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Question Number 137627 by rs4089 last updated on 04/Apr/21

Answered by MrGaster last updated on 16/Feb/25

$$$$$$\mathrm{Prove}:\frac{d}{dt}\left[\overrightarrow{u}\frac{d\overrightarrow{u}}{dt}\right]=\overrightarrow{u}\frac{d^2\overrightarrow{u}}{{dt}^2}+\frac{d\overrightarrow{u}}{dt}\frac{d\overrightarrow{u}}{dt}$$$$\frac{d}{dt}\left[\overrightarrow{u}\frac{d\overrightarrow{u}}{dt}\right]=\overrightarrow{u}\frac{d^2\overrightarrow{u}}{{dt}^2}+{\left(\frac{d\overrightarrow{u}}{dt}\right)}^2$$$$\frac{d}{dt}\left[\overrightarrow{u}\frac{d\overrightarrow{u}}{dt}\frac{d^{2}u}{{dt}^2}\right]=\frac{d}{dt}\left[\overrightarrow{u}\frac{d^2\overrightarrow{u}}{{dt}^2}\right]\frac{d\overrightarrow{u}}{dt}+\overrightarrow{u}\frac{d}{dt}\left[\frac{d\overrightarrow{u}}{dt}\frac{d^2\overrightarrow{u}}{{dt}^2}\right]$$$$\frac{d}{dt}\left[\overrightarrow{u}\frac{d^2\overrightarrow{u}}{{dt}^2}\right]=\overrightarrow{u}\frac{d^3\overrightarrow{u}}{{dt}^3}+\frac{d\overrightarrow{u}}{dt}\frac{d^2\overrightarrow{u}}{{dt}^2}$$$$\frac{d}{dt}\left[\overrightarrow{u}\frac{d\overrightarrow{u}}{dt}\frac{d^2\overrightarrow{u}}{{dt}^2}\right]=(\overrightarrow{u}\frac{d^3\overrightarrow{u}}{{dt}^3}+\frac{d\overrightarrow{u}}{dt}\frac{d^2\overrightarrow{u}}{{dt}^2})\frac{d\overrightarrow{u}}{dt}+\overrightarrow{u}\frac{d}{dt}\left[\frac{d\overrightarrow{u}}{dt}\frac{d^2\overrightarrow{u}}{{dt}^2}\right]$$$$\frac{d}{dt}\left[\frac{d\overrightarrow{u}}{dt}\frac{d^2\overrightarrow{u}}{{dt}^2}\right]=\frac{d^2\overrightarrow{u}}{dt}\frac{d^2\overrightarrow{u}}{{dt}^2}+\frac{d\overrightarrow{u}}{dt}\frac{d^3\overrightarrow{u}}{{dt}^3}$$$$=\overrightarrow{u}\frac{d\overrightarrow{u}}{dt}\frac{d^3\overrightarrow{u}}{dt}+{\left(\frac{d\overrightarrow{u}}{dt}\right)}^2=\frac{d^2\overrightarrow{u}}{{dt}^2}+\overrightarrow{u}\frac{d^2\overrightarrow{u}}{{dt}^2}\frac{d^2\overrightarrow{u}}{{dt}^2}$$$$\begin{array}{c}\frac{d}{dt}\left[\overrightarrow{u}\frac{d\overrightarrow{u}}{dt}\frac{d^2\overrightarrow{u}}{{dt}^2}\right]=\left[\stackrel{\mathrm{\_}}{u}\frac{d\overrightarrow{u}}{dt}\frac{d^3\overrightarrow{u}}{{dt}^3}\right]\\ \frac{d}{dt}\left[u\frac{d\overrightarrow{u}}{dt}\frac{d^2\overrightarrow{u}}{{dt}^2}\right]=\left[u\frac{d\overrightarrow{u}}{dt}\frac{d^3\overrightarrow{u}}{{dt}^3}\right]\end{array}$$

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