# Math Help - differental eq from calc

1. ## differental eq from calc

if x=t^(2)+1 and y=t^3 than d^2y/dt^2=?

2. Originally Posted by AnnaBee123
if x=t^(2)+1 and y=t^3 than d^2y/dt^2=?
Do you need $\frac{d^2y}{dt^2}$ or $\frac{d^2y}{dx^2}~?$

3. Originally Posted by AnnaBee123
if x=t^(2)+1 and y=t^3 than d^2y/dt^2=?

well lets get started....

$\frac{dy}{dt}=3t^2$

Now if we take the derivative of the above with respect to t we get

$\frac{d}{dt}\left(\frac{dy}{dt} \right)=\frac{d}{dt}\left(3t^2 \right)$

$\frac{d^2y}{dt^2}=6t$

4. Originally Posted by Plato
Do you need $\frac{d^2y}{dt^2}$ or $\frac{d^2y}{dx^2}~?$
I was thinking the same as there was an equation given for both

$y=f(t)$ and $x=f(t)$

Therefore a solution could be

$\frac{d^2y}{dx^2} = \frac{\tfrac{d^2y}{dt^2}}{\tfrac{d^2x}{dt^2}}$

$= \frac{6t}{2}$

$= 3t$

5. Originally Posted by pickslides
I was thinking the same as there was an equation given for both

$y=f(t)$ and $x=f(t)$

Therefore a solution could be

$\frac{d^2y}{dx^2} = \frac{\tfrac{d^2y}{dt^2}}{\tfrac{d^2x}{dt^2}}$

$= \frac{6t}{2}$

$= 3t$
We need to be careful...

$x=t^2+1$ $y=t^3$

so $\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}= \frac{3t^2}{2t}=\frac{3}{2}t$

Now $\frac{d^2y}{dx^2}=\frac{\frac{d}{dt}\frac{dy}{dx}} {\frac{dx}{dt}}=\frac{\frac{3}{2}}{2t}=\frac{3}{4t }$

Note we can write y as a function of x as follows

$y=(x-1)^{\frac{3}{2}}$

Now if we take two derivatives we get

$\frac{dy}{dx}=\frac{3}{2}(x-1)^{\frac{1}{2}}$

$\frac{d^2y}{dx^2}=\frac{3}{4}\frac{1}{(x-1)^{\frac{1}{2}}}$

Note that since $x=t^2+1$ we get

$\frac{d^2y}{dx^2}=\frac{3}{4}\frac{1}{(x-1)^{\frac{1}{2}}}=\frac{3}{4}\frac{1}{(t^2+1-1)^{\frac{1}{2}}}=\frac{3}{4t}$