Question

Use Theorem 11.24 to prove that the curvature of a linear function y = mx + b is zero for every value of x.

Short answer

It is proved thatthe curvature of a linear functiony = mx + b is zero for every value ofx.

Step-by-step solution

Step 1. Given Information.

The given linear function is$y=mx+b.$

Step 2. Prove.

To prove that the curvature of a linear function y = mx + b is zero for every value of x,we will use the formula for Curvature in the Plane.

The curvature in the plane is defined as letting y = f(x) be a twice-differentiable function. Then the curvature of the graph of fis given by $k=\frac{\left|f\text{'}\text{'}\left(x\right)\right|}{{\left(1+{\left(f\text{'}\left(x\right)\right)}^2\right)}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}.$

Now,

$$\begin{gathered} y=f\left(x\right)=mx+b \\ f\text{'}\left(x\right)=m \\ f\text{'}\text{'}\left(x\right)=0 \end{gathered}$$

Put all the values in the formula,

$$\begin{gathered} k=\frac{\left|0\right|}{{\left(1+m^2\right)}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}} \\ k=0 \end{gathered}$$

Hence proved.