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# Rewrite the polar equation r = 3 cos(theta) as a Cartesian equation.

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Rewrite the polar equation r = 3 cos(theta) as a Cartesian equation.

Guest May 29, 2015

#4
+93691
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$$\\x^2+y^2=r^2\\ r=3cos\theta\\ x=rcos\theta\\ r^2=3rcos\theta\\ so\\ r^2=3x\\ so\\ x^2+y^2=3x\\ x^2-3x+1.5^2+y^2=0+2.5^2\\ (x-1.5)^2+y^2=(1.5)^2\\  This is a circle centre (1.5,\;0) and radius 1.5$$$NOW I HAVE GOT IT !! Thanks Chris :)) Melody May 29, 2015 #1 +90088 +10 r = 3cosΘ Multiply both sides by r r^2 = 3rcosΘ x^2 + y^2 = 3x Here is the graph of each form.......notice that they overlap perfectly......https://www.desmos.com/calculator/hg8gmfwwu2 CPhill May 29, 2015 #2 +93691 0 Thanks CPhill, I am trying to get my head around this solution. I can see how you have done it but is do not intrisically understand :// Melody May 29, 2015 #3 +90088 +5 Melody....the reason for multiplying by "r" on each side is that we have no Cartesian equivalent for cosΘ...but rcosΘ is easily transformed to "x"....also, multiplying the left side by "r" gets rid of a "root" problem since r = √[x2 + y2]...but.....r2 = x2 + y2.....and the root "disappears"..... CPhill May 29, 2015 #4 +93691 +10 Best Answer Let me think about this $$\\x^2+y^2=r^2\\ r=3cos\theta\\ x=rcos\theta\\ r^2=3rcos\theta\\ so\\ r^2=3x\\ so\\ x^2+y^2=3x\\ x^2-3x+1.5^2+y^2=0+2.5^2\\ (x-1.5)^2+y^2=(1.5)^2\\ This is a circle centre (1.5,\;0) and radius 1.5$$$

NOW I HAVE GOT IT   !!

Thanks Chris  :))

Melody  May 29, 2015