009A Sample Midterm 1, Problem 5 - Revision history

Kayla Murray at 01:37, 7 November 2017

2017-11-07T01:37:55Z

Kayla Murray at 02:37, 6 November 2017

2017-11-06T02:37:39Z

Kayla Murray at 14:20, 3 November 2017

2017-11-03T14:20:43Z

Kayla Murray at 14:20, 3 November 2017

2017-11-03T14:20:22Z

Kayla Murray at 23:52, 26 February 2017

2017-02-26T23:52:35Z

Kayla Murray at 23:38, 26 February 2017

2017-02-26T23:38:46Z

Kayla Murray at 23:59, 18 February 2017

2017-02-18T23:59:03Z

Kayla Murray at 22:00, 18 February 2017

2017-02-18T22:00:28Z

Kayla Murray at 19:55, 18 February 2017

2017-02-18T19:55:38Z

Kayla Murray at 21:00, 16 February 2017

2017-02-16T21:00:02Z

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 <hr>
 [[009A Sample Midterm 1, Problem 5 Detailed Solution|'''<u>Detailed Solution with Background Information</u>''']]
 [[File:9ASM1P5.jpg|600px|thumb|center]]
 [[009A_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]

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 <span class="exam">(c) &nbsp; <math style="vertical-align: -20px">h(x)=\frac{e^{-5x^3}}{\sqrt{x^2+1}}</math>
 <hr>
-[[009A Sample Midterm 1, Problem 5 Detailed Solution|'''<u>Detailed Solution for this Problem</u>''']]
+[[File:9ASM1P5.jpg|600px|thumb|center]]
 [[009A_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]

← Older revision		Revision as of 14:20, 3 November 2017
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	[[009A Sample Midterm 1, Problem 5 Detailed Solution\|'''<u>Detailed Solution for this Problem</u>''']]		[[009A Sample Midterm 1, Problem 5 Detailed Solution\|'''<u>Detailed Solution for this Problem</u>''']]
−

	[[009A_Sample_Midterm_1\|'''<u>Return to Sample Exam</u>''']]		[[009A_Sample_Midterm_1\|'''<u>Return to Sample Exam</u>''']]

@@ Line 1: / Line 1: @@
-<span class="exam">The displacement from equilibrium of an object in harmonic motion on the end of a spring is:
+<span class="exam"> Find the derivatives of the following functions. Do not simplify.
-::<span class="exam"><math>y=\frac{1}{3}\cos(12t)-\frac{1}{4}\sin(12t)</math>
+<span class="exam">(a) &nbsp; <math style="vertical-align: -5px">f(x)=\sqrt{x}(x^2+2)</math>
-<span class="exam">where &nbsp;<math style="vertical-align: -4px">y</math>&nbsp; is measured in feet and &nbsp;<math style="vertical-align: 0px">t</math>&nbsp; is the time in seconds.
+<span class="exam">(b) &nbsp; <math style="vertical-align: -17px">g(x)=\frac{x+3}{x^{\frac{3}{2}}+2}</math> where <math style="vertical-align: 0px">x>0</math>
-<span class="exam">Determine the position and velocity of the object when &nbsp;<math style="vertical-align: -14px">t=\frac{\pi}{8}.</math>
+<span class="exam">(c) &nbsp; <math style="vertical-align: -20px">h(x)=\frac{e^{-5x^3}}{\sqrt{x^2+1}}</math>
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+[[009A Sample Midterm 1, Problem 5 Detailed Solution|'''<u>Detailed Solution for this Problem</u>''']]
 [[009A_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 11: / Line 11: @@
 !Foundations: &nbsp;
 |-
-|What is the relationship between position <math style="vertical-align: -5px">s(t)</math> and velocity <math style="vertical-align: -5px">v(t)</math> of an object?
+|What is the relationship between position &nbsp;<math style="vertical-align: -5px">s(t)</math>&nbsp; and velocity &nbsp;<math style="vertical-align: -5px">v(t)</math>&nbsp; of an object?
 |-
 |&nbsp; &nbsp; &nbsp; &nbsp; <math>v(t)=s'(t)</math>
@@ Line 22: / Line 22: @@
 !Step 1: &nbsp;
 |-
-|To find the position of the object at <math>t=\frac{\pi}{8},</math>
+|To find the position of the object at &nbsp;<math>t=\frac{\pi}{8},</math>
 |-
-|we need to plug <math>t=\frac{\pi}{8}</math> into the equation <math style="vertical-align: -5px">y.</math>
+|we need to plug &nbsp;<math>t=\frac{\pi}{8}</math>&nbsp; into the equation &nbsp;<math style="vertical-align: -5px">y.</math>
 |-
 |Thus, we have
@@ Line 54: / Line 54: @@
 \end{array}</math>
 |-
-|Therefore, the velocity of the object at time <math>t=\frac{\pi}{8}</math> is
+|Therefore, the velocity of the object at time &nbsp;<math>t=\frac{\pi}{8}</math>&nbsp; is
 |-
 |&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
@@ Line 69: / Line 69: @@
 !Final Answer: &nbsp;
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; position is <math>\frac{1}{4} \text{ foot}.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp; position is &nbsp;<math>\frac{1}{4} \text{ foot}.</math>
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; velocity is <math>4 \text{ feet/second}.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp; velocity is &nbsp;<math>4 \text{ feet/second}.</math>
 |}
 [[009A_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]