Difference between revisions of "009B Sample Midterm 2, Problem 3"

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(Created page with "<span class="exam"> Evaluate ::<span class="exam">a) <math style="vertical-align: -14px">\int_1^2\bigg(2t+\frac{3}{t^2}\bigg)\bigg(4t^2-\frac{5}{t}\bigg)~dt</math> ::<span...")
 
 
(4 intermediate revisions by the same user not shown)
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<span class="exam"> Evaluate
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<span class="exam"> A particle moves along a straight line with velocity given by:
  
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::<span class="exam">a) <math style="vertical-align: -14px">\int_1^2\bigg(2t+\frac{3}{t^2}\bigg)\bigg(4t^2-\frac{5}{t}\bigg)~dt</math>  
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::<math>v(t)=-32t+200</math>
  
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::<span class="exam">b) <math style="vertical-align: -14px">\int_0^2 (x^3+x)\sqrt{x^4+2x^2+4}~dx</math>
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<span class="exam">feet per second. Determine the total distance traveled by the particle
 +
 
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<span class="exam">from time &nbsp;<math style="vertical-align: 0px">t=0</math>&nbsp; to time &nbsp;<math style="vertical-align: -1px">t=10.</math>
  
  
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!Foundations: &nbsp;  
 
!Foundations: &nbsp;  
 
|-
 
|-
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|Review <math style="vertical-align: 0px">u</math>-substitution 
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|'''1.''' How are the velocity function &nbsp;<math style="vertical-align: -5px">v(t)</math>&nbsp; and the position function &nbsp;<math style="vertical-align: -5px">s(t)</math>&nbsp; related?
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|-
 +
|
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&nbsp; &nbsp; &nbsp; &nbsp; They are related by the equation &nbsp;<math style="vertical-align: -5px">v(t)=s'(t).</math>
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|-
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|'''2.''' If we calculate &nbsp;<math style="vertical-align: -14px">\int_a^b v(t)~dt,</math>&nbsp; what are we calculating?
 +
|-
 +
|
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&nbsp; &nbsp; &nbsp; &nbsp; We are calculating &nbsp;<math style="vertical-align: -5px">s(b)-s(a).</math>
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|-
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|
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&nbsp; &nbsp; &nbsp; &nbsp; This is the displacement of the particle from &nbsp;<math style="vertical-align: 0px">t=a</math>&nbsp; to &nbsp;<math style="vertical-align: 0px">t=b.</math>
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|-
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|'''3.''' If we calculate &nbsp;<math style="vertical-align: -14px">\int_a^b |v(t)|~dt,</math>&nbsp; what are we calculating?
 +
|-
 +
|
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&nbsp; &nbsp; &nbsp; &nbsp; We are calculating the total distance traveled by the particle from &nbsp;<math style="vertical-align: 0px">t=a</math>&nbsp; to &nbsp;<math style="vertical-align: 0px">t=b.</math>
 
|}
 
|}
 +
  
 
'''Solution:'''
 
'''Solution:'''
  
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'''(a)'''
 
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Step 1: &nbsp;  
 
!Step 1: &nbsp;  
 
|-
 
|-
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|We multiply the product inside the integral to get
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|To calculate the total distance the particle traveled from &nbsp;<math style="vertical-align: -1px">t=0</math>&nbsp; to &nbsp;<math style="vertical-align: -5px">t=10,</math>
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|-
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|we need to calculate
 
|-
 
|-
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| &nbsp;&nbsp; <math style="vertical-align: -16px">\int_1^2\bigg(2t+\frac{3}{t^2}\bigg)\bigg(4t^2-\frac{5}{t}\bigg)~dt=\int_1^2 \bigg(8t^3-10+12-\frac{15}{t^3}\bigg)~dt=\int_1^2 (8t^3+2-15t^{-3})~dt</math>. 
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>\int_0^{10} |v(t)|~dt=\int_0^{10} |-32t+200|~dt.</math>
 
|}
 
|}
  
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!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
|-
 
|-
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|We integrate to get
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|We need to figure out when &nbsp;<math style="vertical-align: -2px">-32t+200</math>&nbsp; is positive and negative in the interval &nbsp;<math style="vertical-align: -6px">[0,10].</math>
 
|-
 
|-
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| &nbsp;&nbsp; <math style="vertical-align: -16px">\int_1^2\bigg(2t+\frac{3}{t^2}\bigg)\bigg(4t^2-\frac{5}{t}\bigg)~dt=\left. 2t^4+2t+\frac{15}{2}t^{-2}\right|_1^2</math>.
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|We set
 
|-
 
|-
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|We now evaluate to get
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|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -2px">-32t+200=0</math>
 
|-
 
|-
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| &nbsp;&nbsp; <math style="vertical-align: -16px">\int_1^2\bigg(2t+\frac{3}{t^2}\bigg)\bigg(4t^2-\frac{5}{t}\bigg)~dt=2(2)^4+2(2)+\frac{15}{2(2)^2}-\bigg(2+2+\frac{15}{2}\bigg)=36+\frac{15}{8}-4-\frac{15}{2}=\frac{211}{8}</math>.
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|and solve for &nbsp;<math style="vertical-align: -1px">t.</math>  
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|}
 
