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Av) 2mied foom lift off point.Angle =\frac{\pi}{3}. changing at 0.1 \mathrm{rad} / \mathrm{min},Let h be the haight of air ballon abou ground 40 be angle between line of bight and horizentalThen \tan \theta=\frac{b}{2}Diffeertiating the abeve eqn wit time tt,\begin{array}{l}\frac{d}{d \theta} \frac{d}{d t}[\tan \theta]=\frac{d}{d t}\left[\frac{h}{2}\right] \\\sec ^{2} \theta \frac{d \theta}{d t}=\frac{1}{2} \frac{d h}{d t} \\\frac{d b}{d t}=2 \sec ^{2} \theta \frac{d \theta}{d t}\end{array}\begin{aligned}\theta=\frac{\pi}{3} \& \frac{d Q}{d t} & =0.1 \mathrm{rad} / \mathrm{min} \\\frac{d h}{d t} & =2 \mathrm{sec}^{2}\left[\frac{\pi}{3}\right][0.1] \\\frac{d b}{d t} & =2[4][0.1]=0.8 \mathrm{miles} / \mathrm{min}\end{aligned}This the ballion is aring at rate of 0.8 miles 1 \mathrm{~min} at that moment.(* Kindly like for bolution) ...