Chapter 9, Problem 47CLT

To determine

Find the equation in the polar form.

Answer to Problem 47CLT

The equation in the polar form is $r=-\frac{1}{4(\cos \theta +\sin \theta )}$ .

Explanation of Solution

Given:

It is given in the question that the rectangular equation is $4x+4y+1=0$ .

Concept Used:

In this, use the concept of the relationship $x^2+y^2=r^2;x=r\cos \theta ,y=r\sin \theta$ .

Calculation:

Here, the equation is $4x+4y+1=0$ .

It has to be use the relationship and put the value of x and y in the equation:

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

Substitute to the rectangular equation:

  $\begin{array}{l}4r\cos \theta +4r\sin \theta +1=0\\ 4r\cos \theta +4r\sin \theta =-1\end{array}$

Factor out $4r$ :

  $4r(\cos \theta +\sin \theta )=-1$

Divide by $4(\cos \theta +\sin \theta )$ :

  $r=-\frac{1}{4(\cos \theta +\sin \theta )}$

Conclusion:

The equation is $r=-\frac{1}{4(\cos \theta +\sin \theta )}$ .