007A Sample Midterm 1, Problem 5 Detailed Solution

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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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.

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

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