Chapter 8, Problem 1PRE

For Exercises 1-8,

1. Match the polar or rectangular equation on the left with the letter of the rectangular or polarequation on the right.
2. Describe the graph.
3. determine if the graph is symmetric with respect to the origin (pole), x-axis (polar axis), or
   y-axis $\left(\text{line }\theta =\frac{\pi }{2}\right).$

1. $x^2+y^2=144$	A. $2x+y=4$
2. $x=-3$	B. $r=-10$
3. $r=\frac{4}{2\cos \theta +\sin \theta }$	C. $r=12$
4. $r=-6\cos \theta$	D. $\theta =\frac{2\pi }{3}$
5. $y=-\sqrt{3x}$	E. $y=5$
6. $x^2+{\left(y+5\right)}^2=25$	F. $r=-3\sec \theta$
7. $r=5\csc \theta$	G. ${\left(x-1\right)}^2+{\left(y-1\right)}^2=2$
8. $r=2\sin \theta +2\cos \theta$	H. ${\left(x+3\right)}^2+y^2=9$

To determine

The correct match either polar or rectangular equation for $x^2+{\left(y\right)}^2=144$ .

Answer to Problem 1PRE

The correct match for $x^2+{\left(y\right)}^2=144$ is $r=12$ .

Explanation of Solution

Given:The list polar and rectangular equation is given as:

1. $2x+y=4$ .
2. $r=-10\sin \theta$ .
3. $r=12$ .
4. $\theta =\frac{2\pi }{3}$ .
5. $y=5$ .
6. $r=-3\sec \theta$ .
7. ${\left(x-1\right)}^2+{\left(y-1\right)}^2=2$ .
8. ${\left(x+3\right)}^2+{\left(y\right)}^2=9$ .

Calculation:

We know that: $\left({\cos }^2\theta +{\sin }^2\theta \right)=1$

The given equation as: $x^2+{\left(y\right)}^2=144$

Put the values as: $x=r\cos \theta$ and $y=r\sin \theta$

Hence, eq. written as:

  $\begin{array}{l}{\left(r\cos \theta \right)}^2+{\left(r\sin \theta \right)}^2=25\\ r^2{\cos }^2\theta +r^2{\sin }^2\theta =144\\ r^2\left({\cos }^2\theta +{\sin }^2\theta \right)=144\\ r^2\left(1\right)=144\\ r=12\end{array}$

Hence, the correct match from given options is C.

Conclusion:

The correct match for $x^2+{\left(y\right)}^2=144$ is C: $r=12$

To determine

To describe the graph of $x^2+{\left(y\right)}^2=144$

Answer to Problem 1PRE

The equation $x^2+{\left(y\right)}^2=144$ represents the circle having points $\left(x,y\right)=\left(0,0\right)$ or origin and radius is $12$ .

Explanation of Solution

Given:

The given equation as: $x^2+{\left(y\right)}^2=144$

Graph:

  

Calculation:

We know that equation as: $x^2+{\left(y\right)}^2=144$

On comparing with standard equation form ${\left(x+a\right)}^2+{\left(y+b\right)}^2=r^2$

We get that: $a=0,b=0,r=5$ .

Hence, the equation $x^2+{\left(y\right)}^2=144$ represents the circle having points $\left(x,y\right)=\left(0,0\right)$ and radius is $12$ .

Conclusion:

The graph is described as above.

To determine

To check:

The symmetry of the equation $x^2+{\left(y\right)}^2=144$ with $x$ -axis and $y$ -axis.

Answer to Problem 1PRE

The function $x^2+{\left(y\right)}^2=144$ is symmetric with $x$ -axis , $y$ -axis and origin.

Explanation of Solution

Given: The given equation as: $x^2+{\left(y\right)}^2=144$

Calculation:

The given equation as: $x^2+{\left(y\right)}^2=144$

To check symmetry, we have to put $x=-x$ and $y=-y$ one by one.

Let’s put $x=-x$ in given equation:

  $\begin{array}{l}{x}^2+{\left(y\right)}^2=144\\ {\left(-x\right)}^2+{\left(y\right)}^2=144\\ x^2+{\left(y\right)}^2=144\end{array}$

We got that, after putting $x=-x$ , the result produces same as an equivalent equation, i.e. the equation is symmetric with $y$ -axis.

Let’s put $y=-y$ in given equation:

  $\begin{array}{l}{x}^2+y^2=144\\ {\left(x\right)}^2+{\left(-y\right)}^2=144\\ x^2+y^2=144\end{array}$

We got that, after putting $y=-y$ , the result produces same as an equivalent equation, i.e. the equation is symmetric with $x$ -axis and origin.

Conclusion:

The function $x^2+{\left(y+5\right)}^2=25$ is symmetricwith $x$ -axis , $y$ -axis and origin.