Final answer:
After 2 hours, the distance between the two boats remains constant at 100 nautical miles. The rate of change of the angle θ after 2 hours is approximately 126.87°.
Step-by-step explanation:
To find the rate at which the distance between the two boats is changing, we can use the concept of relative velocity. Let's denote the northbound boat as Boat A and the eastbound boat as Boat B. The velocity of Boat A is 30 knots per hour towards the north, and the velocity of Boat B is 40 knots per hour towards the east.
After 2 hours, Boat A will have traveled a distance of 60 nautical miles (30 knots/hour x 2 hours) to the north. Boat B will have traveled a distance of 80 nautical miles (40 knots/hour x 2 hours) to the east.
To calculate the distance between the two boats, we can use the Pythagorean theorem: distance = sqrt((60)^2 + (80)^2) = 100 nautical miles.
To find the rate at which the distance is changing, we can calculate the derivative of the distance equation with respect to time. Differentiating, we get:
d(distance)/dt = (d/dt)sqrt((60)^2 + (80)^2) = (1/2)*(1/sqrt((60)^2 + (80)^2))*(0 + 0) = 0 nautical miles per hour.
Therefore, after 2 hours, the distance between the two boats is not changing. They will remain 100 nautical miles apart.
To calculate the rate of change of the angle θ, we can use trigonometry. Let's consider the position of Boat A as the reference point. After 2 hours, Boat A will be 60 nautical miles to the north and Boat B will be 80 nautical miles to the east. We can use the tangent function to find the angle:
tan(θ) = (opposite/adjacent) = (80/60) = 4/3.
Taking the inverse tangent of both sides, we get:
θ = arctan(4/3) ≈ 53.13°.
Since the angle is measured above due west, the actual angle θ will be 180° - 53.13° ≈ 126.87° in radians. Therefore, the rate of change of the angle θ after 2 hours is 126.87°.