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Workbook for Stewart’s "Calculus"

Cory Wilson

Section 10.3 Polar Coordinates

Subsection 10.3.1 Before Class

Figure 74. Pre-Class Video 1
Figure 75. Pre-Class Video 2

Subsubsection 10.3.1.1 The Polar Coordinate System

Question 10.3.1.
The Cartesian coordinate system carries two pieces of information: a horizontal distance, \(x\text{,}\) and a vertical distance, \(y\text{.}\) If polar coordinates deal with circular objects, what two pieces of information should a polar coordinate carry?
Solution.
Radius and angle
Example 10.3.2.
Use the polar grid below to plot the points. Be sure to label the points.
  1. \(\displaystyle \lrpar{2,-\dfrac{\pi}{6}}\)
  2. \(\displaystyle \lrpar{-3,\pi}\)
  3. \(\displaystyle \lrpar{1,\dfrac{\pi}{4}}\)
  4. \(\displaystyle \lrpar{4,-\dfrac{2\pi}{3}}\)
  5. \(\displaystyle \lrpar{-1,\dfrac{5\pi}{6}}\)
  6. \(\displaystyle \lrpar{2,0}\)
A blank polar grid
Solution.
Done in video, need to fix after "image in solution" issue is figured out.
Example 10.3.3.
Consider the polar point \(\lrpar{1,\dfrac{\pi}{4}}\) plotted above. This is not the only way of describing the point.
  1. Give two or three different ways of expressing this point in polar coordinates.
  2. Do you see a pattern in the expressions? What is it?
Solution.
  1. Answers vary
  2. Answers vary
A Point in Polar Coordinates.
Any point \((r,\theta)\) in polar coordinates can also be represented by the points
\begin{equation*} (r,\theta + 2n\pi) \end{equation*}
and
\begin{equation*} (-r,\theta + (2n+1)\pi) \end{equation*}
for any \(n\in \Z\)
There is a connection between Cartesian coordinates and polar coordinates; we can exploit trigonometry to find the relationship.
Example 10.3.4.
  1. Below is a circle of radius \(r\text{,}\) with an inscribed right triangle. Label as much as you can on the figure.
  2. Use the triangle and trigonometric functions to obtain a conversion from polar coordinates to Cartesian coordinates.
A circle of radius \(r\text{,}\) with an inscribed right triangle
Solution.
To be filled out when the image issue is resolved
Polar-to-Cartesian Conversion.
If a point \(P\) has the polar coordinate expression \((r,\theta)\text{,}\) then its Cartesian coordinate expression is given by \(x = r\cos\theta\) and \(y = r\sin\theta\)
Cartesian-to-Polar Conversion.
If a point \(P\) has the Cartesian coordinate expression \((x,y)\text{,}\) then its polar coordinate expression is given by \(r^2 = x^2 + y^2\) and \(\theta = \inv{\tan}\lrpar{\dfrac{y}{x}}\)
Example 10.3.5.
Convert the following points from polar to Cartesian coordinates.
  1. \(\displaystyle \lrpar{2,\dfrac{3\pi}{2}}\)
  2. \(\displaystyle \lrpar{\sqrt{2},\dfrac{\pi}{4}}\)
  3. \(\displaystyle \lrpar{-3,-\dfrac{\pi}{3}}\)
Solution.
Technically there are an infinite number of representations, but here are perhaps the most straightforward.
  1. \(\displaystyle (0,-2)\)
  2. \(\displaystyle (1,1)\)
  3. \(\displaystyle \lrpar{-\dfrac{3}{2},-\dfrac{3\sqrt{3}}{2}}\)
Example 10.3.6.
Convert the following points from Cartesian to polar coordinates.
  1. \(\displaystyle \lrpar{-4,4}\)
  2. \(\displaystyle \lrpar{\sqrt{3},1}\)
  3. \(\displaystyle \lrpar{3,3\sqrt{3}}\)
Solution.
Same comment as above
  1. \(\displaystyle \lrpar{4\sqrt{2},\dfrac{3\pi}{4}}\)
  2. \(\displaystyle \lrpar{2,\dfrac{\pi}{6}}\)
  3. \(\displaystyle \lrpar{6,\dfrac{\pi}{3}}\)
Example 10.3.7.
Convert each equation from polar to rectangular coordinates. If possible, identify the curve.
  1. \(\displaystyle r^2 = 16\)
  2. \(\displaystyle r = 4\sin\theta\)
  3. \(\displaystyle \theta = \dfrac{\pi}{3}\)
  4. \(\displaystyle r^2\cos 2\theta = 1\)
Solution.
  1. \(x^2 + y^2 = 16\text{,}\) a circle of radius 4 centered at the origin.
  2. \(x^2 + (y-2)^2 = 4\text{,}\) a circle of radius 2 centered at \((0,-2)\)
  3. \(y = \sqrt{3}x\text{,}\) a line of slope \(\sqrt{3}\)
  4. \(\displaystyle \)
Example 10.3.8.
Convert each equation from rectangular coordinates to polar coordinates.
  1. \(\displaystyle y = 2\)
  2. \(\displaystyle 4y^2 = x\)
  3. \(\displaystyle y= x\)
Solution.
  1. \(\displaystyle r = 2\csc\theta\)
  2. \(\displaystyle 4r^2\sin^2\theta = r\cos\theta\)
  3. \(\displaystyle \theta = \dfrac{\pi}{4}\)

Subsection 10.3.2 Pre-Class Activities

Example 10.3.9.

