009A Sample Midterm 1, Problem 5 - Revision history

MathAdmin at 21:43, 8 November 2017

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MathAdmin at 15:32, 7 November 2017

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MathAdmin at 02:27, 14 April 2017

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MathAdmin: Created page with "The displacement from equilibrium of an object in harmonic motion on the end of a spring is: :: $y=\frac{1}{3}\cos(12t)-\frac{1}{4}\..."$

2017-04-09T18:36:05Z

Created page with "The displacement from equilibrium of an object in harmonic motion on the end of a spring is: ::<math>y=\frac{1}{3}\cos(12t)-\frac{1}{4}\..."

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The displacement from equilibrium of an object in harmonic motion on the end of a spring is:

::<math>y=\frac{1}{3}\cos(12t)-\frac{1}{4}\sin(12t)</math>

where  <math style="vertical-align: -4px">y</math>  is measured in feet and  <math style="vertical-align: 0px">t</math>  is the time in seconds.

Determine the position and velocity of the object when  <math style="vertical-align: -14px">t=\frac{\pi}{8}.</math>

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|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?
|-
|        <math>v(t)=s'(t)</math>
|}

'''Solution:'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|To find the position of the object at  <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>
|-
|Thus, we have
|-
|        <math>\begin{array}{rcl}
\displaystyle{y\bigg(\frac{\pi}{8}\bigg)} & = & \displaystyle{\frac{1}{3}\cos\bigg(\frac{12\pi}{8}\bigg)-\frac{1}{4}\sin\bigg(\frac{12\pi}{8}\bigg)}\\
&&\\
& = & \displaystyle{\frac{1}{3}\cos\bigg(\frac{3\pi}{2}\bigg)-\frac{1}{4}\sin\bigg(\frac{3\pi}{2}\bigg)}\\
&&\\
& = & \displaystyle{0-\frac{1}{4}(-1)}\\
&&\\
&= & \displaystyle{\frac{1}{4} \text{ foot}.}
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, to find the velocity function, we need to take the derivative of the position function.
|-
|Thus, we have
|-
|        <math>\begin{array}{rcl}
\displaystyle{v(t)} & = & \displaystyle{y'}\\
&&\\
& = & \displaystyle{\frac{-1}{3}\sin(12t)(12)-\frac{1}{4}\cos(12t)(12)}\\
&&\\
& = & \displaystyle{-4\sin(12t)-3\cos(12t).}
\end{array}</math>
|-
|Therefore, the velocity of the object at time  <math>t=\frac{\pi}{8}</math>  is
|-
|        <math>\begin{array}{rcl}
\displaystyle{v\bigg(\frac{\pi}{8}\bigg)} & = & \displaystyle{-4\sin\bigg(\frac{3\pi}{2}\bigg)-3\cos\bigg(\frac{3\pi}{2}\bigg)}\\
&&\\
& = & \displaystyle{-4(-1)+0}\\
&&\\
& = & \displaystyle{4 \text{ feet/second}.}
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|        position is  <math>\frac{1}{4} \text{ foot}.</math>
|-
|        velocity is  <math>4 \text{ feet/second}.</math>
|}
[[009A_Sample_Midterm_1|'''Return to Sample Exam''']]

@@ Line 7: / Line 7: @@
 <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 with Background Information</u>''']]
+[[009A Sample Midterm 1, Problem 5 Solution|'''<u>Solution</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>
 [[009A_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 35: / Line 35: @@
 & = & \displaystyle{0-\frac{1}{4}(-1)}\\
 &&\\
-&= & \displaystyle{\frac{1}{4} \text{ foot}.}
+&= & \displaystyle{\frac{1}{4} \text{ ft}.}
 \end{array}</math>
 |}
@@ Line 61: / Line 61: @@
 & = & \displaystyle{-4(-1)+0}\\
 &&\\
-& = & \displaystyle{4 \text{ feet/second}.}
+& = & \displaystyle{4 \text{ ft/sec}.}
 \end{array}</math>
 |}
@@ Line 69: / Line 69: @@
 !Final Answer: &nbsp;
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; position is &nbsp;<math>\frac{1}{4} \text{ foot}.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp; position is &nbsp;<math>\frac{1}{4} \text{ ft}.</math>
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; velocity is &nbsp;<math>4 \text{ feet/second}.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp; velocity is &nbsp;<math>4 \text{ ft/sec}.</math>
 |}
 [[009A_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]