Chapter 6, Problem 73RE

Solution

a)

To explain: The equation $r=a\sec \theta$ is a polar form for the line $x=a$ .

  $r=a\sec \theta$ is a polar form for the line $x=a$ .

Given information:

The polar form for the line $x=a$ is $r=a\sec \theta$ .

Formula used:

Let $\left(r,\theta \right)$ and $\left(x,y\right)$ are the polar and rectangular coordinates of the point P.

Then,

  $\begin{array}{c}x=r\cos \theta \\ y=r\sin \theta \\ r^2=x^2+y^2\\ \tan \theta =\frac{y}{x}\end{array}$

Calculation:

  $r=a\sec \theta$

Further simplify the equation.

  $\begin{array}{l}r=a\sec \theta \\ a=\frac{r}{\sec \theta }\\ a=r\cos \theta \end{array}$

It is known that $x=r\cos \theta$ . Substitute $x$ for $r\cos \theta$.

  $a=x$

Therefore, $r=a\sec \theta$ is a polar form for the line $x=a$ .

b)

To explain: The equation $r=b\text{cosec}\theta$ is a polar form for the line $y=b$ .

  $r=b\text{cosec}\theta$ is a polar form for the line $y=b$ .

Given information:

The polar form for the line $y=b$ is $r=b\text{cosec}\theta$ .

Formula used:

Let $\left(r,\theta \right)$ and $\left(x,y\right)$ are the polar and rectangular coordinates of the point P.

Then,

  $\begin{array}{c}x=r\cos \theta \\ y=r\sin \theta \\ r^2=x^2+y^2\\ \tan \theta =\frac{y}{x}\end{array}$

Calculation:

  $r=b\text{cosec}\theta$

Further simplify the equation.

  $\begin{array}{l}b=\frac{r}{\text{cosec}\theta }\\ b=r\sin \theta \end{array}$

It is known that $y=r\sin \theta$ . Substitute $y$ for $r\sin \theta$ .

  $b=y$

Therefore, $r=b\text{cosec}\theta$ is a polar form for the line $y=b$ .

c)

To prove: $r=\frac{b}{\sin \theta -m\cos \theta }$ and find the domain of $r$ .

It is proved that $r=\frac{b}{\sin \theta -m\cos \theta }$ for $y=mx+b$ and the domain of $r$ is $\left(-\infty ,\infty \right)$ .

Given information:

The given expression is $y=mx+b$ .

Formula used:

Let $\left(r,\theta \right)$ and $\left(x,y\right)$ are the polar and rectangular coordinates of the point P.

Then,

  $\begin{array}{c}x=r\cos \theta \\ y=r\sin \theta \\ r^2=x^2+y^2\\ \tan \theta =\frac{y}{x}\end{array}$

Calculation:

  $y=mx+b$

Substitute $r\cos \theta$ for $x$ and $r\sin \theta$ for $y$ .

  $\begin{array}{c}r\sin \theta =m\left(r\cos \theta \right)+b\\ r\sin \theta -m\left(r\cos \theta \right)=b\\ r\left(\sin \theta -m\cos \theta \right)=b\\ r=\frac{b}{\left(\sin \theta -m\cos \theta \right)}\end{array}$

The domain of $r$ is $\left(-\infty ,\infty \right)$ .

Therefore, it is proved that $r=\frac{b}{\sin \theta -m\cos \theta }$ for $y=mx+b$ and the domain of $r$ is $\left(-\infty ,\infty \right)$ .

d)

To graph:the polar equation of the line $y=2x+3$ .

It can be observed that the graph of the polar equation $r=\frac{3}{\left(\sin \theta -2\cos \theta \right)}$ is a straight line.

Given information:

  $y=2x+3$

Formula used:

Let $\left(r,\theta \right)$ and $\left(x,y\right)$ are the polar and rectangular coordinates of the point P.

Then,

  $\begin{array}{c}x=r\cos \theta \\ y=r\sin \theta \\ r^2=x^2+y^2\\ \tan \theta =\frac{y}{x}\end{array}$

Calculation:

  $y=2x+3$

Substitute $r\cos \theta$ for $x$ and $r\sin \theta$ for $y$ .

  $\begin{array}{c}r\sin \theta =2\left(r\cos \theta \right)+3\\ r\sin \theta -2\left(r\cos \theta \right)=3\\ r\left(\sin \theta -2\cos \theta \right)=3\\ r=\frac{3}{\left(\sin \theta -2\cos \theta \right)}\end{array}$

Plot the graph of the polar equation.

  

Therefore, it can be observed that the graph of the polar equation $r=\frac{3}{\left(\sin \theta -2\cos \theta \right)}$ is a straight line.