W14 - Notes

More polar curves

01 Theory - Polar limaçons

Theory 2

To draw the polar graph of some function, it can help to first draw the Cartesian graph of the function. (In other words, set $y=r$ and $x=\theta$, and draw the usual graph.) By tracing through the points on the Cartesian graph, one can visualize the trajectory of the polar graph.

This Cartesian graph may be called a graphing tool for the polar graph.

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A limaçon is the polar graph of $r(\theta )=a+b\cos \theta$.

The shape of a limaçon is determined by the value of $c={\displaystyle \frac{b}{a}}$. Any limaçon can be rescaled to have this form:

$$r=1+c\cos \theta$$

$c=0$: Limaçon satisfying $r(\theta )=1$: unit circle.

$c=0.5$: Limaçon satisfying $r(\theta )=2+\cos \theta$: ‘outer loop’ circle with ‘flat spot’, not quite a ‘dimple’:

$c=1$: Limaçon satisfying $r(\theta )=1+\cos \theta$: ‘cardioid’ $=$ ‘outer loop’ circle with ‘dimple’ that creates a cusp:

$c=2$: Limaçon satisfying $r(\theta )=1+2\cos \theta$: ‘dimple’ pushes past cusp to create ‘inner loop’:

$c=\infty$: Limaçon satisfying $r(\theta )=\cos \theta$: ‘inner loop’ only, no outer loop exists:

$c=2$: Limaçon satisfying $r(\theta )=1+2\sin \theta$: ‘inner loop’ and ‘outer loop’ and rotated $\circlearrowleft {90}^{\circ }$:

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Transitions between limaçon types, $r(\theta )=1+c\sin \theta$:

Notice the transition points at $|c|=0.5$ and $|c|=1$:

The flat spot occurs when $c=\pm 0.5$

• Smaller $c$ gives convex shape

The cusp occurs when $c=\pm 1$

• Smaller $c$ gives dimple (assuming $|c|>0.5$)
• Larger $c$ gives inner loop

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02 Theory - Polar roses

Theory 3

Roses are polar graphs of this form:

$$r=\cos (\theta ),r=\sin (2\theta ),r=\sin (3\theta ),r=\cos (4\theta )$$

The pattern of petals:

• $n=2k$ (even): obtain $2n$ petals
  • These petals traversed once
• $n=2k+1$ (odd): obtain $n$ petals
  • These petals traversed twice
• Either way: total-petal-traversals: always $2n$

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03 Illustration

Example - Finding vertical tangents to a limaçon

Finding vertical tangents to a limaçon

Let us find the vertical tangents to the limaçon (the cardioid) given by $r=1+\sin \theta$.

Solution

(1) Convert to Cartesian parametric using $x=r\cos \theta$ and $y=r\sin \theta$:

$$r(\theta )=1+\sin \theta \gg \gg (x,y)=((1+\sin \theta )\cos \theta ,(1+\sin \theta )\sin \theta )$$

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(2) Compute $x^{\prime }$ and $y^{\prime }$:

$$\begin{array}{c}(x^{\prime }(\theta ),y^{\prime }(\theta ))\\ \\ \gg \gg (\cos \theta \cos \theta +(1+\sin \theta )(-\sin \theta ),\cos \theta \sin \theta +(1+\sin \theta )\cos \theta )\\ \\ \gg \gg ({\cos }^2\theta -{\sin }^2\theta -\sin \theta ,\cos \theta (1+2\sin \theta ))\end{array}$$

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(3) The vertical tangents occur when $x^{\prime }(\theta )=0$. We must double check that $y^{\prime }(\theta )\ne 0$ at these points.

$$\begin{array}{c}{x}^{\prime }(\theta )=0\gg \gg {\cos }^2\theta -{\sin }^2\theta -\sin \theta =0\\ \\ \gg \gg (1-{\sin }^2\theta )-{\sin }^2\theta -\sin \theta =0\end{array}$$

