when a foreign object lodged in the trachea forces a person to cough, the diaphragm thrusts upward, causing…

when a foreign object lodged in the trachea forces a person to cough, the diaphragm thrusts upward, causing an increase in pressure in the lungs. this is accompanied by a contraction of the trachea, making a narrower channel for the expelled air to flow through. for a given amount of air to escape in a fixed time, it must move faster through the narrower channel than the wider one. the greater the velocity of the airstream, the greater the force on the foreign object. x - rays show that the radius of the circular tracheal tube contracts to about two - thirds of its normal radius during a cough. according to a mathematical model of coughing, the velocity v of the airstream is related to the radius r of the trachea by the equation\n\n( v(r)=kleft(r_{0}-r\right) r^{2}, quad \frac{1}{2} r_{0} leq r leq r_{0} )\n\nwhere k is a constant and ( r_{0} ) is the normal radius of the trachea. the restriction on r is due to the fact that the tracheal wall stiffens under pressure and a contraction greater than ( \frac{1}{2} r_{0} ) is prevented (otherwise the person would suffocate).\n\n(a) determine the value of r in the interval ( left\frac{1}{2} r_{0}, r_{0}\right ) at which v has an absolute maximum.\n\n( t= )\n\n(b) what is the absolute maximum value of v on the interval?\n\n( v= )\n\n(c) sketch the graph of v on the interval ( left0, r_{0}\right ).

when a foreign object lodged in the trachea forces a person to cough, the diaphragm thrusts upward, causing an increase in pressure in the lungs. this is accompanied by a contraction of the trachea, making a narrower channel for the expelled air to flow through. for a given amount of air to escape in a fixed time, it must move faster through the narrower channel than the wider one. the greater the velocity of the airstream, the greater the force on the foreign object. x - rays show that the radius of the circular tracheal tube contracts to about two - thirds of its normal radius during a cough. according to a mathematical model of coughing, the velocity v of the airstream is related to the radius r of the trachea by the equation\n\n( v(r)=kleft(r_{0}-r\right) r^{2}, quad \frac{1}{2} r_{0} leq r leq r_{0} )\n\nwhere k is a constant and ( r_{0} ) is the normal radius of the trachea. the restriction on r is due to the fact that the tracheal wall stiffens under pressure and a contraction greater than ( \frac{1}{2} r_{0} ) is prevented (otherwise the person would suffocate).\n\n(a) determine the value of r in the interval ( left\frac{1}{2} r_{0}, r_{0}\right ) at which v has an absolute maximum.\n\n( t= )\n\n(b) what is the absolute maximum value of v on the interval?\n\n( v= )\n\n(c) sketch the graph of v on the interval ( left0, r_{0}\right ).

Answer

Explanation:

Step1: Find the derivative of (v(r))

Given (v(r)=k(r_{0}-r)r^{2}=kr_{0}r^{2}-kr^{3}). Using the power rule ((x^{n})^\prime = nx^{n - 1}), the derivative (v^\prime(r)=2kr_{0}r-3kr^{2}=kr(2r_{0}-3r)).

Step2: Find the critical points

Set (v^\prime(r) = 0). Since (kr(2r_{0}-3r)=0) and (k\neq0) (a non - zero constant), we have two cases:

  • Case 1: (r = 0) (but (r\in[\frac{1}{2}r_{0},r_{0}]), so we discard (r = 0)).
  • Case 2: (2r_{0}-3r=0), which gives (r=\frac{2}{3}r_{0}). We also need to check the endpoints of the interval (r=\frac{1}{2}r_{0}) and (r = r_{0}). When (r=\frac{1}{2}r_{0}), (v(\frac{1}{2}r_{0})=k(r_{0}-\frac{1}{2}r_{0})(\frac{1}{2}r_{0})^{2}=k(\frac{1}{2}r_{0})(\frac{1}{4}r_{0}^{2})=\frac{1}{8}kr_{0}^{3}). When (r=\frac{2}{3}r_{0}), (v(\frac{2}{3}r_{0})=k(r_{0}-\frac{2}{3}r_{0})(\frac{2}{3}r_{0})^{2}=k(\frac{1}{3}r_{0})(\frac{4}{9}r_{0}^{2})=\frac{4}{27}kr_{0}^{3}). When (r = r_{0}), (v(r_{0})=k(r_{0}-r_{0})r_{0}^{2}=0).

Step3: Compare the values

Since (\frac{4}{27}kr_{0}^{3}>\frac{1}{8}kr_{0}^{3}>0) (because (\frac{4}{27}\approx0.148) and (\frac{1}{8}=0.125)).

Answer:

(a) (r=\frac{2}{3}r_{0}) (b) (v=\frac{4}{27}kr_{0}^{3})