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

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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
 +
 
 +
<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?
 +
|-
 +
|
 +
&nbsp; &nbsp; &nbsp; &nbsp; They are related by the equation &nbsp;<math style="vertical-align: -5px">v(t)=s'(t).</math>
 +
|-
 +
|'''2.''' If we calculate &nbsp;<math style="vertical-align: -14px">\int_a^b v(t)~dt,</math>&nbsp; what are we calculating?
 +
|-
 +
|
 +
&nbsp; &nbsp; &nbsp; &nbsp; We are calculating &nbsp;<math style="vertical-align: -5px">s(b)-s(a).</math>
 +
|-
 +
|
 +
&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>
 +
|-
 +
|'''3.''' If we calculate &nbsp;<math style="vertical-align: -14px">\int_a^b |v(t)|~dt,</math>&nbsp; what are we calculating?
 +
|-
 +
|
 +
&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>
 +
|-
 +
|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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|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>
 
|-
 
|-
−
|
+
|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}}\\
 +
&&\\
 +
& = & \displaystyle{-16(6.25)^2+200(6.25)+(16(10)^2-200(10))-(16(6.25)^2-200(6.25))}\\
 +
&&\\
 +
& = & \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 12:15, 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  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle t=10.}


Foundations:  
1. How are the velocity function  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle v(t)}   and the position function  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle s(t)}   related?

        They are related by the equation  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle v(t)=s'(t).}

2. If we calculate  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int_a^b v(t)~dt,}   what are we calculating?

        We are calculating  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle s(b)-s(a).}

        This is the displacement of the particle from  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle t=a}   to  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle t=b.}

3. If we calculate  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int_a^b |v(t)|~dt,}   what are we calculating?

        We are calculating the total distance traveled by the particle from  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle t=a}   to  


Solution:

Step 1:  
To calculate the total distance the particle traveled from  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle t=0}   to  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle t=10,}
we need to calculate
        Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int_0^{10} |v(t)|~dt=\int_0^{10} |-32t+200|~dt.}
Step 2:  
We need to figure out when  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -32t+200}   is positive and negative in the interval  
We set
       Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -32t+200=0}
and solve for  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle t.}
We get
       Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle t=6.25.}
Then, we use test points to see that  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -32t+200}   is positive from  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [0,6.25]}
and negative from  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [6.25,10].}
Step 3:  
Therefore, we get

        Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{\int_0^{10} |-32t+200|~dt} & = & \displaystyle{\int_0^{6.25} -32t+200~dt+\int_{6.25}^{10}-(-32t+200)~dt}\\ &&\\ & = & \displaystyle{\left. (-16t^2+200t)\right|_{0}^{6.25}+\left. (16t^2-200t)\right|_{6.25}^{10}}\\ &&\\ & = & \displaystyle{-16(6.25)^2+200(6.25)+(16(10)^2-200(10))-(16(6.25)^2-200(6.25))}\\ &&\\ & = & \displaystyle{850}.\\ \end{array}}


Final Answer:  
        The particle travels  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 850}   feet.

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