Find the polar coordinates of the Cartesian points $(5,-5)$, $(1, \sqrt{3})$ and $(-3, -6)$ with $r>0$ and $0 \leq \theta < 2\pi$.
We have standard formulas for how to convert between these formulas.
We know that $r = \sqrt{x^2 + y^2}$ and $\theta = \tan^{-1}\left(\frac{y}{x}\right)$, but we need to be careful about the quadrant for $\theta$.
When you take inverse tangent, you get an angle between $-\pi/2$ and $\pi/2$, and then need to figure out how to get this into the correct quadrant and the correct range of $\theta$.
These points are at $\left(5\sqrt{2}, \frac{7\pi}{4}\right)$, $\left(2, \frac{\pi}{3}\right)$, and $(\sqrt{45}, \tan^{-1}(2) + \pi)$.
Find the Cartesian coordinates of the polar coordinate points $\left(3, \frac{\pi}{4}\right)$, $(5, \pi)$, and $\left(4, \frac{5\pi}{3}\right)$.
We can again apply the standard formulas for converting these points into rectangular coordinates.
We have that $x = r\cos(\theta)$ and $y = r\sin(\theta)$
These points are at $\left(\frac{3}{\sqrt{2}}, \frac{3}{\sqrt{2}}\right)$, $(-5,0)$ and $(2, -2\sqrt{3})$.
Convert the equation $r = \frac{1}{2 - \cos{\theta}}$ into rectangular coordinates.
We do not want to start converting right away, and should look for good terms first. We can start by clearing the denominator to the other side.
This gives $2r - r\cos{\theta} = 1$. We know that $r\cos{\theta}$ is $x$, so we can replace that and move it to the other side. We still have to deal with that $r$ though.
Once we are at $2r = 1 + x$, we can square both sides to get that $4r^2 = (1+x)^2$, which can then be converted and simplified.
We then have $4x^2 + 4y^2 = x^2 + 2x + 1$ or $3x^2 - 2x + 4y^2 = 1$, which we can leave as is, or simplify further by completing the square.
$\displaystyle 3x^2 - 2x + 4y^2 = 1$ or $\displaystyle 3(x - \frac{1}{3})^2 + 4y^2 = \frac{4}{3}$
Convert the equation $xy = 1$ into polar coordinates of the form $r = f(\theta)$.
When going to polar coordinates, we can just convert directly by plugging in our definitions.
Since we have $x = r\cos{\theta}$ and $y = r\sin{\theta}$, this equation becomes $r\cos{\theta} r\sin{\theta} = 1$.
This becomes $\displaystyle r^2 = \frac{1}{\cos{\theta}\sin{\theta}}$ or $r^2 = \frac{2}{\sin{2\theta}}$.
Then we can take square roots, but be careful where the function is defined.
$\displaystyle r = \sqrt{\frac{2}{\sin{2\theta}}}$ for $0 \leq \theta \leq \pi/2$ and $\pi \leq \theta \leq \frac{3\pi}{2}$.
If a point $P = (x,y)$ has polar coordinates $(r, \theta)$, then what are the polar coordinates of $(x,-y)$? What about $(-x, -y)$?
If we want to analyze the point $(x,-y)$, this involves reflecting over the x-axis. How can we compare the polar coordinates of these points?
To get to the opposite y-coordinate, we could rotate around in the other direction. This means I want to take the negative angle, instead of the positive one.
For $(-x,-y)$, this is in the opposite direction of $(x,y)$. How do we get that in polar coordinates?
For this, we can either take a negative radius, or we can add $\pi$ to the angle.
The point $(x,-y)$ has polar coordinates $(r, -\theta)$ and the point $(-x,-y)$ has polar coordinates $(r, \theta + \pi)$.
Write the equation for the circle of radius $5$ centered at $(3,4)$ in polar coordinates. This should be of the form $r = a\cos{\theta} + b\sin{\theta}$.
What is the Cartesian form of this equation?
This equation is $\displaystyle (x-3)^2 + (y-4)^2 = 25$. We can plug in the polar coordinate versions of $x$ and $y$ and expand everything out.
Expanding out gives $\displaystyle r^2 \cos^2{\theta} - 6r\cos{\theta} + 9 + r^2\sin^2\theta - 8r\sin{\theta} + 16 = 25$.
By cancelling the coefficients and combining the cosine squared and sine squared terms, we get $\displaystyle r^2 - 6r\cos{\theta} - 8r\sin{\theta} = 0$
$\displaystyle r = 6\cos{\theta} + 8 \sin{\theta}$
Convert the equation $(x+2)^2 + y^2 = 4$ into polar coordinates of the form $r = f(\theta)$
Convert the equation $r = 2\sin{\theta}\tan{\theta}$ into rectangular coordinates. Hint: The end result here should be $y^2 = $ an expression in terms of $x$.