My problem states that $\vec{F}$ is a conservative field, ie: $F = \nabla \phi$ for some scalar potential $\phi$. $$\begin{align}\vec{F}&=m\vec{a}\\ &=m\frac{\mathrm{d}^2\vec{r}}{\mathrm{d}t^2} \end{align}$$ I now take the dot product of each side:

$$\vec{F} \cdot \frac{\mathrm{d}\vec{r}}{\mathrm{d}t}= m\frac{\mathrm{d}^2\vec{r}}{\mathrm{d}t^2}\cdot \frac{\mathrm{d}\vec{r}}{\mathrm{d}t}$$

Now, the textbook says that:

$$m\frac{\mathrm{d}\vec{r}}{\mathrm{d}t} \cdot \frac{\mathrm{d}^2\vec{r}}{\mathrm{d}t^2} = \frac{m}{2} \frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\mathrm{d}\vec{r}}{\mathrm{d}t}\right)^2$$

How did the textbook get this answer?: $\frac{m}{2} \frac{d}{dt}\left(\frac{d\vec{r}}{dt}\right)^2$ ?

All of this leads to the result, which I'm OK with:

$$\int \limits_{A}^{B} \vec{F} \cdot \mathrm{d}\vec{r} = \bigg[\frac{m}{2}v^2\bigg]^{B}_{A}$$


So first off the right side of the equation that is confusing you involves the derivative of the square of the speed. The speed is not a vector (which is why squaring it even makes sense). The square of the speed is the dot product of the velocity with itself.

That said, to go from right to left in the equation that is confusing you, you can use the fact that $\frac{d}{dt} (\mathbf{u}(t) \cdot \mathbf{v}(t)) = \mathbf{u}(t) \cdot \mathbf{v}'(t) + \mathbf{u}'(t) \cdot \mathbf{v}(t)$. (This is just the product rule applied to each component.)

Another way to arrive at this is to say $\frac{dv^2}{dt}=2v v'$ using the chain rule and then note that $v'=\frac{\mathbf{v}}{v} \cdot \mathbf{a}$ (i.e. the change in the speed is the tangential component of the acceleration).


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:

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

We can prove the derivative table entry as follows:

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

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

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

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

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


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