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Rus_ich [418]
3 years ago
12

Convert the Cartesian equation x^2 + y^2 = 16 to a polar equation.

Mathematics
1 answer:
RideAnS [48]3 years ago
8 0

Answer:

Problem 1: r=4

Problem 2: r=-2\sin(\theta)

Problem 3: r\sin(\theta)=3

Step-by-step explanation:

Problem 1:

So we are going to use the following to help us:

x=r \cos(\theta)

y=r \sin(\theta)

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

So if we make those substitution into the first equation we get:

x^2+y^2=16

(r\cos(\theta))^2+r\sin(\theta))^2=16

r^2\cos^2(\theta)+r^2\sin^2(\theta)=16

Factor the r^2 out:

r^2(\cos^2(\theta)+\sin^2(\theta))=16

The following is a Pythagorean Identity: \cos^2(\theta)+\sin^2(\theta)=1.

We will apply this identity now:

r^2=16

This implies:

r=4 \text{ or } r=-4

We don't need both because both of include points with radius 4.

Problem 2:

x^2+y^2+2y=0

(r\cos(\theta))^2+(r\sin(\theta))^2+2(r\sin(\theta))=0

r^2\cos^2(\theta)+r^2\sin^2(\theta)+2r\sin(theta)=0

Factoring out r^2 from first two terms:

r^2(\cos^2(\theta)+\sin^2(\theta))+2r\sin(\theta)=0

Apply the Pythagorean Identity I mentioned above from problem 1:

r^2(1)+2r\sin(\theta)=0

r^2+2r\sin(\theta)=0

or if we factor out r:

r(r+2\sin(\theta))=0

r=0 \text{ or } r=-2\sin(\theta)

r=0 is actually included in the other equation since when theta=0, r=0.

Problem 3:

y=3

r\sin(\theta)=3

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