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Topic: Polar Coordinates (Read 2766 times) |
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mikedagr8
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Polar Coordinates
« on: Feb 6th, 2008, 12:54am » |
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Could I get an explanation as to why the answer I get is different to what the intended answer is.
y2=8x
My working is as follows.
y2-8x=0
r2sin2 -8rcos=0
r (rsin2- 8cos)=0
r=0 or rsin2-8cos=0
The answer intended is (8cos)/sin2.
Am I missing something or is it something I have not been taught. If it is somethign I have not been taught like cot and sec (I heard these terms) then just say so and I will be fine.
Thanks guys.
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pex
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Re: Polar Coordinates
« Reply #1 on: Feb 6th, 2008, 1:01am » |
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on Feb 6th, 2008, 12:54am, mikedagr8 wrote:Could I get an explanation as to why the answer I get is different to what the intended answer is.
y2=8x
My working is as follows.
y2-8x=0
r2sin2-8rcos=0
r (rsin2- 8cos)=0
r=0 or rsin2-8cos=0
The answer intended is (8cos)/sin2.
Am I missing something or is it something I have not been taught. If it is somethign I have not been taught like cot and sec (I heard these terms) then just say so and I will be fine.
Thanks guys. |
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As long as is not a multiple of  , your result r sin2 - 8cos = 0 is equivalent with r = (8cos)/(sin2). And the r=0 case is covered by, for instance, = /2.
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pex
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Re: Polar Coordinates
« Reply #2 on: Feb 6th, 2008, 1:04am » |
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And just to clarify, cot and sec are nothing special: cot = 1/tan, sec = 1/cos, and csc = 1/sin.
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mikedagr8
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Re: Polar Coordinates
« Reply #3 on: Feb 6th, 2008, 1:17am » |
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Thanks for that. Is this for all cases that are not multiples of that I can write my second term over my first if they are being subtracted? Would it be multiplied if it were added?
Thanks again.
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pex
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Re: Polar Coordinates
« Reply #4 on: Feb 6th, 2008, 1:27am » |
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on Feb 6th, 2008, 1:17am, mikedagr8 wrote:Thanks for that. Is this for all cases that are not multiples of that I can write my second term over my first if they are being subtracted? Would it be multiplied if it were added?
Thanks again. |
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It is a simple rearranging of terms:
r sin2 - 8cos = 0
 r sin2 = 8cos
r = (8cos)/(sin2).
The condition that is not a multiple of ensures that we are not dividing by zero, that's all.
So addition would not lead to multiplication, but to another division:
r sin2 + 8cos = 0
r sin2 = -8cos
r = (-8cos)/(sin2).
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mikedagr8
Uberpuzzler
A rich man is one who is content; not wealthy.
Gender:
Posts: 1105
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Re: Polar Coordinates
« Reply #5 on: Feb 6th, 2008, 1:31am » |
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Of course!
Sorry about that. Mental blank.
I still appreciate the help.
Thanks.
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"It's not that I'm correct, it's that you're just not correct, and so; I am right." - M.P.E.
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