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

Revision as of 10:59, 9 April 2017

Otis Taylor plots the price per share of a stock that he owns as a function of time

and finds that it can be approximated by the function

s(t)=t(25-5t)+18

where $t$ is the time (in years) since the stock was purchased.

Find the average price of the stock over the first five years.

Foundations:

The average value of a 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 f(x)} on an interval 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 [a,b]} 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 f_{\text{avg}}=\frac{1}{b-a}\int_a^b f(x)~dx.}

Solution:

Step 1:
This problem wants us to find the average value of 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)} over the interval 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 [0,5].}
Using the average value formula, we have
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_{\text{avg}}=\frac{1}{5-0} \int_0^5 t(25-5t)+18~dt.}

Step 2:
First, we distribute to 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 s_{\text{avg}}=\frac{1}{5} \int_0^5 25t-5t^2+18~dt.}
Then, we integrate to 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 s_{\text{avg}}=\left. \frac{1}{5}\bigg[\frac{25t^2}{2}-\frac{5t^3}{3}+18t\bigg]\right\|_0^5.}

Step 3:

We now evaluate to 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{s_{\text{avg}}} & = & \displaystyle{\frac{1}{5}\bigg[\frac{25(5)^2}{2}-\frac{5(5)^3}{3}+18(5)\bigg]-0}\\ &&\\ & = & \displaystyle{\frac{233}{6}}\\ &&\\ & \approx & \displaystyle{$38.83.} \end{array}}

Final Answer:
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 \frac{233}{6}\approx $38.83}

Return to Sample Exam

@@ Line 1: / Line 1: @@
-<span class="exam">Find the average value of the function on the given interval.
+<span class="exam"> Otis Taylor plots the price per share of a stock that he owns as a function of time
-::<math>f(x)=2x^3(1+x^2)^4,~~~[0,2]</math>
+<span class="exam">and finds that it can be approximated by the function
+::<math>s(t)=t(25-5t)+18</math>
+<span class="exam">where &nbsp;<math style="vertical-align: 0px">t</math>&nbsp; is the time (in years) since the stock was purchased.
+<span class="exam">Find the average price of the stock over the first five years.
@@ Line 7: / Line 13: @@
 !Foundations: &nbsp;
 |-
-|The average value of a function <math style="vertical-align: -5px">f(x)</math> on an interval <math style="vertical-align: -5px">[a,b]</math> is given by
+|The average value of a function &nbsp;<math style="vertical-align: -5px">f(x)</math>&nbsp; on an interval &nbsp;<math style="vertical-align: -5px">[a,b]</math>&nbsp; is given by
 |-
-|
+|&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -18px">f_{\text{avg}}=\frac{1}{b-a}\int_a^b f(x)~dx.</math>
-::<math>f_{\text{avg}}=\frac{1}{b-a}\int_a^b f(x)~dx.</math>
 |}
 '''Solution:'''
@@ Line 17: / Line 23: @@
 !Step 1: &nbsp;
 |-
-|Using the formula given in Foundations, we have:
+|This problem wants us to find the average value of &nbsp;<math style="vertical-align: -5px">s(t)</math>&nbsp; over the interval &nbsp;<math style="vertical-align: -5px">[0,5].</math>
+|-
+|Using the average value formula, we have
 |-
-|
+| &nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: 0px">s_{\text{avg}}=\frac{1}{5-0} \int_0^5 t(25-5t)+18~dt.</math>
-::<math>\begin{array}{rcl}
-\displaystyle{f_{\text{avg}}} & = & \displaystyle{\frac{1}{2-0}\int_0^2 2x^3(1+x^2)^4~dx}\\
-&&\\
-& = & \displaystyle{\int_0^2 x^3(1+x^2)^4~dx.}\\
-\end{array}</math>
 |}
@@ Line 30: / Line 33: @@
 !Step 2: &nbsp;
 |-
-|Now, we use <math>u</math>-substitution. Let <math style="vertical-align: -2px">u=1+x^2.</math> Then, <math style="vertical-align: 0px">du=2x dx</math> and <math style="vertical-align: -13px">\frac{du}{2}=xdx.</math> Also, <math style="vertical-align: 0px">x^2=u-1.</math>
+|First, we distribute to get
 |-
-|We need to change the bounds on the integral. We have <math style="vertical-align: -4px">u_1=1+0^2=1</math> and <math style="vertical-align: -3px">u_2=1+2^2=5.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>s_{\text{avg}}=\frac{1}{5} \int_0^5 25t-5t^2+18~dt.</math>
 |-
-|So, the integral becomes
+|Then, we integrate to get
 |-
-|
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>s_{\text{avg}}=\left. \frac{1}{5}\bigg[\frac{25t^2}{2}-\frac{5t^3}{3}+18t\bigg]\right|_0^5.</math>
-::<math>\begin{array}{rcl}
-\displaystyle{f_{\text{avg}}} & = & \displaystyle{\int_0^2 x\cdot x^2 (1+x^2)^4~dx}\\
-&&\\
-& = & \displaystyle{\frac{1}{2}\int_1^5(u-1)u^4~du}\\
-&&\\
-& = & \displaystyle{\frac{1}{2}\int_1^5(u^5-u^4)~du.}\\
-\end{array}</math>
 |}
@@ Line 49: / Line 45: @@
 !Step 3: &nbsp;
 |-
-|We integrate to get
+|We now evaluate to get
 |-
-|
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
-::<math>\begin{array}{rcl}
+\displaystyle{s_{\text{avg}}} & = & \displaystyle{\frac{1}{5}\bigg[\frac{25(5)^2}{2}-\frac{5(5)^3}{3}+18(5)\bigg]-0}\\
-\displaystyle{f_{\text{avg}}} & = & \displaystyle{\left.\frac{u^6}{12}-\frac{u^5}{10}\right|_{1}^5}\\
+&&\\
+& = & \displaystyle{\frac{233}{6}}\\
 &&\\
-& = & \displaystyle{\left.u^5\bigg(\frac{u}{12}-\frac{1}{10}\bigg)\right|_{1}^5.}\\
+& \approx & \displaystyle{$38.83.}
 \end{array}</math>
 |}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 4: &nbsp;
-|-
-|We evaluate to get
-|-
-|
-::<math>\begin{array}{rcl}
-\displaystyle{f_{\text{avg}}} & = & \displaystyle{5^5\bigg(\frac{5}{12}-\frac{1}{10}\bigg)-1^5\bigg(\frac{1}{12}-\frac{1}{10}\bigg)}\\
-&&\\
-& = & \displaystyle{3125\bigg(\frac{19}{60}\bigg)-\frac{-1}{60}}\\
-&&\\
-& = & \displaystyle{\frac{59376}{60}}\\
-&&\\
-& = & \displaystyle{\frac{4948}{5}.}\\
-\end{array}</math>
-|}
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 !Final Answer: &nbsp;
 |-
-|&nbsp;&nbsp; <math>\frac{4948}{5}</math>
+| &nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{233}{6}\approx $38.83</math>
 |-
 |
 |}
 [[009B_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]

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

Revision as of 10:59, 9 April 2017

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