2
$\begingroup$

Can someone help me understand why is this true:

$$\vec{v} \cdot d \vec{v} = v \cdot dv$$

where $v$ is speed? I found somewhere that $\vec{v} \cdot d \vec{v}=|\vec{v}||d \vec{v}| \cos \phi$. And I don't know from where it follows that $|d \vec{v}| \cdot \cos \phi = dv$.

$\endgroup$

closed as off-topic by Jon Custer, John Rennie, Emilio Pisanty, user259412, Qmechanic Sep 13 '17 at 11:48

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – Qmechanic
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 3
    $\begingroup$ What operator do you mean to represent with $*$? If you mean the dot product (inner product) you can use $\vec{v}\cdot d\vec{v}$ (using \cdot for the operator). $\endgroup$ – The Photon Sep 11 '17 at 18:43
  • 1
    $\begingroup$ You want to show $\vec{v} \cdot d \vec{v} = v dv$ where $v = |\vec{v}|$. See my answer here - physics.stackexchange.com/questions/211235/… which addresses a similar question. $\endgroup$ – Prahar Sep 11 '17 at 18:56
  • $\begingroup$ Context may matter here. If your source is talking about a particular case where they know that $\mathrm{d}\vec{v} \parallel \vec{v}$ then the answer is easy: it's only true in certain cases but those include the one your source cares about. $\endgroup$ – dmckee Sep 11 '17 at 19:10
  • $\begingroup$ Is this really a maths question? $\endgroup$ – jim Sep 11 '17 at 21:29
8
$\begingroup$

Let $v=|\mathbf v|$. Then by Leibniz rule $$\text d v^2 = 2v\text dv$$ Now $v^2 = \mathbf v\cdot\mathbf v$ by definition, so we also have $$\text dv^2 = \text d(\mathbf v\cdot\mathbf v)=2\mathbf v\cdot\text d\mathbf v$$ whence $$v\text d v = \mathbf v\cdot\text d\mathbf v.$$

$\endgroup$
2
$\begingroup$

I don't know from where it follows that $|d \vec{v}| \cdot cos \phi = dv$.

Not strict, yet more intuitive answer:

Imagine $d \vec{v}$ as a sum of two components:

  • along $\vec{v}$
  • and perpendicular to it.

The former changes only the magnitude of $\vec{v}$ (i.e. $v$) so it acts as $dv$; and the latter changes only the direction of $\vec{v}$.

Then geometrically, if $\phi$ is the angle between $\vec{v}$ and $d \vec{v}$, the former component magnitude is $|d \vec{v}| \cdot cos \phi$.

$\endgroup$
1
$\begingroup$

Update 2 after @DavidZ said I had not answered the question and quite correctly deleted my answer.
I rewrote my answer but could not undelete it as it was deleted by a moderator so this is the reason for a second answer.

This is an attempt to illustrate the two interpretations of the change in velocity, $d\vec v$ and $|d \vec v|$, which is possibly the reason for the confusion.

Here is a diagram which shows how under the action of a force $\vec F$ for a short interval of time the initial velocity $\vec v$ changes to a final velocity $\vec v + d \vec v$

enter image description here

$\vec v \cdot \Delta \vec v = v \, |\Delta \vec v| \cos \phi$ where $|\Delta \vec v|$ is the magnitude of the change in velocity $\Delta \vec v$

As the time interval over which the acceleration took place tends to zero this equation becomes

$\vec v \cdot d \vec v = v \, |d \vec v| \cos \phi$

The magnitude of the final velocity $|\vec v + \Delta \vec v| = v +\Delta v = v\cos \Delta \phi+ |\Delta \vec v|\cos(\phi - \Delta \phi)$

As the time interval over which the force acts tends to zero $\Delta \phi \rightarrow 0$

$v\cos \Delta \phi \rightarrow v$ and $|\Delta \vec v|\cos(\phi - \Delta \phi) \rightarrow |d \vec v|\cos\phi $ which is $dv$.

Finally the desired result $\vec v \cdot d \vec v = v \, |d \vec v| \cos \phi = v \, dv$

$\endgroup$
1
$\begingroup$

The usual case is in 3-dimensional. In this case we have \begin{equation} \vec v \cdot d \vec v = v_x dv_x + v_y dv_y+ v_z dv_z \;. \tag{1} \end{equation} On the other hand the magnitude of a vector $v$ is given by \begin{equation} v =\sqrt{v_x^2 +v_y^2+v_z^2} \;, \tag{2} \end{equation} or \begin{equation} v^2 =v_x^2 +v_y^2+v_z^2 \;. \tag{3} \end{equation} Differentiate on eqn (3) gives us \begin{equation} 2v \,dv = 2v_x dv_x + 2v_y dv_y+ 2v_z dv_z \;. \tag{4} \end{equation} From eqn(1) and eqn(4) we have $$\vec v \cdot d \vec v = v dv$$

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.