Question

Find the exact length of the curve. x = 9 + 9t2, y = 6 + 6t3, 0 ≤ t ≤ 4

Answer 1

ANSWER

$=102 \sqrt{ 17}$

EXPLANATION

The equation of the curve is:

$x = 9 + 9 {t}^{2}$

$y = 6 + 6 {t}^{3}$

We differentiate to obtain:

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

$\frac{dy}{dt} = 18t$

The length of the arc

$l = \int _0 ^{4} \sqrt{ {( \frac{dx}{dt} )}^{2} + {( \frac{dy}{dt} )}^{2} } dt$

This implies that:

$= \int _0 ^{4} \sqrt{ {( 18t )}^{2} + {( {18 {t}^{2} } )}^{2} } dt$

$= \int _0 ^{4} 18t \sqrt{ 1+ { {18 {t}^{2} } }} dt$

$=102 \sqrt{ 17}$