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I can't find the reason for this simplification, I understand that the dot product of a vector with itself would give the magnitude of that squared, so that explains the v squared. What I don't understand is where did the 2 under the "m" come from.

(The bold v's are vectors.)

$$m\int \frac{d\mathbf{v}}{dt} \cdot \mathbf{v} dt = \frac{m}{2}\int \frac{d}{dt}(\mathbf{v}^2)dt$$

Thanks.

Maybe the book's just wrong and that 2 should't be there...

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First, you are not taking the dot product of a vector with itself, you are taking the dot product of the derivative of a vector with the vector. Second, you don't tell us the relation between $v$ and $\mathbf{v}$. –  Arturo Magidin Jun 17 '12 at 1:47
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What you are probably missing is the product rule for the derivative of a dot product. –  Gerry Myerson Jun 17 '12 at 1:48
    
v is just the magnitude of V –  Raul Jun 17 '12 at 1:49
    
@Raul: Well, shouldn't you have said that when you also told us what $\mathbf{v}$ was? The government really doesn't like it when I read minds without a warrant, so I try not to. –  Arturo Magidin Jun 17 '12 at 1:59
    
So cant I just do the dot product of v . v and put that in the derivative? d(v.v)/dt, this way integrating the derivative would cancel eachother and give mv^2 ...oh but that 2! haha srry about the mind reading thing. –  Raul Jun 17 '12 at 2:01
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2 Answers

The derivative of the dot product is given by the rule $$\frac{d}{dt}\Bigl( \mathbf{r}(t)\cdot \mathbf{s}(t)\Bigr) = \mathbf{r}(t)\cdot \frac{d\mathbf{s}}{dt} + \frac{d\mathbf{r}}{dt}\cdot \mathbf{s}(t).$$

Therefore, $$\begin{align*} \frac{d}{dt} \lVert \mathbf{r}(t)\rVert^2 &= \frac{d}{dt}\left( \mathbf{r}(t)\cdot \mathbf{r}(t)\right)\\ &= 2\mathbf{r}(t)\cdot \frac{d\mathbf{r}}{dt}. \end{align*}$$ Dividing by through by $2$, we get $$\frac{d\mathbf{v}}{dt}\cdot \mathbf{v}(t) = \frac{1}{2}\frac{d}{dt}\lVert \mathbf{v}\rVert^2.$$

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It was so simple, yet I didn't know what to look for. Thanks a lot. –  Raul Jun 17 '12 at 2:06
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If two $n$-dimensional vectors $\mathbf u$ and $\mathbf v$ are functions of time, the derivative of their dot product is given by $$\frac{\mathrm d}{\mathrm dt}(\mathbf u\cdot\mathbf v) = \mathbf u\cdot\frac{\mathrm d\mathbf v}{\mathrm dt} + \mathbf v\cdot\frac{\mathrm d\mathbf u}{\mathrm dt}$$ This is analogous to (and indeed, is easily derived from) the product rule for scalars, $\frac{\mathrm d}{\mathrm dt}(ab) = a\frac{\mathrm db}{\mathrm dt} + b\frac{\mathrm da}{\mathrm dt}$.

Therefore, $$\frac{\mathrm d}{\mathrm dt} \lVert\mathbf v\rVert^2 = \frac{\mathrm d}{\mathrm dt}(\mathbf v\cdot\mathbf v) = \mathbf v\cdot\frac{\mathrm d\mathbf v}{\mathrm dt} + \mathbf v\cdot\frac{\mathrm d\mathbf v}{\mathrm dt} = 2\mathbf v\cdot\frac{\mathrm d\mathbf v}{\mathrm dt}$$ just like $\frac{\mathrm d}{\mathrm dt} a^2 = 2a\frac{\mathrm da}{\mathrm dt}$. Halve that, and you have the result you need.

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Brilliant! thanks a lot. –  Raul Jun 17 '12 at 2:05
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