Q 69. - Calculus

Question

Use your result from Exercise $68$ to show that the arc length formula for a function $y = f \left(x\right)$ is a special case of the arc length formula for a parametric curve.

Step-by-Step Solution

Answer

${\int }_a^b\sqrt{1+{\left(f^{\text{'}}\left(t\right)\right)}^2}dt$

Step 1: Given information

$y = f \left(x\right)$

Step 2: Calculation

Consider the function $y=f\left(x\right)$

For a parameter $t$ the function $y=f\left(t\right)$ for some $t\in [a,b]$

The goal is to determine the curve's arc length.

If the curve $\mathrm{C}$ is expressed by parametric equations $x=f\left(t\right), y=y\left(t\right)$ on the interval $[a, b]$ then the arc length is given by the formula,

${\int }_a^b\sqrt{{\left(f^{\text{'}}\left(t\right)\right)}^2+{\left(g^{\text{'}}\left(t\right)\right)}^2}dt$

Thus, $f\left(t\right)=x\Rightarrow f^{\text{'}}\left(t\right)=\frac{dx}{dt}$

$g\left(t\right)=y\Rightarrow g^{\text{'}}\left(t\right)=\frac{dy}{dt}$

Substituting the values of $f^{\text{'}}\left(t\right),g^{\text{'}}\left(t\right)$ then the arc length is

Arc length $={\int }_a^b\sqrt{{\left(\frac{d}{dt}x\right)}^2+{\left(\frac{d}{dt}y\right)}^2}dt$

$={\int }_a^b\sqrt{{\left(\frac{dx}{dt}\right)}^2+{\left(\frac{dy}{dt}\right)}^2}dt$

Then,

Arc length $={\int }_a^b\sqrt{1+{\left(\frac{dy}{dt}\right)}^2}dt$

Arc length $={\int }_a^b\sqrt{1+{\left(f^{\text{'}}\left(t\right)\right)}^2}dt [$ since , for some parameter $t]$ Therefore the arc length of the curve $y=f\left(t\right)$ is ${\int }_a^b\sqrt{1+{\left(f^{\text{'}}\left(t\right)\right)}^2}dt$