Substitute $A=\sin \theta$ and observe quadratic:

$$\gg \gg 1-2A^2-A=0\gg \gg 2A^2+A-1=0$$

Solve:

$$\begin{array}{c}A=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\gg \gg \\ \\ \frac{-1\pm \sqrt{1-4\cdot 2\cdot (-1)}}{2\cdot 2}\gg \gg \frac{1}{2},-1\end{array}$$

Then find $\theta$:

$$\begin{array}{c}A=\sin \theta \gg \gg \sin \theta =\frac{1}{2},-1\\ \\ \gg \gg \theta =\frac{\pi }{6},\frac{5\pi }{6}(\text{for }1/2)\text{and}\theta =\frac{3\pi }{2}(\text{for }-1)\end{array}$$

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(4) Compute the points. In polar coordinates:

$$\begin{array}{c}(r,\theta )=(1+\sin \theta ,\theta ){|}_{\theta =\frac{\pi }{6},\frac{5\pi }{6},\frac{3\pi }{2}}\\ \\ \gg \gg (\frac{3}{2},\frac{\pi }{6}),(\frac{3}{2},\frac{5\pi }{6}),(0,\frac{3\pi }{2})\end{array}$$

In Cartesian coordinates:

At $\theta =\frac{\pi }{6}$:

$$\begin{array}{c}(x,y){|}_{\theta =\frac{\pi }{6}}\gg \gg ((1+\sin \theta )\cos \theta ,(1+\sin \theta )\sin \theta ){|}_{\theta =\frac{\pi }{6}}\\ \\ \gg \gg ((1+\frac{1}{2})\frac{\sqrt{3}}{2},(1+\frac{1}{2})\frac{1}{2})\gg \gg (\frac{3\sqrt{3}}{4},\frac{3}{4})\end{array}$$

At $\theta =\frac{5\pi }{6}$:

$$\begin{array}{c}(x,y){|}_{\theta =\frac{5\pi }{6}}\gg \gg ((1+\sin \theta )\cos \theta ,(1+\sin \theta )\sin \theta ){|}_{\theta =\frac{5\pi }{6}}\\ \\ \gg \gg ((1+\frac{1}{2})\frac{-\sqrt{3}}{2},(1+\frac{1}{2})\frac{1}{2})\gg \gg (-\frac{3\sqrt{3}}{4},\frac{3}{4})\end{array}$$

At $\theta =\frac{3\pi }{2}$:

$$\begin{array}{c}(x,y){|}_{\theta =\frac{3\pi }{2}}\gg \gg ((1+\sin \theta )\cos \theta ,(1+\sin \theta )\sin \theta ){|}_{\theta =\frac{3\pi }{2}}\\ \\ \gg \gg ((1-1)\cdot 0,(1-1)\cdot (-1))\gg \gg (\mathrm{0,0})\end{array}$$

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(5) Correction: $(\mathrm{0,0})$ is a cusp!

The point $(\mathrm{0,0})$ at $\theta =\frac{3\pi }{2}$ is on the cardioid, but the curve is not smooth there, this is a cusp.

Still, the left- and right-sided tangents exists and are equal, so in a certain sense we could say the curve has vertical tangent at $\theta =\frac{3\pi }{2}$.

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Calculus with polar curves

04 Theory - Polar tangent lines, arclength

Theory 1

Polar arclength formula

The arclength of the polar graph of $r(\theta )$, for $\theta \in [\alpha ,\beta ]$:

$$L={\int }_{\alpha }^{\beta }\sqrt{r^{\prime }{(\theta )}^2+r{(\theta )}^2}d\theta$$

To derive this formula, convert to Cartesian with parameter $\theta$:

$$r=r(\theta )\gg \gg (x,y)=(r\cos \theta ,r\sin \theta )$$

From here you can apply the familiar arclength formula with $\theta$ in the place of $t$.

