Problem 42 - University Calculus: Early Transcendentals

Question

A highway patrol plane flies 3 mi above a level, straight road at a steady \(120 \mathrm{mi} / \mathrm{h}\). The pilot sees an oncoming car and with radar determines that at the instant the line-of-sight distance from plane to car is 5 mi, the line-of-sight distance is decreasing at the rate of \(160 \mathrm{mi} / \mathrm{h}\). Find the car's speed along the highway.

Step-by-Step Solution

Answer

The car's speed along the highway is 200 mph.

Step 1: Understand the Exercise

We need to find the car's speed along the highway when the line-of-sight distance from the plane to the car is 5 miles and is decreasing at a rate of 160 mph. The plane is flying 3 miles above the road at a constant speed of 120 mph.

Step 2: Set Up the Relationship Implicitly

Let the horizontal distance from the plane to the car be \( x \) miles. The line-of-sight distance \( s \) from the plane to the car forms a right triangle where the altitude is 3 miles. From the Pythagorean theorem: \[ s^2 = x^2 + 3^2 \] Substituting \( s = 5 \) miles gives us \[ 5^2 = x^2 + 3^2 \] which simplifies to: \[ 25 = x^2 + 9 \] thus, \[ x^2 = 16 \] and \[ x = 4 \] miles.

Step 3: Differentiate the Relationship Implicitly

Differentiate both sides of the equation \( s^2 = x^2 + 3^2 \) with respect to time \( t \): \[ 2s \frac{ds}{dt} = 2x \frac{dx}{dt} \] Divide through by 2 to simplify: \[ s \frac{ds}{dt} = x \frac{dx}{dt} \] Given \( \frac{ds}{dt} = -160 \) mph (the negative sign because the distance is decreasing), substitute \( s = 5 \) and \( x = 4 \). We can find \( \frac{dx}{dt} \).

Step 4: Solve for \( \frac{dx}{dt} \)

Substitute the known values into the differentiated equation: \[ 5(-160) = 4 \frac{dx}{dt} \] This simplifies to: \[ -800 = 4 \frac{dx}{dt} \] Divide both sides by 4: \[ \frac{dx}{dt} = -200 \] Thus, the car's speed is 200 mph towards the plane.

Key Concepts

Pythagorean theorem

The Pythagorean theorem is a fundamental concept in geometry. It's essential for calculating distances in right triangles. In our scenario, there is a right triangle formed by the highway patrol plane, the car, and their line of sight. Here are the basic principles:

• The theorem states that in a right triangle, the square of the length of the hypotenuse (longest side) is equal to the sum of the squares of the other two sides.
• For this exercise, the line-of-sight distance from the plane to the car acts as the hypotenuse.
• The formula is expressed as: \[ s^2 = x^2 + 3^2 \]
• Where:
  • \( s \) is the line-of-sight distance (hypotenuse),
  • \( x \) is the horizontal distance (one side),
  • 3 miles is the altitude of the plane above the road (other side).

By plugging in the values we know (\( s = 5 \) miles), we can solve for \( x \) and later use this to help find the car's speed.

differentiation

Differentiation is a key mathematical process for understanding how variables change. In the problem, we want to understand how the distance between the plane and the car changes over time. Here's how differentiation helps:

• Differentiate the equation \( s^2 = x^2 + 3^2 \) with respect to time \( t \).
• This results in the equation: \[ 2s \frac{ds}{dt} = 2x \frac{dx}{dt} \]
• The process of differentiation tells us how the line-of-sight distance \( s \) and the horizontal distance \( x \) change as time progresses.
• In this scenario, since \( s \) is decreasing (closer distances), \( \frac{ds}{dt} \) is negative.

Using differentiation, we related the change in the line-of-sight distance to the speed of the car. This relationship is crucial for speed calculations.

speed calculation

Calculating speed involves determining how fast a distance changes over time. In this problem, we're tasked with finding the car's speed along the highway. Here's a breakdown of how it's done:

• First, we have identified that the line-of-sight distance decreases at 160 mph. This forms part of our data in the differentiated equation.
• Using the differentiated equation: \[ s \frac{ds}{dt} = x \frac{dx}{dt} \]we substitute the known values: \( s = 5 \), \( \frac{ds}{dt} = -160 \), and \( x = 4 \).
• Solving this equation gives us the value of \( \frac{dx}{dt} \), which represents the horizontal speed of the car. The negative value indicates the car is traveling towards the plane.
• Upon solving, we find \( \frac{dx}{dt} = -200 \), meaning the car is moving at 200 mph.

This speed calculation is key in understanding relative motion along the line-of-sight and the actual trajectory along the highway.