009B Sample Final 2, Problem 5 - Revision history

Kayla Murray at 00:28, 3 December 2017

2017-12-03T00:28:59Z

Kayla Murray at 21:17, 20 May 2017

2017-05-20T21:17:31Z

Kayla Murray at 21:39, 12 March 2017

2017-03-12T21:39:16Z

Kayla Murray at 21:32, 12 March 2017

2017-03-12T21:32:28Z

Kayla Murray at 00:25, 5 March 2017

2017-03-05T00:25:05Z

Kayla Murray at 00:11, 5 March 2017

2017-03-05T00:11:41Z

Kayla Murray at 21:32, 4 March 2017

2017-03-04T21:32:06Z

Kayla Murray at 21:28, 4 March 2017

2017-03-04T21:28:19Z

Kayla Murray at 20:49, 3 March 2017

2017-03-03T20:49:24Z

Kayla Murray at 02:21, 19 February 2017

2017-02-19T02:21:27Z

@@ Line 11: / Line 11: @@
 <span class="exam">between the points &nbsp;<math style="vertical-align: -5px">(1,10)</math>&nbsp; and &nbsp;<math style="vertical-align: -5px">(4,73).</math>
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+<hr>
-!Foundations: &nbsp;
+[[009B Sample Final 2, Problem 5 Solution|'''<u>Solution</u>''']]
-'''Solution:'''
+[[009B Sample Final 2, Problem 5 Detailed Solution|'''<u>Detailed Solution</u>''']]
 [[009B_Sample_Final_2|'''<u>Return to Sample Exam</u>''']]

@@ Line 101: / Line 101: @@
 |Then, we have
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp;<math>L=\int_1^4 \sqrt{1+\frac{27^2x}{2^2}}~dx.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp;<math>L=\int_1^4 \sqrt{1+\frac{729x}{4}}~dx.</math>
 |-
 |Now, we use &nbsp;<math style="vertical-align: 0px">u</math>-substitution.
 |-
-|Let &nbsp;<math style="vertical-align: -14px">u=1+\frac{27^2x}{2^2}.</math>
+|Let &nbsp;<math style="vertical-align: -14px">u=1+\frac{729x}{4}.</math>
 |-
-|Then, &nbsp;<math style="vertical-align: -14px">du=\frac{27^2}{2^2}dx</math>&nbsp; and &nbsp;<math style="vertical-align: -14px">dx=\frac{2^2}{27^2}du.</math>
+|Then, &nbsp;<math style="vertical-align: -14px">du=\frac{729}{4}~dx</math>&nbsp; and &nbsp;<math style="vertical-align: -14px">dx=\frac{4}{729}~du.</math>
 |-
 |Also, since this is a definite integral, we need to change the bounds of integration.
@@ Line 113: / Line 113: @@
 |We have
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -13px">u_1=1+\frac{27^2(1)}{2^2}=1+\frac{27^2}{2^2}</math>&nbsp; and &nbsp;<math style="vertical-align: -13px">u_2=1+\frac{27^2(4)}{2^2}=1+27^2.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -13px">u_1=1+\frac{729(1)}{4}=\frac{733}{4}</math>&nbsp; and &nbsp;<math style="vertical-align: -13px">u_2=1+\frac{729(4)}{4}=730.</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>
+|&nbsp; &nbsp; &nbsp; &nbsp;<math>L=\int_{\frac{733}{4}}^{730} \frac{4}{729}u^{\frac{1}{2}}~du.</math>
 |}
@@ Line 126: / Line 126: @@
 |-
 |&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{L} & = & \displaystyle{\frac{4}{729} \bigg(\frac{2}{3}u^{\frac{3}{2}}\bigg)\bigg|_{\frac{733}{4}}^{730}}\\
 &&\\
-& = & \displaystyle{\frac{2^3}{3^4} u^{\frac{3}{2}}\bigg|_{1+\frac{27^2}{2^2}}^{1+27^2}}\\
+& = & \displaystyle{\frac{8}{2187} u^{\frac{3}{2}}\bigg|_{\frac{733}{4}}^{730}}\\
 &&\\
-& = & \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}}.}
+& = & \displaystyle{\frac{8}{2187} (730)^{\frac{3}{2}}-\frac{8}{2187} \bigg(\frac{733}{4}\bigg)^{\frac{3}{2}}.}
 \end{array}</math>
 |}
@@ Line 140: / Line 140: @@
 |&nbsp; &nbsp;'''(a)'''&nbsp; &nbsp; <math>\frac{\pi}{27} (10)^{\frac{3}{2}}-\frac{\pi}{27}</math>
 |-
-|&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>
+|&nbsp; &nbsp;'''(b)'''&nbsp; &nbsp; <math>\frac{8}{2187} (730)^{\frac{3}{2}}-\frac{8}{2187} \bigg(\frac{733}{4}\bigg)^{\frac{3}{2}}</math>
 |}
 [[009B_Sample_Final_2|'''<u>Return to Sample Exam</u>''']]