Extra - Derivation of polar arclength formula

Let $r=r(\theta )$ and convert to parametric Cartesian, so:

$$\begin{array}{cc}x(\theta )& =r(\theta )\cos \theta \\ y(\theta )& =r(\theta )\sin \theta \end{array}$$

Then:

$$\begin{array}{c}ds=\sqrt{{(x^{\prime })}^2+{(y^{\prime })}^2}d\theta \\ \\ x^{\prime }={(r\cos \theta )}^{\prime }\gg \gg r^{\prime }\cos \theta -r\sin \theta \\ y^{\prime }={(r\sin \theta )}^{\prime }\gg \gg r^{\prime }\sin \theta +r\cos \theta \end{array}$$

Therefore:

$$\begin{array}{cc}{(x^{\prime })}^2+{(y^{\prime })}^2\gg \gg & \phantom{,+}{r}^{\prime 2}{\cos }^2\theta -2r{r}^{\prime }\cos \theta \sin \theta +r^2{\sin }^2\theta \\ & +r^{\prime 2}{\sin }^2\theta +2r{r}^{\prime }\sin \theta \cos \theta +r^2{\cos }^2\theta \\ & \\ & =r^{\prime 2}+r^2\end{array}$$

Therefore:

$$ds=\sqrt{{(x^{\prime })}^2+{(y^{\prime })}^2}d\theta \gg \gg \sqrt{r^{\prime 2}+r^2}d\theta$$

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05 Illustration

Example - Length of the inner loop

Length of the inner loop

Consider the limaçon given by $r(\theta )=\frac{1}{2}+\cos \theta$.

How long is the inner loop? Set up an integral for this quantity.

Solution

The inner loop is traced by the moving point when $\frac{2\pi }{3}\le \theta \le \frac{4\pi }{3}$. This can be seen from the graph:

Therefore the length of the inner loop is given by this integral:

$$L={\int }_{2\pi /3}^{4\pi /3}\sqrt{{(-\sin \theta )}^2+{(\frac{1}{2}+\cos \theta )}^2}d\theta \gg \gg {\int }_{2\pi /3}^{4\pi /3}\sqrt{5/4+\cos \theta }d\theta$$ Link to original

06 Theory - Polar area

Theory 2

Sectorial area from polar curve

$$A={\int }_{\alpha }^{\beta }\frac{1}{2}r{(\theta )}^{2}d\theta$$

The “area under the curve” concept for graphs of functions in Cartesian coordinates translates to a “sectorial area” concept for polar graphs. To compute this area using an integral, we divide the region into Riemann sums of small sector slices.

To obtain a formula for the whole area, we need a formula for the area of each sector slice.

Area of sector slice

Let us verify that the area of a sector slice is ${\displaystyle \frac{1}{2}{r}^2\theta }$.

Take the angle $\theta$ in radians and divide by $2\pi$ to get the fraction of the whole disk.

Then multiply this fraction by $\pi r^2$ (whole disk area) to get the area of the sector slice.

$$\left(\frac{\theta }{2\pi }\right)\left(\pi r^2\right)\gg \gg \frac{1}{2}{r}^2\theta$$

Now use $d\theta$ and $r(\theta )$ for an infinitesimal sector slice, and integrate these to get the total area formula:

$$A={\int }_{\alpha }^{\beta }\frac{1}{2}r{(\theta )}^{2}d\theta$$

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One easily verifies this formula for a circle.

Let $r(\theta )=R$ be a constant. Then:

$$\text{Area of circle}={\int }_0^{2\pi }\frac{1}{2}{R}^{2}d\theta \gg \gg \frac{1}{2}{R}^2\theta {|}_0^{2\pi }\gg \gg R^2\pi$$

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The sectorial area between curves:

Sectorial area between polar curves

$$A={\int }_{\alpha }^{\beta }\frac{1}{2}(r_1(\theta {)}^2-r_0(\theta {)}^2)d\theta$$

Subtract after squaring, not before!

This aspect is not similar to the Cartesian version: ${\displaystyle \int f-gdx}$

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07 Illustration

Area between circle and limaçon

Area between circle and limaçon

Find the area of the region enclosed between the circle $r_0(\theta )=1$ and the limaçon $r_1(\theta )=1+\cos \theta$.

