Difficulty in plotting polar plot for trigonometric functions

Question

I'm having some difficulty with the following question.

Sketch the graph of $r=\tan(\theta)$,where $-\frac{\pi}{2} < \theta < \frac{\pi}{2}$ and $r=\frac{\sqrt{2}}{2}\csc(\theta)$ and find the area between the two curves.

I'm given the hint that the graph has two vertical asymptotes. This is a polar plot in case that wasn't clear.

I've put this into various software, but I am not really sure how to sketch the curves. I get that in cartesian coordinates, there are vertical asymptotes at $-\frac{\pi}{2}$ and $\frac{\pi}{2}$ but I am not sure how that can help me in plotting the polar curve. Can someone help out here?

For the $\frac{\sqrt{2}}{2} \csc(\theta)$, it's just a horizontal line but I'm not really sure how that's being done either...

I feel like I could get the area if I knew how to plot this but I can't even plot this. Can someone please help out here?

Answer 1

Plotting r=csc(θ) as a horizontal line

One way to plot $r(\theta) = \sqrt{2}/2 \cdot \csc{\theta}$ is by converting to Cartesian coordinates.

If you know the conversion formula $\langle x, y\rangle = \langle r\cos{\theta},\, r\sin{\theta}\rangle$, then you can rewrite:

$$r = A \cdot \csc{\theta} \quad \Longrightarrow\quad r=A\cdot\frac{1}{\sin{\theta}} \quad\Longrightarrow\quad r\sin{\theta} = A \quad \Longrightarrow y=A.$$

Now you can see that for any value of $A$, the polar plot of $r = A\cdot \csc{\theta}$ is a horizontal line of height $A$.

In this specific case, it is a horizontal line of height $y=\sqrt{2}/2$.

Plotting r=tan(θ) as a U shape

After thinking about it for a while, I suggest sketching $r=\tan{\theta}$ like this:

• You know the value of $\tan{\theta}$ at special values of $\theta$. For example, you know that $\tan{0^\circ}=0$, $\tan{45^\circ}=1$, $\tan{-45^\circ} = -1$. You can start by plotting those special points. So:

  • When $\theta=0^\circ$, $r=0$. The graph passes through the origin $\langle 0,0\rangle$.
  • When $\theta=45^\circ$, $r=1$. You can show that graph passes through the Cartesian point $\langle x=\sqrt{2}/2,y=\sqrt{2}/2\rangle$.
  • When $\theta=-45^\circ$, then $r=-1$. You can show that the graph passes through the Cartesian point $\langle x=-\sqrt{2}/2, y=\sqrt{2}/2\rangle$.

• You can use the Cartesian-to-polar conversion formulas again to find the vertical asymptotes. If the equation is $r = \tan{\theta}$, and you know that the cartesian conversion formula is $\langle x, y\rangle = \langle r \cos{\theta}, r\sin{\theta}\rangle$, then you can plug in the formula for $r(\theta)$ to get:

  $$\langle x, y\rangle = \left\langle \frac{\sin{\theta}}{\cos{\theta}} \cdot \cos{\theta}, \frac{\sin{\theta}}{\cos{\theta}}\cdot \sin{\theta}\right\rangle.$$ This simplifies to: $$x=\sin{\theta}, \quad y=\frac{\sin^2{\theta}}{\cos{\theta}}\\~\\\text{\{For this specific function }r(\theta)=\tan{\theta}\}$$

  This formula tells us, for this specific function, what the $x$ and $y$ coordinates of the graph are for every value of $\theta$. And this formula actually tells us a lot about the shape of the graph! Notice:

  1. The $x$ coordinate on this graph is equal to $\sin{\theta}$. This means that the whole graph lies between x=-1 and x=+1. The value of $x$ will approach but never go beyond those values because $\sin$ will never go beyond those values. We've found the two asymptotes, at x=+1 and x=-1.

  2. The $y$ coordinate on this graph is equal to $\sin^2{\theta}/\cos{\theta}$. You know that $\sin^2{\theta}$ is positive because it's a square, and $\cos{\theta}$ is positive as long as $\theta$ is between $-90^\circ$ and $90^\circ$. So the entire $y$ coordinate will always be positive when $\theta$ is in that range. And as $\theta$ gets close to the limits $\pm 90^\circ$, the $y$ coordinate will increase toward infinity.

• This is enough to sketch the graph: plot the three anchor points in the first step, then join them up with a symmetric U shape that increases toward infinity as it approaches the vertical asymptotes on either side at $x=\pm 1$.

Answer 2

It is more convenient to express the two curves in Cartesian coordinates, i.e.

$$y= \frac1{\sqrt2}, \>\>\>\>\> y= \frac{x^2}{\sqrt{1-x^2}}$$ as shown below. Note that they meet at $x=\frac1{\sqrt2}$ and the enclosed area is

$$Area = 2\int_0^{\frac1{\sqrt2} }\left( \frac1{\sqrt2} - \frac{x^2}{\sqrt{1-x^2}}\right)dx=\frac32-\frac\pi4 $$