3) Change the following polar coordinate to a rectangular coordinate (6, 3pi/2)

Question

A: (-6,0)
B: (6,0)
C: (0,6)
D: (0,-6)

4) Change the following rectangular coordinate to a polar coordinate (-4,-4 sqrt 3)

A: (4, 4pi/3)
B: (8, 4pi/3)
C: (4, 7pi/6)
D: (8, 5pi/6)

5) Rewrite r=4 cos theta in rectangular form. Show all your work. Your final answer must be in the standard form of a circle. (x-h)^2+(y-k)^2=r^2

Answer 1

r=4 cos theta
r=4(cos theta, sin theta)
x = r cos theta = 4 cos^2 theta
y = r sin theta = 4 sin theta cos theta = 2 sin 2theta

Substituting sin 2theta = 2 sin theta cos theta,
x = 2(2 cos^2 theta - 1)
y = 4 sin theta cos theta

Simplifying,
x = 4 cos^2 theta - 2
y = 2 sin 2theta

Using double-angle formula, sin 2theta = 2 sin theta cos theta,
y = 4 sin theta cos theta = 2y/x

Substituting into the equation x^2 + y^2 = r^2,
x^2 + (2y/x)^2 = 16 cos^2 theta
x^2 + 4y^2/x^2 = 16 cos^2 theta
Multiplying both sides by x^2,
x^4 + 4y^2 = 16x^2 cos^2 theta

Simplifying,
(x^2 - 8x + 16 cos^2 theta) + 4y^2 = 16 cos^2 theta + 16
(x - 4 cos theta)^2 + 4y^2 = 20

Therefore, the rectangular form is: (x - 4 cos theta)^2 + 4y^2 = 20. This represents a circle centered at (4 cos theta, 0) with a radius of √5.

Answer 2

AAAaannndd the bot gets it wrong yet again!

(6, 3π/2) = (0,-6)
(-4,-4√3) = (8,4π/3)

r = 4cosθ
r^2 = 4rcosθ
x^2+y^2 = 4x
(x-2)^2 + y^2 = 4

Answer 3

I apologize for the mistakes in my previous response. Thank you for correcting me.