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Change the subject

  • Thread starter suppy123
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  • #1
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Homework Statement


make [tex]\theta[/tex]the subject
[tex] y= (\tan {\theta})x - (\frac{g}{2vi^2 \cos^2{\theta}})x^2 [/tex]


Homework Equations





The Attempt at a Solution


not sure how to get start :(
should i change tan to sin/cos first?
 
Last edited:

Answers and Replies

  • #2
tiny-tim
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Welcome to PF!

Homework Statement


make [tex]\theta[/tex]the subject
[tex] y= (\tan {\theta})x - (\frac{g}{2vi^2 \cos^2{\theta}})x^2 [/tex]

not sure how to get start :(
should i change tan to sin/cos first?
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:
 
  • #3
29
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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
 
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  • #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:
 
  • #5
29
0
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]
 
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  • #6
tiny-tim
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yup, it's [tex]\tan^2{\theta}+1[/tex]
ok, so the original equation in terms of tanθ is … ? :smile:
 
  • #7
29
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is that right?
 
  • #8
29
0
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]
 
  • #9
tiny-tim
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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]
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:
 
  • #10
29
0
how about this?
[tex]y = x(\tan{\theta})-(\frac{gx^2}{2vi^2})(\tan{\theta}+1)(\tan{\theta}-1)+2[/tex]
 
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  • #11
tiny-tim
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how about this?
[tex]y = x(\tan{\theta})-(\frac{gx^2}{2vi^2})(\tan{\theta}-\sqrt{1})^2+1[/tex]
ooo … :cry:

I've just noticed that you did produce the right equation (before you went berserk):
[tex]y = x(\tan{\theta})-(\frac{gx^2}{2vi^2})(\tan^2{\theta}+1)[/tex]
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:
 
  • #12
29
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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]
 
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  • #13
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oops, forgot to change the y heh
 
  • #14
tiny-tim
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[tex]\tan{\theta} = \frac{-x \pm \sqrt{x^2-4(\frac{gx^2}{2vi^2})^2}}{-2\frac{gx^2}{2vi^2}}[/tex]
erm …
:cry: What happened to poor little y ? :cry:
 
  • #15
29
0
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
 
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  • #16
tiny-tim
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:biggrin: ooh … there he is … ! :biggrin:

hmm … that last equation doesn't look right, though.
 
  • #17
29
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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]
 
  • #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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