Let Rbe the resistance of${\mathbf{R}}_{\mathbf{1}}\mathbf{=}\mathbf{25}\mathbf{}\mathbf{ohms}$and${\mathbf{R}}_{\mathbf{2}}\mathbf{=}\mathbf{15}\mathbf{}\mathbf{ohms}$ohms in parallel. (See Chapter 2, Problem 16.6.) If$\mathbf{}{\mathbf{R}}_{\mathbf{1}}$is changed to$\mathbf{}\mathbf{25}\mathbf{.}\mathbf{1}\mathbf{}\mathbf{ohms}$, find$\mathbf{}{\mathbf{R}}_{\mathbf{2}}\mathbf{}$so that$\mathbf{R}$is not changed.

Let R be the resistance of$R_1=25\Omega ,R_2=15\Omega$ in parallel.

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

Consider a function$\mathbf{y}\mathbf{=}\mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$, find the value of a function$\mathbf{}\mathbf{}\mathbf{y}\mathbf{=}\mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$whenlocalid="1659324930540" $\mathbf{x}\mathbf{=}\mathbf{x}\mathbf{\text{'}}$.

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

localid="1659324996210" $\frac{\mathbf{d}}{\mathbf{dx}}\mathbf{=}\mathbf{\left(}\mathbf{f}\mathbf{\right(}\mathbf{x}\mathbf{\left)}\mathbf{\right)}$is Change in y with respect to change in x as localid="1659325022795" $\mathbf{dx}\mathbf{\to }\mathbf{0}\mathbf{.}$

If the value of $x=x\text{'}$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.

Differentials can be utilized to construct a relationship that can be used to compute or $d{R}_2$the tiny variation in the second distance, and then the new image distance $\left(R_{2n}\right)$can be determined using the parallel resistors addition formula.

As a result, it is possible to rewrite the differences using differentials:

$\frac{1}{R_1}+\frac{1}{R_2}=\frac{1}{R}$

On differentiate,

$\begin{array}{l}-\frac{1}{R_1^2}d{R}_1-\frac{1}{R_2^2}d{R}_2=0\\ d{R}_2=-\frac{R_2^2}{R_1^2}d{R}_1\end{array}$

Substitute the provided values,

$$\begin{gathered} d{R}_1=25.1-25 \\ =0.1 \end{gathered}$$

Here,

$$\begin{gathered} d{R}_2=\frac{-{\left(15\right)}^2}{{\left(25\right)}^2}\times \left(0.1\right) \\ =-0.036 \end{gathered}$$

Then,

$$\begin{gathered} R_{2n}=R_2+d{R}_2 \\ =15-0.036 \\ =14.964 \end{gathered}$$

Hence,$R_{2n}=14.964$