How to translate from polar to cartesian coordinates:

  • Thread starter sjombol
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  • #1
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How to translate r = 2 /(2 - cos(theta)) to cartesian coordinates:


so far:

r = 2 /(2 - cos(theta))

r = 2 /(2 - cos(theta)) |* (2 - cos(theta)) both sides

r (2 - cos(theta))= 2

2*r - rcos(theta) = 2 | know x = rcos(theta)

2*r - x = 2 | know r^2 = x^2 + y^2

2*(x^2 + y^2)^1/2 - x = 2 | : 2

(x^2 + y^2)^1/2 - x/2 = 1 | opens the parentheses

x^2*1/2 + y^2*1/2 - x/2 = 1

x + y - x/2 = 1

y = 1 - x/2 | is this the solution?


Thank you in advance for help.

Regards
sjombol
 

Answers and Replies

  • #2
LCKurtz
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How to translate r = 2 /(2 - cos(theta)) to cartesian coordinates:


so far:

r = 2 /(2 - cos(theta))

r = 2 /(2 - cos(theta)) |* (2 - cos(theta)) both sides

r (2 - cos(theta))= 2

2*r - rcos(theta) = 2 | know x = rcos(theta)

2*r - x = 2 | know r^2 = x^2 + y^2

2*(x^2 + y^2)^1/2 - x = 2 | : 2

(x^2 + y^2)^1/2 - x/2 = 1 | opens the parentheses

x^2*1/2 + y^2*1/2 - x/2 = 1
Is ##\sqrt{a+b} = \sqrt a + \sqrt b##?
 
Last edited:
  • #3
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It's interesting that you document each step along the way with what you did, and I think that's a good thing. Where you fell down, as LCKurtz notes, is in the step that you describe as "opens the parentheses". The distributive property, which is usually represented something like this -- a(b + c) = ab + ac -- might be what you were thinking. The problem is that that this property applies only to multiplication of a sum. It doesn't apply to expressions raised to powers (including square roots) or other operations.
 
  • #4
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Thank you both very much for your reply :-)

It's been a while since last time I went to school, guess old habits do die sometimes or at least needs refreshing.


Another go then:

r = 2 /(2 - cos(theta))

r = 2 /(2 - cos(theta)) |* (2 - cos(theta)) both sides

r (2 - cos(theta))= 2

2*r - rcos(theta) = 2 | know x = rcos(theta)

2*r - x = 2 | know r^2 = x^2 + y^2

2*(x^2 + y^2)^1/2 - x = 2 | move x to right side and both sides ^2, gives me:

2^2 * ((x^2 + y^2)^1/2)^2 = (x + 2)^2

4 * (x^2 + y^2) = (x + 2)^2

4x^2 + 4y^2 = x^2 + 4x + 4 | move 4x^2 to right side

4y^2 = x^2 + 4x + 4 - 4x^2

4y^2 = -3x^2 + 4x + 4 | divide by 4 and takes the square root, giving me:


y = ((-3x^2 + 4x + 4)/4)^1/2 | the calculator draws this as an half ellipse, above x-axis ?
 
Last edited:
  • #5
LCKurtz
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I didn't check all your steps, but if you only took the positive square root, it wouldn't surprise me if half the graph is missing.
 
  • #6
HallsofIvy
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Thank you both very much for your reply :-)

It's been a while since last time I went to school, guess old habits do die sometimes or at least needs refreshing.


Another go then:

r = 2 /(2 - cos(theta))

r = 2 /(2 - cos(theta)) |* (2 - cos(theta)) both sides

r (2 - cos(theta))= 2

2*r - rcos(theta) = 2 | know x = rcos(theta)

2*r - x = 2 | know r^2 = x^2 + y^2

2*(x^2 + y^2)^1/2 - x = 2 | move x to right side and both sides ^2, gives me:

2^2 * ((x^2 + y^2)^1/2)^2 = (x + 2)^2

4 * (x^2 + y^2) = (x + 2)^2

4x^2 + 4y^2 = x^2 + 4x + 4 | move 4x^2 to right side

4y^2 = x^2 + 4x + 4 - 4x^2

4y^2 = -3x^2 + 4x + 4 | divide by 4 and takes the square root, giving me:
Why take the square root? You understand that you will be losing half the graph that way, don't you?

y = ((-3x^2 + 4x + 4)/4)^1/2 | the calculator draws this as an half ellipse, above x-axis ?
 

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