009C Sample Final 2, Problem 10 Detailed Solution

Find the length of the curve given by

$x=t^{2}$

$y=t^{3}$

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 1\leq t\leq 2}

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Background Information:
The arc length  $L$  of a parametric curve with  $\alpha \leq t\leq \beta$  is given by
$L=\int _{\alpha }^{\beta }{\sqrt {{\bigg (}{\frac {dx}{dt}}{\bigg )}^{2}+{\bigg (}{\frac {dy}{dt}}{\bigg )}^{2}}}~dt.$

Solution:

Step 1:
First, we need to calculate  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{dx}{dt}}   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{dy}{dt}.}
Since  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 x=t^2,~\frac{dx}{dt}=2t.}
Since  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 y=t^3,~\frac{dy}{dt}=3t^2.}
Using the arc length formula, 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 L=\int_1^2 \sqrt{(2t)^2+(3t^2)^2}~dt.}

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

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

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