Question

When you cough,the radius of your trachea (windpipe) decreases,affecting the speed S of the air in the trachea. If r0 is the normal radius of the trachea, the relationship between the speed S of the air and the radius r of the trachea during a cough is given by a function of the form
S(r) = (r0 - r) ar^2

where a is positive constant. Find the radius r for which the speed of the air is greatest.

Answer 1

Answer: 2r(0)/3.

Step-by-step explanation:

So, we are given one Important data or o or parameter in the question above and that is the function of the form which is given below(that is);

S(r) = (r0 - r) ar^2 -----------------------------(1).

We will now have to differentiate S(r) with respect to r, so, check below for the differentiation:

dS/dr = 2ar (r0 - r ) + ar^2 (-1 ) ---------;(2).

dS/dr = 2ar(r0) - 2ar^2 - ar^2.

dS/dr = - 3ar^2 + 2ar(r0) ------------------(3).

Note that dS/dr = 0.

Hence, - 3ar^2 + 2ar(r0) = 0.

Making ra the subject of the formula we have;

ra[ - 3r + 2r(0) ] = 0. -------------------------(4).

Hence, r = 0 and r = 2r(0) / 3.

If we take the second derivative of S(r) too, we will have;

d^2S/dr = -6ar + 2ar(0). -------------------(5).

+ 2ar(0) > 0 for r = 0; and r = 2r(0)/3 which is the greatest.

Answer 2

Answer:

$r =\frac{2r_{0}}{3}$

Step-by-step explanation:

We need to take the derivative of S(r) and equal to zero to maximize the function. In this conditions we will find the radius r for which the speed of the air is greatest.

Let's take the derivative:

$\frac{dS}{dr}=a(2r(r_{0}-r)+r^{2}(-1))$

$\frac{dS}{dr}=a(2r*r_{0}-2r^{2}-r^{2})$

$\frac{dS}{dr}=a(2r*r_{0}-3r^{2})$

$\frac{dS}{dr}=ar(2r_{0}-3r)$

Let's equal it to zero, to maximize S.

$0=ar(2r_{0}-3r)$

We will have two solutions:

$r = 0$

$r =\frac{2r_{0}}{3}$

Therefore the value of r for which the speed of the air is greatest is $r =\frac{2r_{0}}{3}$.

I hope it helps you!