What is the Derivation of Work in Classical Dynamics?

[sloane729]

Homework Statement

I'm working through a derivation for work in Thornton's Classical Dynamics but I'm stuck at one step.
$$\begin{align}<br /> \vec{F} \cdot d\vec{r} &= m\frac{d\vec{v}}{dt} \cdot \frac{d\vec{r}}{dt}dt = m\frac{d\vec{v}}{dt} \cdot \vec{v}dt \\ &= \frac{m}{2}\frac{d}{dt}(\vec{v}\cdot \vec{v})dt \end{align}$$

I'm having trouble getting from the last equality on the first line to the second line.

Homework Equations

The Attempt at a Solution

I don't know what the mathematical reasoning is.

 

[MrWarlock616]

I don't get it too. Might be a mistake.

 

[sloane729]

Okay I understand now from here: http://en.wikipedia.org/wiki/Kinetic_energy#Derivation
But I've never seen differentials like $$d(\vec{v}\cdot\vec{v})$$ before. Why am I not allowed to say $$d(\vec{v}\cdot\vec{v}) = dv^2$$?

For anyone else wondering about this see here also: http://math.stackexchange.com/questions/159284/product-rule-for-the-derivative-of-a-dot-product

 

[HallsofIvy]

Okay I understand now from here: http://en.wikipedia.org/wiki/Kinetic_energy#Derivation
But I've never seen differentials like $$d(\vec{v}\cdot\vec{v})$$ before. Why am I not allowed to say $$d(\vec{v}\cdot\vec{v}) = dv^2$$?

You can't say that because you haven't defined "v^2". Since v is a vector, what you really mean is $\vec{v}\cdot\vec{v}= ||\vec{v}||^2$.
But, the basic idea is correct:
$$\frac{d(\vec{v}\cdot\vec{v})}{dt}= 2\vec{v}\cdot\frac{d\vec{v}}{dt}$$

For anyone else wondering about this see here also: http://math.stackexchange.com/questions/159284/product-rule-for-the-derivative-of-a-dot-product