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'''(b)'''
 
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
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!Step 1: &nbsp;
 
 
|-
 
|-
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|We use <math style="vertical-align: 0px">u</math>-substitution. Let <math style="vertical-align: -2px">u=x^4+2x^2+4</math>. Then, <math style="vertical-align: -5px">du=(4x^3+4x)dx</math> and <math style="vertical-align: -14px">\frac{du}{4}=(x^3+x)dx</math>. Also, we need to change the bounds of integration.
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|We get
 
|-
 
|-
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|Plugging in our values into the equation <math style="vertical-align: -2px">u=x^4+2x^2+4</math>, we get <math style="vertical-align: -5px">u_1=0^4+2(0)^2+4=4</math> and <math style="vertical-align: -4px">u_2=2^4+2(2)^2+4=28</math>.
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|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -1px">t=6.25.</math>
 
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|-
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|Therefore, the integral becomes&nbsp; <math style="vertical-align: -14px">\frac{1}{4}\int_4^{28}\sqrt{u}~du</math>.
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|Then, we use test points to see that &nbsp;<math style="vertical-align: -2px">-32t+200</math>&nbsp; is positive from &nbsp;<math style="vertical-align: -6px">[0,6.25]</math>
 
|-
 
|-
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|
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|and negative from &nbsp;<math>[6.25,10].</math>
 
|}
 
|}
  
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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!Step 2: &nbsp;
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!Step 3: &nbsp;
 
|-
 
|-
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|We now have:
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|Therefore, we get
 
|-
 
|-
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| &nbsp;&nbsp; <math style="vertical-align: -16px">\int_0^2 (x^3+x)\sqrt{x^4+2x^2+4}~dx=\frac{1}{4}\int_4^{28}\sqrt{u}~du=\left.\frac{1}{6}u^{\frac{3}{2}}\right|_4^{28}=\frac{1}{6}(28^{\frac{3}{2}}-4^{\frac{3}{2}})=\frac{1}{6}((\sqrt{28})^3-(\sqrt{4})^3)=\frac{1}{6}((2\sqrt{7})^3-2^3)</math>.
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|
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|-
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&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
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|So, we have
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\displaystyle{\int_0^{10} |-32t+200|~dt} & = & \displaystyle{\int_0^{6.25} -32t+200~dt+\int_{6.25}^{10}-(-32t+200)~dt}\\
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|-
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&&\\
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| &nbsp;&nbsp; <math style="vertical-align: -16px">\int_0^2 (x^3+x)\sqrt{x^4+2x^2+4}~dx=\frac{28\sqrt{7}-4}{3}</math>.
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& = & \displaystyle{\left. (-16t^2+200t)\right|_{0}^{6.25}+\left. (16t^2-200t)\right|_{6.25}^{10}}\\
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&&\\
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& = & \displaystyle{-16(6.25)^2+200(6.25)+(16(10)^2-200(10))-(16(6.25)^2-200(6.25))}\\
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&&\\
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& = & \displaystyle{850}.\\
 +
\end{array}</math>  
 
|}
 
|}
 +
  
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Final Answer: &nbsp;  
 
!Final Answer: &nbsp;  
 
|-
 
|-
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|'''(a)''' &nbsp; <math>\frac{211}{8}</math>
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|&nbsp; &nbsp; &nbsp; &nbsp; The particle travels &nbsp;<math style="vertical-align: -1px">850</math> &nbsp; feet.
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|-
 
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|'''(b)''' &nbsp; <math>\frac{28\sqrt{7}-4}{3}</math>
 
 
|}
 
|}
 
[[009B_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']]
 
[[009B_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']]

Latest revision as of 11:15, 9 April 2017

A particle moves along a straight line with velocity given by:

feet per second. Determine the total distance traveled by the particle

from time    to time  


Foundations:  
1. How are the velocity function    and the position function    related?

        They are related by the equation  

2. If we calculate    what are we calculating?

        We are calculating  

        This is the displacement of the particle from    to  

3. If we calculate    what are we calculating?

        We are calculating the total distance traveled by the particle from    to  


Solution:

Step 1:  
To calculate the total distance the particle traveled from    to  
we need to calculate
       
Step 2:  
We need to figure out when    is positive and negative in the interval  
We set
       
and solve for  
We get
       
Then, we use test points to see that    is positive from  
and negative from  
Step 3:  
Therefore, we get

       


Final Answer:  
        The particle travels     feet.

Return to Sample Exam

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