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

Guest May 29, 2015

Best Answer 

 #4
avatar+91038 
+10

 

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
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4+0 Answers

 #1
avatar+78643 
+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
avatar+91038 
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
avatar+78643 
+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
avatar+91038 
+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

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