Arc Length of Parametric Equation

The arc length of a curve defined by parametric equations $(x(t), y(t))$ for $a \leq t \leq b$ is given by

$\text{Arc Length} = \int\limits _{a} ^{b} \sqrt{ \left( \frac{dx}{dt} \right)^{2} + \left( \frac{dy}{dt} \right)^{2} } \, dt$

In our problem, we can differentiate $x$ and $y$ with respect to $t$ to obtain $\frac{dx}{dt} = 6t$ and $\frac{dy}{dt} = 3t^{2} - 3$ . We want to find the arc length for $1 \leq t \leq 2$ . Therefore the arc length is

$\begin{aligned} \text{Arc Length} & = \int\limits _{1} ^{2} \sqrt{ \left( 6t \right)^{2} + \left( 3t^{2} - 3 \right)^{2} } \, dt \\ & = \int \limits_{1} ^{2} \sqrt{ 36t^{2} + \left( 9t^{4} + 9 - 18 t^{2}\right) } \, dt \\ & = \int \limits _{1} ^{2} \sqrt{ 9t^{4} + 9 + 18 t^{2} } \, dt \\ & = \int \limits _{1} ^{2} \sqrt{ (3t^{2} + 3)^{2} } \, dt \\ & = \int \limits _{1} ^{2} |3t^{2} + 3| \, dt \\ \end{aligned}$

Since $3t^{2} + 3$ is always positive, $|3t^{2} + 3| = 3t^{2} + 3$

$\begin{aligned} \text{Arc Length} & = \int \limits _{1} ^{2} 3t^{2} + 3 \, dt \\ & = [t^{3} + 3t ] _{1} ^{2} \\ & = (2^{3} + 3 \times 2) - (1^{3} + 3 \times 1) \\ & = (8 + 6 ) - (1 + 3) \\ & = 14 - 4 = \boxed{10} & _\square\\ \end{aligned}$