A plane flying horizontally at an altitude of 2 mi and a speed of 540 mi/h passes directly over a radar station. Find the rate at which the distance from the plane to the station is increasing when it has a total distance of 3 mi away from the station. (Round your answer to the nearest whole number.) mi/h
Question
A plane flying horizontally at an altitude of 2 mi and a speed of 540 mi/h passes directly over a radar station. Find the rate at which the distance from the plane to the station is increasing when it has a total distance of 3 mi away from the station. (Round your answer to the nearest whole number.) mi/h
Solution
This problem can be solved using the Pythagorean theorem and the concept of related rates in calculus.
Step 1: Understand the problem The plane is flying horizontally at a constant altitude of 2 miles and a speed of 540 miles per hour. We want to find the rate at which the distance from the plane to the radar station is increasing when the plane is 3 miles away from the station.
Step 2: Set up the equation We can use the Pythagorean theorem to relate the distance of the plane from the radar station (d), the horizontal distance of the plane from the radar station (x), and the altitude of the plane (h). The equation is:
d² = x² + h²
Given that h = 2 miles (constant), we can differentiate both sides of the equation with respect to time (t) to get:
2d(dd/dt) = 2x(dx/dt)
Step 3: Substitute the known values We know that dx/dt = 540 mi/h (the speed of the plane), d = 3 miles (the distance from the plane to the station at the time we are interested in), and x can be found from the Pythagorean theorem when d = 3 and h = 2:
x = sqrt(d² - h²) = sqrt(3² - 2²) = sqrt(5)
Substitute these values into the differentiated equation:
23(dd/dt) = 2*sqrt(5)*540
Step 4: Solve for dd/dt Solving for dd/dt gives:
dd/dt = (sqrt(5)*540) / 3 = 360 mi/h
So, the rate at which the distance from the plane to the station is increasing when it is 3 miles away from the station is approximately 360 miles per hour.
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