Solving a Tricky Substitution with d^2x/dt^2

[strokebow]

Hi,

Firstly: This is not a homework Q. Check my previous posts, you will see the stuff i ask is for my own genuine learning.

Let:
$\vartheta$ = ($\Omega$*t)/2

Now I have: d^2x/dt^2

And I want to sub in for t.

So:

d/dt = d$\vartheta$/dt * d/d$\vartheta$
(Basic chain rule)

I can work out that: d$\vartheta$/dt = $\Omega$/2

So:

d/dt = $\Omega$/2 * d/d$\vartheta$

Now for the second derivative:

d^2/dt^2 = d$\vartheta$/dt * d/d$\vartheta$ * (d/dt)

Now, I already have an expression for d/dt = d$\vartheta$/dt * d/d$\vartheta$

So I can sub this in and get:

d^2/dt^2 = d$\vartheta$/dt * d/d$\vartheta$ * d$\vartheta$/dt * d/d$\vartheta$
Which is:
(correct me if I am wrong):
d^2/dt^2 = d$\vartheta$^2/dt^2 * d^2/d$\vartheta$^2

This is where I need the help of the experts :-)

The text I am trying to understand gives this:
d^2x/dt^2 = $\Omega$^2 / 4 * d^2x/d$\vartheta$^2

Any ideas how they have made that step... it seems like they have simply said:
d$\vartheta$/dt * d$\vartheta$/dt = (d$\vartheta$/dt)^2
Is this acceptable?
If $\Omega$ = 2$\pi$*(1/t). Then its like saying:
d^2/dt^2 (($\Omega$*t)/2) = $\Omega$^2 / 4 Any ideas/help?

thanks

 

[tiny-tim]

hi strokebow!

(try using the X² button just above the Reply box )

Let:
$\vartheta$ = ($\Omega$*t)/2
…
d/dt = d$\vartheta$/dt * d/d$\vartheta$
(Basic chain rule)
…
The text I am trying to understand gives this:
d²x/dt² = $\Omega$^2 / 4 * d^2x/d$\vartheta$^2

i don't really understand your equations

the way they got that is

d²x/dt²

= d/dt (dx/dt)

= {d/dθ (dx/dt)} dθ/dt

= {d/dθ (dθ/dt dx/dθ))} dθ/dt

= {(Ω/2 d²x/dθ²)} Ω/2

 

[strokebow]

= {d/dθ (dθ/dt dx/dθ))} dθ/dt

= {(Ω/2 d²x/dθ²)} Ω/2

Hi,

Thanks you for your reply! :-)

The inbetweener steps would be . . . (?)

= {d/dθ (dθ/dt dx/dθ))} dθ/dt

= {d/dθ dx/dθ (dθ/dt )} dθ/dt

= {d/dθ dx/dθ (Ω/2)} Ω/2

= {d²x/(dθ)² (Ω/2)} Ω/2

= {(Ω/2 d²x/dθ²)} Ω/2

yar?

 

[tiny-tim]

yes.