SOLUTION: Convert the equation to polar form. 5x=5y i have already started the problem to: 5(rcostheta)=5(rsintheta) and don't where to go from there Please help me!

Algebra ->  Coordinate-system -> SOLUTION: Convert the equation to polar form. 5x=5y i have already started the problem to: 5(rcostheta)=5(rsintheta) and don't where to go from there Please help me!      Log On


   

Question 429427: Convert the equation to polar form.
5x=5y
i have already started the problem to:
5(rcostheta)=5(rsintheta)
and don't where to go from there Please help me!
Answer by richard1234(7193) About Me  (Show Source):
You can put this solution on YOUR website!
I presume the linear equation will have a strange polar representation.
From 5x = 5y we obtain x = y and r%2Acos%28theta%29+=+r%2Asin%28theta%29. We wish to isolate r.

Moving all terms to one side we get r%2Acos%28theta%29+-+r%2Asin%28theta%29+=+0 --> r%28cos%28theta%29+-+sin%28theta%29%29+=+0 and r+=+0%2F%28cos%28theta%29-sin%28theta%29%29. What's interesting about this graph is that when cos%28theta%29+-+sin%28theta%29+=+0 (theta+=+pi%2F4+%2B+k%2Api), we get an indeterminate form and r can equal anything. 0%2F0 can theoretically be equal to any number, which is why it is considered indeterminate instead of undefined.