# Why is $\vec{v}\cdot d \vec{v} = v dv$? [closed]

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 $|d \vec{v}| \cdot \cos \phi = dv$.

• 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). – The Photon Sep 11 '17 at 18:43
• 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. – Prahar Sep 11 '17 at 18:56
• 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. – dmckee --- ex-moderator kitten Sep 11 '17 at 19:10
• Is this really a maths question? – jim Sep 11 '17 at 21:29

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 $$v\text d v = \mathbf v\cdot\text d\mathbf v.$$

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 $|d \vec{v}| \cdot cos \phi$.

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 $\vec v + d \vec v$

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

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