## Related questions with answers

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 $\frac{d E}{d v}=\frac{0.31 v^{2}-471.75}{v^{2}}$ 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 $v_{\min }$ that minimizes E. a. What is the most economical speed $v_{\min } ?$ b. Suppose that when the parakeet flies at the most economical speed $v_{\min }$ its energy expenditure is $E_{\min }.$ Use this information to find E(v) for v > 0 in terms of $E_{\min }.$

Solution

Verifieda) \ Find where $\dfrac{dE}{dv } = 0$

$\begin{align*} 0.31v^2 - 471.75 &= 0 \\\\ 0.31v^2 &= 471.75 \\\\ v^2 &= \dfrac{ 471.75}{ 0.31} \\\\ v &= \pm \sqrt{ \dfrac{ 471.75}{ 0.31} } \approx \pm 39 \end{align*}$

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

$-39 < v < 39$: $E$ decreasing

$v < -39$ or $v > 39$: $E$ increasing

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

$v_{min} = 39$ km/hr

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