theta\\ \amp = \int_{\theta=0}^{\theta = \pi/3} \sqrt{4\sin^2(3\theta) + 36\cos^2(3\theta)}\;d\theta \end{align*}

Subsection Areas Bounded by Curves

The last calculus topic we’ll visit in this section will be one to reinforce, for one last time, an important theme: the theme of accumulation.

We’re going to go back to our slice-and-sum process to build a formula for calculating a polar area.

Activity 10.4.4 Slice-and-Sum for Polar Areas.

Remember slice-and-sum? Here’s the general procedure:

We’ll divide up the interval of inputs into \(n\) (equally-sized) subintervals. We’ll pick one value from each subinterval.
One each slice, we’ll use the representative input variable to calculate the measurement we’re interested in. In this case, it’s an area.
Add up the areas of the \(n\) slices, and the sum should be an approximation of the actual area we’re interested in.
Use a limit as \(n\to\infty\) to produce an integral our of a Riemann sum!

(a)

The area of a sector of a circle with angle \(\alpha\) is:

\begin{equation*} A=\frac{1}{2}\alpha r^2\text{.} \end{equation*}

Convince yourselves that this is true, using appropriate test angles for:

A circle.
A semicircle.
A quarter of a circle.

(b)

The reason that we’re thinking about sectors of circles is that when we slice an interval of angles in a polar coordinate system, a bunch of rays extending away from the pole.

When do this for a polar curve, we’ll pick a point in the subinterval and assume that the whole subinterval has a constant function output, just like we did with the rectangle. Take a look at the graphing utility below, and get a feel for how we’ll find the \(k\)th slice.

(c)

Calculate the area of the \(k\)th slice, \(A_k\text{.}\)

Solution.

\(A_k = \frac{1}{2}(r(\theta_k^*))^2\Delta \theta\)

(d)

Add these up to approximate the total area.

Solution.

\(A\approx \frac{1}{2}\displaystyle \sum_{k=1}^n (r(\theta_k^*))^2\Delta \theta\)

(e)

Introduce a limit to find an integral formula for the area trapped inside a polar curve from \(\theta=\alpha\) to \(\theta=\beta\text{.}\)

Solution.

\begin{align*} A \amp = \lim_{n\to\infty}\frac{1}{2}\displaystyle \sum_{k=1}^n (r(\theta_k^*))^2\Delta \theta \\ \amp = \frac{1}{2}\int_{\theta=\alpha}^{\theta=\beta} r^2\;d\theta \end{align*}

Example 10.4.5 Petal Area.

Let’s consider, one last time, the polar flower \(r=2\sin(3\theta)\text{.}\)

Set up and evaluate an integral calculate the area bounded inside one of the petals of the polar flower.

Solution.

\begin{align*} A \amp = \frac{1}{2}\int_{\theta=0}^{\theta=\pi/3} \left(2\sin(3\theta)\right)^2\;d\theta \\ \amp= \frac{1}{2}\int_{\theta=0}^{\theta=\pi/3} \left(4\sin^2(3\theta)\right)\;d\theta\\ \amp = \int_{\theta=0}^{\theta=\pi/3} \left(1-\cos(3\theta)\right)\;d\theta\\ \amp = \left(\theta + \frac{1}{3}\sin(3\theta)\right)\bigg|_{\theta=0}^{\theta=\pi/3}\\ \amp = \frac{\pi}{3} \end{align*}

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