# Section9.5Calculus and Polar Functions¶ permalink

The previous section defined polar coordinates, leading to polar functions. We investigated plotting these functions and solving a fundamental question about their graphs, namely, where do two polar graphs intersect?

We now turn our attention to answering other questions, whose solutions require the use of calculus. A basis for much of what is done in this section is the ability to turn a polar function $r=f(\theta)$ into a set of parametric equations. Using the identities $x=r\cos(\theta)$ and $y=r\sin(\theta)\text{,}$ we can create the parametric equations $x=f(\theta)\cos(\theta)\text{,}$ $y=f(\theta)\sin(\theta)$ and apply the concepts of Section 9.3.

# Subsection9.5.1Polar Functions and $\ds \frac{dy}{dx}$

We are interested in the lines tangent a given graph, regardless of whether that graph is produced by rectangular, parametric, or polar equations. In each of these contexts, the slope of the tangent line is $\frac{dy}{dx}\text{.}$ Given $r=f(\theta)\text{,}$ we are generally not concerned with $r\,'=\fp(\theta)\text{;}$ that describes how fast $r$ changes with respect to $\theta\text{.}$ Instead, we will use $x=f(\theta)\cos(\theta)\text{,}$ $y=f(\theta)\sin(\theta)$ to compute $\frac{dy}{dx}\text{.}$

Using Key Idea 9.3.1 we have \begin{equation*} \frac{dy}{dx} = \frac{dy}{d\theta}\Big/\frac{dx}{d\theta}. \end{equation*}

Each of the two derivatives on the right hand side of the equality requires the use of the Product Rule. We state the important result as a Key Idea.

##### Key Idea9.5.1Finding $\frac{dy}{dx}$ with Polar Functions

Let $r=f(\theta)$ be a polar function. With $x=f(\theta)\cos(\theta)$ and $y=f(\theta)\sin(\theta)\text{,}$ \begin{equation*} \frac{dy}{dx} = \frac{\fp(\theta)\sin(\theta) +f(\theta)\cos(\theta) }{\fp(\theta)\cos(\theta) -f(\theta)\sin(\theta) }. \end{equation*}

##### Example9.5.2Finding $\frac{dy}{dx}$ with polar functions.

Consider the limaçon $r=1+2\sin(\theta)$ on $[0,2\pi]\text{.}$

1. Find the equations of the tangent and normal lines to the graph at $\theta=\pi/4\text{.}$

2. Find where the graph has vertical and horizontal tangent lines.

Solution

When the graph of the polar function $r=f(\theta)$ intersects the pole, it means that $f(\alpha)=0$ for some angle $\alpha\text{.}$ Thus the formula for $\frac{dy}{dx}$ in such instances is very simple, reducing simply to \begin{equation*} \frac{dy}{dx} = \tan\alpha. \end{equation*}

This equation makes an interesting point. It tells us the slope of the tangent line at the pole is $\tan\alpha\text{;}$ some of our previous work (see, for instance, Example 9.4.10) shows us that the line through the pole with slope $\tan\alpha$ has polar equation $\theta=\alpha\text{.}$ Thus when a polar graph touches the pole at $\theta=\alpha\text{,}$ the equation of the tangent line at the pole is $\theta=\alpha\text{.}$

##### Example9.5.4Finding tangent lines at the pole.

Let $r=1+2\sin(\theta)\text{,}$ a limaçon. Find the equations of the lines tangent to the graph at the pole.

Solution

# Subsection9.5.2Area

When using rectangular coordinates, the equations $x=h$ and $y=k$ defined vertical and horzontal lines, respectively, and combinations of these lines create rectangles (hence the name “rectangular coordinates”). It is then somewhat natural to use rectangles to approximate area as we did when learning about the definite integral.

When using polar coordinates, the equations $\theta=\alpha$ and $r=c$ form lines through the origin and circles centered at the origin, respectively, and combinations of these curves form sectors of circles. It is then somewhat natural to calculate the area of regions defined by polar functions by first approximating with sectors of circles.

