009C Sample Final 2, Problem 10 - Revision history

Kayla Murray at 23:03, 3 December 2017

2017-12-03T23:03:01Z

2017-12-03T01:37:54Z

2017-03-10T23:52:10Z

2017-03-05T04:35:11Z

2017-03-05T04:32:42Z

2017-03-05T04:31:42Z

2017-03-05T04:18:49Z

2017-03-05T04:16:18Z

2017-03-05T00:51:17Z

2017-03-05T00:45:30Z

@@ Line 2: / Line 2: @@
 ::<span class="exam"><math>x=t^2</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>
 <hr>

@@ Line 4: / Line 4: @@
 ::<span class="exam"><math>0\leq t \leq 2</math>
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+<hr>
-!Foundations: &nbsp;
+[[009C Sample Final 2, Problem 10 Solution|'''<u>Solution</u>''']]
-'''Solution:'''
+[[009C Sample Final 2, Problem 10 Detailed Solution|'''<u>Detailed Solution</u>''']]
 [[009C_Sample_Final_2|'''<u>Return to Sample Exam</u>''']]

@@ Line 66: / Line 66: @@
 \displaystyle{L} & = & \displaystyle{\frac{1}{18}\int_{13}^{40} \sqrt{u}~du}\\
 &&\\
-& = & \displaystyle{\frac{1}{27} u^{\frac{3}{2}}\bigg|_{13}^{40}}\\
+& = & \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}}.}\\

@@ Line 51: / Line 51: @@
 |Now, we use &nbsp;<math>u</math>-substitution.
 |-
-|Let &nbsp;<math>u=4+9t^2.</math>
+|Let &nbsp;<math style="vertical-align: -2px">u=4+9t^2.</math>
 |-
-|Then, &nbsp;<math>du=18tdt</math>&nbsp; and &nbsp;<math>\frac{du}{18}=tdt.</math>
+|Then, &nbsp;<math style="vertical-align: -2px">du=18tdt</math>&nbsp; and &nbsp;<math style="vertical-align: -15px">\frac{du}{18}=tdt.</math>
 |-
 |Also, since this is a definite integral, we need to change the bounds of integration.
@@ Line 59: / Line 59: @@
 |We have
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp;<math>u_1=4+9(1)^2=13</math>&nbsp; and &nbsp;<math>u_2=4+9(2)^2=40.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -5px">u_1=4+9(1)^2=13</math>&nbsp; and &nbsp;<math style="vertical-align: -5px">u_2=4+9(2)^2=40.</math>
 |-
 |Hence,

@@ Line 51: / Line 51: @@
 |Now, we use &nbsp;<math>u</math>-substitution.
 |-
-|Let &nbsp;<math>u=4+9t^2.</math>&nbsp;
+|Let &nbsp;<math>u=4+9t^2.</math>
 |-
 |Then, &nbsp;<math>du=18tdt</math>&nbsp; and &nbsp;<math>\frac{du}{18}=tdt.</math>
@@ Line 59: / Line 59: @@
 |We have
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp;<math>u_1=4+9(1)^2=13</math> and <math>u_2=4+9(2)^2=40.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp;<math>u_1=4+9(1)^2=13</math>&nbsp; and &nbsp;<math>u_2=4+9(2)^2=40.</math>
 |-
 |Hence,

@@ Line 19: / Line 19: @@
 !Step 1: &nbsp;
 |-
+|First, we need to calculate &nbsp;<math style="vertical-align: -14px">\frac{dx}{dt}</math>&nbsp; and &nbsp;<math style="vertical-align: -14px">\frac{dy}{dt}.</math>
 |-
+|Using the formula in Foundations, we have
 |-
+|
 |}
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 !Step 2: &nbsp;
 |-
+|
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
 |}
@@ Line 38: / Line 76: @@
 !Final Answer: &nbsp;
 |-
-|&nbsp;&nbsp;
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{1}{27}(40)^{\frac{3}{2}}-\frac{1}{27}(13)^{\frac{3}{2}}</math>
 |}
 [[009C_Sample_Final_2|'''<u>Return to Sample Exam</u>''']]

@@ Line 7: / Line 7: @@
 !Foundations: &nbsp;
 |-
-|The formula for the arc length &nbsp;<math style="vertical-align: 0px">L</math>&nbsp; of a polar curve &nbsp;<math style="vertical-align: -5px">r=f(\theta)</math>&nbsp; with &nbsp;<math style="vertical-align: -4px">\alpha_1\leq \theta \leq \alpha_2</math>&nbsp; is
+|The formula for the arc length &nbsp;<math style="vertical-align: 0px">L</math>&nbsp; of a parametric curve with &nbsp;<math style="vertical-align: -4px">\alpha \leq t \leq \beta </math>&nbsp; is
 |-
+|
-&nbsp; &nbsp; &nbsp; &nbsp;<math>L=\int_{\alpha_1}^{\alpha_2} \sqrt{r^2+\bigg(\frac{dr}{d\theta}\bigg)^2}d\theta.</math>
+&nbsp; &nbsp; &nbsp; &nbsp;<math>L=\int_{\alpha}^{\beta} \sqrt{\bigg(\frac{dx}{dt}\bigg)^2+\bigg(\frac{dy}{dt}\bigg)^2}~dt.</math>
 |}

@@ Line 1: / Line 1: @@
 <span class="exam">Find the length of the curve given by
-::::<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>0\leq t \leq 2</math>
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"