Trying to understand following proof: "Show that if the dot product of the velocity and acceleration of a moving particle is positive (or negative), then the speed of the particle is increasing (decreasing)".
That will be the dot-product ($\cdot$) of vector $\vec v(t)$ (velocity) and vector $\vec a(t)$ (acceleration), where $\vec a(t)$ is equal to the derivative ($\frac{d}{dt}$) of vector $\vec v(t)$ ($\frac{d}{dt} \vec v(t) = \vec a(t)$), and the absolute value of the velocity $\vec v(t)=|\vec v(t)|$ is the moving particles speed.
The solution reads:
$ \frac{d}{dt} |\vec v(t)|^2 = \frac{d}{dt} \vec v(t) \cdot \vec v(t) = 2 \vec v(t) \cdot \vec a(t)$
if $\vec v(t) \cdot \vec a(t) > 0$ then $|\vec v(t)|$ is increasing
if $\vec v(t) \cdot \vec a(t) < 0$ then $|\vec v(t)|$ is decreasing
There seems to be something with the notation that I do not quite understand.
When I try myself:
$\vec v(t) \cdot \vec a(t) = \frac{d}{dt} \vec v(t) \cdot \vec v(t) = \frac{d}{dt}(\vec v(t))^2 = 2 \vec v(t) \cdot \vec a(t)$ (Last step according to the chain rule)
Which results in an inequality. If this would be the case then $\vec v(t) \cdot \vec a(t)$ would equal itself times 2 to the power of any natural number.
What is the correlation between $\vec v(t) \cdot \vec v(t)$ and $|\vec v(t)|^2$?
Best Answer
As for your last question, "What is the correlation between $v(t)\cdot v(t)$ and $|v(t)|^2$", they are exactly the same! For any vector u, |u| is defined as "$\sqrt{u\cdot u}$". Squaring both sides, $u\cdot u= |u|^2$.
Now for the main question- you are confusing yourself with poor notation, writing "d/dt v(t)" instead of "dv(t)/dt". Becaus of that notation, you are writing "d/dt v(t)v(t)= d/dt v^2(t)". That last step is wrong.
d/dt is being applied to v(t) THEN the result is multiplied by v(t). v(t) is NOT being multiplie by v(t) and then d/dt applied to the product. Had you written "(dv(t)/dt)v(t)" or, better, "$\frac{dv(t)}{dt}v(t)$" you would have seen that.