009B Sample Final 2, Problem 4 - Revision history

Kayla Murray at 22:05, 26 May 2017

2017-05-26T22:05:55Z

Kayla Murray at 22:00, 26 May 2017

2017-05-26T22:00:45Z

Kayla Murray at 20:30, 26 May 2017

2017-05-26T20:30:41Z

Kayla Murray at 20:26, 26 May 2017

2017-05-26T20:26:26Z

Kayla Murray at 20:20, 26 May 2017

2017-05-26T20:20:09Z

Kayla Murray at 02:21, 19 February 2017

2017-02-19T02:21:50Z

Kayla Murray at 03:14, 18 February 2017

2017-02-18T03:14:56Z

Kayla Murray: Created page with "Consider the region bounded by the following two functions: :::::::: $y=2(-x^2+9)$ and
2017-02-18T02:43:41Z

Created page with "Consider the region bounded by the following two functions: :::::::: <math style="vertical-align: -5px">y=2(-x^2+9)</math> and <math styl..."

New page
Consider the region bounded by the following two functions:
:::::::: <math style="vertical-align: -5px">y=2(-x^2+9)</math> and <math style="vertical-align: -4px">y=0</math>.

a) Using the lower sum with three rectangles having equal width, approximate the area.

b) Using the upper sum with three rectangles having equal width, approximate the area.

c) Find the actual area of the region.

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|
|-
|
|-
|
|-
|
|}

'''Solution:'''

'''(a)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|
|-
|
|-
|
|-
|
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|
|-
|
|-
|
|-
|
|}

'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|
|-
|
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|
|-
|
|}

'''(c)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|
|-
|
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|
|-
|
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|'''(a)'''
|-
|'''(b)'''
|-
|'''(c)'''
|}
[[009B_Sample_Final_2|'''Return to Sample Exam''']]

@@ Line 16: / Line 16: @@
 |&nbsp; &nbsp; &nbsp; &nbsp; <math>\rho(x)=25,000e^{-0.15x}</math>
 |-
-|where &nbsp;<math>x</math>&nbsp; is the distance from the center.
+|where &nbsp;<math style="vertical-align: 0px">x</math>&nbsp; is the distance from the center.
 |-
 |Any slice along a radius gives us a cross section.
@@ Line 41: / Line 41: @@
 |We can treat this as a solid of revolution, and use the shell method.
 |-
-|We are working on a half disk of radius 7, so we can integrate a cross-section where &nbsp;<math>x</math>&nbsp; goes from 0 to 7
+|We are working on a half disk of radius 7, so we can integrate a cross-section where &nbsp;<math style="vertical-align: 0px">x</math>&nbsp; goes from 0 to 7
 |-
-|and the height at each &nbsp;<math>x</math>&nbsp; is our density function, &nbsp;<math>\rho(x).</math>&nbsp;
+|and the height at each &nbsp;<math style="vertical-align: 0px">x</math>&nbsp; is our density function, &nbsp;<math style="vertical-align: -5px">\rho(x).</math>&nbsp;
 |-
-|Normally &nbsp;<math>2\pi R</math>&nbsp; represents once around a circle of radius &nbsp;<math>R,</math>&nbsp;
+|Normally &nbsp;<math style="vertical-align: 0px">2\pi R</math>&nbsp; represents once around a circle of radius &nbsp;<math style="vertical-align: -5px">R,</math>&nbsp;
 |-
 |but in this case we only go half way around.
@@ Line 73: / Line 73: @@
 |To solve this, we need to use integration by parts.
 |-
-|Let &nbsp;<math>u=x</math>&nbsp; and &nbsp;<math>dv=e^{-0.15x}dx.</math>&nbsp;
+|Let &nbsp;<math style="vertical-align: 0px">u=x</math>&nbsp; and &nbsp;<math style="vertical-align: 0px">dv=e^{-0.15x}dx.</math>&nbsp;
 |-
-|Then, &nbsp;<math>du=dx</math>&nbsp; and &nbsp;<math>v=-{\displaystyle \frac{e^{-0.15x}}{0.15}.}</math>&nbsp;
+|Then, &nbsp;<math style="vertical-align: 0px">du=dx</math>&nbsp; and &nbsp;<math style="vertical-align: -14px">v=-{\displaystyle \frac{e^{-0.15x}}{0.15}.}</math>&nbsp;
 |-
 |Thus,
@@ Line 99: / Line 99: @@
 !Final Answer: &nbsp;
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp;<math>{\displaystyle 25,000\pi\left[\left(-\frac{7e^{-1.05}}{0.15}-\frac{e^{-1.05}}{(0.15)^{2}}\right)+\frac{1}{(0.15)^{2}}\right]\ \approx\ 986,556.}</math>
+|&nbsp; &nbsp; &nbsp; &nbsp;<math>{\displaystyle 25,000\pi\left[\left(-\frac{7e^{-1.05}}{0.15}-\frac{e^{-1.05}}{(0.15)^{2}}\right)+\frac{1}{(0.15)^{2}}\right]\ \approx\ 986,556}</math>
 |}
 [[009B_Sample_Final_2|'''<u>Return to Sample Exam</u>''']]