Use this space to write any questions you might have from the videos
Solution.
Answers vary

Example 10.3.10.

Plot each point below on a polar grid. Then, convert each from polar coordinates to Cartesian coordinates.
  1. \(\displaystyle \lrpar{2,\dfrac{5\pi}{6}}\)
  2. \(\displaystyle \lrpar{1,-\dfrac{2\pi}{3}}\)
  3. \(\displaystyle \lrpar{-1,\dfrac{5\pi}{4}}\)
Solution.
Again, the plots will have to wait, but the conversions are below.
  1. \(\displaystyle \lrpar{-\sqrt{3},1}\)
  2. \(\displaystyle \lrpar{-\dfrac{1}{2},-\dfrac{\sqrt{3}{2}}}\)
  3. \(\displaystyle \lrpar{\dfrac{\sqrt{2}}{2},\dfrac{\sqrt{2}}{2}}\)

Example 10.3.11.

Convert \(r=4\csc\theta\) into rectangular coordinates, and describe/identify the graph.
Solution.
This was conveniently done earlier in the before class: \(y = 4\)

Example 10.3.12.

Express the ellipse \(\dfrac{x^2}{9} + \dfrac{y^2}{4} = 1\) in polar coordinates.
Solution.
\(4r^2\cos^2\theta + 9r^2\sin^2\theta = 1\text{,}\) so this simplifies as \(r^2(4 + 5\sin^2\theta) = 1\)

Subsection 10.3.3 In Class

Subsubsection 10.3.3.1 Polar Curves

Example 10.3.13.
Graph the polar equation \(\theta = \dfrac{\pi}{4}\)
Solution.
This ends up the line \(y = x\text{,}\) but again with the graphing issue.
Example 10.3.14.
Let \(r = 2\sin\theta\)
  1. Make a table of values for the curve, for \(0\leq\theta\leq \pi\)
  2. Use the table to sketch the curve
  3. Find the Cartesian equation for the curve
Solution.
  1. ,
    \(\theta\) \(r = 2\sin\theta\)
    \(0\) \(0\)
    \(\dfrac{\pi}{6}\) \(1\)
    \(\dfrac{\pi}{4}\) \(\sqrt{2}\)
    \(\dfrac{\pi}{3}\) \(\sqrt{3}\)
    \(\dfrac{\pi}{2}\) \(2\)
    \(\dfrac{2\pi}{3}\) \(\sqrt{3}\)
    \(\dfrac{3\pi}{4}\) \(\sqrt{2}\)
    \(\dfrac{5\pi}{6}\) \(1\)
    \(\pi\) \(0\)
  2. Graphing issues, but you should see a circle of radius 1 centered at \((0,1)\)
  3. \(\displaystyle x^2 + (y-1)^2 = 1\)
Example 10.3.15.
Sketch the polar curve \(r = 1 + \cos\theta\text{.}\) This shape is called a cardioid.
Solution.
Example 10.3.16.
Sketch the polar curve \(r = \sin 2\theta\text{.}\) This shape is called a rose of four leaves.
Solution.
Symmetry in Polar Graphs.
Consider the polar curve \(r = f(\theta)\)
  • The curve is symmetric about the polar axis if \(f(\theta) = f(-\theta)\)
  • The curve is symmetric about the pole if \(-f(\theta) = f(\theta)\) or \(f(\theta) = f(\theta + \pi)\)
  • The curve is symmetric about \(\theta = \dfrac{\pi}{2}\) if \(f(\theta) = f(\pi - \theta)\)
Example 10.3.17.
What kind of symmetries does the cardioid \(r = 1 + \cos\theta\) possess? What about the cardioid \(r = 1+\sin\theta\text{?}\)
Solution.
Let \(f(\theta) = 1 + \cos\theta\) and \(g(\theta) = 1 + \sin\theta\)
\(f(-\theta) = f(\theta)\text{,}\) so the first cardioid is symmetric about the polar axis. It is clear that \(-f(\theta)\neq f(\theta)\text{,}\) so there is no symmetry about the pole. \(f(\pi - \theta) = 1+\cos\pi\cos\theta - \sin\theta\sin\pi = 1 -\cos\theta \neq f(\theta)\text{,}\) so there is no symmetry about \(\theta = \dfrac{\pi}{2}\)
\(g(-\theta) = -g(\theta)\text{,}\) so the second cardioid is not symmetric about the polar axis. \(-f(\theta) \neq f(\theta)\text{,}\) so there is no symmetry about the pole. \(f(\pi-\theta) = 1 + \sin\pi\cos\theta + \cos\pi\sin\theta = 1-\sin\theta\neq f(\theta)\text{,}\) so there is no symmetry about \(\theta = \dfrac{\pi}{2}\)
Example 10.3.18.
Use symmetries to help you graph \(r^2 = \cos 4\theta\text{.}\)
Solution.
Graphing issues

