If you are outdoors, the wind may make it feel a lot colder than the thermometer reads. You feel the windchill temperature, which if the air temperature is 20^circ F is given in 'F by W(v)=48.17-27.2v^0.16 where vis the wind velocity in mph for 5leqslant vleqslant 60^1 (a) If the air temperature is 20^circ F and the wind is blowing at 30 mph.what is the windchill temperature, to the nearest degree? The windchill temperature is square -F (b) Find W'(30) and explain what this means interms of windchill. Round your answer to three decimal places. W'(30)=boxed (square )square

(a) To find the windchill temperature when the air temperature is $20^{\circ }F$ and the wind is blowing at 30 mph, we need to substitute $v = 30$ into the given formula for $W(v)$.<br /><br />$W(30) = 48.17 - 27.2(30)^{0.16}$<br /><br />First, calculate $(30)^{0.16}$:<br />$(30)^{0.16} \approx 2.238$<br /><br />Now, substitute this value back into the equation:<br />$W(30) = 48.17 - 27.2(2.238)$<br />$W(30) = 48.17 - 60.7856$<br />$W(30) \approx -12.6156$<br /><br />Rounding to the nearest degree, the windchill temperature is approximately $-13^{\circ }F$.<br /><br />(b) To find $W'(30)$, we need to take the derivative of $W(v)$ with respect to $v$ and then evaluate it at $v = 30$.<br /><br />$W(v) = 48.17 - 27.2v^{0.16}$<br /><br />Taking the derivative with respect to $v$:<br />$W'(v) = -27.2 \cdot 0.16v^{0.16 - 1}$<br />$W'(v) = -4.352v^{-0.84}$<br /><br />Now, substitute $v = 30$ into the derivative:<br />$W'(30) = -4.352(30)^{-0.84}$<br />$W'(30) \approx -0.137$<br /><br />Rounding to three decimal places, $W'(30) \approx -0.137$.<br /><br />This means that for every 1 mph increase in wind velocity, the windchill temperature decreases by approximately 0.137 degrees Fahrenheit when the wind is blowing at 30 mph.