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 \(\displaystyle r\cos \theta\),

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

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

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

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

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 \(\displaystyle r\cos \theta\),

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

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

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

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 \(\displaystyle y=\frac{7}{3}x^{2}\), then \(\displaystyle r\sin \theta=\frac{7}{3}{r^{2}\cos^{2} \theta}\). Simplifying accordingly, 

\(\displaystyle \tan \theta=\frac{7}{3}{r\cos\theta}\)

\(\displaystyle \tan \theta\sec \theta =\frac{7}{3}r\)

\(\displaystyle \frac{3}{7}\tan \theta\sec \theta =r\)

Knowing that \(\displaystyle x=3t+15\) and \(\displaystyle y=15t-6\), we can isolate \(\displaystyle t\) in both equations as follows:

\(\displaystyle x=3t+15\rightarrow t=\frac{x-15}{3}\)

\(\displaystyle y=15t-6\rightarrow t=\frac{y+6}{15}\)

Since both of these equations equal \(\displaystyle t\), we can set them equal to each other:

\(\displaystyle \frac{y+6}{15}=\frac{x-15}{3}\)

\(\displaystyle y+6=15(\frac{x-15}{3})\)

\(\displaystyle y+6=5({x-15})\)

\(\displaystyle y+6=5x-75\)

\(\displaystyle y=5x-81\)

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.

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

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

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

To convert from rectangular to polar form, we must use the following formulas:

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

\(\displaystyle \frac{y}{x}=\tan(\theta)\)

It is easier to find our angle \(\displaystyle \theta\) first, which is done by plugging in our x and y into the second formula:

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

Find the angle by taking the inverse of the function:

\(\displaystyle \tan^{-1}(\tan(\theta))=\tan^{-1}(1)\)

\(\displaystyle \theta=\frac{\pi}{4}\)

Now find r by plugging in our angle and x and y into the first formula, and solving for r:

\(\displaystyle 2=r\cos\left(\frac{\pi}{4}\right)\)

\(\displaystyle r=2\sqrt{2}\)

Our final answer is reported in polar coordinate form \(\displaystyle (r, \theta)\):

\(\displaystyle \left(2\sqrt{2}, \frac{\pi}{4}\right)\)