Problem 36 - Calculus - AP Edition

Question

A police helicopter is flying due north at \(100 \mathrm{mi} / \mathrm{h}\) and at a constant altitude of \(\frac{1}{2} \mathrm{mi}\). Below, a car is traveling west on a highway at \(75 \mathrm{mi} / \mathrm{h}\). At the moment the helicopter crosses over the highway the car is 2 mi east of the helicopter. (a) How fast is the distance between the car and helicopter changing at the moment the helicopter crosses the highway? (b) Is the distance between the car and helicopter increasing or decreasing at that moment?

Step-by-Step Solution

Answer

The distance is decreasing at approximately 72.82 mi/h.

Step 1: Understanding the Scenario

We have a helicopter flying north at 100 mi/h and a car traveling west at 75 mi/h. At the given moment, the helicopter is directly over the highway, and the car is 2 miles to the east of the helicopter.

Step 2: Define Variables

Define the position of the helicopter as \( (0,100t, \frac{1}{2}) \) where \( t \) is the time in hours, and for the car: \( (2-75t,0,0) \), since it is moving west.

Step 3: Distance Formula

The distance \( D \) between the car and the helicopter can be expressed using the distance formula: \[ D = \sqrt{(2-75t)^2 + (100t)^2 + (0.5)^2} \]

Step 4: Differentiation

To find the rate of change of the distance \( D \), differentiate the distance function with respect to time: \[ \frac{dD}{dt} = \frac{1}{2D} \cdot \left[2(2-75t)(-75) + 2(100t)(100)\right] \]

Step 5: Evaluate at t=0

At \( t=0 \), substitute the values into the distance formula: \[ D = \sqrt{4 + 0.25} = \sqrt{4.25} \] and then into the derivative: \[ \frac{dD}{dt} = \frac{1}{2 \sqrt{4.25}} \cdot [-300 + 0] \]

Step 6: Simplify the Derivative

Calculate the value: \[ \frac{dD}{dt} = \frac{-300}{2 \sqrt{4.25}} \approx \frac{-300}{4.12} \approx -72.82 \]

Key Concepts

Distance Formula

The distance formula is a crucial tool when determining the distance between two points in a 3D space. In our scenario, the helicopter and the car are moving orthogonally relative to each other. To find the distance between the car and the helicopter, we use:- \[ D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \]This formula computes the straight-line distance by taking into account differences in each of the three dimensions: x, y, and z. For our exercise:- The helicopter is flying at 0 in the x-direction (since it is flying vertically) and 0.5 miles up in the z-direction.- The car is moving in the negative x-direction.Substituting the positions of each at any given time:\[D = \sqrt{(2 - 75t)^2 + (100t)^2 + (0.5)^2}\]Here, \(t\) is the time in hours. At \(t=0\), the car starts 2 miles east of the helicopter. This gives students a fusion of the coordinate geometry concepts within a dynamic context like motion.

Differentiation

Differentiation is the process of finding the derivative, or the rate at which a function is changing at any given point. In problems involving motion, such as this one, we use differentiation to determine how fast distances or positions are changing over time. In our exercise, we differentiate the distance equation with respect to time \(t\) to determine how quickly the distance between the car and helicopter is changing. This is often represented as:- \[ \frac{dD}{dt} \]This derivative gives us the rate of change of distance with respect to time. For our problem, differentiating involves applying the chain rule to the distance equation where:- The components of the car's and helicopter's movements are each considered.- Each part of the distance equation is differentiated, reflecting the change in distance due to velocities in different directions.The differentiation step in our solution gives: - \[ \frac{dD}{dt} = \frac{1}{2D} \cdot \left[2(2-75t)(-75) + 2(100t)(100)\right] \]This derivative is then evaluated further to solve for the exact rate of change at a specific time, unveiling insights into whether the vehicles are moving towards or away from each other.

Speed and Velocity

Speed and velocity are fundamental concepts in the study of motion. They help us understand how fast an object is moving and in what direction. In this exercise: - **Speed** refers to how fast the helicopter and car are moving, namely at 100 mi/h north and 75 mi/h west, respectively. - **Velocity** is speed with a given direction. When solving this type of problem: - Examine each object's velocity vector. For example, the helicopter’s velocity is purely in the y-direction while the car’s is in the x-direction. - Calculate how these velocities affect the rates of change of the relative distance between the car and helicopter. The initial setup gives us the velocities but understanding these in context with their directional nature is key in solving related rates problems. Essential in this scenario is how the northward and westward paths combine and impact the rate of change in the relative distance, as determined by differentiating the distance function. By examining velocity this way, you understand the dynamic interaction of motion in different directions.