## Calculus, 10th Edition (Anton)

$v = 18 \, mi/h$
We need to find the wind speed to the nearest mile per hour. Denote it $v$. We are given that $WCT = 5^\circ F$, $T = 20^\circ F$. Recall the definition of the WCT: $$WCT = \left\{\begin{array}{lr} T, & 0\leq v\leq 3\\ 35.74 + 0.6215T − 35.75v^{0.16} + 0.4275T v^{0.16}, & v>3\\ \end{array}\right\}$$ For the sake of contradiction, suppose $0\leq v\leq 3$. In this case we have $WCT = T$ ,meaning, 20 = 5. A contradiction. Thus, we must have that $v > 3$. In this case we can write: $$WCT = 35.74 + 0.6215T − 35.75v^{0.16} + 0.4275T v^{0.16}$$ Substitute for the value of the $WCT$ to get: $$35.74 + 0.6215T − 35.75v^{0.16} + 0.4275T v^{0.16} = 5$$ Substitude for the value of $T$ to get: $$35.74 + 0.6215 \cdot 20 − 35.75 \cdot v^{0.16} + 0.4275\cdot 20 \cdot v^{0.16} = 5$$ Rewrite the equation as follows: $$v^{0.16} \cdot (0.4275\cdot 20 − 35.75) = 5 - 35.74 -0.6215 \cdot 20$$ $$v^{0.16} = 1.5871$$ $$v^{0.16 \cdot \frac{1}{0.16}} = 1.5871^{\frac{1}{0.16}}$$ $$v \approx 17.94 \, mi/h$$ We are asked to find the wind speed to the nearest mile per hour. Thus, the answer is $v = 18 \, mi/h$.