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  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{du}{18}=tdt.}
Also, since this is a definite integral, we need to change the bounds of integration.
We have
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u_1=4+9(1)^2=13}   and  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u_2=4+9(2)^2=40.}
Hence,
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{L} & = & \displaystyle{\frac{1}{18}\int_{13}^{40} \sqrt{u}~du}\\ &&\\ & = & \displaystyle{\frac{1}{27} u^{\frac{3}{2}}\bigg|_{13}^{40}}\\ &&\\ & = & \displaystyle{\frac{1}{27}(40)^{\frac{3}{2}}-\frac{1}{27}(13)^{\frac{3}{2}}.}\\ \end{array}}

Final Answer:
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1}{27}(40)^{\frac{3}{2}}-\frac{1}{27}(13)^{\frac{3}{2}}}

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