$(\frac{d}{dt}\vec{r})\cdot(\frac{d}{dt}(\frac{d}{dt}\vec{r})) = \frac{1}{2}\frac{d}{dt}(\frac{d}{dt}\vec{r})^2$

Question

prove the following is true:

$(\frac{d}{dt}\vec{r})\bullet(\frac{d}{dt}(\frac{d}{dt}\vec{r})) = \frac{1}{2}\frac{d}{dt}(\frac{d}{dt}\vec{r})^2$

where:

$\bullet \equiv \text{dot product}$

$\vec{r} \equiv x\hat{\text{i}} + y\hat{\text{j}} + z\hat{\text{k}}$

I tried applying chain rule on right hand side. This gave me:

$(\frac{d}{dt}\vec{r})\bullet(\frac{d}{dt}(\frac{d}{dt}\vec{r})) = \frac{1}{2} 2(\frac{d}{dt}\vec{r})\frac{d}{dt}(\frac{d}{dt}\vec{r})$

$(\frac{d}{dt}\vec{r})\bullet(\frac{d}{dt}(\frac{d}{dt}\vec{r})) = (\frac{d}{dt}\vec{r})\frac{d}{dt}(\frac{d}{dt}\vec{r})$

not sure how I can add the dot product back in to the right hand side to get them to be equal...

also, is there a way to prove this by transforming the left-hand side instead of the right-hand side?

because its backwards for the exercise that I took this math snippet from to guess backwards from the answer of the problem...

Answer 1

Just put $$ {\bf v} = \frac{d{\bf r}}{dt}. $$ Then, your identity becomes $$ {\bf v}\cdot\frac{d{\bf v}}{dt}=\frac{1}{2}\frac{d}{dt}{\bf v}^2= \frac{1}{2}\frac{d}{dt}({\bf v}\cdot{\bf v}) $$ that you will recognize as the standard derivative of powers like $x^2/2$ with respect to $x$.

Answer 2

Your manipulation is essentially correct. The thing to remember is that if we have a vector $\vec{v}$, then the notation $\vec{v}^2$ means $\vec{v} \cdot \vec{v}$. Thus, for example we actually have $\frac{d}{dt} (\vec{r}^2 )= 2 \vec{r} \cdot \frac{d}{dt} \vec{r}$. Do you see how this tells fixes your issue?

Answer 3

This is from a standard table of derivatives in vector calculus:

$\vec{v}\cdot \frac{d\vec{v}}{dt} = \frac{1}{2}\frac{d}{dt}v^2$

Subsitute $\vec{v} = \frac{d\vec{r}}{dt}$ into this eqation:

$\frac{d\vec{r}}{dt} \bullet \frac{d^2 \vec{r}}{dt^2} = \frac{1}{2}\frac{d}{dt}(\frac{d\vec{r}}{dt})^2$

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We can prove the derivative table entry as follows:

$\frac{d\vec{v}^2}{dt} = \frac{d}{dt}(\vec{v}\cdot \vec{v})$

$\frac{d}{dt}\vec{v}^2 = \frac{d}{dt}(\vec{v})\cdot \vec{v} + \vec{v}\cdot \frac{d}{dt}(\vec{v})$

$\frac{d}{dt}\vec{v}^2 = \vec{v} \cdot \frac{d}{dt}(\vec{v}) + \vec{v}\cdot \frac{d}{dt}(\vec{v})$

$\frac{d}{dt}\vec{v}^2 = 2(\vec{v} \cdot \frac{d}{dt}(\vec{v}))$

$\boxed{\frac{1}{2}\frac{d}{dt}\vec{v}^2 = \vec{v} \cdot \frac{d}{dt}(\vec{v})}$