Problem 11 - The Calculus with Analytic Geometry

Question

A bomber is flying parallel to the ground at an altitude of \(2 \mathrm{mi}\) and at a speed of \(4 \frac{1}{2} \mathrm{mi} / \mathrm{min}\). If the bomber flies directly over a target, at what rate is the line-of-sight distance between the bomber and the target changing 20 sec later?

Step-by-Step Solution

Answer

\( \frac{dh}{dt} = 0.045 \text{ mi/s} \)

Step 1 - Convert Units

First, convert the given speed of the bomber from miles per minute to miles per second. Speed in miles per minute: \[ 4 \frac{1}{2} \text{ mi/min} = 4.5 \text{ mi/min} \]Speed in miles per second: \[ \frac{4.5 \text{ mi/min}}{60 \text{ seconds/min}} = 0.075 \text{ mi/s} \]

Step 2 - Calculate Distance Travelled in 20 Seconds

Calculate the distance the bomber travels in 20 seconds. Distance travelled: \[ d = \text{speed} \times \text{time} = 0.075 \text{ mi/s} \times 20 \text{ s} = 1.5 \text{ mi} \]

Step 3 - Use Pythagorean Theorem

Set up the problem using the Pythagorean theorem. Let the altitude (2 mi) be one leg of a right triangle, and the distance travelled (1.5 mi) be the other leg. The hypotenuse will be the line-of-sight distance. \[ h = \text{line-of-sight distance} \ h^2 = 2^2 + 1.5^2 = 4 + 2.25 = 6.25 \ h = \text{sqrt}(6.25) = 2.5 \text{ mi} \]

Step 4 - Differentiate with Respect to Time

Differentiate the Pythagorean theorem equation with respect to time. \[ h^2 = 4 + x^2 \ 2h \frac{dh}{dt} = 2x \frac{dx}{dt} \]Given: \( h = 2.5 \text{ mi}, x = 1.5 \text{ mi}, \frac{dx}{dt} = 0.075 \text{ mi/s} \)Substitute: \[ 2(2.5) \frac{dh}{dt} = 2(1.5)(0.075) \ 5 \frac{dh}{dt} = 0.225 \ \frac{dh}{dt} = 0.045 \text{ mi/s} \]

Key Concepts

Pythagorean Theorem

The Pythagorean Theorem is fundamental in this type of problem. It relates the lengths of the sides of a right triangle. Specifically, if you have a right triangle, the theorem states that 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.

In mathematical terms, this is written as: \[ c^2 = a^2 + b^2 \]

In the original exercise, the bomber's altitude (2 miles) and the distance it flies in 20 seconds (1.5 miles) are the legs of the right triangle. When we apply the theorem here: \[ h^2 = 2^2 + 1.5^2 \]

You can see how the line-of-sight distance (hypotenuse) is calculated by first squaring the sides, adding them together, and then taking the square root of the sum. Thus, you get: \( h = \sqrt{6.25} = 2.5 \text{ mi} \)

This theorem makes it easy to find the distance between two points when you know their coordinate differences.

Unit Conversion

Converting units is crucial for solving problems correctly. In our exercise, the speed of the bomber was first given in miles per minute. For our calculations, we needed it in miles per second.

Here's how to do it:

• Start with the speed in miles per minute: \( 4.5 \text{ mi/min} \)
• Convert minutes to seconds by dividing by 60 (since there are 60 seconds in a minute): \( \frac{4.5 \text{ mi/min}}{60 \text{ seconds/min}} = 0.075 \text{ mi/s} \)

Now, we have our speed in the correct units for our calculations. This step is important as using inconsistent units can lead to incorrect answers. Always make sure your units match, otherwise you'll need to convert them!

Differentiation with Respect to Time

Differentiation with respect to time is a technique from calculus used to determine how a particular quantity changes over time. In related rates problems, this method helps us find the rate of change of one variable in relation to another.

Consider our exercise: we needed to find how fast the line-of-sight distance was changing over time, given how fast the bomber was moving.

We used the Pythagorean theorem equation: \[ h^2 = 4 + x^2 \] and differentiated it with respect to time (t): \[ 2h \frac{dh}{dt} = 2x \frac{dx}{dt} \]
By applying the known values for \( h \), \( x \), and \( \frac{dx}{dt} \), we solved for \( \frac{dh}{dt} \), which represents the rate of change of the line-of-sight distance with respect to time.
Differentiation allowed us to understand how the distance is changing in real-time, providing the insight needed to complete the problem.

Distance-Rate-Time Relationship

The distance-rate-time relationship is foundational in problems dealing with motion. It's expressed as: \[ d = r \times t \]

Where:

• \( d \) is the distance traveled
• \( r \) is the rate of speed
• \( t \) is the time taken.

For our bomber problem, we wanted to find the distance (d) the bomber traveled in a given time (t). Given the speed (rate) \( 0.075 \text{ mi/s} \), and time (20 seconds), we used the formula: \( d = 0.075 \text{ mi/s} \times 20 \text{ s} = 1.5 \text{ mi} \).
This step crucially provided us with one side of the triangle needed to apply the Pythagorean theorem. The distance-rate-time formula is simple but incredibly powerful, often acting as the backbone for solving various motion-related problems.