Question

A balloon is released 4 feet away from an observer. The balloon is rising vertically at a rate of 3 ft/sec and at the same time the wind is carrying it horizontally away from the observer at a rate of 4 ft/sec. At what speed is the angle of inclination of the observer’s line of sight changing 4 seconds after the balloon is released?

Answer 1

Answer:0.022 rad/s

Explanation:

Given

balloon is rising at 3 ft/s

wind is blowing horizontally at 4 ft/s

balloon is released 4 feet away from observer

after t sec balloon has moved a distance of 3t vertically and 4t  horizontally

therefore from diagram

$\tan \theta =\frac{3t}{4+4t}$

differentiate w.r.t time

$\sec^2 \theta \frac{\mathrm{d} \theta }{\mathrm{d} t}=\frac{3}{4}\times =\left [ \frac{\left ( 1+t\right )-1\cdot t}{\left ( 1+t\right )^2}\right ]$

now at t=4 s

$\frac{\mathrm{d} \theta }{\mathrm{d} t}=\frac{3}{4}\times \left [ \frac{1}{\left ( 1+t\right )^2}\right ]\times \frac{1}{1+\tan^2\theta }$

at t=4  s $\tan \theta =\frac{3}{5}$

$\frac{\mathrm{d} \theta }{\mathrm{d} t}=\frac{25}{25+9}\times \frac{3}{4}\times \frac{1}{25}$

$\frac{\mathrm{d} \theta }{\mathrm{d} t}=\frac{3}{136}=0.022 rad/s$