007A Sample Midterm 1, Problem 5 Detailed Solution

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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?

Background Information:
Power Rule
${\frac {d}{dx}}(x^{n})=nx^{n-1}$ Solution:

Step 1:
First, we have  $m=85.$ So, we have
${\begin{array}{rcl}\displaystyle {\text{BSA}}&=&\displaystyle {\frac {\sqrt {85h}}{60}}\\&&\\&=&\displaystyle {{\frac {{\sqrt {85}}{\sqrt {h}}}{60}}.}\end{array}}$ Step 2:
Now, using the Power Rule, we get
${\begin{array}{rcl}\displaystyle {\frac {d({\text{BSA}})}{dh}}&=&\displaystyle {{\frac {\sqrt {85}}{60}}\cdot {\frac {1}{2}}h^{-1/2}}\\&&\\&=&\displaystyle {{\frac {\sqrt {85}}{120{\sqrt {h}}}}.}\end{array}}$ Step 3:
For  $h=170,$ we get
${\frac {d({\text{BSA}})}{dh}}={\frac {\sqrt {85}}{120{\sqrt {170}}}}{\text{ m}}^{2}{\text{/cm}}.$ For  $h=190,$ we get
${\frac {d({\text{BSA}})}{dh}}={\frac {\sqrt {85}}{120{\sqrt {190}}}}{\text{ m}}^{2}{\text{/cm}}.$ Step 4:
Since  $h$ is in the denominator of the formula for  ${\frac {d({\text{BSA}})}{dh}},$ ${\text{BSA}}$ increases more rapidly with respect to height at lower heights.

${\frac {d({\text{BSA}})}{dh}}={\frac {\sqrt {85}}{120{\sqrt {h}}}}$ ${\frac {\sqrt {85}}{120{\sqrt {170}}}}{\text{ m}}^{2}{\text{/cm}}$ ${\frac {\sqrt {85}}{120{\sqrt {190}}}}{\text{ m}}^{2}{\text{/cm}}$ ${\text{BSA}}$ increases more rapidly with respect to height at lower heights.