W15 - Examples

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:

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

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

Complex multiplication

Complex multiplication

Compute the products:

(a) $(1-i)(4-7i)$ (b) $(2+5i)(2-5i)$

Solution

(a) $(1-i)(4-7i)$

Expand:

$$(1-i)(4-7i)\gg \gg 4-7i-4i+7i^2$$

Simplify $i^2$:

$$\gg \gg 4-7i-4i+7(-1)\gg \gg -3-11i$$

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(b) $(2+5i)(2-5i)$

Expand:

$$(2+5i)(2-5i)\gg \gg 4-10i+10i-25i^2$$

Simplify $i^2$:

$$\gg \gg 4-10i+10i-25(-1)\gg \gg 29$$ Link to original

Complex division

Complex division

Compute the following divisions of complex numbers:

(a) $\frac{1}{-3+i}$ (b) $\frac{1}{i}$ (c) $\frac{1}{7i}$ (d) $\frac{2+5i}{2-5i}$

Solution

(a) $\frac{1}{-3+i}$

Conjugate is $-3-i$:

$$\frac{1}{-3+i}\gg \gg \frac{1}{-3+i}\cdot \frac{-3-i}{-3-i}$$

Simplify:

$$\gg \gg \frac{-3-i}{9+1}\gg \gg \frac{-3}{10}+\frac{-1}{10}i$$

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(b) $\frac{1}{i}$

Conjugate is $-i$:

$$\frac{1}{i}\gg \gg \frac{1}{i}\cdot \frac{-i}{-i}\gg \gg -i$$

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(c) $\frac{1}{7i}$

Factor out the $1/7$:

$$\frac{1}{7i}\gg \gg \frac{1}{7}\cdot \frac{1}{i}$$

Use $\frac{1}{i}=-i$:

$$\gg \gg \frac{1}{7}\cdot (-i)\gg \gg \frac{-1}{7}i$$

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(d) $\frac{2+5i}{2-5i}$

Denominator conjugate is $2+5i$:

$$\frac{2+5i}{2-5i}\gg \gg \frac{2+5i}{2-5i}\cdot \frac{2+5i}{2+5i}$$

Simplify:

$$\gg \gg \frac{4+20i+25i^2}{4+25}\gg \gg \frac{-21}{29}+\frac{20}{29}i$$Link to original