@@ Line 73: / Line 73: @@
 |To solve this, we need to use integration by parts.
 |-
+|Let &nbsp;<math>u=x</math>&nbsp; and &nbsp;<math>dv=e^{-0.15x}dx.</math>&nbsp;
-|}
+|-
+|Then, &nbsp;<math>du=dx</math>&nbsp; and &nbsp;<math>v=-{\displaystyle \frac{e^{-0.15x}}{0.15}.}</math>&nbsp;
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+|-
-!Step 3: &nbsp;
+|Thus,
 |-
+|Note that in a calculator-prohibited test, no one would expect the actual numerical answer.
 |-
+|However, you would likely need the line above it to receive full credit.
 |}
@@ Line 88: / Line 99: @@
 !Final Answer: &nbsp;
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp;<math>{\displaystyle 25,000\pi\left[\left(-\frac{7e^{-1.05}}{0.15}-\frac{e^{-1.05}}{(0.15)^{2}}\right)+\frac{1}{(0.15)^{2}}\right]\ \approx\ 986,556.}</math>
 |}
 [[009B_Sample_Final_2|'''<u>Return to Sample Exam</u>''']]

@@ Line 61: / Line 61: @@
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 !Step 2: &nbsp;
 |-
+|
 |-
+|To solve this, we need to use integration by parts.
 |-
+|

@@ Line 35: / Line 35: @@
 '''Solution:'''
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 !Step 1: &nbsp;
 |-
+|We can treat this as a solid of revolution, and use the shell method.
 |-
+|We are working on a half disk of radius 7, so we can integrate a cross-section where &nbsp;<math>x</math>&nbsp; goes from 0 to 7
 |-
+|and the height at each &nbsp;<math>x</math>&nbsp; is our density function, &nbsp;<math>\rho(x).</math>&nbsp;
 |-
+|Normally &nbsp;<math>2\pi R</math>&nbsp; represents once around a circle of radius &nbsp;<math>R,</math>&nbsp;
 |-
+|but in this case we only go half way around.
 |-
+|Therefore, we '''adjust''' our usual shell method formula to find the population as
 |-
+|
 |}
@@ Line 77: / Line 64: @@
+|
 |-
-−
+|
 |-
+|
@@ Line 91: / Line 72: @@
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 2: &nbsp;
+!Step 3: &nbsp;
 |-
+|
@@ Line 102: / Line 83: @@
 !Final Answer: &nbsp;
 |-
-|'''(a)'''
 |}
 [[009B_Sample_Final_2|'''<u>Return to Sample Exam</u>''']]

@@ Line 1: / Line 1: @@
 <span class="exam"> A city bordered on one side by a lake can be approximated by a semicircle of radius 7 miles, whose city center is on the shoreline. As we move away from the center along a radius the population density of the city can be approximated by:
-::::<math>\rho(x)=25000e^{-0.15x}</math>
+::<math>\rho(x)=25000e^{-0.15x}</math>
 <span class="exam">people per square mile. What is the population of the city?
@@ Line 16: / Line 16: @@
+|
 |}
 '''Solution:'''
@@ Line 80: / Line 81: @@
+|
 |}
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"

@@ Line 8: / Line 8: @@
 !Foundations: &nbsp;
 |-
+|Many word problems can be confusing, and this is a good example.
 |-
+|We know that we are going to integrate over a half-disk of radius 7, but how do we construct the integral?
 |-
+|One key could be the expression of our density,
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>\rho(x)=25,000e^{-0.15x}</math>
 |}