We are given several Cartesian equations and asked to find their corresponding polar equations. Problem 13: $y = -2$ Problem 15: $x^2 + y^2 = 4$ Problem 16: $x^2 = 4py$

1. Problem Description

We are given several Cartesian equations and asked to find their corresponding polar equations.

Problem 13: $y = -2$

Problem 15: $x^2 + y^2 = 4$

Problem 16: $x^2 = 4py$

2. Solution Steps

Problem 13: $y = -2$

To convert to polar coordinates, we use the relationship $y = r \sin(\theta)$.

Therefore, we have $r \sin(\theta) = -2$.

Solving for $r$, we get $r = -\frac{2}{\sin(\theta)}$.

Alternatively, we can write $r = -2 \csc(\theta)$.

Problem 15: $x^2 + y^2 = 4$

To convert to polar coordinates, we use the relationship $x^2 + y^2 = r^2$.

Therefore, we have $r^2 = 4$.

Taking the square root of both sides gives $r = 2$.

Problem 16: $x^2 = 4py$

To convert to polar coordinates, we use the relationships $x = r \cos(\theta)$ and $y = r \sin(\theta)$.

Substituting these into the equation, we get $(r \cos(\theta))^2 = 4p(r \sin(\theta))$.

This simplifies to $r^2 \cos^2(\theta) = 4pr \sin(\theta)$.

If $r \neq 0$, we can divide by $r$ to get $r \cos^2(\theta) = 4p \sin(\theta)$.

Solving for $r$, we get $r = \frac{4p \sin(\theta)}{\cos^2(\theta)}$.

We can also write this as $r = 4p \tan(\theta) \sec(\theta)$.

3. Final Answer

Problem 13: $r = -2 \csc(\theta)$

Problem 15: $r = 2$

Problem 16: $r = 4p \tan(\theta) \sec(\theta)$