The rate of change of the angle of elevation when the firework is 40 feet above the ground is 0.12 radians/second.
First we will draw a right angle triangle ΔABC, where ∠B = 90°
Lets, assume the height(AB) = h and base(BC)= x
If the angle of elevation, ∠ACB = α, then
tan(α) = [tex] \frac{AB}{BC} = \frac{h}{x} [/tex]
Taking inverse trigonometric function, α = tan⁻¹ ([tex] \frac{h}{x} [/tex]) .............(1)
As we need to find the rate of change of the angle of elevation, so we will differentiate both sides of equation (1) with respect to time (t) :
[tex] \frac{d\alpha}{dt}=[\frac{1}{1+ \frac{h^2}{x^2}}]*(\frac{1}{x})\frac{dh}{dt} [/tex]
Here, the firework is launched from point B at the rate of 10 feet/second and when it is 40 feet above the ground it reaches point A,
that means h = 40 feet and [tex] \frac{dh}{dt} [/tex] = 10 feet/second.
C is the observer's position which is 50 feet away from the point B, so x = 50 feet.
[tex] \frac{d\alpha}{dt}= [\frac{1}{1+ \frac{40^2}{50^2}}] *\frac{1}{50} *10\\ \\ \frac{d\alpha}{dt} = [\frac{1}{1+\frac{16}{25}}] *\frac{1}{5}\\ \\ \frac{d\alpha}{dt} = [\frac{25}{41}] *\frac{1}{5}\\ \\ \frac{d\alpha}{dt}= \frac{5}{41} =0.1219512 [/tex]
= 0.12 (Rounding up to two decimal places)
So, the rate of change of the angle of elevation is 0.12 radians/second.
