Question

Q8.(10 points) a)When you cough,the radius of your trachea (windpipe) decreases,affecting the speed of the air in the trachea. If 0 r 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 2 0 S r ar r r ( ) ( ) = − , where a is positive constant. Find the radius r for which the speed of the air is greatest

Answer 1

Answer:

$\bf{r=\frac{2r_o}{3}}$ is greatest

Explanation:

Given:

$S(r)=ar^2(r_{o}-r)$

$\frac{ds}{dr}=2ar(r_{o}-r)+ar^2(-1)$

$\frac{ds}{dr}=2ar_{o}-2ar^2-ar^2$

$\frac{ds}{dr}=2ar_{o}-3ar^2$

Substitute $\frac{ds}{dr} = 0$, so

$2ar_{o}-3ar^2=0$

Then, get the common value from the equation.

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

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

$(\frac{d^2s}{dr^2})_{r=0}=2ar_{o}-6ar$

$(\frac{d^2s}{dr^2})_{r=0}=2ar_{o} >0\;and, \\(\frac{d^2s}{dr^2})_{r}=\frac{2r_o}{3} <0$

So, the speed of the air is greatest.

$\bf{r=\frac{2r_o}{3}}\;is\;greatest$