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HELP!!! when a person coughs, the trachea(windpipe contracts, allowing air to be expelled at max velocity. it can be shown that during a cough the velocity v of airflow is given by the function v=f(r)=kx^2(Rr) where r is the radius of the trachea in cm during a cough, R is the trachea's normal radius in cm, and k is a positive constant that depends on the length of the trachea. find the radius r for which the velocity of airflow is greatest.
 2 years ago
 2 years ago
HELP!!! when a person coughs, the trachea(windpipe contracts, allowing air to be expelled at max velocity. it can be shown that during a cough the velocity v of airflow is given by the function v=f(r)=kx^2(Rr) where r is the radius of the trachea in cm during a cough, R is the trachea's normal radius in cm, and k is a positive constant that depends on the length of the trachea. find the radius r for which the velocity of airflow is greatest.
 2 years ago
 2 years ago

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mathqBest ResponseYou've already chosen the best response.0
Hey ask it in other greoups
 2 years ago

suju101Best ResponseYou've already chosen the best response.0
this is a maths problem.
 2 years ago

omar_86Best ResponseYou've already chosen the best response.2
i think u have to differentiate f(r) to have df(r)/dr, but i can't see what's ment by the "x" in f(r)?
 2 years ago

suju101Best ResponseYou've already chosen the best response.0
sorry...thats "r". v=f(r)=kr^2(Rr)
 2 years ago

omar_86Best ResponseYou've already chosen the best response.2
aha, then: v=kr^2Rkr^2r, where k, R are constants dv/dr=2kRr3kr^2 finding the maxima by equaling to zero 0=2kRr3kr^2
 2 years ago

suju101Best ResponseYou've already chosen the best response.0
any particular reason for finding the maxima??
 2 years ago
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