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 December 10th, 2009, 05:23 PM #1 Newbie   Joined: Dec 2009 Posts: 1 Thanks: 0 derivatives I am going through a quiz I took a while back and can't figure out this problem. V=3-(.05)(cos((200)(3.14)(t)) Where V is the volume of the lungs in liters and t is the time in minutes. 1) What are the the maximum and minimum volumes of air in the lungs? 2) How do you determine the answer to part 1? 3) find the maximum rate (in Liter/min) of air flow during inspiration (i.e. breathing)
 December 10th, 2009, 09:55 PM #2 Senior Member   Joined: Dec 2009 From: Las Vegas Posts: 209 Thanks: 0 Re: derivatives Hi Calavan11; Did you graph function? It is really wild.
December 11th, 2009, 04:42 AM   #3
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Re: derivatives

Hello, Calavan11!

Quote:
 $V\:=\:3\,-\,\frac{1}{20}\,\cos(200\pi t)$ [color=beige]. . [/color]where $V$ is the volume of the lungs in liters and $t$ is the time in minutes. 1) What are the the maximum and minimum volumes of air in the lungs?

$\text{Solve }V' \,=\,0$

[color=beige]. . [/color]$V' \:=\:-\frac{1}{20}\,\sin(200\pi t)\,\cdot\,(200\pi) \:=\:10\pi\,\sin(200\pi t)$

[color=beige]. . [/color]$10\pi\,\sin(200\pi t) \:=\:0 \qquad\quad\Rightarrow\qquad\quad \sin(200\pi t) \:=\:0 \qquad\quad\Rightarrow\qquad\quad 200\pi t \:=\:\pi n$

[color=beige]. . [/color]$t \:=\:\frac{1}{200}\,n \qquad\quad\Rightarrow\qquad\quad n \:=\:0,\;\frac{1}{200},\;\frac{1}{100},\:\text{ . . .}$

$V(0) \;=\;3\,-\,\frac{1}{20}\cos(0) \;=\;3 \,-\, \frac{1}{20} \;=\;2\,\frac{19}{20}\quad \text{ minimum}$

$V\left(\frac{1}{200}\right) \;=\;3\,-\,\frac{1}{20}\cos(\pi) \;=\;3\,+\,\frac{1}{20} \;=\;3\,\frac{1}{20} \quad \text{ maximum}$

Quote:
 3) Find the maximum rate (in Liter/min) of air flow during inspiration (breathing in).

$\text{The rate of air flow is: }\;R \;=\;V#39; \;=\;10\pi\sin(200\pi t)$

$\text{For max/min }R\text{, solve: }\:R' \,=\,0$

[color=beige]. . [/color]$R' \;=\;10\pi\cos(200\pi t)\,\cdot\,(200\pi) \;=\;2000\pi^2\cos(200\pi t)$

$\text{Then: }\;2000\pi^2\cos(200\pi t) \;=\;0 \qquad\quad\Rightarrow\qquad\quad \cos(200\pi t) \;=\;0$

[color=beige]. . [/color]$200\pi t \;=\;\frac{\pi}{2} \qquad\quad\Rightarrow\qquad\quad t \:=\:\frac{1}{400}$

$\text{Therefore, maximum air flow is:}$

[color=beige]. . [/color]$R\left(\frac{1}{400}\right) \;=\;10\pi\sin\left(200\pi\,\cdot\,\frac{1}{400}\r ight) \;=\;
10\pi\sin\left(\frac{\pi}{2}\right) \;=\;10\pi\text{ Liters/min}$

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