Difference between revisions of "009A Sample Midterm 2, Problem 4"

Latest revision as of 11:08, 11 November 2017

To determine drug dosages, doctors estimate a person's body surface area (BSA) (in meters squared) using the formula:

{\text{BSA}}={\frac {\sqrt {hm}}{60}}

where $h$ is the height in centimeters and $m$ is the mass in kilograms. Calculate the rate of change of BSA with respect to height for a person of a constant mass of $m=85.$ What is the rate at $h=170$ and $h=190?$ Express your results in the correct units. Does the BSA increase more rapidly with respect to height at lower or higher heights?

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Detailed Solution

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-<span class="exam">Find the derivatives of the following functions. Do not simplify.
+<span class="exam"> To determine drug dosages, doctors estimate a person's body surface area (BSA) (in meters squared) using the formula:
-<span class="exam">(a) &nbsp; <math style="vertical-align: -5px">f(x)=x^3(x^{\frac{4}{3}}-1)</math>
+::<span class="exam"><math>\text{BSA}=\frac{\sqrt{hm}}{60}</math>
-<span class="exam">(b) &nbsp; <math style="vertical-align: -14px">g(x)=\frac{x^3+x^{-3}}{1+6x}</math>&nbsp; where &nbsp;<math style="vertical-align: 0px">x>0</math>
+<span class="exam">where &nbsp;<math style="vertical-align: 0px">h</math>&nbsp; is the height in centimeters and &nbsp;<math style="vertical-align: 0px">m</math>&nbsp; is the mass in kilograms. Calculate the rate of change of BSA with respect to height for a person of a constant mass of &nbsp;<math style="vertical-align: 0px">m=85.</math>&nbsp; What is the rate at &nbsp;<math style="vertical-align: -1px">h=170</math>&nbsp; and &nbsp;<math style="vertical-align: -1px">h=190?</math>&nbsp; Express your results in the correct units. Does the BSA increase more rapidly with respect to height at lower or higher heights?
+<hr>
+[[009A Sample Midterm 2, Problem 4 Solution|'''<u>Solution</u>''']]
+[[009A Sample Midterm 2, Problem 4 Detailed Solution|'''<u>Detailed Solution</u>''']]
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Foundations: &nbsp;
-|-
-|'''1.''' '''Product Rule'''
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{d}{dx}(f(x)g(x))=f(x)g'(x)+f'(x)g(x)</math>
-|-
-|'''2.''' '''Quotient Rule'''
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{d}{dx}\bigg(\frac{f(x)}{g(x)}\bigg)=\frac{g(x)f'(x)-f(x)g'(x)}{(g(x))^2}</math>
-|-
-|'''3.''' '''Power Rule'''
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{d}{dx}(x^n)=nx^{n-1}</math>
-|}
-'''Solution:'''
-'''(a)'''
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 1: &nbsp;
-|-
-|Using the Product Rule, we have
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math>f'(x)=x^3(x^{\frac{4}{3}}-1)'+(x^3)'(x^{\frac{4}{3}}-1).</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{f'(x)} & = & \displaystyle{x^3(x^{\frac{4}{3}}-1)'+(x^3)'(x^{\frac{4}{3}}-1)}\\
-&&\\
-& = & \displaystyle{x^3\bigg(\frac{4}{3}x^{\frac{1}{3}}\bigg)+(3x^2)(x^{\frac{4}{3}}-1).}
-\end{array}</math>
-|}
-'''(b)'''
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Step 1: &nbsp;
-|-
-|Using the Quotient Rule, we have
-|-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math>g'(x)=\frac{(1+6x)(x^3+x^{-3})'-(x^3+x^{-3})(1+6x)'}{(1+6x)^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{g'(x)} & = & \displaystyle{\frac{(1+6x)(x^3+x^{-3})'-(x^3+x^{-3})(1+6x)'}{(1+6x)^2}}\\
-&&\\
-& = & \displaystyle{\frac{(1+6x)(3x^2-3x^{-4})-(x^3+x^{-3})(6)}{(1+6x)^2}.}
-\end{array}</math>
-|}
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-!Final Answer: &nbsp;
-|-
-|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; <math>x^3\bigg(\frac{4}{3}x^{\frac{1}{3}}\bigg)+(3x^2)(x^{\frac{4}{3}}-1)</math>
-|-
-|&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp; <math>\frac{(1+6x)(3x^2-3x^{-4})-(x^3+x^{-3})(6)}{(1+6x)^2}</math>
-|}
 [[009A_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']]

Difference between revisions of "009A Sample Midterm 2, Problem 4"

Latest revision as of 11:08, 11 November 2017

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