# Thread: 2nd order parametric differentiation problem

1. ## 2nd order parametric differentiation problem

I'm slightly confused with this problem:
$\displaystyle x(t) = t^3 + 3t + 1$ & $\displaystyle y(t) = ln(3t^2 - 3)$

Find the function $\displaystyle p(t)$ such that:

$\displaystyle \frac{d^2y}{dx^2} = p(t)\frac{d^2y}{dt^2}\frac{d^2t}{dx^2}$

I'm new to this subject so any help would be really appreciated.

Dojo

2. Originally Posted by dojo
I'm slightly confused with this problem:
$\displaystyle x(t) = t^3 + 3t + 1$ & $\displaystyle y(t) = ln(3t^2 - 3)$

Find the function $\displaystyle p(t)$ such that:

$\displaystyle \frac{d^2y}{dx^2} = p(t)\frac{d^2y}{dt^2}\frac{d^2t}{dx^2}$

I'm new to this subject so any help would be really appreciated.

Dojo
First note that

$\displaystyle \displaystyle \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$

and that

$\displaystyle \displaystyle \frac{d^2y}{dx^2}=\frac{\frac{d}{dt}\left(\frac{dy }{dx} \right)}{\frac{dx}{dt}}$

Now just compute all of the derivatives and solve for $\displaystyle p(t)$

3. Thanks for your help. I end up with this monster! $\displaystyle 1/9\, \left( 3\,{t}^{2}+3 \right) \left( t-1 \right) ^{-1} \left( t+1 \right) ^{-1} \left( {t}^{2}+1 \right) ^{-1} \left( 6\, \left( 3\,{t} ^{2}-3 \right) ^{-1}-36\,{\frac {{t}^{2}}{ \left( 3\,{t}^{2}-3 \right) ^{2}}} \right) ^{-1}$

4. is my answer right?