Answer:
We have the velocity function given as follows:
v ( r ) = a r 2 ( r 0 − r ) = a r 0 r 2 − a r 3... ( 1 )
To find the value of r that makes the function v(r) a maximum we seek its critical points. To that end we need to calculate the first derivative of v(r) which will be:
v ′ ( r ) = 2 a r 0 r − 3 a r 2 ...( 2 )
Since v'(r) exists for all values of r from (2) we seek critical points of v(r) by solving v'(r) = 0 to get
a r ( 2 r 0 − 3 r ) = 0 ⇒ r = 0 , r = 2 r 0 ÷3
Since r = 0 will give us a trachea with zero radius which is of no use in a real-world setting we disregard r = 0 and only consider
r = 2 r 0÷ 3
To make sure this value of r maximizes v(r) we calculate the second derivative of v(r) to get
v ′′ ( r ) = 2 a r 0 − 6 a r... ( 3 )
From (3),
v ′′ ( 2 r 0 3 ) = 2 a r 0 − 6 a ( 2 r 0 3 ) = 2 a r 0 − 4 a r 0 = − 2 a r 0 < 0.
Hence from the second derivative test,
r = 2 r 0 ÷3
maximizes the velocity of air in the trachea during a cough.
Answer: your question is incomplete, as you have not shown the formula of how velocity is related the radius. Please let me assume your question to be this;
According to B. F. Visser, the velocity(V) of air in the trachea during a cough is related to the radius(r) of the trachea according to the following, where "A" is a constant and R is the radius of the trachea in a relaxed state.
V= Ar^2(R-r)
Find the radius (r) that produces the maximum velocity of air in the trachea during a cough.
THE ANSWER IS r=2R/3
Explanation: To solve this we have to make "r" the subject of the formula in the equation.
The equation states
V=Ar^2(R-r) .....................1
We have to open bracket
V=ARr^2 - Ar^3................2
From equation two we can see that r has two equation, so we need to find the critical point of equation 2. Remember, to find a critical point we have to find the first order differential.
Therefore differentiating Eqn 2 in respect to r
V'(r)=2ARr - 3Ar^2…....…......3
Simplifying Eqn 3 to get
V'(r)=Ar(2R - 3r).......................4
To get the critical point of V(r) we have to set Eqn 4 to be equal to zero. Therefore
Ar(2R - 3r)= 0......................5
Then we solve Eqn 5 using quadratic method.
2R -3r = 0
Using formula method which is
(-b+(√b^2 - √4ac))/2a.........6
AND
(-b-(√b^2 - √4ac))/2a..........7
Eqn 6 and 7 are the almighty formula method for solving quadratic Eqn
Therefore dividing the equation using the formula rule: The part with the highest value of the variable to find should replace "a" in the equation, and it should be followed in accordance to replace "b" and "c" in the formula method.
Therefore:
a= -3,. b= 2R. c= 0
Substituting "a", "b" and "c" into the Eqn 6
(-2R+(√[2R]^2 - [-3×2R×0]))/2×[-3]
Solving out gives
0/-6
Therefore r=0
This is the formula of radius that will produce the minimum velocity of air during a cough.
Substituting a,b and c into Eqn 7
(-2R-(√[2R]^2 - [-3×2R×0]))/2×[-3]
Solving out gives
-4R/-6
using -2 to divide the numerator and denominator gives
r= 2R/3
This is the formula for radius that will produce the maximum velocity of air in the trachea during a cough
