009B Sample Final 3, Problem 3 - Revision history

Kayla Murray at 21:46, 12 March 2017

2017-03-12T21:46:20Z

Kayla Murray at 20:38, 3 March 2017

2017-03-03T20:38:24Z

Kayla Murray at 20:35, 3 March 2017

2017-03-03T20:35:54Z

Kayla Murray at 20:28, 3 March 2017

2017-03-03T20:28:30Z

Kayla Murray at 20:24, 3 March 2017

2017-03-03T20:24:33Z

Kayla Murray at 20:15, 3 March 2017

2017-03-03T20:15:39Z

Kayla Murray at 20:13, 3 March 2017

2017-03-03T20:13:49Z

Kayla Murray at 19:58, 3 March 2017

2017-03-03T19:58:01Z

Kayla Murray at 19:31, 3 March 2017

2017-03-03T19:31:31Z

Kayla Murray at 00:59, 1 March 2017

2017-03-01T00:59:18Z

@@ Line 69: / Line 69: @@
 |We need to find the absolute maximum and minimum of &nbsp;<math style="vertical-align: -5px">\rho(x).</math>
 |-
-|We begin by finding the critical points of &nbsp;<math style="vertical-align: -2px">-x^2+6x+16.</math>
+|We begin by finding the critical points of
 |-
-|Taking the derivative, we have &nbsp;<math style="vertical-align: -2px">-2x+6.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -2px">-x^2+6x+16.</math>
 |-
-|Solving &nbsp;<math style="vertical-align: -4px">-2x+6=0,</math>&nbsp; we get a critical point at &nbsp;<math style="vertical-align: 0px">x=3.</math>
+|Taking the derivative, we get
 |-
 |Now, we calculate &nbsp;<math style="vertical-align: -5px">\rho(0),~\rho(3),~\rho(12).</math>
@@ Line 120: / Line 126: @@
 !Final Answer: &nbsp;
 |-
-|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; The minimum of &nbsp;<math style="vertical-align: -5px">\rho(x)</math>&nbsp; is &nbsp;<math style="vertical-align: -1px">16</math>&nbsp; and the maximum of &nbsp;<math style="vertical-align: -5px">\rho(x)</math>&nbsp; is &nbsp;<math style="vertical-align: -1px">56.</math>
+|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; The minimum of &nbsp;<math style="vertical-align: -5px">\rho(x)</math>&nbsp; is &nbsp;<math style="vertical-align: -1px">16</math>&nbsp; and the maximum of &nbsp;<math style="vertical-align: -5px">\rho(x)</math>&nbsp; is &nbsp;<math style="vertical-align: -1px">56.</math> (See above for graph.)
 |-
 |&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp; There are approximately &nbsp;<math style="vertical-align: -1px">251</math>&nbsp; trout.

@@ Line 75: / Line 75: @@
 |Solving &nbsp;<math style="vertical-align: -4px">-2x+6=0,</math>&nbsp; we get a critical point at &nbsp;<math style="vertical-align: 0px">x=3.</math>
 |-
-|Now, we calculate &nbsp;<math style="vertical-align: -5px">\rho(0),\rho(3),\rho(12).</math>
+|Now, we calculate &nbsp;<math style="vertical-align: -5px">\rho(0),~\rho(3),~\rho(12).</math>
 |-
 |We have
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp;<math>\rho(0)=16,\rho(3)=25,\rho(12)=56.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp;<math>\rho(0)=16,~\rho(3)=25,~\rho(12)=56.</math>
 |-
 |Therefore, the minimum of &nbsp;<math style="vertical-align: -5px">\rho(x)</math>&nbsp; is &nbsp;<math style="vertical-align: -1px">16</math>&nbsp; and the maximum of &nbsp;<math style="vertical-align: -5px">\rho(x)</math>&nbsp; is &nbsp;<math style="vertical-align: -1px">56.</math>
@@ Line 95: / Line 95: @@
 |Using the information from Step 1 of (a), we have
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp;<math> \int_0^{12} \rho(x)~dx.=\int_0^8 (-x^2+6x+16)~dx+\int_8^{12} (x^2-6x-16)~dx.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp;<math> \int_0^{12} \rho(x)~dx=\int_0^8 (-x^2+6x+16)~dx+\int_8^{12} (x^2-6x-16)~dx.</math>
 |}
@@ Line 113: / Line 113: @@
 \end{array}</math>
 |-
-|Thus, there are approximately &nbsp;<math>251</math>&nbsp; trout.
+|Thus, there are approximately &nbsp;<math style="vertical-align: -1px">251</math>&nbsp; trout.
 |}
@@ Line 122: / Line 122: @@
 |&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; The minimum of &nbsp;<math style="vertical-align: -5px">\rho(x)</math>&nbsp; is &nbsp;<math style="vertical-align: -1px">16</math>&nbsp; and the maximum of &nbsp;<math style="vertical-align: -5px">\rho(x)</math>&nbsp; is &nbsp;<math style="vertical-align: -1px">56.</math>
 |-
-|&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp; There are approximately &nbsp;<math>251</math>&nbsp; trout.
+|&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp; There are approximately &nbsp;<math style="vertical-align: -1px">251</math>&nbsp; trout.
 |-
+|
 |}
 [[009B_Sample_Final_3|'''<u>Return to Sample Exam</u>''']]