Solution

First draw the region:

The two curves intersect at $\theta =\pm \frac{\pi }{2}$. Therefore the area enclosed is given by integrating over $-\frac{\pi }{2}\le \theta \le +\frac{\pi }{2}$:

$$\begin{array}{c}A={\int }_{\alpha }^{\beta }\frac{1}{2}(r_1^2-r_0^2)d\theta \gg \gg {\int }_{-\pi /2}^{\pi /2}\frac{1}{2}((1+\cos \theta {)}^2-1^2)d\theta \\ \\ \gg \gg \frac{1}{2}{\int }_{-\pi /2}^{\pi /2}2\cos \theta +{\cos }^2\theta d\theta \gg \gg {\int }_{-\pi /2}^{\pi /2}\cos \theta +\frac{1}{4}(1+\cos (2\theta ))d\theta \\ \\ \gg \gg {\sin \theta +\frac{\theta }{4}+\frac{1}{8}\sin (2\theta )|}_{-\pi /2}^{\pi /2}\gg \gg 2+\frac{\pi }{4}\end{array}$$ Link to original

Area of small loops

Area of small loops

Consider the following polar graph of $r(\theta )=1+2\cos (4\theta )$:

Find the area of the shaded region.

Solution

Find bounds for one small loop. Lower left loop occurs first. This loop is when $1+2\cos (4\theta )\le 0$.

$$\begin{array}{c}1+2\cos (4\theta )\le 0\gg \gg \cos (4\theta )\le -\frac{1}{2}\\ \\ \gg \gg \frac{2\pi }{3}\le 4\theta \le \frac{4\pi }{3}\gg \gg \frac{\pi }{6}\le \theta \le \frac{\pi }{3}\end{array}$$

Now set up area integral:

$$\begin{array}{c}A=4{\int }_{\alpha }^{\beta }\frac{1}{2}r(\theta {)}^{2}d\theta \gg \gg 4{\int }_{\pi /6}^{\pi /3}\frac{1}{2}(1+2\cos (4\theta ){)}^{2}d\theta \\ \\ \gg \gg 2{\int }_{\pi /6}^{\pi /3}1+4\cos (4\theta )+4{\cos }^2(4\theta )d\theta \end{array}$$

Power-to-frequency conversion: ${\cos }^{2}A⇝\frac{1}{2}(1+\cos (2A))$ with $A=4\theta$:

$$\begin{array}{c}\gg \gg 2{\int }_{\pi /6}^{\pi /3}1+4\cos (4\theta )+4\cdot \frac{1}{2}(1+\cos (8\theta ))d\theta \\ \\ \gg \gg {6\theta +2\sin (4\theta )+\frac{1}{4}\sin (8\theta )|}_{\pi /6}^{\pi /3}\gg \gg \pi -\frac{3\sqrt{3}}{2}\end{array}$$ Link to original

Overlap area of circles

Overlap area of circles

Compute the area of the overlap between crossing circles. For concreteness, suppose one of the circles is given by $r(\theta )=\sin \theta$ and the other is given by $r(\theta )=\cos \theta$.

Solution

Drawing of the overlap:

Notice: total overlap area = $2\times$ area of red region. Bounds for red region: $0\le \theta \le \frac{\pi }{4}$.

Area formula applied to $r(\theta )=\sin \theta$:

$$A=2{\int }_{\alpha }^{\beta }\frac{1}{2}r{(\theta )}^{2}d\theta \gg \gg 2{\int }_0^{\pi /4}\frac{1}{2}{\sin }^2\theta d\theta$$

Power-to-frequency: ${\sin }^2\theta ⇝\frac{1}{2}(1-\cos (2\theta ))$:

$$\begin{array}{c}\gg \gg 2{\int }_0^{\pi /4}\frac{1}{4}(1-\cos (2\theta ))d\theta \\ \\ \gg \gg \frac{2}{4}\theta -{\frac{2}{8}\sin (2\theta )|}_0^{\pi /4}\gg \gg \frac{\pi }{8}-\frac{1}{4}\end{array}$$Link to original