Problem 60 - Single Variable Calculus

Question

Distance Between Two Aircraft Two aircraft approach each other, each flying at a speed of \(500 \mathrm{mph}\) and at an altitude of \(35,000 \mathrm{ft}\). Their paths are straight lines that intersect at an angle of \(120^{\circ}\). At a certain instant of time, one aircraft is \(200 \mathrm{mi}\) from the point of intersection of their paths, while the other is \(300 \mathrm{mi}\) from it. At what time will the aircraft be closest to each other, and what will that distance be?

Step-by-Step Solution

Answer

The two aircraft will be closest to each other after approximately \(t \approx 0.447 \, \text{h} \, (26.82 \, \text{min})\) of flying, and their distance at that time will be approximately \(z \approx 4724.56\, \text{ft}\).

Step 1: Draw a diagram

First, let's draw a diagram to visualize the problem. Label the point of intersection of the paths as A, and the positions of the two aircraft at a given time as B and C, with B being 200 mi from A and C being 300 mi from A. Let D be the orthogonal projection of B onto AC. So, ∠ADC = 90°. Also, let x be the distance BD, y be the distance DC, and z be the distance BC.

Step 2: Write the relation for the Distance BC using the Law of Cosines

We can use the Law of Cosines to find a relation for the distance BC (z) in terms of x, y, and the angle between their paths, which is 120°. In triangle BAC, we have: \[z^2 = x^2 +y^2 - 2xy \cos(120^{\circ})\] Since \(\cos (120^{\circ}) = -\frac{1}{2}\), the above equation can be simplified to: \[z^2= x^2 + y^2 + xy\]

Step 3: Write the expression in terms of the given distances and speed

Using the Pythagorean theorem for the altitude, we can find x and y in terms of their distances from the point of intersection A. \[x^2 = (200\, \text{mi})^2 - (35,000\, \text{ft})^2\] \[y^2 = (300\, \text{mi})^2 - (35,000\, \text{ft})^2\] Now, we need to relate distances x and y to time, t. Since both aircraft have the same speed: \(x = 500t\) and \(y = 500t + 100\) Now substitute the expressions for x and y in the equation for z from Step 2: \[z^2 = (500t)^2 + (500t+100)^2 + 500t(500t+100)\]

Step 4: Differentiate z with respect to time t

Now, we'll find the derivative of z^2 with respect to time t: \[\frac{d(z^2)}{dt} = 4000000t^2 + 1000000t + 10000000\] To find the time of closest approach, set this derivative to zero and solve for t: \[4000000t^2 + 1000000t + 10000000 = 0\] This is a quadratic equation, and we can now solve for t: \[t = \frac{-1000000 \pm \sqrt{(-1000000)^2 - 4(4000000)(10000000)}}{2(4000000)}\] Since we're considering a real problem with aircraft, we'll only look at the positive value of t as our solution: \[t \approx 0.447\, \text{h}\]

Step 5: Find the distance between aircraft at the time of closest approach

Now that we have the time, we can find the distance between the aircraft when they are closest. Substitute the calculated value of t into the expression for z^2: \[z^2 = (500 \cdot 0.447)^2 + (500 \cdot 0.447 + 100)^2 + 500 \cdot 0.447 (500 \cdot 0.447 + 100)\] \[z^2 \approx 22312795\] Now, find the distance z: \[z \approx \sqrt{22312795} \approx 4724.56\, \text{ft}\]

Step 6: Final Answer

So, the two aircraft will be closest to each other after approximately 0.447 hours (about 26.82 minutes) of flying, and their distance will be approximately 4724.56 ft. Time of closest approach: \(t \approx 0.447 \, \text{h} \, (26.82 \, \text{min})\) Distance at closest approach: \(z \approx 4724.56\, \text{ft}\)

Key Concepts

Distance Between Two Aircraft

When considering the distance between two aircraft, the scenario often involves objects with straight-line motion intersecting at an angle. As two aircraft approach each other at constant speeds and altitudes, their minimum distance occurs when they are the closest to one another. This situation not only needs an understanding of geometrical concepts but also an application of calculus to find out when this minimum distance occurs.

Illustrating the positions of the aircrafts as points in a triangular formation with the intersection point serves as a practical first step. This intersection point helps create a sense of orientation and forms a critical part of the triangle used to apply the Law of Cosines. Understanding how to transform real-world motion into a static geometric framework is essential for solving distance-related problems in physics and navigation.

Law of Cosines Application

The Law of Cosines is instrumental in solving problems where the traditional right triangle trigonometry doesn't suffice, especially when a triangle doesn't include a right angle. This law relates the lengths of the sides of any triangle to the cosine of one of its angles. The formula \[z^2 = x^2 + y^2 - 2xy \cos(\theta)\] becomes a vital tool in calculating the distance between two aircraft at any given point.

In the given problem, the angle is given as 120°, which is not within a right angle triangle. By substituting the speed and distances of each aircraft into the relation, and considering their rate of change with respect to time, the Law of Cosines helps to provide a square of the distance (z²) as a function of time (t). This abstract algebraic approach allows an easy transition into finding the time of closest approach by leveraging calculus.

Rate of Change Calculus

Calculus allows us to calculate the rate of change of one variable with respect to another; in aviation problems, this often means distance versus time. By differentiating the square of the distance between two aircraft with respect to time, we can determine when this rate of change is zero - this corresponds to the moment when the distance between the two aircraft starts to increase again, signifying the closest approach.

Using the derivative of the square of the distance, we set up a quadratic equation which, upon solving, yields the exact time the aircraft come nearest to each other. While it may seem counterintuitive to differentiate z² instead of z, it streamlines the mathematics involved, avoiding the complexity of square roots and simplifies finding the time at minimum distance. The critical part to grasp is that, in practice, the time at which the derivative of z² is zero corresponds to when the actual distance z is at its minimum, which is why calculus is so powerful in solving such real-life problems.