@@ Line 27: / Line 27: @@
 !Step 1: &nbsp;
 |-
-|To graph &nbsp;<math>\rho(x),</math>&nbsp; we need to find out when &nbsp;<math>-x^2+6x+16</math>&nbsp; is negative.
+|To graph &nbsp;<math style="vertical-align: -5px">\rho(x),</math>&nbsp; we need to find out when &nbsp;<math style="vertical-align: -2px">-x^2+6x+16</math>&nbsp; is negative.
 |-
 |To do this, we set
@@ Line 43: / Line 43: @@
 \end{array}</math>
 |-
-|Hence, we get &nbsp;<math>x=-2</math>&nbsp; and &nbsp;<math>x=8.</math>&nbsp;
+|Hence, we get &nbsp;<math style="vertical-align: 0px">x=-2</math>&nbsp; and &nbsp;<math style="vertical-align: 0px">x=8.</math>&nbsp;
 |-
-|But, &nbsp;<math>x=-2</math>&nbsp; is outside of the domain of &nbsp;<math>\rho(x).</math>
+|But, &nbsp;<math style="vertical-align: 0px">x=-2</math>&nbsp; is outside of the domain of &nbsp;<math style="vertical-align: -5px">\rho(x).</math>
 |-
-|Using test points, we can see that &nbsp;<math>-x^2+6x+16</math>&nbsp; is positive in the interval &nbsp;<math>[0,8]</math>
+|Using test points, we can see that &nbsp;<math style="vertical-align: -2px">-x^2+6x+16</math>&nbsp; is positive in the interval &nbsp;<math style="vertical-align: -5px">[0,8]</math>
 |-
-|and negative in the interval &nbsp;<math>[8,12].</math>
+|and negative in the interval &nbsp;<math style="vertical-align: -5px">[8,12].</math>
 |-
 |Hence, we have
@@ Line 61: / Line 61: @@
 </math>
 |-
-|The graph of &nbsp;<math>\rho(x)</math>&nbsp; is displayed below.
+|The graph of &nbsp;<math style="vertical-align: -5px">\rho(x)</math>&nbsp; is displayed below.
 |}
@@ Line 67: / Line 67: @@
 !Step 2: &nbsp;
 |-
-|We need to find the absolute maximum and minimum of &nbsp;<math>\rho(x).</math>
+|We need to find the absolute maximum and minimum of &nbsp;<math style="vertical-align: -5px">\rho(x).</math>
 |-
-|We begin by finding the critical points of &nbsp;<math>-x^2+6x+16.</math>
+|We begin by finding the critical points of &nbsp;<math style="vertical-align: -2px">-x^2+6x+16.</math>
 |-
-|Taking the derivative, we have &nbsp;<math>-2x+6.</math>
+|Taking the derivative, we have &nbsp;<math style="vertical-align: -2px">-2x+6.</math>
 |-
-|Solving &nbsp;<math>-2x+6=0,</math>&nbsp; we get a critical point at &nbsp;<math>x=3</math>.
+|Solving &nbsp;<math style="vertical-align: -4px">-2x+6=0,</math>&nbsp; we get a critical point at &nbsp;<math style="vertical-align: 0px">x=3.</math>
 |-
-|Now, we calculate &nbsp;<math>\rho(0),\rho(3),\rho(12).</math>
+|Now, we calculate &nbsp;<math style="vertical-align: -5px">\rho(0),\rho(3),\rho(12).</math>
 |-
 |We have
@@ Line 81: / Line 81: @@
 |&nbsp; &nbsp; &nbsp; &nbsp;<math>\rho(0)=16,\rho(3)=25,\rho(12)=56.</math>
 |-
-|Therefore, the minimum of &nbsp;<math>\rho(x)</math>&nbsp; is &nbsp;<math>16</math>&nbsp; and the maximum of &nbsp;<math>\rho(x)</math>&nbsp; is &nbsp;<math>56.</math>
+|Therefore, the minimum of &nbsp;<math style="vertical-align: -5px">\rho(x)</math>&nbsp; is &nbsp;<math style="vertical-align: -1px">16</math>&nbsp; and the maximum of &nbsp;<math style="vertical-align: -5px">\rho(x)</math>&nbsp; is &nbsp;<math style="vertical-align: -1px">56.</math>
 |}
@@ Line 120: / Line 120: @@
 !Final Answer: &nbsp;
 |-
-|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; The minimum of &nbsp;<math>\rho(x)</math>&nbsp; is &nbsp;<math>16</math>&nbsp; and the maximum of &nbsp;<math>\rho(x)</math>&nbsp; is &nbsp;<math>56.</math>&nbsp; (See Step 1 for graph)
+|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; The minimum of &nbsp;<math style="vertical-align: -5px">\rho(x)</math>&nbsp; is &nbsp;<math style="vertical-align: -1px">16</math>&nbsp; and the maximum of &nbsp;<math style="vertical-align: -5px">\rho(x)</math>&nbsp; is &nbsp;<math style="vertical-align: -1px">56.</math>
 |-
 |&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp; There are approximately &nbsp;<math>251</math>&nbsp; trout.

