# How do you determine the length of a parametric curve?

Aug 29, 2015

${\int}_{f \left({t}_{1}\right)}^{f \left({t}_{2}\right)} \sqrt{1 + {\left(\frac{g ' \left(t\right)}{f ' \left(t\right)}\right)}^{2}} f ' \left(t\right) \mathrm{dt}$ (with respect to $x$) OR ${\int}_{g \left({t}_{1}\right)}^{g \left({t}_{2}\right)} \sqrt{1 + {\left(\frac{f ' \left(t\right)}{g ' \left(t\right)}\right)}^{2}} g ' \left(t\right) \mathrm{dt}$ (with respect to $y$)

#### Explanation:

Let the curve $C$ be defined as $x = f \left(t\right)$ and $y = g \left(t\right)$

Then taking the derivative with respect to $t$: $\frac{\mathrm{dx}}{\mathrm{dt}} = f ' \left(t\right)$ and $\frac{\mathrm{dy}}{\mathrm{dt}} = g ' \left(t\right)$

$\setminus R i g h t a r r o w \int \mathrm{dx} = \int f ' \left(t\right) \mathrm{dt}$ and $\int \mathrm{dy} = \int g ' \left(t\right) \mathrm{dt}$

and $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{g ' \left(t\right)}{f ' \left(t\right)}$ and $\frac{\mathrm{dx}}{\mathrm{dy}} = \frac{f ' \left(t\right)}{g ' \left(t\right)}$

The length of the arc $L$ between $\left({x}_{1} , {y}_{1}\right)$ and $\left({x}_{2} , {y}_{2}\right)$ is given by the formula(e):
$L = {\int}_{{x}_{1}}^{{x}_{2}} \sqrt{1 + {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2}} \mathrm{dx} = {\int}_{{y}_{1}}^{{y}_{2}} \sqrt{1 + {\left(\frac{\mathrm{dx}}{\mathrm{dy}}\right)}^{2}} \mathrm{dy}$

Substituting accordingly:

$L = {\int}_{f \left({t}_{1}\right)}^{f \left({t}_{2}\right)} \sqrt{1 + {\left(\frac{g ' \left(t\right)}{f ' \left(t\right)}\right)}^{2}} f ' \left(t\right) \mathrm{dt} = {\int}_{g \left({t}_{1}\right)}^{g \left({t}_{2}\right)} \sqrt{1 + {\left(\frac{f ' \left(t\right)}{g ' \left(t\right)}\right)}^{2}} g ' \left(t\right) \mathrm{dt}$

Where $f \left({t}_{i}\right) = {x}_{i}$ and $g \left({t}_{i}\right) = {y}_{i} , i = 1 , 2$