g A balloon rises at a rate of4 metersper second from a point on the ground 50 meters from an observer. Find the rate of change of the angle

Answer :

Answer:

0.04 rad/sec

Step-by-step explanation:

A balloon rises at a rate of 4 meters per second from a point on the ground 50 meters from an observer. Find the rate of change of the angle of elevation of the balloon from the observer when the balloon is 50 meters above the ground.

The rate of rise = dh/ dt = 4 m/s

h is the height of the balloon = 50 meters

Tanθ = opposite/ adjacent

tanθ = h / 50

Differentiating both sides of the equation with respect to t:

d(tanθ) / dt = (d/dt)(h/50)

dtanθ / dt = (dh/dt)(1/50)

But dh/dt = 4 m/s, dtanθ/dt = sec²θ

[tex]sec^2\theta\frac{d\theta}{dt} = (4)(1/50)\\\\sec^2\theta=\frac{1}{cos^2\theta}\\\\\frac{1}{cos^2\theta} \frac{d\theta}{dt} =0.08\\\frac{d\theta}{dt} =cos^2\theta(0.08)\\\\At\ h=50/ m\\\\tan\theta=\frac{h}{50}\\ \\tan\theta=\frac{50}{50}=1\\\\\\\theta=tan^{-1}1=45^o\\\\Substituting\ \theta=45^o:\\\\\frac{d\theta}{dt} =cos^2(45)*(0.08)\\\\\frac{d\theta}{dt}=0.04\ rad/sec[/tex]

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