@@ Line 27: / Line 27: @@
 !Step 1: &nbsp;
 |-
-|To graph <math>\rho(x),</math> we need to find out when <math>-x^2+6x+16</math> is negative.
+|To graph &nbsp;<math>\rho(x),</math>&nbsp; we need to find out when &nbsp;<math>-x^2+6x+16</math>&nbsp; is negative.
 |-
 |To do this, we set
@@ Line 43: / Line 43: @@
 \end{array}</math>
 |-
-|Hence, we get <math>x=-2</math> and <math>x=8.</math> But, <math>x=-2</math> is outside of the domain of <math>\rho(x).</math>
+|Hence, we get &nbsp;<math>x=-2</math>&nbsp; and &nbsp;<math>x=8.</math>&nbsp;
 |-
-|Using test points, we can see that <math>-x^2+6x+16</math> is positive in the interval <math>[0,8]</math>
+|But, &nbsp;<math>x=-2</math>&nbsp; is outside of the domain of &nbsp;<math>\rho(x).</math>
 |-
-|and negative in the interval <math>[8,12].</math>
+|Using test points, we can see that &nbsp;<math>-x^2+6x+16</math>&nbsp; is positive in the interval &nbsp;<math>[0,8]</math>
 |-
 |Hence, we have
 |-
-|<math>\rho(x) = \left\{
+|&nbsp; &nbsp; &nbsp; &nbsp;<math>\rho(x) = \left\{
       \begin{array}{lr}
         -x^2+6x+16 &  \text{if }0\le x \le 8\\
@@ Line 59: / Line 61: @@
 </math>
 |-
-|The graph of <math>\rho(x)</math> is displayed below.
+|The graph of &nbsp;<math>\rho(x)</math>&nbsp; is displayed below.
 |}
@@ Line 65: / Line 67: @@
 !Step 2: &nbsp;
 |-
-|We need to find the absolute maximum and minimum of <math>\rho(x).</math>
+|We need to find the absolute maximum and minimum of &nbsp;<math>\rho(x).</math>
 |-
-|We begin by finding the critical points of <math>-x^2+6x+16.</math>
+|We begin by finding the critical points of &nbsp;<math>-x^2+6x+16.</math>
 |-
-|Taking the derivative, we have <math>-2x+6.</math>
+|Taking the derivative, we have &nbsp;<math>-2x+6.</math>
 |-
-|Solving <math>-2x+6=0,</math> we get a critical point at <math>x=3</math>.
+|Solving &nbsp;<math>-2x+6=0,</math>&nbsp; we get a critical point at &nbsp;<math>x=3</math>.
 |-
-|Now, we calculate <math>\rho(0),\rho(3),\rho(12).</math>
+|Now, we calculate &nbsp;<math>\rho(0),\rho(3),\rho(12).</math>
 |-
 |We have
 |-
-|<math>\rho(0)=16,\rho(3)=25,\rho(12)=56.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp;<math>\rho(0)=16,\rho(3)=25,\rho(12)=56.</math>
 |-
-|Therefore, the minimum of <math>\rho(x)</math> is <math>16</math> and the maximum of <math>\rho(x)</math> is <math>56.</math>
+|Therefore, the minimum of &nbsp;<math>\rho(x)</math>&nbsp; is &nbsp;<math>16</math>&nbsp; and the maximum of &nbsp;<math>\rho(x)</math>&nbsp; is &nbsp;<math>56.</math>
 |}
@@ Line 89: / Line 91: @@
 |To calculate the total number of trout, we need to find
 |-
-|<math> \int_0^{12} \rho(x)~dx.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp;<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>
+|&nbsp; &nbsp; &nbsp; &nbsp;<math> \int_0^{12} \rho(x)~dx.=\int_0^8 (-x^2+6x+16)~dx+\int_8^{12} (x^2-6x-16)~dx.</math>
 |}
@@ Line 111: / Line 113: @@
 \end{array}</math>
 |-
-|So there are approximately <math>251</math> trout.
+|Thus, there are approximately &nbsp;<math>251</math>&nbsp; trout.
 |}
@@ Line 118: / Line 120: @@
 !Final Answer: &nbsp;
 |-
-|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; The minimum of <math>\rho(x)</math> is <math>16</math> and the maximum of <math>\rho(x)</math> is <math>56.</math> (See Step 1 for graph)
+|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; The minimum of &nbsp;<math>\rho(x)</math>&nbsp; is &nbsp;<math>16</math>&nbsp; and the maximum of &nbsp;<math>\rho(x)</math>&nbsp; is &nbsp;<math>56.</math>&nbsp; (See Step 1 for graph)
 |-
-|&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp; There are approximately <math>251</math> trout.
+|&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp; There are approximately &nbsp;<math>251</math>&nbsp; trout.
 |-
+|
 |}
 [[009B_Sample_Final_3|'''<u>Return to Sample Exam</u>''']]

