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| | ::<math>y^3=x</math> | | ::<math>y^3=x</math> |
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| − | <span class="exam">between <math style="vertical-align: -5px">(0,0)</math> and <math style="vertical-align: -5px">(1,1)</math> about the <math style="vertical-align: -4px">y</math>-axis. | + | <span class="exam">between <math style="vertical-align: -5px">(0,0)</math> and <math style="vertical-align: -5px">(1,1)</math> about the <math style="vertical-align: -4px">y</math>-axis. |
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| | <span class="exam">(b) Find the length of the arc | | <span class="exam">(b) Find the length of the arc |
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| | ::<math>y=1+9x^{\frac{3}{2}}</math> | | ::<math>y=1+9x^{\frac{3}{2}}</math> |
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| − | <span class="exam">between the points <math style="vertical-align: -5px">(1,10)</math> and <math style="vertical-align: -5px">(4,73).</math> | + | <span class="exam">between the points <math style="vertical-align: -5px">(1,10)</math> and <math style="vertical-align: -5px">(4,73).</math> |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| + | <hr> |
| − | !Foundations:
| + | [[009B Sample Final 2, Problem 5 Solution|'''<u>Solution</u>''']] |
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| − | |'''1.''' The formula for the length <math style="vertical-align: 0px">L</math> of a curve <math style="vertical-align: -5px">y=f(x)</math> where <math style="vertical-align: -3px">a\leq x \leq b</math> is | |
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| − | <math>L=\int_a^b \sqrt{1+\bigg(\frac{dy}{dx}\bigg)^2}~dx.</math>
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| − | |-
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| − | |'''2.''' The surface area <math style="vertical-align: 0px">S</math> of a function <math style="vertical-align: -5px">y=f(x)</math> rotated about the <math style="vertical-align: -4px">y</math>-axis is given by
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| − | <math style="vertical-align: -13px">S=\int 2\pi x\,ds,</math> where <math style="vertical-align: -18px">ds=\sqrt{1+\bigg(\frac{dx}{dy}\bigg)^2}dy.</math>
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| − | |}
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| − | '''Solution:''' | + | [[009B Sample Final 2, Problem 5 Detailed Solution|'''<u>Detailed Solution</u>''']] |
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| − | '''(a)'''
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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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| − | |We start by calculating <math>\frac{dx}{dy}.</math>
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| − | |Since <math style="vertical-align: -13px">x=y^3,~ \frac{dx}{dy}=3y^2.</math>
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| − | |Now, we are going to integrate with respect to <math>y.</math>
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| − | |-
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| − | |Using the formula given in the Foundations section,
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| − | |-
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| − | |we have
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| − | |-
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| − | | <math>\begin{array}{rcl}
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| − | \displaystyle{S} & = & \displaystyle{\int_0^1 2\pi x \sqrt{1+(3y^2)^2}~dy}\\
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| − | &&\\
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| − | & = & \displaystyle{2\pi \int_0^1 y^3 \sqrt{1+9y^4}~dy.}
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| − | |-
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| − | |where <math>S</math> is the surface area.
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| − | \end{array}</math>
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| − | |}
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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 use <math>u</math>-substitution.
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| − | |Let <math>u=1+9y^4.</math>
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| − | |Then, <math>du=36y^3dy</math> and <math>\frac{du}{36}=y^3dy.</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>u_1=1+9(0)^4=1</math> and <math>u_2=1+9(1)^4=10.</math>
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| − | |-
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| − | |Thus, we get
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| − | | <math>\begin{array}{rcl}
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| − | \displaystyle{S} & = & \displaystyle{\frac{2\pi}{36} \int_1^{10} \sqrt{u}~du}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{\pi}{27} u^{\frac{3}{2}}\bigg|_1^{10}}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{\pi}{27} (10)^{\frac{3}{2}}-\frac{\pi}{27}.}
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| − | \end{array}</math>
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| − | |}
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| − |
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| − | '''(b)'''
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| − |
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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 calculate <math>\frac{dy}{dx}.</math>
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| − | |Since <math>y=1+9x^{\frac{3}{2}},</math> we have
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| − | | <math>\frac{dy}{dx}=\frac{27\sqrt{x}}{2}.</math>
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| − | |Then, the arc length <math>L</math> of the curve is given by
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| − | |-
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| − | | <math>L=\int_1^4 \sqrt{1+\bigg(\frac{27\sqrt{x}}{2}\bigg)^2}~dx.</math>
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| − | |}
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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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| − | |-
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| − | |Then, we have
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| − | |-
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| − | | <math>L=\int_1^4 \sqrt{1+\frac{27^2x}{2^2}}~dx.</math>
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| − | |-
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| − | |Now, we use <math>u</math>-substitution.
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| − | |-
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| − | |Let <math>u=1+\frac{27^2x}{2^2}.</math>
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| − | |-
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| − | |Then, <math>du=\frac{27^2}{2^2}dx</math> and <math>dx=\frac{2^2}{27^2}du.</math>
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| − | |-
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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 <math>u_1=1+\frac{27^2(1)}{2^2}=1+\frac{27^2}{2^2}</math>
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| − | |-
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| − | |and <math>u_2=1+\frac{27^2(4)}{2^2}=1+27^2.</math>
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| − | |-
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| − | |Hence, we now have
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| − | | <math>L=\int_{1+\frac{27^2}{2^2}}^{1+27^2} \frac{2^2}{27^2}u^{\frac{1}{2}}~du.</math>
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| − | |}
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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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| − | |-
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| − | |Therefore, we have
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| − | |-
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| − | | <math>\begin{array}{rcl}
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| − | \displaystyle{L} & = & \displaystyle{\frac{2^2}{27^2} \bigg(\frac{2}{3}u^{\frac{3}{2}}\bigg)\bigg|_{1+\frac{27^2}{2^2}}^{1+27^2}}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{2^3}{3^4} u^{\frac{3}{2}}\bigg|_{1+\frac{27^2}{2^2}}^{1+27^2}}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{2^3}{3^4} (1+27^2)^{\frac{3}{2}}-\frac{2^3}{3^4} \bigg(1+\frac{27^2}{2^2}\bigg)^{\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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| − | |-
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| − | | '''(a)''' <math>\frac{\pi}{27} (10)^{\frac{3}{2}}-\frac{\pi}{27}</math>
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| − | |-
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| − | | '''(b)''' <math>\frac{2^3}{3^4} (1+27^2)^{\frac{3}{2}}-\frac{2^3}{3^4} \bigg(1+\frac{27^2}{2^2}\bigg)^{\frac{3}{2}}</math>
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| − | |}
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| | [[009B_Sample_Final_2|'''<u>Return to Sample Exam</u>''']] | | [[009B_Sample_Final_2|'''<u>Return to Sample Exam</u>''']] |
