Problem 51 - Calculus Early Transcendentals: Pearson New International Edition

Question

Starting at noon, airplane A flies due north at 400 miles per hour. Starting 1 hour later, airplane \(\mathrm{B}\) flies due east at 300 miles per hour. Neglecting the curvature of the Earth and assuming that they fly at the same altitude, find a formula for \(D(t)\), the distance between the two airplanes \(t\) hours after noon. Hint: There will be two formulas for \(D(t)\), one if \(0

Step-by-Step Solution

Answer

For \\(0 < t < 1\\), \\(D(t) = 400t\\), and for \\(t \geq 1\\), \\(D(t) = \sqrt{250000t^2 - 180000t + 90000}\\).

Step 1: Define Coordinate System

Since airplane A is flying due north, set it on the y-axis moving upwards. Airplane B starts 1 hour later moving due east, so it will be on the x-axis heading to the right. Let the coordinates of airplane A be \(0, y_A\) and those of airplane B be \(x_B, 0\).

Step 2: Determine Airplane A's Position for Given Timings

Airplane A starts at noon moving north at 400 mph, so its position at any time \(t \text{ hours}\) after noon is given by \[y_A = 400t.\] This will hold for any \(t > 0\).

Step 3: Determine Airplane B's Position for Different Time Intervals

Airplane B starts flying due east at 300 mph but begins 1 hour after noon. Therefore, the position function is different depending on the time intervals:- For \(0 < t < 1\): Airplane B has not started flying yet, so \[x_B = 0.\]- For \(t \geq 1\): Airplane B starts moving, so from \(t = 1\) onward, it covers a distance \(300(t - 1)\) given by \[x_B = 300(t - 1).\]

Step 4: Calculate Distance for Different Time Intervals

The distance between the two airplanes, \(D(t)\), is the hypotenuse of a right triangle formed by their position coordinates. Apply the Pythagorean theorem:- For \(0 < t < 1\): \[D(t) = \sqrt{(0 - 400t)^2 + (0)^2} = 400t.\]- For \(t \geq 1\): \[D(t) = \sqrt{(0 - 400t)^2 + (300(t - 1))^2}.\] Expanding the distance: \[D(t) = \sqrt{(400t)^2 + (300(t - 1))^2}.\]

Step 5: Derive and Write the Formulas for D(t)

For each time interval, derive the distance formula:- If \(0 < t < 1\): \[D(t) = 400t.\]- If \(t \geq 1\): \[D(t) = \sqrt{(400t)^2 + (300(t - 1))^2} = \sqrt{160000t^2 + 90000(t^2 - 2t + 1)} = \sqrt{250000t^2 - 180000t + 90000}.\]

Key Concepts

Coordinate System

The coordinate system is a crucial tool used to represent the positions of objects in space. In this scenario, the coordinate system helps us visualize the paths of airplanes A and B. Here's how it works:

• Airplane A travels due north, so we place it on the y-axis moving upwards.
• Airplane B travels due east, therefore it moves along the x-axis to the right.

We use this system to easily determine each airplane's position at any given time, setting a point of reference. In this context, airplane A's position will be given as coordinate \(0, y_A\), and airplane B’s position as \(x_B, 0\). This simplifies the problem as it allows us to track movement along two straight, perpendicular paths.

Distance Formula

The distance formula derives from the Pythagorean theorem and is used here to calculate the distance between two points in a plane. Understanding the formula is essential in finding how far apart two objects in motion are at any given time.

• The general distance formula between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given as \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\).
• This formula helps us calculate the changing distance between airplanes A and B as they fly north and east, respectively.

By applying this formula, we can calculate their distance for different time intervals, as demonstrated when \( 0 < t < 1 \) and \( t \geq 1 \).