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
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    What you are probably missing is the product rule for the derivative of a dot product.2012-06-17
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    _v_ is just the magnitude of **V**2012-06-17
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    @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

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