Problem 34 - Calculus - AP Edition

Question

An aircraft is flying at a constant altitude with a constant speed of \(600 \mathrm{mi} / \mathrm{h}\). An antiaircraft missile is fired on a straight line perpendicular to the flight path of the aircraft so that it will hit the aircraft at a point \(P\) (see the accompanying figure). At the instant the aircraft is 2 mi from the impact point \(P\) the missile is 4 mi from \(P\) and flying at 1200 mi/h. At that instant, how rapidly is the distance between missile and aircraft decreasing? PICTURE CANT COPY

Step-by-Step Solution

Answer

The distance between the missile and the aircraft decreases at approximately 1209.54 mi/h.

Step 1: Set Up the Coordinate System

Consider a coordinate plane where the aircraft is traveling along the x-axis toward a point of impact P at x=0. At the same time, the missile travels along the y-axis toward point P. The aircraft starts from x=2 at the instant of observation, while the missile starts from y=4.

Step 2: Analyze the Positions Over Time

At time t=0, the aircraft is 2 miles from P, so its position along the x-axis is x(t) = 2 - 600t. Similarly, the missile is 4 miles from P along the y-axis, so its position is y(t) = 4 - 1200t.

Step 3: Express Distance Between Aircraft and Missile

The distance between the aircraft and the missile, D, is given by the equation \( D = \sqrt{x(t)^2 + y(t)^2} \). Substituting the positions from Step 2, we get: \( D = \sqrt{(2 - 600t)^2 + (4 - 1200t)^2} \).

Step 4: Differentiate with Respect to Time

To find how rapidly the distance D is changing, differentiate D with respect to time t. Use the chain rule: \( \frac{dD}{dt} = \frac{1}{2\sqrt{(2 - 600t)^2 + (4 - 1200t)^2}} \left[ 2(2 - 600t)(-600) + 2(4 - 1200t)(-1200) \right] \).

Step 5: Substitute Values at Instant

Substitute t=0 into the derivative to find the rate at that specific instant. This gives \( \frac{dD}{dt} = \frac{1}{2 \times \sqrt{(2)^2 + (4)^2}} [-1200 + (-9600)] \).

Step 6: Simplify the Rate of Change

Calculate the values: the expression simplifies to \( \frac{dD}{dt} = \frac{-10800}{2 \times \sqrt{20}} \). Further simplification provides the rate: \( \frac{dD}{dt} = -540 \sqrt{5} \).

Step 7: Calculate the Exact Value

Finally, approximate the answer by calculating \( -540 \sqrt{5} \) to find the rate. This gives approximately \( -1209.54 \). The negative sign indicates the distance is decreasing.

Key Concepts

Calculus

Calculus is the branch of mathematics that focuses on change. It provides tools to study how quantities vary over time or within a system. Such tools are critical when dealing with dynamic systems in motion, like an aircraft and a missile. In this exercise, we apply calculus to determine how the distance between two moving objects changes at a specific instant.

We use concepts like differentiation to find rates of change, which tells us how fast something is changing. In our example, we specifically want to know how quickly the distance between the aircraft and the missile is decreasing at a particular moment.

When dealing with such problems:

• Identify the dynamic quantities you are interested in.
• Set up equations that describe these quantities over time.
• Use calculus techniques to find the rates of change of these quantities.

With calculus, dynamic scenarios become understandable and manageable, allowing us to solve real-world problems.

Differentiation

Differentiation is a fundamental concept in calculus. It is the process of finding a derivative, which represents how a function is changing at any point. In the exercise, we used differentiation to find how quickly the distance between the missile and aircraft decreases.

Differentiation has several steps:

• First, you need a function that describes the quantity you're interested in. Here, it was the distance function, \( D = \sqrt{(2 - 600t)^2 + (4 - 1200t)^2} \).
• Then, apply the differentiation rules (like the chain rule) to find its rate of change. This gives you \, \( \frac{dD}{dt} \).
• You substitute the specific time \( t=0 \) to determine the instantaneous rate of change.

By differentiating, you can precisely calculate how quickly variables change, essential for scenarios where time and change matter, such as tracking moving objects.

Coordinate System

A coordinate system is an essential tool for visualizing and solving geometric problems in mathematics. In this scenario, we used a Cartesian coordinate plane to track the positions of the aircraft and missile.

Setting up the coordinate system:

• We positioned the aircraft's path along the x-axis, indicating that it moves horizontally.
• The missile's trajectory was along the y-axis, indicating vertical movement towards point P.

This setup allowed us to describe the positions of each object relative to the point P as functions of time:

• For the aircraft: \( x(t) = 2 - 600t \).
• For the missile: \( y(t) = 4 - 1200t \).

Once a problem is translated into a coordinate system, geometric relationships become solvable with algebraic equations.

Chain Rule

The chain rule is a method for differentiating composite functions. It allows us to handle situations where one variable depends on another, which in turn depends on a third variable. In our exercise, it was crucial for differentiating the distance function.

Using the chain rule involves:

• Breaking down the function you want to differentiate into its component parts.
• Taking the derivative of the outer function with respect to the inner function. Then multiply it by the derivative of the inner function.

In our situation, we differentiated the composite function \( D \), incorporating both \( (2 - 600t) \) and \( (4 - 1200t) \) with respect to time. The chain rule helped us compute how the distance between the aircraft and missile changes by taking into account each part's rate of change. This powerful tool is vital for handling problems involving nested relationships and dependencies.