Difference between revisions of "009A Sample Midterm 1, Problem 2"

Revision as of 09:45, 7 November 2017

Suppose the size of a population at time $t$ is given by

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 N(t)=\frac{1000t}{5+t},~t\ge 0.}

(a) Determine the size of the population as 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\rightarrow \infty.} We call this the limiting population size.

(b) Show that at 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=5,} the size of the population is half its limiting size.

Detailed Solution with Background Information

Return to Sample Exam

@@ Line 6: / Line 6: @@
 <span class="exam">(b) Show that at time &nbsp;<math style="vertical-align: -4px">t=5,</math>&nbsp; the size of the population is half its limiting size.
+<hr>
+[[009A Sample Midterm 1, Problem 2 Detailed Solution|'''<u>Detailed Solution with Background Information</u>''']]
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+[[File:9ASM1P2.jpg|600px|thumb|center]]
-!Foundations: &nbsp;
-|-
-|'''1.''' If &nbsp;<math style="vertical-align: -15px">\lim_{x\rightarrow a^-} f(x)=\lim_{x\rightarrow a^+} f(x)=c,</math>
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp; then &nbsp;<math style="vertical-align: -12px">\lim_{x\rightarrow a} f(x)=c.</math>
-|-
-|'''2.''' &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; is continuous at &nbsp;<math style="vertical-align: 0px">x=a</math>&nbsp; if
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -14px">\lim_{x\rightarrow a^+}f(x)=\lim_{x\rightarrow a^-}f(x)=f(a).</math>
-|}
-'''Solution:'''
-'''(a)'''
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 1: &nbsp;
-|-
-|Notice that we are calculating a left hand limit.
-|-
-|Thus, we are looking at values of &nbsp;<math style="vertical-align: 0px">x</math>&nbsp; that are smaller than &nbsp;<math style="vertical-align: -1px">1.</math>
-|-
-|Using the definition of &nbsp;<math style="vertical-align: -5px">f(x),</math>&nbsp; we have
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math>\lim_{x\rightarrow 1^-} f(x)=\lim_{x\rightarrow 1^-} x^2.</math>
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 2: &nbsp;
-|-
-|Now, we have
-|-
-|
-&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
-\displaystyle{\lim_{x\rightarrow 1^-} f(x)} & = & \displaystyle{\lim_{x\rightarrow 1^-} x^2}\\
-&&\\
-& = & \displaystyle{\lim_{x\rightarrow 1} x^2}\\
-&&\\
-& = & \displaystyle{1^2}\\
-&&\\
-& = & \displaystyle{1.}\\
-\end{array}</math>
-|}
-'''(b)'''
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 1: &nbsp;
-|-
-|Notice that we are calculating a right hand limit.
-|-
-|Thus, we are looking at values of &nbsp;<math style="vertical-align: 0px">x</math>&nbsp; that are bigger than &nbsp;<math style="vertical-align: -2px">1.</math>
-|-
-|Using the definition of &nbsp;<math style="vertical-align: -5px">f(x),</math>&nbsp; we have
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math>\lim_{x\rightarrow 1^+} f(x)=\lim_{x\rightarrow 1^+} \sqrt{x}.</math>
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 2: &nbsp;
-|-
-|Now, we have
-|-
-|
-&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
-\displaystyle{\lim_{x\rightarrow 1^+} f(x)} & = & \displaystyle{\lim_{x\rightarrow 1^+} \sqrt{x}}\\
-&&\\
-& = & \displaystyle{\lim_{x\rightarrow 1} \sqrt{x}}\\
-&&\\
-& = & \displaystyle{\sqrt{1}}\\
-&&\\
-& = & \displaystyle{1.}\\
-\end{array}</math>
-|}
-'''(c)'''
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 1: &nbsp;
-|-
-|From (a) and (b), we have
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math>\lim_{x\rightarrow 1^-}f(x)=1</math>
-|-
-|and
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math>\lim_{x\rightarrow 1^+}f(x)=1.</math>
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 2: &nbsp;
-|-
-|Since
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math>\lim_{x\rightarrow 1^-}f(x)=\lim_{x\rightarrow 1^+}f(x)=1,</math>
-|-
-|we have
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math>\lim_{x\rightarrow 1}f(x)=1.</math>
-|}
-'''(d)'''
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 1: &nbsp;
-|-
-|From (c), we have
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math>\lim_{x\rightarrow 1}f(x)=1.</math>
-|-
-|Also,
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math>f(1)=\sqrt{1}=1.</math>
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 2: &nbsp;
-|-
-|Since
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math>\lim_{x\rightarrow 1}f(x)=f(1),</math>
-|-
-|<math style="vertical-align: -5px">f(x)</math> &nbsp;is continuous at &nbsp;<math style="vertical-align: -1px">x=1.</math>
-|-
-|
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Final Answer: &nbsp;
-|-
-|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; <math>1</math>
-|-
-|&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp; <math>1</math>
-|-
-|&nbsp; &nbsp; '''(c)''' &nbsp; &nbsp; <math>1</math>
-|-
-|&nbsp; &nbsp; '''(d)''' &nbsp; &nbsp; <math style="vertical-align: -5px">f(x)</math>&nbsp; is continuous at &nbsp;<math style="vertical-align: -1px">x=1</math>&nbsp; since &nbsp;<math style="vertical-align: -12px">\lim_{x\rightarrow 1}f(x)=f(1).</math>
-|}
 [[009A_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]

Difference between revisions of "009A Sample Midterm 1, Problem 2"

Revision as of 09:45, 7 November 2017

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