How do I convert the polar equation r - 3cos(theta) = 5sin to cartesian equation form?
For your equation
r - 3cosØ = 5sinØ
√(x^2 + y^2) - 3x/r = 5y/r
√(x^2 + y^2) = 5y/r + 3x/r
√(x^2 + y^2) = (5y + 3x)/r
√(x^2 + y^2) = (5y + 3x)/√(x^2 + y^2)
cross multiply
x^2 + y^2 = 5y + 3x
which looks like the equation of a circle to me
see the bottom part of my reply to
http://www.jiskha.com/display.cgi?id=1269266887
To convert the given polar equation, r - 3cos(theta) = 5sin, into Cartesian equation form, we can use the following trigonometric relationships:
r = sqrt(x^2 + y^2)
cos(theta) = x / sqrt(x^2 + y^2)
sin(theta) = y / sqrt(x^2 + y^2)
Replacing these in the given equation, we have:
sqrt(x^2 + y^2) - 3(x / sqrt(x^2 + y^2)) = 5(y / sqrt(x^2 + y^2))
Next, let's simplify the equation:
Multiply both sides by sqrt(x^2 + y^2) to eliminate the square roots:
x^2 + y^2 - 3x = 5y
Finally, rearrange the terms to obtain the Cartesian equation form:
x^2 - 3x + y^2 - 5y = 0
So, the Cartesian equation form of the given polar equation is:
x^2 - 3x + y^2 - 5y = 0
To convert a polar equation into Cartesian equation form, you can use the following relationships between polar and Cartesian coordinates:
x = r * cos(theta)
y = r * sin(theta)
In your polar equation, r - 3cos(theta) = 5sin(theta), we need to express r in terms of x and y to convert it into Cartesian equation form.
Given:
r = 3cos(theta) + 5sin(theta)
Replace r in terms of x and y:
x = (3cos(theta) + 5sin(theta)) * cos(theta)
y = (3cos(theta) + 5sin(theta)) * sin(theta)
Simplify:
x = 3cos^2(theta) + 5sin(theta)cos(theta)
y = 3cos(theta)sin(theta) + 5sin^2(theta)
To convert this into Cartesian equation form, we can use the following trigonometric identity:
sin(2θ) = 2sin(theta)cos(theta)
Applying this identity, we get:
x = 3cos^2(theta) + (5/2)sin(2theta)
y = (5/2)sin^2(theta) + 3sin(theta)cos(theta)
And there you have it! The Cartesian equation form of the polar equation r - 3cos(theta) = 5sin(theta) is:
x = 3cos^2(theta) + (5/2)sin(2theta)
y = (5/2)sin^2(theta) + 3sin(theta)cos(theta)