Problem 5 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

Two planes are traveling at the same altitude toward an airport. One plane is flying at 500 kilometers per hour from a position due North of the airport and the other plane is traveling at 300 kilometers per hour from a position due East of the airport. At what rate is the distance between the planes decreasing when the first plane is \(8 \mathrm{~km}\) North of the airport and the second plane is \(5 \mathrm{~km}\) East of the airport?

Step-by-Step Solution

Answer

The distance between the planes is decreasing at approximately 583.1 km/h.

Step 1: Introduction and Understanding the Problem

Two planes are approaching an airport: one from the North and one from the East. We need to find the rate at which the distance between the two planes is decreasing when plane A is 8 km north of the airport and plane B is 5 km east of the airport.

Step 2: Set Up Variables and Equations

Let plane A's distance from the airport be denoted by \(y\) (from the North) and plane B's distance be \(x\) (from the East). The rate of change for plane A is \(-500 \text{ km/h}\) (since it's approaching, this is a negative value), while for plane B it's \(-300 \text{ km/h}\). We are interested in finding the rate of change of the distance between the planes, \(s\), when \(y = 8\) km and \(x = 5\) km.

Step 3: Use the Distance Formula

The distance \(s\) between the planes forms a right triangle with legs \(x\) and \(y\): \[ s = \sqrt{x^2 + y^2} \]. We need to find \(\frac{ds}{dt}\), the rate at which the distance between the planes is changing.

Step 4: Differentiate the Distance Function

To find \(\frac{ds}{dt}\), differentiate both sides of the equation \(s = \sqrt{x^2 + y^2}\) with respect to time \(t\). Using the chain rule, we have: \[ \frac{ds}{dt} = \frac{1}{2\sqrt{x^2 + y^2}} \cdot 2x \frac{dx}{dt} + 2y \frac{dy}{dt} \]. This simplifies to: \[ \frac{ds}{dt} = \frac{x \frac{dx}{dt} + y \frac{dy}{dt}}{\sqrt{x^2 + y^2}} \].

Step 5: Substitute Known Values

Substitute the given values into the equation: \(x = 5\), \(y = 8\), \(\frac{dx}{dt} = -300\), and \(\frac{dy}{dt} = -500\). Calculate \(s\) when \(x = 5\) and \(y = 8\): \[ s = \sqrt{5^2 + 8^2} = \sqrt{25 + 64} = \sqrt{89} \]. Then, substitute these into the differentiated equation: \[ \frac{ds}{dt} = \frac{5(-300) + 8(-500)}{\sqrt{89}} = \frac{-1500 - 4000}{\sqrt{89}} = \frac{-5500}{\sqrt{89}} \].

Step 6: Calculate the Final Rate

Finally, simplify to find \(\frac{ds}{dt}\): \[\frac{ds}{dt} = \frac{-5500}{\sqrt{89}} \text{ km/h} \]. Calculating the numerical value gives approximately \(-583.1 \text{ km/h}\). This means the distance between the planes is decreasing at approximately 583.1 kilometers per hour.

Key Concepts

Pythagorean Theorem

The Pythagorean Theorem is a fundamental principle in geometry that connects the lengths of the sides in a right triangle. It is expressed as:

• For a triangle with right angle, the sum of the squares of the two shorter sides (known as the legs) equals the square of the longest side (known as the hypotenuse).
• Mathematically, this relationship is \(a^2 + b^2 = c^2\).

In the context of the problem, the two planes and the airport form a right triangle, with the line from the North plane to the airport as one side, the line from the East plane to the airport as the other side, and the direct line between the two planes as the hypotenuse. When we calculate the distance between the planes, we are essentially solving for the hypotenuse, using \(s = \sqrt{x^2 + y^2}\). Understanding this setup is crucial in applying related rates to find how fast the distance between them is changing.

Differentiation

Differentiation is a key concept in calculus that deals with finding the rate at which a function is changing at any given point. In this exercise, we use differentiation to find how the distances and speeds of the planes change over time.

• To differentiate means to calculate the derivative, which represents the instantaneous rate of change.
• For functions of time, such as the position of each plane, the derivative tells us how quickly this position changes as time progresses.

In the problem, we use the derivative to express how the horizontal and vertical distances of the planes change as they move towards the airport. Notably, negative values indicate the planes are approaching the airport. By finding the derivatives of these distances with respect to time, we are able to compute the rate at which the overall distance between the planes is decreasing.

Chain Rule

The chain rule is a fundamental technique in calculus used to find the derivative of composite functions. It is crucial for understanding how different rates are related to one another. Here, we apply the chain rule to determine how quickly the distance between the planes changes over time.

• A basic instance of the chain rule is expressed as: if a function \(z\) depends on \(y\) and \(y\) depends on \(x\), we can express a change in \(z\) relative to \(x\) as \(\frac{dz}{dx} = \frac{dz}{dy} \cdot \frac{dy}{dx}\).
• In our problem, the distance \(s\) between the planes is linked through the distances \(x\) and \(y\) they travel.

To find \(\frac{ds}{dt}\), the rate of change of \(s\) with respect to time, we differentiate the function \(s = \sqrt{x^2 + y^2}\) using the chain rule: \(\frac{ds}{dt} = \frac{x \cdot \frac{dx}{dt} + y \cdot \frac{dy}{dt}}{\sqrt{x^2 + y^2}}\). This formula allows us to analyze how the position changes specifically impact the rate at which the distance between the two planes alters.