Answer:
Problem 1: [tex]r=4[/tex]
Problem 2: [tex]r=-2\sin(\theta)[/tex]
Problem 3: [tex]r\sin(\theta)=3[/tex]
Step-by-step explanation:
Problem 1:
So we are going to use the following to help us:
[tex]x=r \cos(\theta)[/tex]
[tex]y=r \sin(\theta)[/tex]
[tex]\frac{y}{x}=\tan(\theta)[/tex]
So if we make those substitution into the first equation we get:
[tex]x^2+y^2=16[/tex]
[tex](r\cos(\theta))^2+r\sin(\theta))^2=16[/tex]
[tex]r^2\cos^2(\theta)+r^2\sin^2(\theta)=16[/tex]
Factor the [tex]r^2[/tex] out:
[tex]r^2(\cos^2(\theta)+\sin^2(\theta))=16[/tex]
The following is a Pythagorean Identity: [tex]\cos^2(\theta)+\sin^2(\theta)=1[/tex].
We will apply this identity now:
[tex]r^2=16[/tex]
This implies:
[tex]r=4 \text{ or } r=-4[/tex]
We don't need both because both of include points with radius 4.
Problem 2:
[tex]x^2+y^2+2y=0[/tex]
[tex](r\cos(\theta))^2+(r\sin(\theta))^2+2(r\sin(\theta))=0[/tex]
[tex]r^2\cos^2(\theta)+r^2\sin^2(\theta)+2r\sin(theta)=0[/tex]
Factoring out [tex]r^2[/tex] from first two terms:
[tex]r^2(\cos^2(\theta)+\sin^2(\theta))+2r\sin(\theta)=0[/tex]
Apply the Pythagorean Identity I mentioned above from problem 1:
[tex]r^2(1)+2r\sin(\theta)=0[/tex]
[tex]r^2+2r\sin(\theta)=0[/tex]
or if we factor out r:
[tex]r(r+2\sin(\theta))=0[/tex]
[tex]r=0 \text{ or } r=-2\sin(\theta)[/tex]
r=0 is actually included in the other equation since when theta=0, r=0.
Problem 3:
[tex]y=3[/tex]
[tex]r\sin(\theta)=3[/tex]
