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Homework Help: Change the subject

  1. Jun 9, 2008 #1
    1. The problem statement, all variables and given/known data
    make [tex]\theta[/tex]the subject
    [tex] y= (\tan {\theta})x - (\frac{g}{2vi^2 \cos^2{\theta}})x^2 [/tex]


    2. Relevant equations



    3. The attempt at a solution
    not sure how to get start :(
    should i change tan to sin/cos first?
     
    Last edited: Jun 9, 2008
  2. jcsd
  3. Jun 9, 2008 #2

    tiny-tim

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    Welcome to PF!

    Hi suppy123! Welcome to PF! :smile:

    You ultimately want to find an equation beginning "θ =".

    But the last-step-but-one will be an equation beginning "cosθ =" or "tanθ =".

    So the technique you need is to change the given equation so that it either has only cosθ (and no tanθ) or only tanθ (and no cosθ).

    In other words: either rewrite tanθ in terms of cosθ, or cosθ in terms of tanθ (whichever seems easier). :smile:
     
  4. Jun 9, 2008 #3
    hi,
    ok,
    [tex]y = x(\frac{\sin{\theta}}{\cos{\theta}})-(\frac{gx^2}{2vi^2})(\frac{1}{\cos^2{\theta}})[/tex]

    [tex]y = x(\frac{\cos{\theta+\frac{\pi}{2}}}{\cos{\theta}})-(\frac{gx^2}{2vi^2})(\frac{2}{1+\cos{2{\theta}}})[/tex]

    is this correct? :S
     
    Last edited: Jun 9, 2008
  5. Jun 9, 2008 #4

    tiny-tim

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    Nooo …

    You need the whole of "cosθ", not just the "cos" part.

    An equation with both cosθ and cos(π/2 - θ) won't do. :frown:

    Hint: tanθ is easier. How can you express 1/cos²θ in terms of tanθ? :smile:
     
  6. Jun 9, 2008 #5
    yup, it's [tex]\tan^2{\theta}+1[/tex]
    [tex]y = x(\tan{\theta})-(\frac{gx^2}{2vi^2})(\tan^2{\theta}+1)[/tex]
    [tex]\tan{2\theta} = \frac{2\tan{\theta}}{1-\tan^2{\theta}}[/tex]
    [tex]\Rightarrow \tan^2{\theta} = \frac{-2\tan{\theta}}{\tan{2\theta}}[/tex]

    so
    [tex]y = x(\tan{\theta})-(\frac{gx^2}{2vi^2})(\frac{2\tan{\theta}}{\tan{2\theta}}+2)[/tex]
     
    Last edited: Jun 9, 2008
  7. Jun 9, 2008 #6

    tiny-tim

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    ok, so the original equation in terms of tanθ is … ? :smile:
     
  8. Jun 9, 2008 #7
    is that right?
     
  9. Jun 9, 2008 #8
    sry it should be
    [tex]\Rightarrow \tan^2{\theta} = \frac{-2\tan{\theta}}{\tan{2\theta}}+1[/tex]

    [tex]y = x(\tan{\theta})-(\frac{gx^2}{2vi^2})(\frac{-2\tan{\theta}}{\tan{2\theta}}+2)[/tex]
     
  10. Jun 9, 2008 #9

    tiny-tim

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    oh nooo … you've gone completely off the rails … :cry:

    It's [tex]\tan^2{\theta} = \frac{-2\tan{\theta}}{\sin{2\theta}}+1[/tex]

    Your original equation 1/cos²θ = tan²θ +1 was right.

    But don't start using tan2θ … having both tan2θ and tanθ is as bad as having both cosθ and cos(π/2 - θ). :frown:

    You want an equation with only tanθ, tan²θ, tan³θ, and so on. :smile:
     
  11. Jun 9, 2008 #10
    how about this?
    [tex]y = x(\tan{\theta})-(\frac{gx^2}{2vi^2})(\tan{\theta}+1)(\tan{\theta}-1)+2[/tex]
     
    Last edited: Jun 9, 2008
  12. Jun 9, 2008 #11

    tiny-tim

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    ooo … :cry:

    I've just noticed that you did produce the right equation (before you went berserk):
    That's the equation you want!

    Now that's a quadratic equation in tanθ, so use the usual -b ± √etc formula, to get an equation starting "tanθ =" :smile:
     
  13. Jun 9, 2008 #12
    lol...sweet, so
    [tex]y = x(\tan{\theta})-(\frac{gx^2}{2vi^2})(\tan^2{\theta}+1)[/tex]
    [tex]0 = (-\frac{gx^2}{2vi^2})(\tan^2{\theta}) + x(\tan{\theta}) - (\frac{gx^2}{2vi^2} - y)[/tex]
    [tex]\tan{\theta} = \frac{-x \pm \sqrt{x^2-4(\frac{gx^2}{2vi^2}-y)^2}}{-2\frac{gx^2}{2vi^2-y}}[/tex]
     
    Last edited: Jun 9, 2008
  14. Jun 9, 2008 #13
    oops, forgot to change the y heh
     
  15. Jun 9, 2008 #14

    tiny-tim

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    erm …
    :cry: What happened to poor little y ? :cry:
     
  16. Jun 9, 2008 #15
    there you go
    [tex]\tan{\theta} = \frac{-x \pm \sqrt{x^2-4(-\frac{gx^2}{2vi^2})(\frac{-gx^2}{2vi^2}-y)^2}}{-2(\frac{gx^2}{2vi^2})}[/tex]
    ;0
     
    Last edited: Jun 9, 2008
  17. Jun 9, 2008 #16

    tiny-tim

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    :biggrin: ooh … there he is … ! :biggrin:

    hmm … that last equation doesn't look right, though.
     
  18. Jun 9, 2008 #17
    should be right this time ;)
    [tex]\tan{\theta} = \frac{-x \pm \sqrt{x^2-4(-\frac{gx^2}{2vi^2})(\frac{-gx^2}{2vi^2}-y)}}{2(-\frac{gx^2}{2vi^2})}[/tex]
     
  19. Jun 9, 2008 #18

    tiny-tim

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    :biggrin: Woohoo! :biggrin:

    And now tidy it up a bit (get rid of some of the fractions)! :smile:
     
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