Suppose we're given the position $x$ , velocity $\dot{x}$ , and acceleration $\ddot{x}$ at $t_1 $.
We need to approximate the position $x$ at $t_2$.
In below picture, I understand how $x_1$ represents the vertical black length, and how $ \color{blue}{\dot{x_1}(t_1)}h$ represents the vertical blue length. However I don't get how $\color{green}{\ddot{x}(t_1)}\frac{h^2}{2}$ represents the green length. I'm wondering if there a simple derivation for proving that the green vertical length equals $\color{green}{\ddot{x}(t_1)}\frac{h^2}{2}$ . Any ideas ? Thanks !
- aylo
You are assuming that the accelearation is constant with data at $t_1$ according to $x_1,x'(t_1),x''(t_1)$ and it results in the quadratic curve,
$$ \text{quadratic}(h) = x_1 + x'(t_1)h + x''(t_1)\frac{h^2}{2} $$
Now you did understand the black and blue part. The green is then what remains to get to the quadratic approximation, which is $x''(t_1) h^2/2$ according to the formula and also according to the diagram.