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

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

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

\(\displaystyle \left ( r^{2} \right )^{\frac{3}{2}} =\left ( r \sin \theta \right ) ^{2}\)

\(\displaystyle \left (x^{2} +y^{2} \right )^{\frac{3}{2}} =y ^{2}\)

\(\displaystyle \left [ \left (x^{2} +y^{2} \right )^{\frac{3}{2}} \right ] ^{2} =\left ( y ^{2} \right ) ^{2}\)

\(\displaystyle \left (x^{2} +y^{2} \right )^{3} = y ^{4}\)

or \(\displaystyle y ^{4} = \left (x^{2} +y^{2} \right )^{3}\)

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

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

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

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

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

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

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

\(\displaystyle r ^{2} + 1 = \cos \theta\)

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

\(\displaystyle \sqrt{x^{2} + y ^{2}} \left ( {x^{2} + y ^{2}} +1 \right ) = x\)

\(\displaystyle \sqrt{x^{2} + y ^{2}} =\frac{ x}{{x^{2} + y ^{2}} +1}\)

This line has slope \(\displaystyle \frac{4-0}{0- (-6)} = \frac{2}{3}\) and \(\displaystyle y\)-intercept \(\displaystyle (0.4)\), so its Cartesian equation is \(\displaystyle y = \frac{2}{3}x + 4\).

By substituting, we can rewrite this:

\(\displaystyle r \sin \theta = \frac{2}{3}r \cos \theta + 4\)

\(\displaystyle r \sin \theta - \frac{2}{3}r \cos \theta = 4\)

\(\displaystyle 3 r \sin \theta - 2 r \cos \theta = 12\)

\(\displaystyle r \left ( 3 \sin \theta - 2 \cos \theta \right )= 12\)

\(\displaystyle r = \frac{12}{3 \sin \theta - 2 \cos \theta }\)

\(\displaystyle x = r \cos \theta = -4 \cos \frac{\pi}{3} = -4 \cdot \frac{1}{2} = -2\)

\(\displaystyle y = r \sin \theta = -4 \sin \frac{\pi}{3} = -4 \cdot \frac{ \sqrt{3}}{2} = -2 \sqrt{3}\)

The point will have rectangular coordinates \(\displaystyle \left (- 2, -2 \sqrt{3} \right )\).

We know \(\displaystyle \small y=rsin\theta\) and \(\displaystyle \small x=rcos\theta\).

Subbing that in to the equation \(\displaystyle \small y=x^{2}\) will give us \(\displaystyle \small rsin\theta=r^{2}cos^{2}\theta\).

Multiplying both sides by \(\displaystyle \small 1/(rcos^{2}\theta)\) gives us 

\(\displaystyle \small \small \small \small \small r=sin\theta /cos^{2}\theta=tan\theta/cos\theta=tan\theta\cdot sec\theta\).

To go from polar form to cartesion coordinates, use the following two relations.

\(\displaystyle x=rcos(\theta)\)

\(\displaystyle y=rsin(\theta)\)

In this case, our \(\displaystyle r\) is \(\displaystyle 3\) and our \(\displaystyle \theta\) is \(\displaystyle \frac{\pi}{3}\).

Plugging those into our relations we get 

\(\displaystyle x=\frac{3}{2}\)

\(\displaystyle y=\frac{3\sqrt3}{2}\), 

which gives us our \(\displaystyle (x,y)\) coordinate.

Although not explicitly stated, the problem is asking for the polar coordinates of the point \(\displaystyle (3,4)\). To calculate the magnitude, \(\displaystyle r\), calculate the following:

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

\(\displaystyle r=5\)

To calculate \(\displaystyle \theta\), do the following:

\(\displaystyle \theta=tan^{-1}\left(\frac{y}{x}\right)\) in radians. (The problem asks for radians)

\(\displaystyle \theta = 0.927\)

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 \(\displaystyle r\cos \theta\),

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

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

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

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