how do you obtain the cartesian equation for r^2=16cos(3theata)?? I've tried looking on many website and at my notes but have had no luck.
i know that r^2=x^2+y^2 and x=rcos(theta) but don't know were to go from here?
the messy part is the cos (3Ø)
we know cos 3Ø = 4cos^3 Ø - 3cosØ
and since you know cosØ = x/r
we get
x^2 + y^2 = 16x^3/r^3 - 3x/r
x^2 + y^2 = 16x/((x^2+y^2)√(x^2 + y^2)) - 3x/√(x^2+y^2)
see:
http://www.wolframalpha.com/input/?i=plot+x%5E2+%2B+y%5E2+%3D+16x%2F%28%28x%5E2%2By%5E2%29√%28x%5E2+%2B+y%5E2%29%29+-+3x%2F√%28x%5E2%2By%5E2%29+
thanks so much ill have a look
forgot the 4 in front of the 4cos^3 Ø
should be
x^2 + y^2 = 16x^3/(64r^3) - 3x/r
x^2 + y^2 = (1/4)x/((x^2+y^2)√(x^2 + y^2)) - 3x/√(x^2+y^2)
To obtain the Cartesian equation for the polar equation r^2 = 16cos(3θ), we need to convert the equation from polar coordinates (r, θ) to Cartesian coordinates (x, y). Here's how you can do it step by step:
1. Start with the given polar equation: r^2 = 16cos(3θ).
2. Substitute r^2 with its equivalent Cartesian expression, x^2 + y^2: x^2 + y^2 = 16cos(3θ).
3. Convert the polar coordinates (r, θ) to Cartesian coordinates (x, y). Recall that x = rcos(θ) and y = rsin(θ).
4. Replace r^2 with (x^2 + y^2), and replace cos(θ) with x/r and sin(θ) with y/r in the equation:
x^2 + y^2 = 16cos(3θ)
x^2 + y^2 = 16cos(3arctan(y/x)) (using the relationship θ = arctan(y/x))
x^2 + y^2 = 16cos(arctan(3y/x)) (using the identity cos(3θ) = cos(arctan(3y/x)))
5. Simplify the expression further if possible, but note that this equation in terms of x and y might not be the most practical form. You may consider simplifying or expanding the cos(3arctan(3y/x)) term, but it could get quite complicated.
That's how you obtain the Cartesian equation for the given polar equation. Keep in mind that some polar equations might not easily simplify to Cartesian form, and it might be more convenient to work with them in polar coordinates.