@@ Line 35: / Line 35: @@
 |We start by calculating &nbsp;<math style="vertical-align: -16px">\frac{dx}{dy}.</math>
 |-
-|Since &nbsp;<math style="vertical-align: -17px">x=y^3,~ \frac{dx}{dy}=3y^2.</math>
+|Since &nbsp;<math style="vertical-align: -4px">x=y^3,</math>
 |-
 |Now, we are going to integrate with respect to &nbsp;<math style="vertical-align: -3px">y.</math>
@@ Line 65: / Line 67: @@
 |We have
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp;<math>u_1=1+9(0)^4=1</math>&nbsp; and &nbsp;<math>u_2=1+9(1)^4=10.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -5px">u_1=1+9(0)^4=1</math>&nbsp; and &nbsp;<math style="vertical-align: -5px">u_2=1+9(1)^4=10.</math>
 |-
 |Thus, we get
@@ Line 111: / Line 113: @@
 |We have
 |-
-|&nbsp; &nbsp; &nbsp; &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>
+|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -13px">u_1=1+\frac{27^2(1)}{2^2}=1+\frac{27^2}{2^2}</math>&nbsp; and &nbsp;<math style="vertical-align: -13px">u_2=1+\frac{27^2(4)}{2^2}=1+27^2.</math>
 |-
 |Hence, we now have

@@ Line 3: / Line 3: @@
 ::<math>y^3=x</math>
-<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 &nbsp;<math style="vertical-align: -5px">(0,0)</math>&nbsp; and &nbsp;<math style="vertical-align: -5px">(1,1)</math>&nbsp; about the &nbsp;<math style="vertical-align: -4px">y</math>-axis.
 <span class="exam">(b) Find the length of the arc
@@ Line 9: / Line 9: @@
 ::<math>y=1+9x^{\frac{3}{2}}</math>
-<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 &nbsp;<math style="vertical-align: -5px">(1,10)</math>&nbsp; and &nbsp;<math style="vertical-align: -5px">(4,73).</math>
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"

