Proving that $\frac{\text{d}\vec{v}}{\text{dt}}=\frac{\text{d}|\vec{v}|}{\text{dt}}\hat{v}+\frac{\text{d}\hat{v}}{\text{dt}}|\vec{v}|$.

Question

We were taught the following equation on a physics lecture:

$$\frac{\text{d}\vec{v}}{\text{d}t}=\frac{\text{d}|\vec{v}|}{\text{d}t}\hat{v}+\frac{\text{d}\hat{v}}{\text{d}t}|\vec{v}|$$

where $\vec{v}$ is a velocity vector dependent on time $t$. I was wondering how the equation could be proven.

Answer 1

For scalar $\phi$ and vector $A$,$$\frac{\text{d}}{\text{d}t}(\phi A_i)=\frac{\text{d}\phi}{\text{d}t}A_i+\frac{\text{d}A_i}{\text{d}t}\phi\implies\frac{\text{d}}{\text{d}t}(\phi A)=\frac{\text{d}\phi}{\text{d}t}A+\frac{\text{d}A}{\text{d}t}\phi,$$by contraction with the standard basis $e^i$. Now take $\phi:=|v|,\,A:=\hat{v}$.

Answer 2

Just write

$$ \vec{v} = |\vec{v}| \hat{v} \tag{1} $$

that is to say: vector $\vec{v}$ is of magnitude $|\vec{v}|$ and pointing in the direction $\hat{u}$. After that, apply the product rule on Eq. (1)

Answer 3

If $\hat{v}$ is the unit vector with the same direction as $v$, then this should be true by the fact that $$ \vec{v}=|\vec{v}|\hat{v} $$ and by using the product rule for derivative.