Difference between revisions of "009B Sample Final 2, Problem 5"

Revision as of 14:32, 4 March 2017

(a) Find the area of the surface obtained by rotating the arc of the curve

y^{3}=x

between $(0,0)$ and $(1,1)$ about the $y$ -axis.

(b) Find the length of the arc

y=1+9x^{\frac {3}{2}}

between the points $(1,10)$ and $(4,73).$

Foundations:
1. The formula for the length $L$ of a curve $y=f(x)$ where $a\leq x\leq b$ is
$L=\int _{a}^{b}{\sqrt {1+{\bigg (}{\frac {dy}{dx}}{\bigg )}^{2}}}~dx.$
2. The surface area $S$ of a function $y=f(x)$ rotated about the $y$ -axis is given by
$S=\int 2\pi x\,ds,$ where $ds={\sqrt {1+{\bigg (}{\frac {dy}{dx}}{\bigg )}^{2}}}.$

Solution:

(a)

Step 1:

Step 2:

(b)

Step 1:
First, we calculate ${\frac {dy}{dx}}.$
Since $y=1+9x^{\frac {3}{2}},$ we have
${\frac {dy}{dx}}={\frac {27{\sqrt {x}}}{2}}.$
Then, the arc length $L$ of the curve is given by
$L=\int _{1}^{4}{\sqrt {1+{\bigg (}{\frac {27{\sqrt {x}}}{2}}{\bigg )}^{2}}}~dx.$

Step 2:
Then, we have
$L=\int _{1}^{4}{\sqrt {1+{\frac {27^{2}x}{2^{2}}}}}~dx.$
Now, we use $u$ -substitution.
Let $u=1+{\frac {27^{2}x}{2^{2}}}.$
Then, $du={\frac {27^{2}}{2^{2}}}dx$ and $dx={\frac {2^{2}}{27^{2}}}du.$
Also, since this is a definite integral, we need to change the bounds of integration.
We have $u_{1}=1+{\frac {27^{2}(1)}{2^{2}}}=1+{\frac {27^{2}}{2^{2}}}$
and $u_{2}=1+{\frac {27^{2}(4)}{2^{2}}}=1+27^{2}.$
Hence, we now have
$L=\int _{1+{\frac {27^{2}}{2^{2}}}}^{1+27^{2}}{\frac {2^{2}}{27^{2}}}u^{\frac {1}{2}}~du.$

Step 3:

Therefore, we have

{\begin{array}{rcl}\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}}}\\&&\\&=&\displaystyle {{\frac {2^{3}}{3^{4}}}u^{\frac {3}{2}}{\bigg |}_{1+{\frac {27^{2}}{2^{2}}}}^{1+27^{2}}}\\&&\\&=&\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}}.}\end{array}}

Final Answer:
(a)
(b) ${\frac {2^{3}}{3^{4}}}(1+27^{2})^{\frac {3}{2}}-{\frac {2^{3}}{3^{4}}}{\bigg (}1+{\frac {27^{2}}{2^{2}}}{\bigg )}^{\frac {3}{2}}$

Return to Sample Exam

@@ Line 59: / Line 59: @@
 !Step 1: &nbsp;
 |-
-|
+|First, we calculate &nbsp;<math>\frac{dy}{dx}.</math>
+|-
+|Since <math>y=1+9x^{\frac{3}{2}},</math> we have
+|-
+|&nbsp; &nbsp; &nbsp; &nbsp;<math>\frac{dy}{dx}=\frac{27\sqrt{x}}{2}.</math>
+|-
+|Then, the arc length &nbsp;<math>L</math>&nbsp; of the curve is given by
 |-
-|
+|&nbsp; &nbsp; &nbsp; &nbsp;<math>L=\int_1^4 \sqrt{1+\bigg(\frac{27\sqrt{x}}{2}\bigg)^2}~dx.</math>
 |}
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 !Step 2: &nbsp;
+|-
+|Then, we have
+|-
+|&nbsp; &nbsp; &nbsp; &nbsp;<math>L=\int_1^4 \sqrt{1+\frac{27^2x}{2^2}}~dx.</math>
+|-
+|Now, we use &nbsp;<math>u</math>-substitution.
+|-
+|Let &nbsp;<math>u=1+\frac{27^2x}{2^2}.</math>
+|-
+|Then, &nbsp; <math>du=\frac{27^2}{2^2}dx</math>&nbsp; and &nbsp;<math>dx=\frac{2^2}{27^2}du.</math>
+|-
+|Also, since this is a definite integral, we need to change the bounds of integration.
+|-
+|We have &nbsp; <math>u_1=1+\frac{27^2(1)}{2^2}=1+\frac{27^2}{2^2}</math>
+|-
+|and &nbsp; <math>u_2=1+\frac{27^2(4)}{2^2}=1+27^2.</math>
+|-
+|Hence, we now have
+|-
+|&nbsp; &nbsp; &nbsp; &nbsp;<math>L=\int_{1+\frac{27^2}{2^2}}^{1+27^2} \frac{2^2}{27^2}u^{\frac{1}{2}}~du.</math>
 |-
 |
+|}
+{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+!Step 3: &nbsp;
 |-
-|
+|Therefore, we have
+|-
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
+\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}}\\
+&&\\
+& = & \displaystyle{\frac{2^3}{3^4} u^{\frac{3}{2}}\bigg|_{1+\frac{27^2}{2^2}}^{1+27^2}}\\
+&&\\
+& = & \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}}.}
+\end{array}</math>
 |}
@@ Line 78: / Line 116: @@
 |'''(a)'''
 |-
-|'''(b)'''
+|&nbsp; &nbsp;'''(b)'''&nbsp; &nbsp; <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>
 |}
 [[009B_Sample_Final_2|'''<u>Return to Sample Exam</u>''']]

Difference between revisions of "009B Sample Final 2, Problem 5"

Revision as of 14:32, 4 March 2017

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