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Consider some vector field $\underline{u}(\underline{r}(t),t).$ Find the total derivative $\frac{du}{dt}$.

What I have done so far is $$\frac{du}{dt} = \frac{∂u}{∂r} \frac{dr}{dt} + \frac{∂u}{∂t} \frac{dt}{dt}.$$

This is a 'show that' question so what I need to show is that $\frac{∂u}{∂r} \frac{dr}{dt} =(u \cdot ∇)u.$

I wrote this in components: $$\left[ \frac{∂u}{∂x_1} \frac{dx_1}{dt} e_1 + \frac{∂u}{∂x_2}\frac{dx_2}{dt} e_2 + \frac{∂u}{∂x_3} \frac{dx_3}{dt}e_3 \right]$$

From this it seems that I have $(∇\cdot u)u$ instead? Any ideas?

Oh, and for the partial derivative with respect to r, I believe this should be with respect to |r|.

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