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Online solved math and physics exercises applied to medicine with complete detailed solution - maximize the velocity of exhalation (Bernouilli) in your trachea while you are coughing - calculate the trachea radius and diameter.

### Derivative : solution exercise 4.5 (Advanced math level)

4.5 Derivatives applied to biology, medicine and economy : The diameter of your trachea is reduced by 33% of its original diameter when you are coughing

When you are coughing, the airway (trachea) narrows to increase the speed of exhalation and to dislodge foreign bodies (see Bernouilli). This fact leads us to wonder: what is the required reduction in trachea diameter (in centimeters) during a cough to maximize the velocity of exhalation ? The following formula describes the relationship between v (velocity of exhalation in the trachea during a cough) and r (tracheal radius).

v = c . (r0 - r) . r2

v in cm/s,

c is a constant greater than zero. Its value depends on the lenght of the trachea,

r0 is the tracheal radius at rest ( = when you are not coughing),

r is the tracheal radius while coughing.

You are asked to:

a) Demonstrate that the velocity of the exhaled air, v, is maximum when
r = (2/3).r0 , which means when the tracheal radius is reduced by 33,33% compared to its normal size at rest.

The most remarkable thing is that the pictures taken with an X-ray device confirms the reduction of 33,33% calculated numerically.

The velocity v is a function of the variable r.

v(r) = c . (r0 - r) . r2

c and r0 are real constants.

We know (thanks to what we have learned at the basic math level and intermediate math level) that if v'(r1) = 0 and v''(r1) < 0 then we have a maximu at r = r1.

Let us figure out the first and second derivative of the function
v(r).

First let us find the value of r that makes the first derivative zero :
v'(r) = c . ( [r0 - r] . r2 )'
= 2 c r0 r - 3 c r2

v'(r) = 0 = 2 c r0 r - 3 c r2
0 = cr (2 r0 - 3 r)
Thus 0 = 2 r0 - 3 r because c and r cannot equal 0
and we get 2 r0 = 3 r

hence r = (2/3) . r0

Next, let us find out the value of r that makes the second derivative negative :
v''(r) = 2 c r0 - 6 c r

v''(r) < 0 then 2 c r0 - 6 c r < 0 hence

r > r0/3 (this means that r must be greater than r0/3 to get the maximum value of v).

Conclusion: the second derivative confirms that (2/3) . r0 is the value of the tracheal radius for which we have the maximum speed of exhalation because (2/3) r0 is greater than r0/3.

b) Sketch the graph of v(r) and its derivative v'(r) using our free graph plotter online.

Use the following values for r0 and c : r0 = 2 cm and c = 1. Do not forget to replace the variable r by the letter x in the equation you will enter in the graph plotter. Next, check graphically that the value of the radius r corresponding to the maximum exhaled velocity is two thirds of the value r0. Check graphically as well the value of the derivative, v'(r), when the exhaled velocity is maximum (the value of the derivative must be zero).

I cannot do anything more for you here. Just go to our free function tracer online and draw the functions.