3) Change the following polar coordinate to a rectangular coordinate (6, 3pi/2)
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
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
I apologize for the mistakes in my previous response. Thank you for correcting me.
To change polar coordinates to rectangular coordinates, you can use the following conversions:
x = r * cos(theta)
y = r * sin(theta)
For the polar coordinate (6, 3pi/2):
x = 6 * cos(3pi/2) = 6 * 0 = 0
y = 6 * sin(3pi/2) = 6 * (-1) = -6
So the rectangular coordinate is (0, -6).
For the rectangular coordinate (-4, -4 sqrt 3), to change it to polar coordinates, we can use the following formulas:
r = sqrt(x^2 + y^2)
theta = arctan(y/x)
r = sqrt((-4)^2 + (-4 sqrt 3)^2) = sqrt(16 + 48) = sqrt(64) = 8
theta = arctan((-4 sqrt 3)/(-4)) = arctan(sqrt(3)) = pi/3
So the polar coordinate is (8, pi/3) or (8, 4pi/3).
For the equation r = 4cos(theta), we need to rewrite it in rectangular form by substituting the values of x and y using the formulas:
x = r * cos(theta)
y = r * sin(theta)
x = 4cos(theta)
y = 4sin(theta)
Now, we can square both x and y and rearrange the equation:
x^2 = (4cos(theta))^2 = 16cos^2(theta)
y^2 = (4sin(theta))^2 = 16sin^2(theta)
We also know that cos^2(theta) + sin^2(theta) = 1 (from the trigonometric identity). So we can rearrange that equation to solve for cos^2(theta) and substitute it into the equation:
cos^2(theta) = 1 - sin^2(theta)
Substituting into the equation:
x^2 = 16(1 - sin^2(theta))
x^2 + y^2 = 16(1 - sin^2(theta)) + 16sin^2(theta)
Simplifying:
x^2 + y^2 = 16
This is the equation of a circle with radius 4 and center (0, 0) in standard form.
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.