ATAR Notes: Forum
VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: Homer J on January 16, 2011, 11:53:26 am
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--- question: convert x^2 + 4x + y^2 - 2y = 0 into a polar equation
thanks
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x^2 + 4x + y^2 -2y = 0
Know that the conversion formulas that relate the point (x,y) and (r, θ) are: x= rcos θ, y=rsin θ, r²= x² + y² and tan θ= y/x. These are important for any type of conversion between the two forms as well as some trigonometric identities
(x^2 + y^2) +4x - 2y = 0
subsitute r^2 = x^2 + y^2 aswell as y= rsin(t) , x= rcos(t)
so becomes
r^2 - 2rsin(t) +4rcos(t) = 0
factorise
r [r -2sin(t) + 4cos(t) ] = 0
r= 0 and (r -2sin(t) + 4cos(t)) = 0
The equation R = 0 is the pole. But the pole is included in the graph of the second equation r - 2sin(t) + 4cos(t) = 0 .
We therefore can keep only the second equation.
now solve for r to get polar equation
r - 2 sin(t) + 4cos(t) = 0
r = 2 sin(t) - 4 cos(t)
correct me if i am wrong
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converting cartesian to polar is normally done in a systematic way, substitute x = rcos(theta), y = rsin(theta), substitute the in the cartesan and solve for r.
convince yourself why we have the relationship x = f(theta,r) and y = f(theta, r)