@@ Line 89: / Line 89: @@
 |To calculate the total number of trout, we need to find
 |-
-|<math> \int_0^12 \rho(x)~dx.</math>
+|<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+12~dx+\int_8^{12} x^2-6x-16~dx.</math>
+|<math> \int_0^{12} \rho(x)~dx.=\int_0^8 (-x^2+6x+16)~dx+\int_8^{12} (x^2-6x-16)~dx.</math>
 |}
@@ Line 99: / Line 99: @@
 !Step 2: &nbsp;
 |-
+|We integrate to get
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 |}
@@ Line 110: / Line 120: @@
 |&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; The minimum of <math>\rho(x)</math> is <math>16</math> and the maximum of <math>\rho(x)</math> is <math>56.</math> (See Step 1 for graph)
 |-
-|'''(b)'''
+|&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp; There are approximately <math>251</math> trout.
 |-
+|
 |}
 [[009B_Sample_Final_3|'''<u>Return to Sample Exam</u>''']]

@@ Line 12: / Line 12: @@
 !Foundations: &nbsp;
 |-
+|What is the relationship between population density &nbsp;<math style="vertical-align: -5px">\rho(x)</math>&nbsp; and the total populations?
 |-
-|'''1.''' Absolute minimum and maximum
+|&nbsp; &nbsp; &nbsp; &nbsp;The total population is equal to &nbsp;<math style="vertical-align: -15px">\int_a^b \rho(x)~dx</math>
 |-
-|'''2.''' Relationship between population density and total population
+|&nbsp; &nbsp; &nbsp; &nbsp;for appropriate choices of &nbsp;<math style="vertical-align: -5px">a,b.</math>
 |}

@@ Line 51: / Line 51: @@
 '''(b)'''
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
@@ Line 94: / Line 76: @@
 |'''(b)'''
 |-
-|'''(c)'''
 |}
 [[009B_Sample_Final_3|'''<u>Return to Sample Exam</u>''']]