Subsubsection 10.3.3.2 Tangents to Polar Curves

Derivative of a Polar Curve.
Given the polar curve \(r = f(\theta)\text{,}\) consider the parametric equations \(x = r\cos\theta = f(\theta)\cos\theta\) and \(y = r\sin\theta = f(\theta)\sin\theta\text{.}\) Then,
\begin{equation*} \dfrac{dy}{dx} = \dfrac{dy/d\theta}{dx/d\theta} = \dfrac{\dfrac{dr}{d\theta}\sin\theta + r\cos\theta}{\dfrac{dr}{d\theta}\cos\theta - r\sin\theta} \end{equation*}
Example 10.3.19.
Consider the cardioid \(r = 1 + \sin\theta\text{.}\)
  1. Find the slope of the tangent line when \(\theta = \dfrac{\pi}{6}\text{.}\)
  2. Find the points where the tangent line is horizontal or vertical.
Solution.
  1. \(\dfrac{dy}{dx} = \dfrac{\cos\theta(1+2\sin\theta)}{(1+\sin\theta)(1-2\sin\theta)}\text{,}\) so
    \begin{equation*} \dfrac{dy}{dx}\bigg\rvert_{\theta = \pi/6} = \dfrac{\sqrt{3}}{0} \end{equation*}
    which is undefined.
  2. The tangent line is horizontal when \(\cos\theta(1+2\sin\theta) = 0\iff \theta = \dfrac{\pi}{2},\dfrac{7\pi}{6},\dfrac{11\pi}{6}\text{.}\)
    The tangent line is vertical when \((1+\sin\theta)(1-2\sin\theta) = 0\iff \theta = \dfrac{3\pi}{2},\dfrac{\pi}{6},\dfrac{5\pi}{6}\)
Example 10.3.20.
Find the slope of the tangent line to the polar curve \(r = 2\cos\theta\) at \(\theta = \dfrac{\pi}{3}\text{.}\)
Solution.
\(\dfrac{dr}{d\theta}\bigg\rvert_{\pi/3} = -2\sin\theta\bigg\rvert_{\pi/3} = -\sqrt{3}\text{,}\) \(\sin\lrpar{\dfrac{\pi}{3}} = \sqrt{3}{2}\) and \(\cos\lrpar{\dfrac{\pi}{3} = \dfrac{1}{2}}\text{,}\) so
\begin{equation*} \dfrac{dy}{dx} = \dfrac{(-\sqrt{3})\lrpar{\dfrac{\sqrt{3}{2}+2\lrpar{\dfrac{1}{2}}{\dfrac{1}{2}}}}}{(-\sqrt{3})\lrpar{\dfrac{1}{2}} - 2\lrpar{\dfrac{1}{2}}\lrpar{\dfrac{\sqrt{3}}{2}}} = \dfrac{1-3\sqrt{3}}{-3\sqrt{3}} \end{equation*}
Example 10.3.21.
Find the points on the curve \(r = 1-\sin\theta\) where the tangent line is horizontal or vertical.
Solution.
\(\dfrac{dr}{d\theta} = -\cos\theta\text{,}\) so the tangent line is horizontal when
\begin{equation*} -\cos^2\theta - \sin\theta - \sin^2\theta = 0 \end{equation*}
which gives \(\theta = \dfrac{\3pi}{2}\text{.}\) The tangent line is vertical when
\begin{equation*} -\cos\theta\sin\theta - \cos\theta + \sin\theta\cos\theta = 0 \end{equation*}
i.e. when \(\cos\theta = 0\iff \theta = \dfrac{\pi}{2},\dfrac{3\pi}{2}\text{.}\)
This means that the tangent line is never horizontal.

Subsection 10.3.4 After Class Activities

Example 10.3.22.

Sketch the curve \(r = 2(1+\cos\theta)\text{.}\) Then, convert the equation to Cartesian coordinates.
Solution.

Example 10.3.23.

The graph below shows \(r\) as a function of \(\theta\text{,}\) in Cartesian coordinates. Use the graph to sketch the corresponding polar curve.
Solution.

Example 10.3.24.

Find the slope of the tangent line to the polar curve \(r = 2 + \sin 3\theta\) at \(\theta = \dfrac{\pi}{4}\text{.}\)
Solution.
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