# 009C Sample Final 2, Problem 10

Find the length of the curve given by

$x=t^{2}$ $y=t^{3}$ $0\leq t\leq 2$ Foundations:
The formula for the arc length  $L$ of a parametric curve with  $\alpha \leq t\leq \beta$ is

$L=\int _{\alpha }^{\beta }{\sqrt {{\bigg (}{\frac {dx}{dt}}{\bigg )}^{2}+{\bigg (}{\frac {dy}{dt}}{\bigg )}^{2}}}~dt.$ Solution:

Step 1:
First, we need to calculate  ${\frac {dx}{dt}}$ and  ${\frac {dy}{dt}}.$ Since  $x=t^{2},~{\frac {dx}{dt}}=2t.$ Since  $y=t^{3},~{\frac {dy}{dt}}=3t^{2}.$ Using the formula in Foundations, we have

$L=\int _{1}^{2}{\sqrt {(2t)^{2}+(3t^{2})^{2}}}~dt.$ Step 2:
Now, we have

${\begin{array}{rcl}\displaystyle {L}&=&\displaystyle {\int _{1}^{2}{\sqrt {4t^{2}+9t^{4}}}~dt}\\&&\\&=&\displaystyle {\int _{1}^{2}{\sqrt {t^{2}(4+9t^{2})}}~dt}\\&&\\&=&\displaystyle {\int _{1}^{2}t{\sqrt {4+9t^{2}}}~dt.}\\\end{array}}$ Step 3:
Now, we use  $u$ -substitution.
Let  $u=4+9t^{2}.$ Then,  $du=18tdt$ and  ${\frac {du}{18}}=tdt.$ Also, since this is a definite integral, we need to change the bounds of integration.
We have
$u_{1}=4+9(1)^{2}=13$ and  $u_{2}=4+9(2)^{2}=40.$ Hence,
${\begin{array}{rcl}\displaystyle {L}&=&\displaystyle {{\frac {1}{18}}\int _{13}^{40}{\sqrt {u}}~du}\\&&\\&=&\displaystyle {{\frac {1}{18}}\cdot {\frac {2}{3}}u^{\frac {3}{2}}{\bigg |}_{13}^{40}}\\&&\\&=&\displaystyle {{\frac {1}{27}}(40)^{\frac {3}{2}}-{\frac {1}{27}}(13)^{\frac {3}{2}}.}\\\end{array}}$ ${\frac {1}{27}}(40)^{\frac {3}{2}}-{\frac {1}{27}}(13)^{\frac {3}{2}}$ 