Find the Cartesian form of the equation, r3 = 3r cosØ.
r^3 = 3rcosθ
(x^2 + y^2)^(3/2) = 3x
or
x^2+y^2 = (3x)^(2/3)
To find the Cartesian form of the polar equation r^3 = 3r*cos(θ), we need to express r in terms of x and y.
First, we can replace r with √(x^2 + y^2), since r represents the distance from the origin to a point (x, y) in the Cartesian coordinate system.
Now our equation becomes (√(x^2 + y^2))^3 = 3√(x^2 + y^2)*cos(θ).
Simplifying further, we expand (√(x^2 + y^2))^3:
(x^2 + y^2)^(3/2) = 3√(x^2 + y^2)*cos(θ).
To eliminate the square roots, we square both sides of the equation:
[(x^2 + y^2)^(3/2)]^2 = [3√(x^2 + y^2)*cos(θ)]^2.
This gives us:
x^2 + y^2 = 9(x^2 + y^2)cos^2(θ).
Expanding further, we have:
x^2 + y^2 = 9x^2cos^2(θ) + 9y^2cos^2(θ).
Now we can use the identity cos^2(θ) = (x^2 / (x^2 + y^2)) and plug it in:
x^2 + y^2 = 9x^2[x^2 / (x^2 + y^2)] + 9y^2[x^2 / (x^2 + y^2)].
Simplifying the expression, we get:
x^2 + y^2 = 9x^4 / (x^2 + y^2) + 9xy^2 / (x^2 + y^2).
To remove the denominator, we multiply both sides of the equation by (x^2 + y^2):
(x^2 + y^2)^2 = 9x^4 + 9xy^2.
Expanding further, we have:
x^4 + 2x^2y^2 + y^4 = 9x^4 + 9xy^2.
Simplifying,
8x^4 - 2x^2y^2 + y^4 = 0.
So, the Cartesian form of the polar equation r^3 = 3r*cos(θ) is 8x^4 - 2x^2y^2 + y^4 = 0.