# How do you find the exact arc length of a parametric equation ? x(t)=2 + 9t^2 y(t)=9 + 6t^3

May 5, 2018

The length
$L = {\int}_{0}^{1} \sqrt{{\left(18 t\right)}^{2} + {\left(18 {t}^{2}\right)}^{2}} \cdot \mathrm{dt} = {\left[6 {\left({t}^{2} + 1\right)}^{\frac{3}{2}}\right]}_{0}^{1} = 10.97056$

#### Explanation:

if a curve is given by a parametric equations

$x \left(t\right) = 2 + 9 {t}^{2}$

$y \left(t\right) = 9 + 6 {t}^{3}$

where 0 ≤ t ≤ 1

the length of the curve is given by

$L = {\int}_{a}^{b} \sqrt{{\left(\frac{\mathrm{dx}}{\mathrm{dt}}\right)}^{2} + {\left(\frac{\mathrm{dy}}{\mathrm{dt}}\right)}^{2}} \cdot \mathrm{dt}$

$\frac{\mathrm{dx}}{\mathrm{dt}} = 18 t$

$\frac{\mathrm{dy}}{\mathrm{dt}} = 18 {t}^{2}$

$L = {\int}_{0}^{1} \sqrt{{\left(18 t\right)}^{2} + {\left(18 {t}^{2}\right)}^{2}} \cdot \mathrm{dt} = {\int}_{0}^{1} \sqrt{324 {t}^{2} + 324 {t}^{4}} \cdot \mathrm{dt}$

$L = {\int}_{0}^{1} \sqrt{324 {t}^{2} + 324 {t}^{4}} \cdot \mathrm{dt} = {\int}_{0}^{1} \sqrt{\left(324 {t}^{2}\right) \cdot \left(1 + {t}^{2}\right)} \cdot \mathrm{dt} = {\int}_{0}^{1} \left(18 \cdot t\right) \cdot \sqrt{1 + {t}^{2}} \cdot \mathrm{dt}$

$L = {\int}_{0}^{1} \left(18 \cdot t\right) \cdot \sqrt{1 + {t}^{2}} \cdot \mathrm{dt}$

$= {\left[6 \cdot {\left({t}^{2} + 1\right)}^{\frac{3}{2}}\right]}_{0}^{1} = 18 \cdot \left({2}^{\frac{3}{2}} / 3 - \frac{1}{3}\right) = 3 \cdot {2}^{\frac{5}{2}} - 6 = 10.97056$