# Classical physics Time dependent vector calculation

1. Jul 9, 2014

### wavecaster

1. The problem statement, all variables and given/known data
If A is a time dependent vector, calculate

[itex] \int_{t1}^{t2} dtA(t) \times \frac{d^2A}{dt^2} [\itex]

2. Relevant equations

3. The attempt at a solution

I think we should somehow relate it with something's derivative.

\int_{t1}^{t2}A(t)\frac{d^2A(t)}{dt^2}dt=
A(t)\int_{t1}^{t2}\frac{d^2A(t)}{dt^2}dt-\int_{t1}^{t2}\frac{dA(t)}{dt}\left ( \int \frac{d^2A(t)}{dt^2} \right )dt=
A(t)\frac{dA(t)}{dt}\left.\right|_a^b\int_{t1}^{t2}\left ( \frac{dA(t)}{dt} \right )^2dt=
A(t2)\frac{dA(t)}{dt}\left.\right|_t_{2}-A(t1)\frac{dA(t)}{dt}\left.\right|_t_{1}-\int_{t1}^{t2}\left ( \frac{dA(t)}{dt} \right )^2dt

Last edited: Jul 9, 2014
2. Jul 9, 2014

### SteamKing

Staff Emeritus
Your Latex seems to have blown up. Would you like to edit the OP and try asking your question again?

Hint: before posting, use the Preview Button to check the appearance of your post and correct as required.

3. Jul 9, 2014

### Orodruin

Staff Emeritus
Your notation is a bit horrible, you really should not write $A(t)$ outside of the integrals and you are missing the $\times$ from the original expression - making it unclear whether or not it is a cross product or a scalar product. The approach with using partial integration does make sense (and in case of a cross product, what is $\dot A \times \dot A$?). However, it is not necessary to use partial integration as you can rewrite $A\times \ddot A$ directly as the total derivative of something. What total derivative would have the given term as one of its terms? What would be the value of the other term?