# Doubt in differentiation

#### logearav

Joined Aug 19, 2011
243
dW = -ΔV(r,t).dr = -(dV/dt - ∂V/∂t)dt,
This is how given in my book. I understand the first part ie dV/dt, but can't understand how ∂V/∂t comes. Please help, revered members.

#### logearav

Joined Aug 19, 2011
243
Revered members,
why none has answered my query. Just give some tips, so that i could try something. Please help.

Joined Jul 7, 2009
1,583
You don't define your terms, so I for one don't know the context. Don't assume there is any standardization of notation (there isn't). An easy way to give the context is to include a digital photo of the relevant book material.

Are these mathematical equations or do they represent some physical process? If they represent a physical process and the variables r and t represent some physical parameters, then the equation is not dimensionally consistent. Here's why.

Suppose V is in units of "vees", r is in units of "ars", and t is in units of "tees". Then the ΔV*dr has units vees*ars and the right hand side has units vees. This is an incorrect equation. Hint: always check the units of your equations, even if they are purely mathematical -- it can help uncover math or assumption mistakes.

#### logearav

Joined Aug 19, 2011
243
dW = -ΔV(r,t).dr = -(dV/dt - ∂V/∂t)dt
Thanks for the reply someonesdad. Here W is work done, so for some infinitesimal work done, dW = -F.dr, where F is the force and dr is displacement. So i presume, here Δ stands for differentiation. dV/dt is ok, but how ∂V/dt arises?

#### Zazoo

Joined Jul 27, 2011
114
If ΔV represents ∂V/∂r. Working backwards from the end result:

-(dV/dt - ∂V/∂t) dt = -( [∂V/∂t + (∂V/∂r)(dr/dt)] - ∂V/∂t) dt

= -(∂V/∂r)(dr/dt)) dt = -(∂V/∂r) dr = -ΔV dr