Without using a computer or a calculator, estimate the change in length of a space diagonal of a box whose dimensions are changed from $\mathbf{200}\mathbf{\times }\mathbf{200}\mathbf{\times }\mathbf{100}$to $\mathbf{}\mathbf{201}\mathbf{\times }\mathbf{202}\mathbf{\times }\mathbf{99}\mathbf{.}$

Provided:

The change in length of a space diagonal of a box whose dimensions are changed from $200\times 200\times 100$to$201\times 202\times 99$ .

This method is based on the derivatives of functions whose values must be calculated at certain locations, as the name implies.

Consider a function $\mathit{y}\mathbf{=}\mathit{f}\left(x\right)$, find the value of a function$\mathit{y}\mathbf{=}\mathit{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$when$\mathit{x}\mathbf{=}\mathit{x}\mathbf{\text{'}}$.

As an example, the derivative of a function $\mathit{y}\mathbf{=}\mathit{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$with respect to xwill be employed.

$\frac{\mathbf{d}}{\mathbf{d}\mathbf{x}}\mathbf{=}\left(f\left(x\right)\right)$is Change in y with respect to change in x as $\mathit{d}\mathit{x}\mathbf{\Rightarrow }\mathbf{0}$.

If the value of role="math" localid="1659172184296" $x=x\text{'}$from a value of x near it, such that the difference in the two values, dx, is vanishingly small, one can derive the change in the value of the function $y=f\left(x\right)$corresponding to the change dx in x. In practice, however, the concept of vanishingly small is not possible.

Provide the differential of the function,$f(x,y,z)=\sqrt{x^2+y^2+z^2}$to get an estimate of the variation where x, y and z define the box's dimensions and the function denotes the length of the diagonals.

$f(x,y,z)=\sqrt{x^2+y^2+z^2}for\sqrt{(201{)}^2+({202}^2)+(99{)}^2}$

Provided $x=200;dx=1.0,y=200;dy=2.0,z=100;dz=-1.0$

Put all the provided values in above equation,

$$\begin{gathered} f\left(200,200,100\right)=\sqrt{{\left(200\right)}^2+{\left(200\right)}^2+{\left(100\right)}^2} \\ =\sqrt{90000} \\ =300 \end{gathered}$$

On differentiate,

$$\begin{gathered} df=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy+\frac{\partial f}{\partial z}dz \\ =\frac{xdx}{{\left(x^2+y^2+z^2\right)}^{\frac{1}{2}}}+\frac{ydy}{{\left(x^2+y^2+z^2\right)}^{\frac{1}{2}}}+\frac{zdx}{{\left(x^2+y^2+z^2\right)}^{\frac{1}{2}}} \end{gathered}$$

Substitute the provide values in above equation.

role="math" localid="1659172533369" $$\begin{gathered} df=\frac{200\times 1.0}{{\left(x^2+y^2+z^2\right)}^{{\displaystyle \frac{1}{2}}}}+\frac{200\times 2.0}{{\left(x^2+y^2+z^2\right)}^{{\displaystyle \frac{1}{2}}}}+\frac{100\times \left(-1.0\right)}{{\left(x^2+y^2+z^2\right)}^{{\displaystyle \frac{1}{2}}}} \\ =\frac{200\times 1.0+200\times 2.0+100\times \left(-1.0\right)}{300} \\ =\frac{500}{300} \end{gathered}$$

Therefore,

$df=\frac{5}{3}$