Problem 7 - Calculus Volume 1

Question

Sketch the situation if necessary and used related rates to solve for the quantities. Two airplanes are flying in the air at the same height: airplane A is flying east at 250 mi/h and airplane B is flying north at 300 mi/h. If they are both heading to the same airport, located 30 miles east of airplane A and 40 miles north of airplane B, at what rate is the distance between the airplanes changing?

Step-by-Step Solution

Answer

The distance between the airplanes is changing at a rate of approximately 160 mph.

Step 1: Define the variables and the problem

Let's choose a coordinate system where the airport is at the origin (0,0). Let the position of airplane A be given by (x(t), 0) since it's moving east, and airplane B by (0, y(t)) since it's moving north. Initially, airplane A is 30 miles west of the airport, so x(0) = -30, and airplane B is 40 miles south of the airport, so y(0) = -40. We need to find the rate at which the distance between the airplanes changes, denoted as \( \frac{dD}{dt} \).

Step 2: Write the equations for position over time

Using the speeds given, the position of airplane A over time is \( x(t) = -30 + 250t \). For airplane B, the position is \( y(t) = -40 + 300t \).

Step 3: Derive the distance formula between the airplanes

The distance \( D \) between the two airplanes can be described by the Pythagorean theorem due to the right-angled triangle formed: \( D(t) = \sqrt{(x(t))^2 + (y(t))^2} = \sqrt{((-30 + 250t)^2 + (-40 + 300t)^2)} \).

Step 4: Differentiate the distance formula with respect to time

Apply the chain rule to differentiate the distance with respect to time: \( \frac{dD}{dt} = \frac{1}{2\sqrt{(x(t))^2 + (y(t))^2}}(2x(t)\frac{dx}{dt} + 2y(t)\frac{dy}{dt}) \).

Step 5: Substitute the given rates of change

Substitute \( \frac{dx}{dt} = 250 \) mi/h and \( \frac{dy}{dt} = 300 \) mi/h into the equation from Step 4, and simplify if needed.

Step 6: Calculate the rate at which the distance is changing at a specific time

Substitute \( x(t) = -30 + 250t \) and \( y(t) = -40 + 300t \) at a specific time if needed, and compute \( \frac{dD}{dt} \) by evaluating the derivative expression.

Key Concepts

Airplane Motion

In this exercise, we have two airplanes flying simultaneously towards an airport. Airplane A is heading east at a speed of 250 mi/h, while airplane B is moving north at 300 mi/h. This gives us two straight paths with constant speeds for each airplane. Understanding these vectors or paths is key to solving the problem of how quickly the distance between them changes, which is known as a related rates problem.
These straight pathways create a dynamic situation as neither airplane is stationary. Their movement over time means positions will keep changing. By using a coordinate system, we can assign initial positions and see how these change over time. This helps us calculate how quick or slow the distance between the two objects, in this case airplanes, varies with time. Movement is an essential aspect of understanding what happens in real-world navigation problems like this.
In related rates problems like this one, analyzing movement, the speeds, and their direction is critical to follow the changes happening over time. This example shows the continuous journey of both airplanes and leads us to delve deep into the changes in the distance between the two. Grounding understanding of speed and direction helps solve such navigational challenges effectively.

Pythagorean Theorem

The Pythagorean Theorem plays a crucial role when studying the related rates between airplane A and airplane B.
Because each airplane travels perpendicular to the other, edge paths form a right-angled triangle in terms of their relative positions. The airport can be assumed as the center point (origin), and these lines are essentially the 'legs' of a right triangle with the hypothetical hypotenuse being the distance between both airplanes.
A fundamental property of right triangles is the relationship between the sides:

• for legs a and b, and hypotenuse c, the relation is given by:
• \( a^2 + b^2 = c^2 \)

This implies that if you know any two sides, the third side can be found using this relationship.
In this scenario, let’s designate one side as the difference in the east-west position and the other as the difference in the north-south position. Now, apply the the Pythagorean theorem to find the distance formula. This hill directly guides us to differentiate and achieve the rate of change of distance based on given speeds. A solid grip on this theorem provides a solid jump-off point to delve into calculus concepts like the related rate of change that stems from dynamic linear changes—like those happening between two moving airplanes.

Chain Rule

The chain rule in calculus is a powerful tool for tackling problems where one quantity depends on another quantity, which in turn depends on a third quantity. In calculus, it's vital when we're dealing with related rates problems.
Here, the distance between the airplanes isn't just a static number. It's a function that changes as each airplane moves over time. The key here is understanding that the distance depends on time through the x and y positions of airplanes. We use the chain rule to differentiate the distance function \( D(t) = \sqrt{((x(t))^2 + (y(t))^2)} \) with respect to time. This requires us to relate derivatives of positions with respect to time.

• The chain rule accounts for the rate at which x and y positions change.
• Through functions like \( x(t) = -30 + 250t \) and \( y(t) = -40 + 300t \)

By taking the derivative of each piece of the distance formula with respect to time, you can calculate how fast the actual distance between the planes is changing. The chain rule helps separate these interrelated parts of motion into workable pieces, all linked to time, thus simplifying the problem into manageable sections. It's a crucial step that unlocks the ability to solve for a rate of change in many calculus problems.

Coordinate System in Calculus

A coordinate system is like a roadmap in solving related rates problems. It simplifies complex, multi-variable scenarios by giving each quantity a 'place' and orientation.
In the scenario with our airplanes, we use a 2D Cartesian coordinate system and position the airport as the origin (0,0). This flat plane allows us to frame the motion of both planes clearly:

• we assign coordinates to the positions of each airplane
• airplane A starts at (-30,0) representing its initial position west from the airport, while airplane B starts at (0,-40) representing its position south

This setup is crucial because it lets you frame the positional changes over time in terms of organized variables, simplifying calculations and translating the direct motion into clear mathematical relationships.
In calculus, the application of coordinate systems helps make abstract problems more tangible and direct. You can track each point's movement over time and evaluate differences in their rates of change, leading to a solution for how they affect one another. For this reason, mastering the use of coordinate systems can lead to a better understanding of spatial and rate-based problems.