Differential calculus - Error calculation

Question

Let $V_0$ be the volume of a cuboid with the edges $a_0, \ b_0, \ c_0> 0$.

A measurement of the edge lengths yields the values ​​$a, \ b, \ c$ with the absolute errors $\Delta a=a-a_0, \ \Delta b=b-b_0 , \ \Delta c=c-c_0$.

The volume of the cuboid is thus determined with an absolute error $\Delta V = V-V_0$.

How can we explain the following approximation formula for the relative error by means of differential calculus? $$\frac{\Delta V}{V_0}\sim \frac{\Delta a}{a_0}+\frac{\Delta b}{b_0}+\frac{\Delta c}{c_0}$$

$$$$

We have that $V(a,b,c)=V(a_0+\Delta a,b_0+\Delta b, c_0+\Delta c)$

Do we use the Taylor expansion? Or is an other meant by "by means of differential calculus" ?

Answer 1

It suffices to use the concept of differential by the gradient, that is

$$f(x+h)=f(x)+\nabla f(x)\cdot h +o(|h|)\implies f(x+h)\approx f(x)+\nabla f(x)\cdot h\\\implies \Delta f=f(x+h)-f(x) \approx \nabla f(x)\cdot h$$

Indeed in this case we have

• $V=abc\implies \nabla V=(bc,ac,ab)$
• $h=(\Delta a,\Delta b,\Delta c)$

thus

$$\Delta V\approx \nabla V(a_0,b_0,c_0)\cdot h=b_0c_0\Delta a+a_0c_0\Delta b+a_0b_0\Delta c$$

and

$$\frac{\Delta V}{V_0}\approx\frac{\Delta a}{a_0}+\frac{\Delta b}{b_0}+\frac{\Delta c}{c_0}$$