MathAdmin: Created page with "Find the length of the curve given by :: $x=t^2$ :: $y=t^3$ :: $0\leq t \l..."$

2017-03-12T16:56:38Z

Created page with "Find the length of the curve given by ::<math>x=t^2</math> ::<math>y=t^3</math> ::<math>0\leq t \l..."

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Find the length of the curve given by
::<math>x=t^2</math>
::<math>y=t^3</math>
::<math>0\leq t \leq 2</math>

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|The formula for the arc length  <math style="vertical-align: 0px">L</math>  of a parametric curve with  <math style="vertical-align: -4px">\alpha \leq t \leq \beta </math>  is
|-
|
       <math>L=\int_{\alpha}^{\beta} \sqrt{\bigg(\frac{dx}{dt}\bigg)^2+\bigg(\frac{dy}{dt}\bigg)^2}~dt.</math>
|}

'''Solution:'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First, we need to calculate  <math style="vertical-align: -14px">\frac{dx}{dt}</math>  and  <math style="vertical-align: -14px">\frac{dy}{dt}.</math>
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|Since  <math style="vertical-align: -14px">x=t^2,~\frac{dx}{dt}=2t.</math>
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|Since  <math style="vertical-align: -14px">y=t^3,~\frac{dy}{dt}=3t^2.</math>
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|Using the formula in Foundations, we have
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|
       <math>L=\int_1^2 \sqrt{(2t)^2+(3t^2)^2}~dt.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, we have
|-
|
       <math>\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}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 3:  
|-
|Now, we use  <math>u</math>-substitution.
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|Let  <math style="vertical-align: -2px">u=4+9t^2.</math>
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|Then,  <math style="vertical-align: -2px">du=18tdt</math>  and  <math style="vertical-align: -15px">\frac{du}{18}=tdt.</math>
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|Also, since this is a definite integral, we need to change the bounds of integration.
|-
|We have
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|       <math style="vertical-align: -5px">u_1=4+9(1)^2=13</math>  and  <math style="vertical-align: -5px">u_2=4+9(2)^2=40.</math>
|-
|Hence,
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|       <math>\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}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
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|        <math>\frac{1}{27}(40)^{\frac{3}{2}}-\frac{1}{27}(13)^{\frac{3}{2}}</math>
|}
[[009C_Sample_Final_2|'''Return to Sample Exam''']]

009C Sample Final 2, Problem 10 - Revision history

MathAdmin: Created page with "Find the length of the curve given by ::x=t^2 ::y=t^3 ::0\leq t \l..."

MathAdmin: Created page with "Find the length of the curve given by :: $x=t^2$ :: $y=t^3$ :: $0\leq t \l..."$