Consider Figure 9.5.6 (a) where a region defined by $r=f(\theta)$ on $[\alpha,\beta]$ is given. (Note how the “sides” of the region are the lines $\theta=\alpha$ and $\theta=\beta\text{,}$ whereas in rectangular coordinates the “sides” of regions were often the vertical lines $x=a$ and $x=b\text{.}$)

Recall that the area of a sector of a circle with radius $r$ subtended by an angle $\theta$ is $A = \frac12\theta r^2\text{.}$

Partition the interval $[\alpha,\beta]$ into $n$ equally spaced subintervals as $\alpha = \theta_1 \lt \theta_2 \lt \cdots \lt \theta_{n+1}=\beta\text{.}$ The length of each subinterval is $\Delta\theta = (\beta-\alpha)/n\text{,}$ representing a small change in angle. The area of the region defined by the $i^\text{ th }$ subinterval $[\theta_i,\theta_{i+1}]$ can be approximated with a sector of a circle with radius $f(c_i)\text{,}$ for some $c_i$ in $[\theta_i,\theta_{i+1}]\text{.}$ The area of this sector is $\frac12f(c_i)^2\Delta\theta\text{.}$ This is shown in part (b) of the figure, where $[\alpha,\beta]$ has been divided into 4 subintervals. We approximate the area of the whole region by summing the areas of all sectors: \begin{equation*} \text{ Area } \approx \sum_{i=1}^n \frac12f(c_i)^2\Delta\theta. \end{equation*}

This is a Riemann sum. By taking the limit of the sum as $n\to\infty\text{,}$ we find the exact area of the region in the form of a definite integral.

The theorem states that $0\leq \beta-\alpha\leq 2\pi\text{.}$ This ensures that region does not overlap itself, which would give a result that does not correspond directly to the area.

##### Example9.5.10Area of a polar region

Find the area of the circle defined by $r=\cos(\theta)\text{.}$ (Recall this circle has radius $1/2\text{.}$)

Solution

Example 9.5.10 requires the use of the integral $\ds\int \cos^2(\theta) \ d\theta\text{.}$ This is handled well by using the power reducing formula as found in the back of this text. Due to the nature of the area formula, integrating $\cos^2(\theta)$ and $\sin^2(\theta)$ is required often. We offer here these indefinite integrals as a time–saving measure. \begin{equation*} \int\cos^2(\theta) \ d\theta = \frac12\theta+\frac14\sin(2\theta)+C \end{equation*} \begin{equation*} \int\sin^2(\theta) \ d\theta = \frac12\theta-\frac14\sin(2\theta)+C \end{equation*}

##### Example9.5.11Area of a polar region

Find the area of the cardiod $r=1+\cos(\theta)$ bound between $\theta=\pi/6$ and $\theta=\pi/3\text{,}$ as shown in Figure 9.5.12.

Solution

Area Between Curves

Our study of area in the context of rectangular functions led naturally to finding area bounded between curves. We consider the same in the context of polar functions.

Consider the shaded region shown in Figure 9.5.13. We can find the area of this region by computing the area bounded by $r_2=f_2(\theta)$ and subtracting the area bounded by $r_1=f_1(\theta)$ on $[\alpha,\beta]\text{.}$ Thus \begin{equation*} \text{ Area } \ = \ \frac12\int_\alpha^\beta r_2^{\,2}\ d\theta - \frac12\int_\alpha^\beta r_1^{\,2}\ d\theta = \frac12\int_\alpha^\beta \big(r_2^{\,2}-r_1^{\,2}\big)\ d\theta. \end{equation*}

##### Key Idea9.5.14Area Between Polar Curves

The area $A$ of the region bounded by $r_1=f_1(\theta)$ and $r_2=f_2(\theta)\text{,}$ $\theta=\alpha$ and $\theta=\beta\text{,}$ where $f_1(\theta)\leq f_2(\theta)$ on $[\alpha,\beta]\text{,}$ is \begin{equation*} A = \frac12\int_\alpha^\beta \big(r_2^{\,2}-r_1^{\,2}\big)\ d\theta. \end{equation*}

##### Example9.5.15Area between polar curves

Find the area bounded between the curves $r=1+\cos(\theta)$ and $r=3\cos(\theta)\text{,}$ as shown in Figure 9.5.16.

