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 3 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.
 3 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.

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mathq
 3 years ago
Best ResponseYou've already chosen the best response.0Hey ask it in other greoups

suju101
 3 years ago
Best ResponseYou've already chosen the best response.0this is a maths problem.

omar_86
 3 years ago
Best ResponseYou've already chosen the best response.2i 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)?

suju101
 3 years ago
Best ResponseYou've already chosen the best response.0sorry...thats "r". v=f(r)=kr^2(Rr)

omar_86
 3 years ago
Best ResponseYou've already chosen the best response.2aha, then: v=kr^2Rkr^2r, where k, R are constants dv/dr=2kRr3kr^2 finding the maxima by equaling to zero 0=2kRr3kr^2

suju101
 3 years ago
Best ResponseYou've already chosen the best response.0any particular reason for finding the maxima??
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