@@ Line 14: / Line 14: @@
 !Foundations: &nbsp;
 |-
-|'''1.''' The formula for the length &nbsp;<math style="vertical-align: 0px">L</math>&nbsp; of a curve &nbsp;<math style="vertical-align: -5px">y=f(x)</math>&nbsp; where &nbsp;<math style="vertical-align: -3px">a\leq x \leq b</math>&nbsp; is
+|'''1.''' The surface area &nbsp;<math style="vertical-align: 0px">S</math>&nbsp; of a function &nbsp;<math style="vertical-align: -5px">y=f(x)</math>&nbsp; rotated about the &nbsp;<math style="vertical-align: -4px">y</math>-axis is given by
 |-
+|
-&nbsp; &nbsp; &nbsp; &nbsp;<math>L=\int_a^b \sqrt{1+\bigg(\frac{dy}{dx}\bigg)^2}~dx.</math>
+&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -13px">S=\int 2\pi x\,ds,</math>&nbsp; where &nbsp;<math style="vertical-align: -19px">ds=\sqrt{1+\bigg(\frac{dx}{dy}\bigg)^2}dy.</math>
 |-
-|'''2.''' The surface area &nbsp;<math style="vertical-align: 0px">S</math>&nbsp; of a function &nbsp;<math style="vertical-align: -5px">y=f(x)</math>&nbsp; rotated about the &nbsp;<math style="vertical-align: -4px">y</math>-axis is given by
+|'''2.''' The formula for the length &nbsp;<math style="vertical-align: 0px">L</math>&nbsp; of a curve &nbsp;<math style="vertical-align: -5px">y=f(x)</math>&nbsp; where &nbsp;<math style="vertical-align: -3px">a\leq x \leq b</math>&nbsp; is
 |-
+|
-&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -13px">S=\int 2\pi x\,ds,</math>&nbsp; where <math style="vertical-align: -18px">ds=\sqrt{1+\bigg(\frac{dx}{dy}\bigg)^2}dy.</math>
+&nbsp; &nbsp; &nbsp; &nbsp;<math>L=\int_a^b \sqrt{1+\bigg(\frac{dy}{dx}\bigg)^2}~dx.</math>
 |}
@@ Line 33: / Line 33: @@
 !Step 1: &nbsp;
 |-
-|We start by calculating &nbsp;<math>\frac{dx}{dy}.</math>
+|We start by calculating &nbsp;<math style="vertical-align: -16px">\frac{dx}{dy}.</math>
 |-
-|Since &nbsp;<math style="vertical-align: -13px">x=y^3,~ \frac{dx}{dy}=3y^2.</math>
+|Since &nbsp;<math style="vertical-align: -17px">x=y^3,~ \frac{dx}{dy}=3y^2.</math>
 |-
-|Now, we are going to integrate with respect to <math>y.</math>
+|Now, we are going to integrate with respect to &nbsp;<math style="vertical-align: -3px">y.</math>
 |-
 |Using the formula given in the Foundations section,
@@ Line 47: / Line 47: @@
 &&\\
 & = & \displaystyle{2\pi \int_0^1 y^3 \sqrt{1+9y^4}~dy.}
 |-
-|where &nbsp;<math>S</math>&nbsp; is the surface area.
+|where &nbsp;<math style="vertical-align: 0px">S</math>&nbsp; is the surface area.
 |}
@@ Line 55: / Line 55: @@
 !Step 2: &nbsp;
 |-
-|Now, we use <math>u</math>-substitution.
+|Now, we use &nbsp;<math style="vertical-align: 0px">u</math>-substitution.
 |-
-|Let <math>u=1+9y^4.</math>
+|Let &nbsp;<math style="vertical-align: -5px">u=1+9y^4.</math>
 |-
-|Then, <math>du=36y^3dy</math> and <math>\frac{du}{36}=y^3dy.</math>
+|Then, &nbsp;<math style="vertical-align: -5px">du=36y^3dy</math>&nbsp; and &nbsp;<math style="vertical-align: -14px">\frac{du}{36}=y^3dy.</math>
 |-
 |Also, since this is a definite integral, we need to change the bounds of integration.
@@ Line 65: / Line 65: @@
 |We have
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp;<math>u_1=1+9(0)^4=1</math> and <math>u_2=1+9(1)^4=10.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp;<math>u_1=1+9(0)^4=1</math>&nbsp; and &nbsp;<math>u_2=1+9(1)^4=10.</math>
 |-
 |Thus, we get
@@ Line 83: / Line 83: @@
 !Step 1: &nbsp;
 |-
-|First, we calculate &nbsp;<math>\frac{dy}{dx}.</math>
+|First, we calculate &nbsp;<math style="vertical-align: -15px">\frac{dy}{dx}.</math>
 |-
-|Since <math>y=1+9x^{\frac{3}{2}},</math> we have
+|Since &nbsp;<math style="vertical-align: -5px">y=1+9x^{\frac{3}{2}},</math>&nbsp; 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
+|Then, the arc length &nbsp;<math style="vertical-align: 0px">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>
@@ Line 101: / Line 101: @@
 |&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.
+|Now, we use &nbsp;<math style="vertical-align: 0px">u</math>-substitution.
 |-
-|Let &nbsp;<math>u=1+\frac{27^2x}{2^2}.</math>
+|Let &nbsp;<math style="vertical-align: -14px">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>
+|Then, &nbsp;<math style="vertical-align: -14px">du=\frac{27^2}{2^2}dx</math>&nbsp; and &nbsp;<math style="vertical-align: -14px">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>
+|We have
 |-
-|and &nbsp; <math>u_2=1+\frac{27^2(4)}{2^2}=1+27^2.</math>
+|&nbsp; &nbsp; &nbsp; &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>
 |}

@@ Line 1: / Line 1: @@
-::<span class="exam">a) Find the area of the surface obtained by rotating the arc of the curve
+<span class="exam">(a) Find the area of the surface obtained by rotating the arc of the curve
-::::<math>y^3=x</math>
+::<math>y^3=x</math>
-::<span class="exam">between <math>(0,0)</math> and <math>(1,1)</math> about the <math>y</math>-axis.
+<span class="exam">between <math>(0,0)</math> and <math>(1,1)</math> about the <math>y</math>-axis.
-::<span class="exam">b) Find the length of the arc
+<span class="exam">(b) Find the length of the arc
-::::<math>y=1+9x^{\frac{3}{2}}</math>
+::<math>y=1+9x^{\frac{3}{2}}</math>
-::<span class="exam">between the points <math>(1,10)</math> and <math>(4,73).</math>
+<span class="exam">between the points <math>(1,10)</math> and <math>(4,73).</math>
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
@@ Line 22: / Line 22: @@
+|
 |}
 '''Solution:'''
@@ Line 86: / Line 87: @@
+|
 |}
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"