Chapter H.1, Problem 48E

To determine

To match: The polar equations with the graphs labeled I − VI

Answer to Problem 48E

The matching is $\left(a\right)\to \text{V},\left(b\right)\to \text{II},\left(c\right)\to \text{VI},\left(d\right)\to \text{III},\left(e\right)\to \text{I},\left(f\right)\to \text{IV}$

Explanation of Solution

Given information:

Polar equations are given as,

  $\begin{array}{l}\left(a\right)r=\sqrt{\theta },0\le \theta \le 16\pi \left(b\right)r={\theta }^2,0\le \theta \le 16\pi \left(c\right)r=\cos \left(\frac{\theta }{3}\right)\\ \left(d\right)r=1+2\cos \theta \left(e\right)r=2+\sin 3\theta \left(f\right)r=1+2\sin 3\theta \end{array}$

And graphs are given as,

  

Graph of $\left(a\right)r=\sqrt{\theta },0\le \theta \le 16\pi$ :

Since, the graph $r=\sqrt{\theta }$ is a spiral and $16\pi =8\left(2\pi \right)$ that is 8 revolutions.

If $\theta =2\pi$ then $r=\sqrt{2\pi }$ and if $\theta =16\pi$ then $r=4\sqrt{\pi }$

Then the difference between the origin to the point at which the last revolution that intersects the polar axis is $\frac{4\sqrt{\pi }}{\sqrt{2\pi }}\approx 3$ times the distance from the origin to the point at which the first revolution that intersects the polar axis. Hence, this is graph V.

  

Graph of $\left(b\right)r={\theta }^2,0\le \theta \le 16\pi$ :

Since, the graph $r={\theta }^2$ is a spiral and $16\pi =8\left(2\pi \right)$ that is 8 revolutions.

If $\theta =2\pi$ then $r={\left(2\pi \right)}^2\approx 4{\pi }^2$ and if $\theta =16\pi$ then $r={\left(16\pi \right)}^2=256{\pi }^2$

Then the difference between the origin to the point at which the last revolution that intersects the polar axis is $\frac{256{\pi }^2}{4{\pi }^2}=64$ times the distance from the origin to the point at which the first revolution that intersects the polar axis. Hence, this is graph II.

  

Graph of $\left(c\right)r=\cos \left(\frac{\theta }{3}\right)$ :

Since, the graph $r=\cos \left(\frac{\theta }{3}\right)$ is a limacon with inner loop. Hence, this is graph VI.

  

Graph of $\left(d\right)r=1+2\cos \theta$ :

Since, the graph $r=1+2\cos \theta$ is a limacon and $r=1$ , then inner loop will be there. Hence, this is graph VI.

  

Graph of $\left(e\right)r=2+\sin 3\theta$ :

Since, the graph $r=2+\sin 3\theta$ is a limacon with 3 leaves. Hence, this is graph I.

  

Graph of $\left(f\right)r=1+2\sin 3\theta$ :

Since, the graph $r=1+2\sin 3\theta$ is a limacon with $2\times 3=6$ leaves. Also $1<2$ , then 3 inner leaves are there. Hence, this is graph IV.

  

Therefore, the correct matching is $\left(a\right)\to \text{V},\left(b\right)\to \text{II},\left(c\right)\to \text{VI},\left(d\right)\to \text{III},\left(e\right)\to \text{I},\left(f\right)\to \text{IV}$