009B Sample Final 1, Problem 3 - Revision history

Kayla Murray at 20:55, 20 May 2017

2017-05-20T20:55:50Z

Kayla Murray at 19:41, 18 March 2017

2017-03-18T19:41:01Z

Kayla Murray at 22:42, 4 March 2017

2017-03-04T22:42:10Z

Kayla Murray at 16:23, 28 February 2017

2017-02-28T16:23:31Z

Kayla Murray at 16:20, 28 February 2017

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Kayla Murray at 15:58, 28 February 2017

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Kayla Murray at 20:08, 27 February 2017

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Kayla Murray at 02:11, 19 February 2017

2017-02-19T02:11:15Z

Grad at 06:16, 26 February 2016

2016-02-26T06:16:10Z

Grad: /* 3 */

2016-02-26T06:16:00Z

@@ Line 36: / Line 36: @@
 !Step 2: &nbsp;
 |-
-|Setting &nbsp;<math style="vertical-align: -4px">\cos x=1-\cos x,</math>&nbsp; we get &nbsp;<math style="vertical-align: 0px">2\cos x=2.</math>
+|Setting &nbsp;<math style="vertical-align: -4px">\cos x=2-\cos x,</math>&nbsp; we get &nbsp;<math style="vertical-align: 0px">2\cos x=2.</math>
 |-
 |Therefore, we have

@@ Line 87: / Line 87: @@
 !Final Answer: &nbsp;
 |-
-|&nbsp; &nbsp;'''(a)''' &nbsp; &nbsp;<math>(0,1),(2\pi,1)</math>
+|&nbsp; &nbsp;'''(a)''' &nbsp; &nbsp;<math>(0,1),(2\pi,1)</math>&nbsp; (See Step 1 above for graph)
 |-
 |&nbsp; &nbsp;'''(b)''' &nbsp; &nbsp;<math>4\pi</math>
 |}
 [[009B_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 8: / Line 8: @@
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 !Foundations: &nbsp;
 |-
 |'''1.''' You can find the intersection points of two functions, say &nbsp;<math style="vertical-align: -5px">f(x),g(x),</math>

@@ Line 44: / Line 44: @@
 |&nbsp; &nbsp; &nbsp; &nbsp;<math>\cos x=1.</math>
 |-
-|In the interval <math>0\le x\le 2\pi,</math> the solutions to this equation are
+|In the interval &nbsp;<math style="vertical-align: -4px">0\le x\le 2\pi,</math>&nbsp; the solutions to this equation are
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math>x=0</math> and <math>x=2\pi.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: 0px">x=0</math>&nbsp; and &nbsp;<math style="vertical-align: 0px">x=2\pi.</math>
 |-
 |Plugging these values into our equations,
 |-
-|we get the intersection points &nbsp;<math>(0,1)</math> and <math>(2\pi,1).</math>
+|we get the intersection points &nbsp;<math style="vertical-align: -4px">(0,1)</math>&nbsp; and &nbsp;<math style="vertical-align: -4px">(2\pi,1).</math>
 |-
 |You can see these intersection points on the graph shown in Step 1.

