How do you convert #r = 15 / (6sin(theta) + 43cos(theta)) # into cartesian form?

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Answer 1 · 2016-06-24T07:39:56

#6 y + 43 x = 15#

Use (the transformation #(r cos theta, r sin theta) = (x, y)#.

The given equation becomes

# r = 15/(6(y/r)+43(x/r))#

#=(15 r)/(6y+43x)#

As #|6 sin theta + 43 cos theta|<=sqrt(6^2+43^2)#.,

#r>=15/sqrt(6^2+43^2)#. So, r does not become 0.

Cancelling the common factor r and cross multiplying,

#6y+43x=15#.