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| | ::<span class="exam"><math>x=t^2</math> | | ::<span class="exam"><math>x=t^2</math> |
| | ::<span class="exam"><math>y=t^3</math> | | ::<span class="exam"><math>y=t^3</math> |
| − | ::<span class="exam"><math>0\leq t \leq 2</math> | + | ::<span class="exam"><math>1\leq t \leq 2</math> |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| + | <hr> |
| − | !Foundations:
| + | [[009C Sample Final 2, Problem 10 Solution|'''<u>Solution</u>''']] |
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| − | |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
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| − | <math>L=\int_{\alpha}^{\beta} \sqrt{\bigg(\frac{dx}{dt}\bigg)^2+\bigg(\frac{dy}{dt}\bigg)^2}~dt.</math>
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| − | |}
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| − | '''Solution:''' | + | [[009C Sample Final 2, Problem 10 Detailed Solution|'''<u>Detailed Solution</u>''']] |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 1:
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| − | |-
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| − | |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>
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| − | |}
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 2:
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| − | |Now, we have
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| − | <math>\begin{array}{rcl}
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| − | \displaystyle{L} & = & \displaystyle{\int_1^2 \sqrt{4t^2+9t^4}~dt}\\
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| − | &&\\
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| − | & = & \displaystyle{\int_1^2 \sqrt{t^2(4+9t^2)}~dt}\\
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| − | &&\\
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| − | & = & \displaystyle{\int_1^2 t\sqrt{4+9t^2}~dt.}\\
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| − | \end{array}</math>
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| − | |}
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 3:
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| − | |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.
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| − | |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>
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| − | |Hence,
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| − | | <math>\begin{array}{rcl}
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| − | \displaystyle{L} & = & \displaystyle{\frac{1}{18}\int_{13}^{40} \sqrt{u}~du}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{1}{27} u^{\frac{3}{2}}\bigg|_{13}^{40}}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{1}{27}(40)^{\frac{3}{2}}-\frac{1}{27}(13)^{\frac{3}{2}}.}\\
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| − | \end{array}</math>
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| − | |}
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Final Answer:
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| − | | <math>\frac{1}{27}(40)^{\frac{3}{2}}-\frac{1}{27}(13)^{\frac{3}{2}}</math>
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| − | |}
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| | [[009C_Sample_Final_2|'''<u>Return to Sample Exam</u>''']] | | [[009C_Sample_Final_2|'''<u>Return to Sample Exam</u>''']] |
