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?

[tex]\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 ][/tex]

now at t=4 s

[tex]\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 }[/tex]

at t=4 s [tex]\tan \theta =\frac{3}{5}[/tex]

[tex]\frac{\mathrm{d} \theta }{\mathrm{d} t}=\frac{25}{25+9}\times \frac{3}{4}\times \frac{1}{25}[/tex]

[tex]\frac{\mathrm{d} \theta }{\mathrm{d} t}=\frac{3}{136}=0.022 rad/s[/tex]

onyebuchinnaji onyebuchinnaji · Answer 2 · 2021-11-11T11:00:42+01:00

The rate at which the angle of inclination of the observer’s line of sight is changing after 4 seconds is 7 ft/s.

The given parameters;

distance between the balloon and the observer, d = 4 ft
change in vertical position, da/dt = 3 ft/s
change in the horizontal position, db/dt = 4 ft/s
time of motion, t = 4 s

The displacement of the balloon after 4 seconds is calculated by applying Pythagoras theorem;

a = 3 x 4 = 12 ft

b = (4 x 4) - 4 = 12 ft

[tex]c^2 = a^2 + b^2\\\\c = \sqrt{(12)^2 + (12)^2} \\\\c = 12 \ ft[/tex]

find the rate of change of the displacement;

[tex]c^2 = a^2 + b^2\\\\2c\frac{dc}{dt} = 2a\frac{da}{dt} + 2b\frac{db}{dt} \\\\c\frac{dc}{dt} = a\frac{da}{dt} + b\frac{db}{dt}\\\\\frac{dc}{dt} = 12(3) + 12(4)\\\\12\frac{dc}{dt} = 84\\\\\frac{dc}{dt} =\frac{84}{12} \\\\\frac{dc}{dt} = 7 \ ft/s[/tex]

Thus, the rate at which the angle of inclination of the observer’s line of sight is changing after 4 seconds is 7 ft/s.

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