Final answer:
Using the concept of related rates in calculus, the increase in the angle of elevation of a balloon, 30 seconds after it has started rising at a constant rate of 4 m/s, is found to be approximately 0.000833 radians per second.
Step-by-step explanation:
To solve the mathematical problem completely regarding how fast the angle of elevation of a balloon is increasing, we will use the concept of related rates in calculus. Let's denote the angle of elevation as θ and use the trigonometric relationship to relate the vertical height and the horizontal distance from the observer to the balloon. Since the observer is 200 meters from the launch site, we can use the tangent function: tan(θ) = opposite/adjacent = height/200.
30 seconds after launch, the balloon has risen 4 m/s × 30 s = 120 meters vertically. We have:
tan(θ) = 120/200 = 0.6
To find the rate at which the angle of elevation is increasing, we differentiate both sides of the equation with respect to time (t):
d/dt [tan(θ)] = d/dt [height/200]
Given that d[height]/dt = 4 m/s (constant rate of ascent), we can substitute to find:
dθ/dt = 1/(cos^2(θ)) × (4/200)
Using a calculator, we find the value of cos(θ) when θ such that tan(θ) = 0.6, and then calculate dθ/dt.
We convert our tangent to an angle, θ ≈ 30.96 degrees, find the cosine value, and then square it. After doing the calculations, the increase in the angle of elevation is approximately 0.000833 radians per second at 30 seconds after the launch.