@@ Line 38: / Line 38: @@
 !Step 2: &nbsp;
 |-
-|Setting <math style="vertical-align: -14px">\sin x=\frac{2}{\pi}x</math>, we get three solutions: <math>x=0,\frac{\pi}{2},\frac{-\pi}{2}.</math>
+|Setting &nbsp;<math style="vertical-align: -4px">\cos x=1-\cos x,</math>&nbsp; we get &nbsp;<math style="vertical-align: 0px">2\cos x=2.</math>
 |-
-|So, the three intersection points are <math style="vertical-align: -15px">(0,0),\bigg(\frac{\pi}{2},1\bigg),\bigg(\frac{-\pi}{2},-1\bigg)</math>.
+|Therefore, we have
 |-
 |You can see these intersection points on the graph shown in Step 1.
@@ Line 50: / Line 60: @@
 !Step 1: &nbsp;
 |-
-|Using symmetry of the graph, the area bounded by the two functions is given by
+|The area bounded by the two functions is given by
 |-
+|
-::<math>2\int_0^{\frac{\pi}{2}}\bigg(\sin(x)-\frac{2}{\pi}x\bigg)~dx.</math>
+&nbsp; &nbsp; &nbsp; &nbsp; <math>\int_0^{2\pi} (2-\cos x)-\cos x~dx.</math>
 |-
+|
@@ Line 64: / Line 74: @@
 |-
+|
-::<math>\begin{array}{rcl}
+&nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
-\displaystyle{2\int_0^{\frac{\pi}{2}}\bigg(\sin (x)-\frac{2}{\pi}x\bigg)~dx} & {=} & \displaystyle{2\bigg(-\cos (x)-\frac{x^2}{\pi}\bigg)\bigg|_0^{\frac{\pi}{2}}}\\
+\displaystyle{\int_0^{2\pi} (2-\cos x)-\cos x~dx} & {=} & \displaystyle{\int_0^{2\pi} 2-2\cos x~dx}\\
 &&\\
-& = & \displaystyle{2\bigg(-\cos \bigg(\frac{\pi}{2}\bigg)-\frac{1}{\pi}\bigg(\frac{\pi}{2}\bigg)^2\bigg)}-2(-\cos(0))\\
+& = & \displaystyle{(2x-2\sin x)\bigg|_0^{2\pi}}\\
 &&\\
-& = & \displaystyle{2\bigg(-\frac{\pi}{4}\bigg)+2}\\
+& = & \displaystyle{(4\pi-2\sin(2\pi))-(0-2\sin(0))}\\
 &&\\
-& = & \displaystyle{-\frac{\pi}{2}+2}.\\
+& = & \displaystyle{4\pi.}\\
 \end{array}</math>
 |}
@@ Line 79: / Line 89: @@
 !Final Answer: &nbsp;
 |-
-|'''(a)''' &nbsp;<math>(0,0),\bigg(\frac{\pi}{2},1\bigg),\bigg(\frac{-\pi}{2},-1\bigg)</math>
+|&nbsp; &nbsp;'''(a)''' &nbsp; &nbsp;<math>(0,1),(2\pi,1)</math>
 |-
-|'''(b)''' &nbsp;<math>-\frac{\pi}{2}+2</math>
+|&nbsp; &nbsp;'''(b)''' &nbsp; &nbsp;<math>4\pi</math>
 |}
 [[009B_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 11: / Line 11: @@
 |Recall:
 |-
-|'''1.''' You can find the intersection points of two functions, say <math style="vertical-align: -5px">f(x),g(x),</math>
+|'''1.''' You can find the intersection points of two functions, say &nbsp;<math style="vertical-align: -5px">f(x),g(x),</math>
 |-
+|
-::by setting <math style="vertical-align: -5px">f(x)=g(x)</math> and solving for <math style="vertical-align: 0px">x</math>.
+&nbsp; &nbsp; &nbsp; &nbsp;by setting &nbsp;<math style="vertical-align: -5px">f(x)=g(x)</math>&nbsp; and solving for &nbsp;<math style="vertical-align: 0px">x.</math>
 |-
-|'''2.''' The area between two functions, <math style="vertical-align: -5px">f(x)</math> and <math style="vertical-align: -5px">g(x)</math>, is given by <math>\int_a^b f(x)-g(x)~dx</math>
+|'''2.''' The area between two functions, &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; and &nbsp;<math style="vertical-align: -5px">g(x),</math>&nbsp; is given by &nbsp;<math>\int_a^b f(x)-g(x)~dx</math>
 |-
+|
-::for <math style="vertical-align: -3px">a\leq x\leq b</math>, where <math style="vertical-align: -5px">f(x)</math> is the upper function and <math style="vertical-align: -5px">g(x)</math> is the lower function.
+&nbsp; &nbsp; &nbsp; &nbsp;for &nbsp;<math style="vertical-align: -3px">a\leq x\leq b,</math>&nbsp; where &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; is the upper function and &nbsp;<math style="vertical-align: -5px">g(x)</math>&nbsp; is the lower function.
 |}

@@ Line 1: / Line 1: @@
 <span class="exam">Consider the area bounded by the following two functions:
-::<math>y=\sin x</math> and <math style="vertical-align: -13px">y=\frac{2}{\pi}x.</math>
+::<span class="exam"><math style="vertical-align: -4px">y=\cos x</math>&nbsp; and &nbsp;<math style="vertical-align: -4px">y=2-\cos x,~0\le x\le 2\pi.</math>
-<span class="exam">(a) Find the three intersection points of the two given functions. (Drawing may be helpful.)
+<span class="exam">(a) Sketch the graphs and find their points of intersection.
 <span class="exam">(b) Find the area bounded by the two functions.

← Older revision		Revision as of 06:16, 26 February 2016
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	\|}		\|}

−	~~== 3 ==~~
	'''(b)'''		'''(b)'''