0

Trying to derive the work equation in a conservative field after a few years of zero practice had me stump for an hour with simple derivatives.

I know that: $\mathrm d(\vec{v} \cdot \vec{v}) = (\mathrm d\vec{v})\cdot\vec{v} + \vec{v}\cdot(\mathrm d\vec{v}) = 2 (\vec{v}\cdot\mathrm d\vec{v})$

How do I show that: $\vec{v}\cdot\mathrm d(m\vec{v}) = \frac{m}{2}\mathrm d(\vec{v}\cdot \vec{v})$?

In this case, $m$ is a constant value.

This is in reference to Derivation in the link kinetic_energy

Graham Kemp
  • 129,094
Mathematicing
  • 6,253
  • 1
    I'm not really sure what kind of derivative you are doing, but it looks like you can multiply both sides of $d(v\cdot v)=2(v\cdot dv)$ by $\frac m 2$ and put the $m$ inside the derivative because it's constant. Does that work? – Jackozee Hakkiuz May 25 '21 at 02:36
  • @JackozeeHakkiuz I have added a link in the OP – Mathematicing May 25 '21 at 02:41
  • 1
    $\vec{v}.d(m\vec{v})=m\vec{v}.d(\vec{v})$ from the equation $\vec{v}d\vec{v}=\frac{d(\vec{v}.\vec{v})}{2}$ we get,the required result? – IITM May 25 '21 at 02:42
  • you're just multiplying the first equation by $m/2$ and then moving the $m$ inside the derivative because it's a constant – peek-a-boo May 25 '21 at 02:50

1 Answers1

1

Begin as before: $$\begin{align}\mathrm d(\vec v\cdot m\vec v) = (\mathrm d \vec v)\cdot (m\vec v)+\vec v\cdot\mathrm d (m\vec v)\end{align}$$

Now $m$ is a scalar constant, so for any $\vec\mu,\vec\nu$ we have: $${\vec \mu\cdot(m\vec \nu)=m\,(\vec \mu\cdot\vec \nu)\\\mathrm d(m\vec \mu)=m\,\mathrm d \vec \nu}$$

So apply these rules ...

Graham Kemp
  • 129,094