When a person coughs, the trachea (windpipe) contracts and the velocity of the airflow in the windpipe is proportional to the product of the increase in the air pressure and the cross sectional area of the trachea. This can be modelled by v = APπr^2 where r is the radius of the windpipe during the cough, P is the increase in the air pressure during the cough, and A is a positive constant (proportionality constant). During the cough, the increase in the air pressure in the trachea is proportional to the change in the radius of the trachea. This can be modelled by P = B(R − r) where R is the normal radius of the windpipe and B is some positive constant (proportionality constant). Using this information determine the radius of the windpipe such that the velocity of the airflow is maximized.
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That seems to be an excercise. Did you mean to ask a question about it, perhaps? – hmakholm left over Monica Nov 11 '14 at 22:38
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Henning Makholm, I'm not sure how to solve for the radius so that the velocity is maximized. – Amanda Nov 11 '14 at 22:54
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x @Amanda: A good start would be to write down the function you want to maximize explicitly. – hmakholm left over Monica Nov 11 '14 at 22:56
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which would be v+APπr^2? – Amanda Nov 11 '14 at 22:59
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x@Amanda: Why do you add the two sides of that equation? What you want is $v$ as a function of $r$. – hmakholm left over Monica Nov 11 '14 at 23:31
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Substitute $P$ with $B(R-r)$ in the first equation to get
$$v = AB(R-r) \pi r^2=ABR \pi r^2-AB \pi r^3$$
The derivative of the last function is
$$v'=2ABR \pi r - 3AB \pi r^2$$
This derivative has a zero in $\displaystyle r=\frac{2}{3}R$, is positive for $\displaystyle 0<r< \frac{2}{3}R$, and is negative for $\displaystyle r>\frac{2}{3}R$. Hence, the value
$$\displaystyle r=\frac{2}{3}R$$
is a maximum of the original function expressing flow velocity, and therefore corresponds to the radius of the windpipe such that the velocity of the airflow is maximized.
Anatoly
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