Difference between revisions of "009A Sample Midterm 1, Problem 5"

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<span class="exam">The displacement from equilibrium of an object in harmonic motion on the end of a spring is:
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<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>
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<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.
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<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>
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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>
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<hr>
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[[009A Sample Midterm 1, Problem 5 Solution|'''<u>Solution</u>''']]
  
  
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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[[009A Sample Midterm 1, Problem 5 Detailed Solution|'''<u>Detailed Solution</u>''']]
!Foundations: &nbsp;
 
|-
 
|What is the relationship between the position &nbsp;<math style="vertical-align: -5px">s(t)</math>&nbsp; and the velocity &nbsp;<math style="vertical-align: -5px">v(t)</math>&nbsp; of an object?
 
|-
 
|&nbsp; &nbsp; &nbsp; &nbsp; <math>v(t)=s'(t)</math>
 
|}
 
  
  
'''Solution:'''
 
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Step 1: &nbsp;
 
|-
 
|To find the position of the object at &nbsp;<math>t=\frac{\pi}{8},</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
 
|-
 
|&nbsp; &nbsp; &nbsp; &nbsp; <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: &nbsp;
 
|-
 
|Now, to find the velocity function, we need to take the derivative of the position function.
 
|-
 
|Thus, we have
 
|-
 
|&nbsp; &nbsp; &nbsp; &nbsp; <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 &nbsp;<math>t=\frac{\pi}{8}</math>&nbsp; is
 
|-
 
|&nbsp; &nbsp; &nbsp; &nbsp; <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: &nbsp;
 
|-
 
|&nbsp; &nbsp; &nbsp; &nbsp; position is &nbsp;<math>\frac{1}{4} \text{ foot}.</math>
 
|-
 
|&nbsp; &nbsp; &nbsp; &nbsp; velocity is &nbsp;<math>4 \text{ feet/second}.</math>
 
|}
 
 
[[009A_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]
 
[[009A_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]

Latest revision as of 14:43, 8 November 2017

Find the derivatives of the following functions. Do not simplify.

(a)  

(b)   where

(c)  


Solution


Detailed Solution


Return to Sample Exam