text: Rotation of Ellipse to FIND the Bxy term, from horizontal to an angle of pi/4
  
  • May 28th 2011, 01:34 PM
  Lizziepaquette
  Rotation of Ellipse to FIND the Bxy term, from horizontal to an angle of pi/4
  In essence I'm trying to take a horizontal ellipse and put it at an angle, rotating it around the origin. It's for my conics project, I'm making the elliptical orbits of the planets.
  
  So I have the a, b, and c terms of the ellipse, (as in the distance from vertice to center and such) which are 28, 21, and $7\sqrt{7}$ respectively. When plugged into the equation they give:
  
  ${x}^{2} / 784 +{y}^{2} /441 =1$
  
  which turns into
  
  $441 {x}^{2} + 784{y}^{2 } - 345,744 = 0$
  
  from staring at my math book i've assumed that these x's and y's are actually x prime and y prime in the grand scheme of how we normally look at rotations.
  
  so actually
  
  $441{x'}^{2 } + 784{y'}^{2 } - 345,744 = 0$
  
  with this is mind I tried to apply the transformation equations knowing that I was trying to acheive a rotation of $\pi / 4$
  
  $x = x' \cos \pi /4 - y' \sin \pi /4$
  
  $x = (x' - y')/\sqrt{2}$
  
  $\sqrt{2}x = x' - y'$
  
  $\sqrt{2}y = x' + y'$
  
  But that's where I get stuck, because I can't think of how to substitute into my prime equation to get ride of the prime variable and create a Bxy term.
  
  So please, its due Wednesday... help?
  • May 29th 2011, 12:20 AM
  Opalg
  Quote:
  
  Originally Posted by Lizziepaquette
  I was trying to achieve a rotation of $\pi / 4$
  
  $x = x' \cos \pi /4 - y' \sin \pi /4$
  
  $x = (x' - y')/\sqrt{2}$
  
  $\sqrt{2}x = x' - y'$
  
  $\sqrt{2}y = x' + y'$
  
  But that's where I get stuck, because I can't think of how to substitute into my prime equation to get ride of the prime variable
  
  If you get from $(x',y')$ to $(x,y)$ by rotating through $\pi/4$, then you can get back from $(x,y)$ to $(x',y')$ by rotating through $-\pi/4$. Therefore if $x = x' \cos \pi /4 - y' \sin \pi /4$ then $x' = x \cos(- \pi /4) - y \sin(- \pi /4) = \tfrac1{\sqrt2}x +\tfrac1{\sqrt2}y.$
  • May 29th 2011, 06:47 AM
  Lizziepaquette
  Hmm let me plug it in
  • May 29th 2011, 06:48 AM
  Lizziepaquette
  I did the math out with that concept and my end result is
  
  $597.5{x}^{2} -373xy + 597.5{y}^{2} -345,744 = 0$
  
  I haven't done the forward math to check but i graphed it and it looks about right.
  
  Thanks (Smile)
explanation: The document contains a detailed math question about rotating an ellipse to find the \( Bxy \) term when the ellipse is rotated by an angle of \( \pi/4 \). The user provides the initial equation of the ellipse and the steps they have taken so far, including the transformation equations. The user then gets stuck and asks for help. Another user responds with a solution, and the original user confirms that the solution works.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: Rotation of Ellipse to FIND the Bxy term, from horizontal to an angle of pi/4
  
  • May 28th 2011, 01:34 PM
  Lizziepaquette
  Rotation of Ellipse to FIND the Bxy term, from horizontal to an angle of pi/4
  In essence I'm trying to take a horizontal ellipse and put it at an angle, rotating it around the origin. It's for my conics project, I'm making the elliptical orbits of the planets.
  
  So I have the a, b, and c terms of the ellipse, (as in the distance from vertice to center and such) which are 28, 21, and $7\sqrt{7}$ respectively. When plugged into the equation they give:
  
  ${x}^{2} / 784 +{y}^{2} /441 =1$
  
  which turns into
  
  $441 {x}^{2} + 784{y}^{2 } - 345,744 = 0$
  
  from staring at my math book i've assumed that these x's and y's are actually x prime and y prime in the grand scheme of how we normally look at rotations.
  
  so actually
  
  $441{x'}^{2 } + 784{y'}^{2 } - 345,744 = 0$
  
  with this is mind I tried to apply the transformation equations knowing that I was trying to acheive a rotation of $\pi / 4$
  
  $x = x' \cos \pi /4 - y' \sin \pi /4$
  
  $x = (x' - y')/\sqrt{2}$
  
  $\sqrt{2}x = x' - y'$
  
  $\sqrt{2}y = x' + y'$
  
  But that's where I get stuck, because I can't think of how to substitute into my prime equation to get ride of the prime variable and create a Bxy term.
  
  So please, its due Wednesday... help?
  • May 29th 2011, 12:20 AM
  Opalg
  Quote:
  
  Originally Posted by Lizziepaquette
  I was trying to achieve a rotation of $\pi / 4$
  
  $x = x' \cos \pi /4 - y' \sin \pi /4$
  
  $x = (x' - y')/\sqrt{2}$
  
  $\sqrt{2}x = x' - y'$
  
  $\sqrt{2}y = x' + y'$
  
  But that's where I get stuck, because I can't think of how to substitute into my prime equation to get ride of the prime variable
  
  If you get from $(x',y')$ to $(x,y)$ by rotating through $\pi/4$, then you can get back from $(x,y)$ to $(x',y')$ by rotating through $-\pi/4$. Therefore if $x = x' \cos \pi /4 - y' \sin \pi /4$ then $x' = x \cos(- \pi /4) - y \sin(- \pi /4) = \tfrac1{\sqrt2}x +\tfrac1{\sqrt2}y.$
  • May 29th 2011, 06:47 AM
  Lizziepaquette
  Hmm let me plug it in
  • May 29th 2011, 06:48 AM
  Lizziepaquette
  I did the math out with that concept and my end result is
  
  $597.5{x}^{2} -373xy + 597.5{y}^{2} -345,744 = 0$
  
  I haven't done the forward math to check but i graphed it and it looks about right.
  
  Thanks (Smile)
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document contains a detailed math question about rotating an ellipse to find the \( Bxy \) term when the ellipse is rotated by an angle of \( \pi/4 \). The user provides the initial equation of the ellipse and the steps they have taken so far, including the transformation equations. The user then gets stuck and asks for help. Another user responds with a solution, and the original user confirms that the solution works.
  
  Label: question-and-answer