Solution
##### Example9.5.17Area defined by polar curves

Find the area bounded between the polar curves $r=1$ and $r=2\cos(2\theta)\text{,}$ as shown in Figure 9.5.18 (a).

Solution

# Subsection9.5.3Arc Length

As we have already considered the arc length of curves defined by rectangular and parametric equations, we now consider it in the context of polar equations. Recall that the arc length $L$ of the graph defined by the parametric equations $x=f(t)\text{,}$ $y=g(t)$ on $[a,b]$ is $$L = \int_a^b \sqrt{\fp(t)^2 + g'(t)^2}\ dt = \int_a^b \sqrt{x'(t)^2+y'(t)^2}\ dt. \label{eq_polar_arclength}\tag{9.5.1}$$

Now consider the polar function $r=f(\theta)\text{.}$ We again use the identities $x=f(\theta)\cos(\theta)$ and $y=f(\theta)\sin(\theta)$ to create parametric equations based on the polar function. We compute $x'(\theta)$ and $y'(\theta)$ as done before when computing $\frac{dy}{dx}\text{,}$ then apply Equation (9.5.1).

The expression $x'(\theta)^2+y'(\theta)^2$ can be simplified a great deal; we leave this as an exercise and state that \begin{equation*} x'(\theta)^2+y'(\theta)^2 = \fp(\theta)^2+f(\theta)^2. \end{equation*}

This leads us to the arc length formula.

##### Key Idea9.5.19Arc Length of Polar Curves

Let $r=f(\theta)$ be a polar function with $\fp$ continuous on an open interval $I$ containing $[\alpha,\beta]\text{,}$ on which the graph traces itself only once. The arc length $L$ of the graph on $[\alpha,\beta]$ is \begin{equation*} L = \int_\alpha^\beta \sqrt{\fp(\theta)^2+f(\theta)^2}\ d\theta = \int_\alpha^\beta\sqrt{(r\,')^2+ r^2}\ d\theta. \end{equation*}

##### Example9.5.20Arc length of a limaçon

Find the arc length of the limaçon $r=1+2\sin(t)\text{.}$

Solution

# Subsection9.5.4Surface Area

The formula for arc length leads us to a formula for surface area. The following Key Idea is based on Key Idea 9.3.20.

##### Key Idea9.5.22Surface Area of a Solid of Revolution

Consider the graph of the polar equation $r=f(\theta)\text{,}$ where $\fp$ is continuous on an open interval containing $[\alpha,\beta]$ on which the graph does not cross itself.

1. The surface area of the solid formed by revolving the graph about the initial ray ($\theta=0$) is: \begin{equation*} \text{ Surface Area } = 2\pi\int_\alpha^\beta f(\theta)\sin(\theta) \sqrt{\fp(\theta)^2+f(\theta)^2}\ d\theta. \end{equation*}

2. The surface area of the solid formed by revolving the graph about the line $\theta=\pi/2$ is: \begin{equation*} \text{ Surface Area } = 2\pi\int_\alpha^\beta f(\theta)\cos(\theta) \sqrt{\fp(\theta)^2+f(\theta)^2}\ d\theta. \end{equation*}

##### Example9.5.23Surface area determined by a polar curve

Find the surface area formed by revolving one petal of the rose curve $r=\cos(2\theta)$ about its central axis (see Figure 9.5.24).

Solution

This chapter has been about curves in the plane. While there is great mathematics to be discovered in the two dimensions of a plane, we live in a three dimensional world and hence we should also look to do mathematics in 3D — that is, in space. The next chapter begins our exploration into space by introducing the topic of vectors, which are incredibly useful and powerful mathematical objects.

# Subsection9.5.5Exercises

Terms and Concepts

In the following exercises, find $\lz{y}{x}$ (in terms of $\theta$). Then find the equations of the tangent and normal lines to the curve at the indicated $\theta$-value.

In the following exercises, find the values of $\theta$ in the given interval where the graph of the polar function has horizontal and vertical tangent lines.

In the following exercises, find the equation of the lines tangent to the graph at the pole.

In the following exercises, find the area of the described region.

In the following exercises, answer the questions involving arc length.

In the following exercises, answer the questions involving surface area.