Difference between revisions of "009A Sample Final 1, Problem 10"

Latest revision as of 16:17, 2 December 2017

If a resistor 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 R} ohms is connected across a battery 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 E} volts with internal resistance 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 r} ohms, then the power (in watts) in the external resistor is

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 P=\frac{E^2R}{(R+r)^2}.}

If 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 E} and 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 r} are fixed but 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 R} varies, what is the maximum value of the power?

Solution

Detailed Solution

Return to Sample Exam

@@ Line 1: / Line 1: @@
-<span class="exam">Consider the following continuous function:
+<span class="exam">If a resistor of &nbsp;<math style="vertical-align: 0px">R</math>&nbsp; ohms is connected across a battery of &nbsp;<math style="vertical-align: 0px">E</math>&nbsp; volts with internal resistance &nbsp;<math style="vertical-align: 0px">r</math>&nbsp; ohms, then the power (in watts) in the external resistor is
-::::::<math>f(x)=x^{1/3}(x-8)</math>
-<span class="exam">defined on the closed, bounded interval <math style="vertical-align: -5px">[-8,8]</math>.
+::<math>P=\frac{E^2R}{(R+r)^2}.</math>
-<span class="exam">a) Find all the critical points for <math style="vertical-align: -5px">f(x)</math>.
+<span class="exam">If &nbsp;<math style="vertical-align: 0px">E</math>&nbsp; and &nbsp;<math style="vertical-align: 0px">r</math>&nbsp; are fixed but &nbsp;<math style="vertical-align: 0px">R</math>&nbsp; varies, what is the maximum value of the power?
-<span class="exam">b) Determine the absolute maximum and absolute minimum values for <math style="vertical-align: -5px">f(x)</math> on the interval <math style="vertical-align: -5px">[-8,8]</math>.
+<hr>
+[[009A Sample Final 1, Problem 10 Solution|'''<u>Solution</u>''']]
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Foundations: &nbsp;
-|-
-|Recall:
-|-
-|'''1.''' To find the critical points for <math style="vertical-align: -5px">f(x),</math> we set <math style="vertical-align: -5px">f'(x)=0</math> and solve for <math style="vertical-align: -1px">x.</math>
-|-
-|
-::Also, we include the values of <math style="vertical-align: -1px">x</math> where <math style="vertical-align: -5px">f'(x)</math> is undefined.
-|-
-|'''2.''' To find the absolute maximum and minimum of <math style="vertical-align: -5px">f(x)</math> on an interval <math>[a,b],</math>
-|-
-|
-::we need to compare the <math style="vertical-align: -5px">y</math> values of our critical points with <math style="vertical-align: -5px">f(a)</math> and <math style="vertical-align: -5px">f(b).</math>
-|}
-'''Solution:'''
+[[009A Sample Final 1, Problem 10 Detailed Solution|'''<u>Detailed Solution</u>''']]
-== Temp1 ==
-'''(a)'''
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 1: &nbsp;
-|-
-|To find the critical points, first we need to find <math style="vertical-align: -5px">f'(x).</math>
-|-
-|Using the Product Rule, we have
-|-
-|
-::<math>\begin{array}{rcl}
-\displaystyle{f'(x)} & = & \displaystyle{\frac{1}{3}x^{-\frac{2}{3}}(x-8)+x^{\frac{1}{3}}}\\
-&&\\
-& = & \displaystyle{\frac{x-8}{3x^{\frac{2}{3}}}+x^{\frac{1}{3}}.}\\
-\end{array}</math>
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 2: &nbsp;
-|-
-|Notice <math style="vertical-align: -5px">f'(x)</math> is undefined when <math style="vertical-align: -1px">x=0.</math>
-|-
-|Now, we need to set <math style="vertical-align: -5px">f'(x)=0.</math>
-|-
-|So, we get
-|-
-|
-::<math>-x^{\frac{1}{3}}\,=\,\frac{x-8}{3x^{\frac{2}{3}}}.</math>
-|-
-|We cross multiply to get <math style="vertical-align: 1px">-3x=x-8.</math>
-|-
-|Solving, we get <math style="vertical-align: -1px">x=2.</math>
-|-
-|Thus, the critical points for <math style="vertical-align: -5px">f(x)</math> are <math style="vertical-align: -5px">(0,0)</math> and <math style="vertical-align: -4px">(2,2^{\frac{1}{3}}(-6)).</math>
-|}
-== Temp2 ==
-'''(b)'''
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 1: &nbsp;
-|-
-|We need to compare the values of <math style="vertical-align: -5px">f(x)</math>&thinsp; at the critical points and at the endpoints of the interval.
-|-
-|Using the equation given, we have <math style="vertical-align: -5px">f(-8)=32</math>&thinsp; and <math style="vertical-align: -5px">f(8)=0.</math>
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 2: &nbsp;
-|-
-|Comparing the values in Step 1 with the critical points in '''(a)''', the absolute maximum value for <math style="vertical-align: -5px">f(x)</math>&thinsp; is <math style="vertical-align: -1px">32</math>
-|-
-|and the absolute minimum value for <math style="vertical-align: -5px">f(x)</math>&thinsp; is <math style="vertical-align: -5px">2^{\frac{1}{3}}(-6).</math>
-|}
-== Temp3 ==
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Final Answer: &nbsp;
-|-
-|'''(a)'''&thinsp; <math style="vertical-align: -4px">(0,0)</math> and <math style="vertical-align: -4px">(2,2^{\frac{1}{3}}(-6))</math>
-|-
-|'''(b)'''&thinsp; The absolute minimum value for <math style="vertical-align: -5px">f(x)</math> is <math style="vertical-align: -5px">2^{\frac{1}{3}}(-6).</math>
-|}
 [[009A_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

Difference between revisions of "009A Sample Final 1, Problem 10"

Latest revision as of 16:17, 2 December 2017

Navigation menu

Search