Problem 19 - College Algebra and Calculus: An Applied Approach

Question

An air traffic controller spots two airplanes at the same altitude converging to a point as they fly at right angles to each other. One airplane is 150 miles from the point and has a speed of 450 miles per hour. The other is 200 miles from the point and has a speed of 600 miles per hour. (a) At what rate is the distance between the planes changing? (b) How much time does the controller have to get one of the airplanes on a different flight path?

Step-by-Step Solution

Answer

The distance between the planes is changing at a rate of -529.15 miles per hour. The controller has 30 minutes to change the course of one of the planes.

Step 1: Set Up the Initial Conditions

Let \( d_1 \) and \( d_2 \) be the distances of the planes from the convergence point, the relative velocities be \( v_1 = 450 \) mph and \( v_2 = 600 \) mph, and \( d \) be the distance between the planes. From the Pythagorean theorem, we can express \( d \) as: \[ d^2 = d_1^2 + d_2^2 \]

Step 2: Differentiating With Respect to Time

Differentiating both sides of equation with respect to time \( t \), we get: \[ 2d \cdot \frac{dd}{dt} = 2d_1 \cdot \frac{d_1}{dt} + 2d_2 \cdot \frac{d_2}{dt} \] As we know \( \frac{d_1}{dt} = -v_1 \) and \( \frac{d_2}{dt} = -v_2 \), because the distances are decreasing as time increases.

Step 3: Substitute Values and Solve for dd/dt

Substituting all the values into the equation, we get: \[ \frac{dd}{dt} = \frac{(d_1 \cdot v_1 + d_2 \cdot v_2)}{d} = \frac{(150 \cdot 450 + 200 \cdot 600)}{ \sqrt{150^2 + 200^2}} = -529.15 \: mph \] This is the rate at which the distance between the planes is decreasing at this moment.

Step 4: Calculate the Time

The controller can change the course of the airplane when they are still not collided. Hence he has until they would collide if they stayed on their existing courses, i.e., when the distance between them becomes 0. We have the changing rate of distance so we can use the equation \( time = \frac{distance}{speed} \), which gives us \[ time = \frac{d}{|-dd/dt|} = \frac{\sqrt{150^2 + 200^2}}{529.15} = 0.5 \: hours \]

Step 5: Convert Time to Minutes

Since there are 60 minutes in 1 hour, the controller has \( 0.5 \times 60 = 30 \: minutes \) to get one of the airplanes on a different flight path before they collide.

Key Concepts

Related Rates Calculus Problems

Related rates problems are a practical application of derivatives in calculus. They deal with the rate at which one quantity changes in relation to another quantity. In these problems, we know the rate of change of one variable, and we want to find out the rate of change of another variable that is related to the first one through some equation. These problems often involve real-life scenarios like moving objects, changing areas, or varying volumes.

For example, in the case of the converging airplanes from the textbook exercise, we are interested in finding out how fast the distance between the airplanes is changing. We have the speed at which each plane is approaching a convergence point, and we need to find out how quickly the gap between them is closing. The key to solving related rates problems is to:

• Identify the known rates of change (i.e., the speeds of the airplanes).
• Set up an equation that relates the various quantities involved (typically using geometry).
• Differentiate this equation with respect to time to relate the rates of change of the various quantities.
• Plug in the known values and solve for the unknown rate.

By doing so, we can determine how variables change with respect to one another over time, which is incredibly useful in fields such as physics, engineering, and economics.

Pythagorean Theorem Application

The Pythagorean theorem is a fundamental principle in geometry that states that in a right-angled triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides. This theorem is expressed as \( c^2 = a^2 + b^2 \), where \( c \) is the length of the hypotenuse, and \( a \) and \( b \) are the lengths of the other two sides.

In the context of the related rates problem involving the two airplanes, the Pythagorean theorem is used to relate the distances of the airplanes from the point of convergence (\(d_1\) and \(d_2\)) to the distance between the airplanes (\(d\)). Here, \(d\) acts as the hypotenuse of a right triangle, while \(d_1\) and \(d_2\) are the perpendicular sides. Mathematically, this relationship is depicted as \(d^2 = d_1^2 + d_2^2\).

Applying the Pythagorean theorem allows us to create an equation that represents the geometrical situation, which we can then differentiate with respect to time to find the rate of change of the distance between the planes. Understanding and applying this theorem is crucial for finding solutions to a variety of problems in mathematics, physics, and engineering where right-angled triangles are involved.

Differentiation with Respect to Time

Differentiation is a fundamental concept in calculus that measures how a function changes as its input changes. It’s like taking a snapshot of the moment when something is happening and measuring the rate at which that event is occurring. When we differentiate with respect to time, denoted by \( \frac{d}{dt} \), we’re interested in finding out how some quantity is changing as time progresses.

In our scenario involving the airplanes converging towards a point, differentiating with respect to time allows us to find the rate at which the distance between the planes is decreasing, denoted by \( \frac{dd}{dt} \). To do this, we need to apply the rules of differentiation to the equation that represents the relationship between the changing quantities - in this case, the Pythagorean theorem as it relates to the distances of the airplanes from the convergence point and the distance between the airplanes themselves.

The process of differentiation yields a new equation that gives us the rate of change of the distance with respect to time, using the known rates of change of the individual distances, which are the negative of the speeds of the airplanes toward the convergence point. The negative sign indicates that the distances \(d_1\) and \(d_2\) are decreasing over time. This differentiation step is crucial to solve not just related rates problems but any problem in calculus where we need to understand the dynamics of changing quantities.