# Strain Gauge Formula Derivation

1. Apr 1, 2015

### yosimba2000

Using R = ρL/A,
I am trying to get this formula, which is the change in resistance due to change in length, area, and resistivity.

ΔR = (∂R/∂L)ΔL + (∂R/∂A)ΔA + (∂R/∂ρ)Δρ

I understand the above conceptually, but I am confused about why we are multiplying by terms ΔL, ΔA, and Δρ.
Intuitively, I think it is just ΔR = (∂R/∂L) + (∂R/∂A) + (∂R/∂ρ), as it says "adding up the changes in resistance by changes in length, area, and resistivity equals total resistance change". And by unit analysis, I agree that multiplying by ΔL, ΔA, and Δρ is correct, but can anyone show me how to derive this mathematically?

Thanks.

*edit*
Just thought about it some more, and while it doesn't satisfy me, it makes a bit more sense.
You can just take R = ρL/A and take the partial of it WRT L and get ∂R/∂L = ρ/A.
Multiply denominator over to get ∂R = (ρ/A)∂L
Rewrite as ΔR = (ρ/A)ΔL
Realize that (ρ/A) is ∂R/∂L and substitute back in.
ΔR = (∂R/∂L)ΔL, then do the same with A and ρ.
However, this seems kind of backwards...

Last edited: Apr 1, 2015
2. Apr 1, 2015

### SteamKing

Staff Emeritus
It's a math concept known as the total derivative:

http://en.wikipedia.org/wiki/Total_derivative

You are linearizing the change in resistance R in terms of the changes in length, area, and resistivity of the wire