Question

Coughing forces the trachea (windpipe) to contract, which affects the velocity \(v\) of the air passing through the trachea. The velocity of the air during coughing is \(v=k(R-r) r^{2}, \quad 0 \leq r

Short answer

The maximum air velocity occurs when the radius of the trachea during a cough is \(r = (1/3)R\).

Step-by-step solution

Differentiate the given function

\First, derive the function \(v = k(R - r) r^2\). Apply the product rule, which is (uv)' = u'v + uv', where u = (\(R - r)\) and v = \(r^{2}\).\Let's differentiate it:\\\(\frac{dv}{dr} = -2kr + 2kr(R - r)\)

Find the critical points

\Maximums or minimums occur at critical points, which are points where the derivative of a function is either zero or undefined. Set \(\frac{dv}{dr} = 0\) and solve for r:\\\(-2kr + 2kr(R - r) = 0\)\Solving the equation for r, we get \(r = (1/3)R\)

Second derivative test to confirm the maximum

\Calculate the second derivative of v, make sure whether the radius \(r=(1/3)R\) found indeed yields a maximum. The second derivative of v with respect to r is:\\\(\frac{d^2v}{dr^2} = -2k + 2k = 0\).\\Since the second derivative is zero, the second derivative test tells us nothing. We must use either the first derivative test. So, the test fails and leaves us uncertain whether the point is a maximum, a minimum or possibly a point of inflection. However, given the physical context of the problem, a maximum air velocity does exist, and it occurs when \(r = (1/3)R\).