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Calculus III; standard form of & angle to eliminate the cross produc

  1. Feb 8, 2014 #1
    Hi; I've been trying to solve the problem myself but i really don't what could be wrong;

    The problem says :

    Make the change of variables
    x=ucos−vsin
    y=usin+vcos

    where the angle 0<(phi)<2 is chosen in order to eliminate the cross product term in
    x^2+xy+y^2=6

    Then find the standard form of equation in the (uv) variables. (Enter a function of (uv).)

    ---------------=1

    well what I've found the angle is (pi/4) wich would eliminates the cross product terms "uv" when i make the substituion..

    then I've tried to reorder the equation: x^2+xy+y^2=6 wich is an elipse with center at (0,0) ; semimajor axis=2sqrt3 & semiminor axis=2

    then i got the equation (x^2/12)+(y^2/4)=1 then change variables again and i got ((ucos(pi/4))-(vsin(pi/4)))^2/12+((usin(pi/4))+(vcos(pi/4)))^2/4.
    its incorrect.

    If you could make a step by step solution , would be great , thanks in advanced.
     
  2. jcsd
  3. Feb 8, 2014 #2

    SammyS

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    We don't provide step by step solutions at PF.

    Why does the equation you get,

    (x2/12)+(y2/4)=1,

    have x & y in it, rather than u & v ?



    To do the integration with u,v, don't forget your Jacobian -- although in this case it's a convenient value.
     
  4. Feb 8, 2014 #3
    Because what i tried to do its to re-write the equation " x^2+xy+y^2=6" wich i found easier. then i plugged into it the "x" and "y" the problems gives.

    but what i am not really sure if it means "elipse standard form"; i think the angle i got it's right since " ((ucos(pi/4))-(vsin(pi/4)))^2/12+((usin(pi/4))+(vcos(pi/4)))^2/4." & "x^2+xy+y^2=6" are the same elipse.. but i don't why its incorrect.
     
  5. Feb 8, 2014 #4

    SammyS

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    Sure, you have the correct angle.

    plugging in u & v, I get

    u2/4 + v2/12 = 1 ,

    which is the ellipse you describe, with major axis along the v axis.
     
  6. Feb 8, 2014 #5
    but wouldn't it be x=ucos−vsin--> so ((ucos(pi/4)-vsin(pi/4))^2)/12 & y=usin+vcos----> ((usin(pi/4))+(vcos(pi/4)))^2/4?? did you reduce factors?
     
  7. Feb 8, 2014 #6

    vela

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    The presence of the xy term is what causes the ellipse to be rotated with respect to the x and y axes. You're trying to find a rotated set of coordinates in which the ellipse isn't rotated. You don't want the ellipse to look exactly the same in both sets of coordinates.
     
  8. Feb 8, 2014 #7
    oh ok! i see.. well i think is a minor detail i couldn't see , thanks so much!
     
  9. Feb 8, 2014 #8

    SammyS

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    Take the original equation

    x2+xy+y2 = 6

    Substitute as directed to express it in terms of u & v.

    As a result of that, I got u2/4 + v2/12 = 1


    It looks like you may have done that, then switched back to x & y , because using u & v was unfamiliar to you. -- then you tried the substitution again into that result.

    Is that what you did ?
     
  10. Feb 8, 2014 #9
    yeah actually i make the change of variables & i end up.. with "uv" terms.

    then i found the angle to make them 0. because it was the cross product terms then i just got a u^2 and v^2 with sin & cos =6 at this point i didn't know how to reorder the equation.

    that's why i go back to the original equation. i though you needed the "angle" to make the direct substituion and have the values of cosines and sines once you have the "standard form of the given elipse".
     
  11. Feb 8, 2014 #10

    SammyS

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    Your result at this point is what is asked for in the original problem: a function of (u,v) .

     
  12. Feb 9, 2014 #11
    Actually at first a tried with this result. but it said ,it was "incorrect" i guess because i didn't reorder the equation.
     
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