Question

Is the following linear function? Prove your conclusions by showing that f(r)satisfies both of the equations $\mathbf{f}\mathbf{(}{\mathbf{r}}_{\mathbf{1}}\mathbf{+}{\mathbf{r}}_{\mathbf{2}}\mathbf{)}\mathbf{=}\mathbf{f}\mathbf{\left(}{\mathbf{r}}_{\mathbf{1}}\mathbf{\right)}\mathbf{+}\mathbf{f}\mathbf{\left(}{\mathbf{r}}_{\mathbf{2}}\mathbf{\right)}$ , and f ( ar ) = af ( r ), where is a scalar, or that it does not satisfy at least one of these.

f (r) = r . r

Short answer

The function f(r) is not linear.

Step-by-step solution

Equations for linear function:

A function of a vector, sayf (r), is called linear if,$\mathbf{f}\mathbf{(}{\mathbf{r}}_{\mathbf{1}}\mathbf{+}{\mathbf{r}}_{\mathbf{2}}\mathbf{)}\mathbf{=}\mathbf{f}\mathbf{\left(}{\mathbf{r}}_{\mathbf{1}}\mathbf{\right)}\mathbf{+}\mathbf{f}\mathbf{\left(}{\mathbf{r}}_{\mathbf{2}}\mathbf{\right)}$, andf ( ar ) = af ( r ), whereis a scalar.

Check for the first equation:

The given function are as follow.

$$\begin{gathered} \mathrm{f}\left(\mathrm{r}\right)=\mathrm{r}.\mathrm{r} \\ \mathrm{f}({\mathrm{r}}_1+{\mathrm{r}}_2)=\mathrm{f}\left({\mathrm{r}}_1\right)+\mathrm{f}\left({\mathrm{r}}_2\right) \end{gathered}$$

And it is to be checked whether the function is linear or not.

The Left-Hand Side (LHS) of the equation is as given below.

$$\begin{gathered} \mathrm{f}({\mathrm{r}}_1+{\mathrm{r}}_2)=({\mathrm{r}}_1+{\mathrm{r}}_2)\times ({\mathrm{r}}_1+{\mathrm{r}}_2) \\ =({\mathrm{r}}_1\times {\mathrm{r}}_1)+({\mathrm{r}}_2\times {\mathrm{r}}_2)+(2{\mathrm{r}}_1\times {\mathrm{r}}_2) \end{gathered}$$

The Right-Hand Side (RHS) of the equation isas given below.

$\mathrm{f}\left({\mathrm{r}}_1\right)+f\left({\mathrm{r}}_2\right)=({\mathrm{r}}_1+{\mathrm{r}}_1)\times ({\mathrm{r}}_2+{\mathrm{r}}_2)$

As, LHS is not equal to RHS, so is does not satisfy the first equation. Therefore, the function is not linear.

Check for the second equation:

The Left-Hand Side (LHS) of the equation f ( ar ) = af ( r ) .

$$\begin{gathered} \mathrm{f}\left(\mathrm{ar}\right)=\mathrm{a}\times \mathrm{r}+\mathrm{a}\times \mathrm{r} \\ ={\mathrm{a}}^2(\mathrm{r}\times \mathrm{r}) \end{gathered}$$

The Right-Hand Side (RHS) of the equation f ( ar ) = af ( r ) .

$\mathrm{af}\left(\mathrm{r}\right)=\mathrm{ar}.\mathrm{r}$

As, LHS is not equal to RHS, so is does not satisfy the second equation.