Question

In an investigation by V. A. Tucker and K. Schmidt-Koenig, it was determined that the energy E expended by a bird in flight varies with the speed v (km/hr) of the bird. For a particular kind of parakeet, the energy expenditure changes at a rate given by dEdv=0.31v2471.75v2\frac{d E}{d v}=\frac{0.31 v^{2}-471.75}{v^{2}} for v > 0 where E is given in joules per gram mass per kilometer. Observations indicate that the parakeet tends to fly at the speed vminv_{\min } that minimizes E. a. What is the most economical speed vmin?v_{\min } ? b. Suppose that when the parakeet flies at the most economical speed vminv_{\min } its energy expenditure is Emin.E_{\min }. Use this information to find E(v) for v > 0 in terms of Emin.E_{\min }.

Solution

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a) \ Find where dEdv=0\dfrac{dE}{dv } = 0

0.31v2471.75=00.31v2=471.75v2=471.750.31v=±471.750.31±39\begin{align*} 0.31v^2 - 471.75 &= 0 \\\\ 0.31v^2 &= 471.75 \\\\ v^2 &= \dfrac{ 471.75}{ 0.31} \\\\ v &= \pm \sqrt{ \dfrac{ 471.75}{ 0.31} } \approx \pm 39 \end{align*}

Since 0.31v2471.750.31v^2 - 471.75 is an upward opening parabola, it is negative only between the zeros.

39<v<39-39 < v < 39:          EE decreasing

v<39v < -39 or v>39v > 39:          EE increasing

EE changes from decreasing to increasing at v=39v = 39 so it is a local minimum of EE.

vmin=39v_{min} = 39 km/hr

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