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 \(B\) 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 \(0 < t < 1\), \(D(t) = 400t\). For \(t \geq 1\), \(D(t) = \sqrt{900(t-1)^2 + 160000t^2}\).

Step 1: Determine the position of Airplane A

Airplane A flies due north starting at noon. For any time \( t \) in hours after noon, Airplane A has traveled a distance of \( 400t \) miles. When \( 0 < t < 1 \), only Airplane A is flying, so its position is at \( (0, 400t) \) in miles, where the y-coordinate represents the northward direction.

Step 2: Determine the position of Airplane B

Airplane B starts flying due east at 1 pm, or 1 hour after noon. For time \( t \geq 1 \), Airplane B has been flying for \( t - 1 \) hours. The distance it travels is \( 300(t-1) \) miles, so its position is \( (300(t-1), 0) \), with the x-coordinate representing the eastward direction.

Step 3: Write the distance formula for 0<t<1

When \( 0 < t < 1 \), only Airplane A is flying, making Airplane B's position the origin \((0, 0)\). Thus, the distance \( D(t) \) between the airplanes is simply the distance Airplane A has traveled north: \[ D(t) = 400t. \]

Step 4: Write the distance formula for t>=1

When \( t \geq 1 \), both airplanes are flying. Using the coordinates \((300(t-1), 0)\) for Airplane B and \((0, 400t)\) for Airplane A, apply the distance formula: \[ D(t) = \sqrt{(300(t-1))^2 + (400t)^2}. \] Simplifying, the formula becomes: \[ D(t) = \sqrt{900(t-1)^2 + 160000t^2}. \]

Key Concepts

Parametric Equations

Parametric equations are a fantastic tool in calculus for describing complex motion or paths over time. They provide a way to chart an object's position by defining coordinates based on an independent variable, often time. Consider our problem with the airplanes: rather than describing the position in just x or y terms, we use time as a reference.

For Airplane A, its movement is described by the parametric equation given by

• Position along the north: (0, 400t) as it flies at 400 miles per hour northward if time, t, is between 0 and 1.
• For Airplane B, its journey eastward begins at t = 1 and is thus represented by the parametric equation (300(t-1), 0) for t ≥ 1.

This use of parametric equations simplifies the understanding of motion in distinct axes and allows solving for complex systems where paths might not align or start simultaneously.

Vector Mathematics

In vector mathematics, understanding the movement of objects includes describing their positions as vectors. A vector has both magnitude and direction, which makes it useful in physics and calculus.

For our scenario with the airplanes, the positions can be interpreted as vectors:

• Airplane A's position as a vector is \( \vec{a} = (0, 400t) \), primarily in the northward (y) direction."
• Airplane B's position vector begins after t=1 and is \( \vec{b} = (300(t-1), 0) \), solely in the eastward (x) direction.

The distance between two points (or vectors) can be found using the distance formula, which leverages their horizontal and vertical distance differences. When both planes are flying after t=1, the calculation involves the Pythagorean theorem, transforming these vectors into a measure of straight-line distance.

Kinematics in Physics

Kinematics is a branch of physics mostly concerned with the motion of objects. The fundamentals include understanding displacement, velocity, and acceleration. These concepts apply to our airplanes, each with a unique pattern of motion.

In the problem, both planes have constant velocities: 400 miles per hour north (Airplane A) and 300 miles per hour east (Airplane B). Since they maintain these speeds:

• Their respective speeds translate simply to geographic position changes over time.
• Kinematics equations help us to write these travels in mathematical form, where displacement depends linearly on velocity and time.

Ultimately, by applying the constant speed to distance calculations—especially after t=1 when both are moving—we can visually map their respective paths, deduce at what points they might be closest or furthest, and use these calculations to determine the exact distance at any given time post-noon.