# 009C Sample Final 2, Problem 10

Find the length of the curve given by

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

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

Solution:

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

${\displaystyle L=\int _{1}^{2}{\sqrt {(2t)^{2}+(3t^{2})^{2}}}~dt.}$

Step 2:
Now, we have

${\displaystyle {\begin{array}{rcl}\displaystyle {L}&=&\displaystyle {\int _{1}^{2}{\sqrt {4t^{2}+9t^{4}}}~dt}\\&&\\&=&\displaystyle {\int _{1}^{2}{\sqrt {t^{2}(4+9t^{2})}}~dt}\\&&\\&=&\displaystyle {\int _{1}^{2}t{\sqrt {4+9t^{2}}}~dt.}\\\end{array}}}$

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

${\displaystyle {\frac {1}{27}}(40)^{\frac {3}{2}}-{\frac {1}{27}}(13)^{\frac {3}{2}}}$