Classical physics Time dependent vector calculation

[wavecaster]

Homework Statement

If A is a time dependent vector, calculate

\int_{t1}^{t2} dtA(t) \times \frac{d^2A}{dt^2} [\itex]<br /> <br /> <h2>Homework Equations</h2><br /> <br /> <br /> <br /> <h2>The Attempt at a Solution</h2><br /> <br /> I think we should somehow relate it with something&#039;s derivative.<br /> <br /> \int_{t1}^{t2}A(t)\frac{d^2A(t)}{dt^2}dt=<br /> 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=<br /> A(t)\frac{dA(t)}{dt}\left.\right|_a^b\int_{t1}^{t2}\left ( \frac{dA(t)}{dt} \right )^2dt=<br /> 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

 

[SteamKing]

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[Orodruin]

$$\int_{t1}^{t2}A(t)\frac{d^2A(t)}{dt^2}dt=
A(t)\int_{t1}^{t2}\frac{d^2A(t)}{dt^2}dt-\int_{t1}^{t2}\frac{dA(t)}{dt}\left ( \int \frac{d^2A(t)}{dt^2} \right )dt
=
A(t)\frac{dA(t)}{dt}\left.\right|_a^b\int_{t1}^{t2}\left ( \frac{dA(t)}{dt} \right )^2dt
=
A(t2)\frac{dA(t)}{dt}\left.\right|_{t_2}-A(t1)\frac{dA(t)}{dt}\left.\right|_{t_1}-\int_{t1}^{t2} \left ( \frac{dA(t)}{dt} \right )^2dt$$

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?