Subtract -5x2 5x2 – 9x + 8 from -6x2 + 5. Answer: Submit Answer what’s the answer
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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) =
Taking inverse trigonometric function, α = tan⁻¹ () .............(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) :
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 = 10 feet/second.
C is the observer's position which is 50 feet away from the point B, so x = 50 feet.
= 0.12 (Rounding up to two decimal places)
So, the rate of change of the angle of elevation is 0.12 radians/second.
