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MathAdmin: Created page with "The population density of trout in a stream is :: $\rho(x)=|-x^2+6x+16|$ where $\rho<..."$

2017-04-10T16:53:20Z

Created page with "The population density of trout in a stream is ::<math>\rho(x)=|-x^2+6x+16|</math> where <math style="vertical-align: -5px">\rho<..."

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The population density of trout in a stream is

::<math>\rho(x)=|-x^2+6x+16|</math>

where  <math style="vertical-align: -5px">\rho</math>  is measured in trout per mile and  <math style="vertical-align: 0px">x</math>  is measured in miles.  <math style="vertical-align: 0px">x</math>  runs from 0 to 12.

(a) Graph  <math style="vertical-align: -5px">\rho(x)</math>  and find the minimum and maximum.

(b) Find the total number of trout in the stream.

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|What is the relationship between population density  <math style="vertical-align: -5px">\rho(x)</math>  and the total populations?
|-
|       The total population is equal to  <math style="vertical-align: -15px">\int_a^b \rho(x)~dx</math>
|-
|       for appropriate choices of  <math style="vertical-align: -5px">a,b.</math>
|}

'''Solution:'''

'''(a)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|To graph  <math style="vertical-align: -5px">\rho(x),</math>  we need to find out when  <math style="vertical-align: -2px">-x^2+6x+16</math>  is negative.
|-
|To do this, we set
|-
|       <math>-x^2+6x+16=0.</math>
|-
|So, we have
|-
|        <math>\begin{array}{rcl}
\displaystyle{0} & = & \displaystyle{-x^2+6x+16}\\
&&\\
& = & \displaystyle{-(x^2-6x-16)}\\
&&\\
& = & \displaystyle{-(x+2)(x-8).}
\end{array}</math>
|-
|Hence, we get  <math style="vertical-align: 0px">x=-2</math>  and  <math style="vertical-align: 0px">x=8.</math> 
|-
|But,  <math style="vertical-align: 0px">x=-2</math>  is outside of the domain of  <math style="vertical-align: -5px">\rho(x).</math>
|-
|Using test points, we can see that  <math style="vertical-align: -2px">-x^2+6x+16</math>  is positive in the interval  <math style="vertical-align: -5px">[0,8]</math>
|-
|and negative in the interval  <math style="vertical-align: -5px">[8,12].</math>
|-
|Hence, we have
|-
|       <math>\rho(x) = \left\{
\begin{array}{lr}
-x^2+6x+16 & \text{if }0\le x \le 8\\
x^2-6x-16 & \text{if }8<x\le 12
\end{array}
\right.
</math>
|-
|The graph of  <math style="vertical-align: -5px">\rho(x)</math>  is displayed below.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|We need to find the absolute maximum and minimum of  <math style="vertical-align: -5px">\rho(x).</math>
|-
|We begin by finding the critical points of
|-
|       <math style="vertical-align: -2px">-x^2+6x+16.</math>
|-
|Taking the derivative, we get
|-
|       <math style="vertical-align: -2px">-2x+6.</math>
|-
|Solving  <math style="vertical-align: -4px">-2x+6=0,</math>  we get a critical point at
|-
|       <math style="vertical-align: 0px">x=3.</math>
|-
|Now, we calculate  <math style="vertical-align: -5px">\rho(0),~\rho(3),~\rho(12).</math>
|-
|We have
|-
|       <math>\rho(0)=16,~\rho(3)=25,~\rho(12)=56.</math>
|-
|Therefore, the minimum of  <math style="vertical-align: -5px">\rho(x)</math>  is  <math style="vertical-align: -1px">16</math>  and the maximum of  <math style="vertical-align: -5px">\rho(x)</math>  is  <math style="vertical-align: -1px">56.</math>
|}

'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|To calculate the total number of trout, we need to find
|-
|       <math> \int_0^{12} \rho(x)~dx.</math>
|-
|Using the information from Step 1 of (a), we have
|-
|       <math> \int_0^{12} \rho(x)~dx=\int_0^8 (-x^2+6x+16)~dx+\int_8^{12} (x^2-6x-16)~dx.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|We integrate to get
|-
|        <math>\begin{array}{rcl}
\displaystyle{\int_0^{12} \rho(x)~dx} & = & \displaystyle{\bigg(\frac{-x^3}{3}+3x^2+16x\bigg)\bigg|_0^8+\bigg(\frac{x^3}{3}-3x^2-16x\bigg)\bigg|_8^{12}}\\
&&\\
& = & \displaystyle{\bigg(\frac{-8^3}{3}+3(8)^2+16(8)\bigg)-0+\bigg(\frac{(12)^3}{3}-3(12)^2-16(12)\bigg)-\bigg(\frac{8^3}{3}-3(8)^2-16(8)\bigg)}\\
&&\\
& = & \displaystyle{8\bigg(\frac{56}{3}\bigg)+12\bigg(\frac{12}{3}\bigg)+8\bigg(\frac{56}{3}\bigg)}\\
&&\\
& = & \displaystyle{\frac{752}{3}.}
\end{array}</math>
|-
|Thus, there are approximately  <math style="vertical-align: -1px">251</math>  trout.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|    '''(a)'''     The minimum of  <math style="vertical-align: -5px">\rho(x)</math>  is  <math style="vertical-align: -1px">16</math>  and the maximum of  <math style="vertical-align: -5px">\rho(x)</math>  is  <math style="vertical-align: -1px">56.</math> (See above for graph.)
|-
|    '''(b)'''     There are approximately  <math style="vertical-align: -1px">251</math>  trout.
|-
|
|}
[[009B_Sample_Final_3|'''Return to Sample Exam''']]

@@ Line 63: / Line 63: @@
 |The graph of &nbsp;<math style="vertical-align: -5px">\rho(x)</math>&nbsp; is displayed below.
 |-
-|[[File:009B_SF3_3.png|center|700px]]
+|[[File:009B_SF3_3.jpg|center|700px]]
 |}

@@ Line 63: / Line 63: @@
 |The graph of &nbsp;<math style="vertical-align: -5px">\rho(x)</math>&nbsp; is displayed below.
 |-
-|[[File:009B_SF3_3.png|center|600px]]
+|[[File:009B_SF3_3.png|center|700px]]
 |}

@@ Line 54: / Line 54: @@
 |-
 |&nbsp; &nbsp; &nbsp; &nbsp;<math>\rho(x) = \left\{
-      \begin{array}{lr}
+      \begin{array}{ll}
         -x^2+6x+16 &  \text{if }0\le x \le 8\\
-       x^2-6x-16 & \text{if }8<x\le 12
+       \,\,\,\,x^2-6x-16\qquad & \text{if }8<x\le 12
       \end{array}
     \right.

009B Sample Final 3, Problem 3 - Revision history

MathAdmin at 17:54, 25 May 2017

MathAdmin at 23:09, 23 May 2017

MathAdmin at 21:26, 23 May 2017

MathAdmin at 21:26, 23 May 2017

MathAdmin: Created page with "The population density of trout in a stream is ::\rho(x)=|-x^2+6x+16| where \rho<..."

MathAdmin: Created page with "The population density of trout in a stream is :: $\rho(x)=|-x^2+6x+16|$ where $\rho<..."$