3
$\begingroup$

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...

  • 2
    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 $\mathbf{v}$.2012-06-17
  • 1
    What you are probably missing is the product rule for the derivative of a dot product.2012-06-17
  • 0
    _v_ is just the magnitude of **V**2012-06-17
  • 0
    @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.2012-06-17
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    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.2012-06-17
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    @Arturo, it's fairly common in physics, or at least in the physics textbooks I've seen, to let *italic letter* implicitly denote the magnitude of the **bold letter** vector.2012-06-17
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    @Rahul: And since the OP took the time to tell us that $\mathbf{v}$ was a vector, and did not assume we would guess that from notation, why not take the time to also tell us what $v$ was?2012-06-17
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    @Arturo: Sometimes I don't even do the former when I'm feeling lazy, so I'm not going to say anything! :)2012-06-17

2 Answers 2

9

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 $$\frac{d\mathbf{v}}{dt}\cdot \mathbf{v}(t) = \frac{1}{2}\frac{d}{dt}\lVert \mathbf{v}\rVert^2.$$

  • 0
    It was so simple, yet I didn't know what to look for. Thanks a lot.2012-06-17
  • 0
    @Raul if it helped you, mark this as an answer :)2016-12-18
7

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.

  • 0
    Brilliant! thanks a lot.2012-06-17