Taking the graph of \(\displaystyle y=sin(2x)\), we only want the areas in the positive first quadrant because the radius is squared and cannot be negative.

This leaves us with the areas from \(\displaystyle 0\) to \(\displaystyle \frac{\pi }{2}\) and \(\displaystyle \pi\) to \(\displaystyle \frac{3\pi }{2}\). 

Then, when we take the square root of the radius, we get both a positive and negative answer with a maximum and minimum radius of \(\displaystyle \pm 1\).

To draw the graph, the radius is 0 at \(\displaystyle 0\) and traces to 1 at \(\displaystyle \frac{\pi }{4}\). As well, the negative part of the radius starts at 0 and traces to-1 in the opposite quadrant, the third quadrant.

From \(\displaystyle \frac{\pi }4{}\) to \(\displaystyle \frac{\pi }{2}\), the curves are traced from 1 to 0 and -1 to 0 in the third quadrant.

Following this pattern, the graph is redrawn again from the areas included in \(\displaystyle \pi\) to \(\displaystyle 2\pi\).    

\(\displaystyle x^{2} = 3xy + 1\)

\(\displaystyle \left (r \cos \theta \right ) ^{2} = 3\left (r \cos \theta \right )\left (r \sin \theta \right ) + 1\)

\(\displaystyle r ^{2} \cos ^{2} \theta = 3 r^{2} \cos \theta\; \sin \theta \right ) + 1\)

\(\displaystyle r ^{2} \cos ^{2} \theta - 3 r^{2} \cos \theta\; \sin \theta \right ) = 1\)

\(\displaystyle r ^{2} \left ( \ \cos ^{2} \theta - 3 \cos \theta\; \sin \theta \right ) = 1\)

\(\displaystyle r ^{2} = \frac{1}{\cos ^{2} \theta - 3 \cos \theta\; \sin \theta }\)

\(\displaystyle r =\sqrt{ \frac{1}{\cos ^{2} \theta - 3 \cos \theta\; \sin \theta }}\)

To calculate the polar coordinate, use

\(\displaystyle r=\sqrt{x^2+y^2}\)

\(\displaystyle r=\sqrt{29}\)

\(\displaystyle \theta=tan^{-1}\left(\frac{y}{x}\right)=68\)

However, keep track of the angle here. 68 degree is the mathematical equivalent of the expression, but we know the point (-2,-5) is in the 3rd quadrant, so we have to add 180 to it to get 248.

Some calculators might already have provided you with the correct answer.

\(\displaystyle \theta=248\).

We can convert from rectangular form to polar form by using the following identities: \(\displaystyle y=r\sin \theta\) and \(\displaystyle x=r\cos \theta\). Given \(\displaystyle y=2x^{2}\), then \(\displaystyle r\sin \theta=2(r\cos \theta )^{2}=2r^{2}\cos^{2} \theta\).

\(\displaystyle r\sin \theta=2(r\cos \theta )^{2}=2r^{2}\cos^{2} \theta\). Dividing both sides by \(\displaystyle r\cos \theta\),

\(\displaystyle \tan \theta=2r\cos \theta\)

\(\displaystyle \frac{\tan \theta}{\cos \theta}=2r\)

\(\displaystyle \tan \theta}{\sec \theta=2r\)

\(\displaystyle \frac{1}{2}\tan \theta}{\sec \theta=r\)

We can convert from rectangular form to polar form by using the following identities: \(\displaystyle y=r\sin \theta\) and \(\displaystyle x=r\cos \theta\). Given \(\displaystyle y=\frac{6}{x}\), then \(\displaystyle r\sin \theta=\frac{6}{r\cos \theta}\). Multiplying both sides by \(\displaystyle r\cos \theta\),

\(\displaystyle r^{2}\sin \theta cos \theta=6\)

\(\displaystyle r^{2}=\frac{6}{\sin \theta cos \theta}\)

\(\displaystyle r=\sqrt\frac{6}{\sin \theta cos \theta}\)

The following formulas were used to convert the function from polar to Cartestian coordinates:

\(\displaystyle x=r\cos(\theta), y=r\sin(\theta), x^2+y^2=r^2\)

Note that the last formula is a manipulation of a trignometric identity.

Simply replace these with x and y in the original function.

\(\displaystyle f(x)=2x^2+5x+2+2y^2\)

\(\displaystyle f(x)=(x^2+y^2)2+5x+2\)

\(\displaystyle f(x)=2r^2+5rcos(\theta)+2\)

We can convert from rectangular to polar form by using the following trigonometric identities: \(\displaystyle y=r\sin\theta\) and \(\displaystyle x=r\cos\theta\). Given \(\displaystyle y=-\frac{1}{2}x^{2}\), then:

\(\displaystyle r\sin\theta=-\frac{1}{2}(r\cos \theta)^{2}\)

\(\displaystyle r\sin\theta=-\frac{1}{2}r^{2}\cos^{2} \theta\)

 Dividing both sides by \(\displaystyle r\cos\theta\), we get:

\(\displaystyle \tan \theta=-\frac{1}{2}r\cos\theta\)

\(\displaystyle \frac{\tan \theta}{\cos\theta}=-\frac{1}{2}r\)

\(\displaystyle \tan \theta\sec\theta=-\frac{1}{2}r\)

\(\displaystyle r=-2\tan \theta\sec\theta\)

We can convert from rectangular to polar form by using the following trigonometric identities: and . Given \(\displaystyle y=-7x^{2}\), then:

\(\displaystyle r\sin\theta=-7(r\cos\theta)^{2}\)

\(\displaystyle r\sin\theta=-7r^{2}\cos^{2}\theta\)

Dividing both sides by , we get:

\(\displaystyle \tan\theta=-7r\cos\theta\)

\(\displaystyle \frac{\tan\theta}{\cos\theta}=-7r\)

\(\displaystyle \tan\theta\sec\theta=-7r\)

\(\displaystyle r=-\frac{1}{7}\tan\theta\sec\theta\)

We can convert from rectangular to polar form by using the following trigonometric identities: and . Given \(\displaystyle y=3-4x\), then:

\(\displaystyle r\sin\theta=3-4(r\cos\theta)\)

\(\displaystyle r\sin\theta+4(r\cos\theta)=3\)

\(\displaystyle r(\sin\theta+4\cos\theta)=3\)

\(\displaystyle r=\frac{3}{\sin\theta+4\cos\theta}\)

We can convert from rectangular to polar form by using the following trigonometric identities: and . Given \(\displaystyle y=-x^{2}\), then:

\(\displaystyle r\sin\theta=-(r\cos\theta)^{2}\)

Dividing both sides by , we get:

\(\displaystyle \tan\theta=-r\cos\theta\)

\(\displaystyle \frac{\tan\theta}{\cos\theta}=-r\)

\(\displaystyle \tan\theta\sec\theta=-r\)

\(\displaystyle r=-\tan\theta\sec\theta\)