Solving y= (tan θ)x - (g/(2vi2cos2θ))x2

[suppy123]

Homework Statement

make \thetathe subject
y= (\tan {\theta})x - (\frac{g}{2vi^2 \cos^2{\theta}})x^2

Homework Equations

The Attempt at a Solution

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

 

[tiny-tim]

Welcome to PF!

Homework Statement

make \thetathe subject
y= (\tan {\theta})x - (\frac{g}{2vi^2 \cos^2{\theta}})x^2

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

Hi suppy123! Welcome to PF!

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).

 

[suppy123]

hi,
ok,
y = x(\frac{\sin{\theta}}{\cos{\theta}})-(\frac{gx^2}{2vi^2})(\frac{1}{\cos^2{\theta}})

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

is this correct? :S

 

[tiny-tim]

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.

Hint: tanθ is easier. How can you express 1/cos²θ in terms of tanθ?

 

[suppy123]

yup, it's \tan^2{\theta}+1
y = x(\tan{\theta})-(\frac{gx^2}{2vi^2})(\tan^2{\theta}+1)
\tan{2\theta} = \frac{2\tan{\theta}}{1-\tan^2{\theta}}
\Rightarrow \tan^2{\theta} = \frac{-2\tan{\theta}}{\tan{2\theta}}

so
y = x(\tan{\theta})-(\frac{gx^2}{2vi^2})(\frac{2\tan{\theta}}{\tan{2\theta}}+2)

 

[tiny-tim]

yup, it's \tan^2{\theta}+1

ok, so the original equation in terms of tanθ is … ?

 

[suppy123]

is that right?

 

[suppy123]

sry it should be
\Rightarrow \tan^2{\theta} = \frac{-2\tan{\theta}}{\tan{2\theta}}+1

y = x(\tan{\theta})-(\frac{gx^2}{2vi^2})(\frac{-2\tan{\theta}}{\tan{2\theta}}+2)

 

[tiny-tim]

sry it should be
\Rightarrow \tan^2{\theta} = \frac{-2\tan{\theta}}{\tan{2\theta}}+1

y = x(\tan{\theta})-(\frac{gx^2}{2vi^2})(\frac{-2\tan{\theta}}{\tan{2\theta}}+2)

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

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

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).

You want an equation with only tanθ, tan²θ, tan³θ, and so on.

 

[suppy123]

how about this?
y = x(\tan{\theta})-(\frac{gx^2}{2vi^2})(\tan{\theta}+1)(\tan{\theta}-1)+2

 

[tiny-tim]

how about this?
y = x(\tan{\theta})-(\frac{gx^2}{2vi^2})(\tan{\theta}-\sqrt{1})^2+1

ooo …

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

y = x(\tan{\theta})-(\frac{gx^2}{2vi^2})(\tan^2{\theta}+1)

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 ="

 

[suppy123]

lol...sweet, so
y = x(\tan{\theta})-(\frac{gx^2}{2vi^2})(\tan^2{\theta}+1)
0 = (-\frac{gx^2}{2vi^2})(\tan^2{\theta}) + x(\tan{\theta}) - (\frac{gx^2}{2vi^2} - y)
\tan{\theta} = \frac{-x \pm \sqrt{x^2-4(\frac{gx^2}{2vi^2}-y)^2}}{-2\frac{gx^2}{2vi^2-y}}

 

[suppy123]

oops, forgot to change the y heh

 

[tiny-tim]

\tan{\theta} = \frac{-x \pm \sqrt{x^2-4(\frac{gx^2}{2vi^2})^2}}{-2\frac{gx^2}{2vi^2}}

erm …

What happened to poor little y ? ​

 

[suppy123]

there you go
\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})}
;0

 

[tiny-tim]

ooh … there he is … ! ​

hmm … that last equation doesn't look right, though.

 

[suppy123]

should be right this time ;)
\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})}

 

[tiny-tim]

Woohoo!
​

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