How to differentiate an equation

[GBA13]

Homework Statement

Hi Everyone,

This seems like a very simple question but I' a bit confused.

In maths if I wanted to differentiate y = x^2 then it would just be 2x but I'm not sure about what you would do in physics. If you had the equation C = m/V (Concentration = mass/volume) how can you differenicate that with respect to time or something?

Homework Equations

The Attempt at a Solution

Surely what ever type of differentiation you did (normal, partial etc.) all the parts would just end up being 0 so the differential is zero.

Can someone please set me straight.

Thanks,

 

[mfb]

In maths if I wanted to differentiate y = x^2 then it would just be 2x

If you calculate the derivative with respect to x.

If you had the equation C = m/V (Concentration = mass/volume) how can you differenicate that with respect to time or something?

Do m or V depend on time?
If yes (how?), you have to take this into account. Simple example: m=c*t with some constant t leads to a non-zero time derivative.
If no, they are just constants.

 

[rude man]

Homework Statement

Hi Everyone,

This seems like a very simple question but I' a bit confused.

In maths if I wanted to differentiate y = x^2 then it would just be 2x but I'm not sure about what you would do in physics. If you had the equation C = m/V (Concentration = mass/volume) how can you differenicate that with respect to time or something?

Homework Equations

The Attempt at a Solution

Surely what ever type of differentiation you did (normal, partial etc.) all the parts would just end up being 0 so the differential is zero.

Can someone please set me straight.

Thanks,

Your example:
C = mV^-1
dC = ∂C/∂m dm + ∂C/∂V dV
so dC/dt = ∂C/∂m dm/dt + ∂C/∂V dV/dt.
But ∂C/∂m = 1/V and ∂C/∂V = -m/V²
So if you know how m and V vary with time you can compute dC/dt.

 

[Chestermiller]

You differentiate it using the quotient rule:

$$\frac{dC}{dt}=\frac{V\frac{dm}{dt}-m\frac{dV}{dt}}{V^2}=\frac{\frac{dm}{dt}-C\frac{dV}{dt}}{V}$$

Chet