A rocket travels vertically at a speed of 1300 km/hr.1300 km/hr. The rocket is tracked through a telescope by an observer located 13 km13 km from the launching pad. Find the rate at which the angle between the telescope and the ground is increasing 3 min3 min after the lift-off.

Final answer:

This calculus problem can be solved by defining the appropriate variables, constructing a relationship using trigonometry, taking the derivative of both sides with respect to time, and solving for the rate of change of the angle with respect to time. The initial distances, rocket's speed, and angle are used to determine the rocket's position after 3 minutes and thus the rate at which the angle is changing.

Explanation:

This problem involves the concept of related rates in calculus and the understanding of trigonometric relationships. Let's denote the rocket's altitude as y and the angle between the ground and the telescope as θ. We know Δy/Δt = 1300 km/hr, we're given the initial distance (13 km), and we want to find Δθ/Δt at 3 min after lift-off.

From trigonometry, we know that tan(θ) = y/x, where x is the horizontal distance (which remains constant at 13 km) and y is the vertical distance (which is changing). Differentiating both sides with respect to t gives sec²(θ) * Δθ/Δt = Δy/Δt / x. Assuming that the speed of the rocket remains constant, we find that y = (1300 km/hr * 3min)*(1hr/60min) = 65 km at 3 min after launch. Plugging x = 13 km and y = 65 km into the equation tan(θ) = y/x, we get θ = atan(65/13) = 78.69°. Now we can solve for Δθ/Δt using the differentiated equation: Δθ/Δt = ( Δy/Δt / x ) / sec²(θ) = (1300 km/hr / 13 km) / sec²(78.69°). Solving this gives the rate of change of θ with respect to time.

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