009C Sample Final 2, Problem 10

Find the length of the curve given by

$x=t^{2}$

$y=t^{3}$

$0\leq t\leq 2$

ExpandFoundations:
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:

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

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

ExpandStep 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,
${\begin{array}{rcl}\displaystyle {L}&=&\displaystyle {{\frac {1}{18}}\int _{13}^{40}{\sqrt {u}}~du}\\&&\\&=&\displaystyle {{\frac {1}{18}}\cdot {\frac {2}{3}}u^{\frac {3}{2}}{\bigg |}_{13}^{40}}\\&&\\&=&\displaystyle {{\frac {1}{27}}(40)^{\frac {3}{2}}-{\frac {1}{27}}(13)^{\frac {3}{2}}.}\\\end{array}}$

ExpandFinal Answer:
${\frac {1}{27}}(40)^{\frac {3}{2}}-{\frac {1}{27}}(13)^{\frac {3}{2}}$

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