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#1 (permalink) |
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Support Team
Expert Zuner Join Date: Nov 2007
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Lul, this probably isn't the right place in the forums to post this... so any mod move this to where it should be if there's a better spot... but anyone know how to do this problem? I have to convert 2+2sin2theta to rectangular form.. then graph.. and I have no idea where to begin.
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#2 (permalink) |
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What do you mean, 'rectangular form?'
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#3 (permalink) |
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The Phenomenon
An army of many
Support Team Jr. Staff Premium Member Zune Freak |
Hey no fair deleting my intelligently funny post
hax.
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#4 (permalink) |
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Friend of the Ood
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Randy, please keep in mind that this section does fall under Serious Discussion.
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#5 (permalink) |
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The Phenomenon
An army of many
Support Team Jr. Staff Premium Member Zune Freak |
oh i didnt realize that :S sry.
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#6 (permalink) |
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Expert Zuner Join Date: Nov 2007
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Rectangular form as in... in terms of X and Y
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#7 (permalink) |
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I'll assume the full equation is R = 2 + 2Θ ? Polar equation, yes?
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#8 (permalink) |
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Zune Freak
Join Date: Jan 2008
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He said R=2+2sin(2theta) I believe. If I'm correct, then you can't convert it to rectangular form considering the fact that that equation is a two leafed clover. You'd need a given value for theta (a number) in order to do so.
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To view links or images in signatures your post count must be 0 or greater. You currently have 0 posts. To view links or images in signatures your post count must be 0 or greater. You currently have 0 posts. Last edited by Darth Camel : 07-23-2008 at 02:24 AM.
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#10 (permalink) |
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Experienced Zuner
Join Date: Jul 2008
Location: Chicago
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Why do you care its the middle of summer. Summer school already ended, atleast for me it did(and no im not retarted im a freshman getting ahead, my mom made me take it).
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#11 (permalink) | |
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Purger of Ignorance
Jr. Staff
zB Programmer Zune Freak Join Date: Jun 2008
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Quote:
Polar Relationships: x = r cos(t) y = r sin(t) x^2 + y^2 = r^2. So: sin(t) = x / r cos(t) = y / r sin(t) = x / sqrt(x^2 + y^2) r = 2 + 2 sin(2t) (Assuming this is what is meant, as opposed to 2 + 2 sin^2(t)) sqrt(x^2 + y^2) = 2 + 2(sin(2t)) sqrt(x^2 + y^2) = 2 + 2(2 sin(t) + cos(t)) sqrt(x^2 + y^2) = 2 + 2(2(x / sqrt(x^2 + y^2)) + (y / sqrt(x^2 + y^2)) sqrt(x^2 + y^2) = 2 + 4x / sqrt(x^2 + y^2) + y / sqrt(x^2 + y^2) Multiply by (x^2 + y^2)sqrt(x^2 + y^2) (x^2 + y^2)^2 = 2((x^2 + y^2)sqrt(x^2 + y^2) + 4x(x^2 + y^2) + y(x^2 + y^2) x^4 + y^4 + 2(x^2)(y^2) = (2x^2 + 2y^2)(x^2 + y^2)^(1/2) + 4x^3 + 4x(y^2) + (x^2)y + y^3 x^4 + y^4 - 4x^3 - y^3 + 2(x^2)(y^2) - (x^2)y - 4x(y^2) - (2x^2 + 2y^2)(x^2 + y^2)^(1/2) = 0 I think that is right...
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#12 (permalink) | |
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Experienced Zuner
Join Date: Jul 2008
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Quote:
HOLY &#$!, good work!
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