Can you show me how to find the rectangular equation of r=2costheta + 3 sin theta
All I need is an example to work off I was absent today from Pre-Calc and I have homework to do and no explanation with them-Please help with this one Thank you
what the heck is a rectangular equation?
hmm. you have given a polar coordinates equation, and want it y = f(x)?
cos theta = x/r
sin theta = y/r
r^2 = x^2 + y^2
r^2 = (2 cos t + 3 sin t)^2
= 4 cos^2 t + 12 sin t cos t + 9 sin^2 t
= 4 x^2/r^2 + 12xy/r^2 + 9y^2/r^2
r^4 = 4x^2 + 12xy + 9y^2
so, *x^2 + y^2)^2 = 4x^2 + 12xy + 9y^2
Not a pretty sight. The curve is cool, though
Certainly! I'd be happy to help you with that.
To find the rectangular equation of a polar equation, you can use the relationships between polar coordinates (r, θ) and rectangular coordinates (x, y), which are given by:
x = r * cos(θ)
y = r * sin(θ)
In your case, the polar equation is r = 2cos(θ) + 3sin(θ). To convert it to rectangular form, we'll substitute the values of x and y using the above relationships.
First, let's find x:
x = r * cos(θ)
x = (2cos(θ) + 3sin(θ)) * cos(θ)
Next, let's simplify it:
x = 2cos²(θ) + 3sin(θ) * cos(θ)
Now, let's find y:
y = r * sin(θ)
y = (2cos(θ) + 3sin(θ)) * sin(θ)
Again, let's simplify it:
y = 2sin(θ) * cos(θ) + 3sin²(θ)
Therefore, the rectangular equation is:
x = 2cos²(θ) + 3sin(θ) * cos(θ)
y = 2sin(θ) * cos(θ) + 3sin²(θ)
I hope this helps! Let me know if you have any more questions.