009A Sample Midterm 2, Problem 4 Detailed Solution

From Grad Wiki
Revision as of 10:32, 11 November 2017 by Kayla Murray (talk | contribs) (Created page with "<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"><math>\text{BSA...")
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
Jump to navigation Jump to search

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

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\text{BSA}}={\frac {\sqrt {hm}}{60}}}

where  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 h}   is the height in centimeters 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 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  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 m=85.}   What is the rate at  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 h=170}   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 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:  
Recall
       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)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}}


Solution:

Step 1:  
Let  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)=\frac{1+x}{3x}.}
Using the limit definition of derivative, 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 \begin{array}{rcl} \displaystyle{f'(x)} & = & \displaystyle{\lim_{h\rightarrow 0} \frac{f(x+h)-f(x)}{h}}\\ &&\\ & = & \displaystyle{\lim_{h\rightarrow 0} \frac{(\frac{1+(x+h)}{3(x+h)})-(\frac{1+x}{3x})}{h}}\\ &&\\ & = & \displaystyle{\lim_{h\rightarrow 0} \frac{(\frac{1+x+h}{3x+3h})-(\frac{1+x}{h})}{h}.} \end{array}}
Step 2:  
Now, we get a common denominator for the fractions in the numerator.
Hence, 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 \begin{array}{rcl} \displaystyle{f'(x)} & = & \displaystyle{\lim_{h\rightarrow 0}\frac{\frac{(1+x+h)3x}{(3x+3h)(3x)}-\frac{(1+x)(3x+3h)}{(3x+3h)(3x)}}{h}}\\ &&\\ & = & \displaystyle{\lim_{h\rightarrow 0}\frac{\frac{3x+3x^2+3xh-(3x+3h+3x^2+3hx)}{(3x+3h)(3x)}}{h}}\\ &&\\ & = & \displaystyle{\lim_{h\rightarrow 0} \frac{-3h}{h(3x+3h)(3x)}}\\ &&\\ & = & \displaystyle{\lim_{h\rightarrow 0} \frac{-3}{(3x+3h)(3x)}}\\ &&\\ & = & \displaystyle{\frac{-3}{(3x)(3x)}}\\ & = & \displaystyle{-\frac{1}{3x^2}.} \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{dy}{dx}=-\frac{1}{3x^2}}

Return to Sample Exam