Q. 77 - Calculus

Question

Prove, in two ways, that the arc length of a linear function $f\left(x\right)=mx+c$ on an interval $\left[a,b\right]$ is equal to $(b-a)\sqrt{1+m^2}$: (a) by using the distance formula; (b) by using Theorem 6.7. 

Step-by-Step Solution

Answer

Arc length of a linear function $f\left(x\right)=mx+c$ on an interval $\left[a,b\right]$is equal to $(b-a)\sqrt{1+m^2}$.

Part (a) Step 1. Given information

A linear function, $f\left(x\right)=mx+c$.

Part (a) Step 2. Proof by using the distance formula

$$\begin{gathered} \mathrm{The} \mathrm{distance} \mathrm{from} (a, f(a\left)\right)=(a, ma+c) \mathrm{to} (b, f(b\left)\right)=(b, mb+c) \mathrm{is} \mathrm{given} \mathrm{by}, \\ D=\sqrt{{(b-a)}^2 + \left(\right(mb+c) - {(ma+c))}^2} \\ D=\sqrt{{(b-a)}^2 + m^{2 }(b-{a)}^2} \\ D=\sqrt{ {(b-a)}^2 \left(1+m^2\right)} \\ D=(b-a)\sqrt{1+m^2} \\ \mathrm{Since} \mathrm{the} \mathrm{graph} \mathrm{of} f\left(x\right) = mx + c \mathrm{is} \mathrm{a} \mathrm{line}, \mathrm{this} \mathrm{is} \mathrm{the} \mathrm{exact} \mathrm{arc} \mathrm{length} \end{gathered}$$

Part (b) step 1. Proof by using Theorem 6.7

$$\begin{gathered} \mathrm{The} \mathrm{arc} \mathrm{length} \mathrm{of} f\left(x\right) \mathrm{from} x=a \mathrm{to} x=b \mathrm{can} \mathrm{be} \mathrm{represented} \mathrm{by} \mathrm{the} \\ \mathrm{definite} \mathrm{integral}:{\int }_{\mathrm{a}}^{\mathrm{b}}\sqrt{1 + ( \mathrm{f}\text{'}{\left(\mathrm{x}\right))}^2}d\mathrm{x} \\ \mathrm{For} \mathrm{the} \mathrm{given} \mathrm{linear} \mathrm{function}, \mathrm{f}\left(\mathrm{x}\right)=\mathrm{mx}+\mathrm{c}, \mathrm{f}\text{'}\left(\mathrm{x}\right)=\mathrm{m} \\ \mathrm{Therefore}, \mathrm{D} = {\int }_{\mathrm{a}}^{\mathrm{b}}\sqrt{1 + {\left(\mathrm{m}\right)}^2}d\mathrm{x} \\ \mathrm{D}=\left(\sqrt{1 + {\mathrm{m}}^2}\right) {\int }_{\mathrm{a}}^{\mathrm{b}}1 d\mathrm{x} \\ \mathrm{D}=\left(\sqrt{1 + {\mathrm{m}}^2}\right) \left[ {\overline{)\mathrm{x}}}_{\mathrm{a}}^{\mathrm{b}} \right] \\ \mathrm{D}=\left(\sqrt{1 + {\mathrm{m}}^2}\right)(\mathrm{b}-\mathrm{a}) \end{gathered}$$