A cop sits in a parked car 10 feet from a straight road. As you drive along the road, the cop aims a radar gun at your car. Let s be the distance from your car to the cop in feet. The radar gun measures the rate at which your distance from the cop is changing with respect to time, which is ds/dt. Let x be the distance, in feet, of your car from the point on the road that is closest to the cop, so your car's velocity is dx/dt. dx (a) Draw and label a diagram of this scenario, and write dt ds in terms of s, and x. dt' ds dt ds dx (b) Use part a to explain why < Thus the radar gun's measurement of dt dt always underestimates your car's velocity. (This is why if the radar gun measures a speed greater than the speed limit on the road, the driver deserves a ticket.) (c) Does the conclusion in part (b) depend on the cop's car being 10 feet, rather than some other (positive) distance, from the road?

Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
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Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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### Understanding Radar Gun Measurements

A police officer sits in a parked car 10 feet from a straight road. As you drive along the road, the officer aims a radar gun at your car. In this scenario, we'll examine the mathematics behind how the radar gun measures your speed differently than your actual velocity.

Let \( s \) be the distance from your car to the cop in feet. The radar gun measures the rate at which your distance from the cop is changing with respect to time, which is \( \frac{ds}{dt} \). Let \( x \) be the distance, in feet, of your car from the point on the road that is closest to the cop, so your car’s velocity is \( \frac{dx}{dt} \).

#### Task Breakdown
1. **Diagram and Relationship Derivation:**
   - **Diagram:** Draw and label a diagram of the scenario, and write \(\frac{dx}{dt}\) in terms of \(\frac{ds}{dt}\), \(s\), and \(x\).
2. **Explanation using the Diagram:** Use part (a) to explain why \(\frac{ds}{dt} < \frac{dx}{dt}\). Thus, the radar gun's measurement of \(\frac{ds}{dt}\) always underestimates your car’s velocity.
3. **Dependence on Distance:** Examine whether the conclusion in part (b) depends on the cop’s car being 10 feet from the road, rather than some other positive distance.

#### Part (a): Diagram and Relationships

**Diagram Description:**
- Draw a right triangle where the cop is at one vertex 10 feet away from the road (adjacent side).
- Your car is a point moving along the opposite side of the triangle.
- The hypotenuse \( s \) represents the distance from the cop to your car.
- The base \( x \) represents the distance along the road from the point closest to the cop to your car.

**Mathematical Relationships:**
Using the Pythagorean theorem in the right triangle:

\[ s^2 = x^2 + 10^2 \]

Differentiate both sides with respect to time \( t \):

\[ 2s \frac{ds}{dt} = 2x \frac{dx}{dt} \]

Simplify:

\[ \frac{dx}{dt} = \frac{s}{x} \frac{ds}{
Transcribed Image Text:### Understanding Radar Gun Measurements A police officer sits in a parked car 10 feet from a straight road. As you drive along the road, the officer aims a radar gun at your car. In this scenario, we'll examine the mathematics behind how the radar gun measures your speed differently than your actual velocity. Let \( s \) be the distance from your car to the cop in feet. The radar gun measures the rate at which your distance from the cop is changing with respect to time, which is \( \frac{ds}{dt} \). Let \( x \) be the distance, in feet, of your car from the point on the road that is closest to the cop, so your car’s velocity is \( \frac{dx}{dt} \). #### Task Breakdown 1. **Diagram and Relationship Derivation:** - **Diagram:** Draw and label a diagram of the scenario, and write \(\frac{dx}{dt}\) in terms of \(\frac{ds}{dt}\), \(s\), and \(x\). 2. **Explanation using the Diagram:** Use part (a) to explain why \(\frac{ds}{dt} < \frac{dx}{dt}\). Thus, the radar gun's measurement of \(\frac{ds}{dt}\) always underestimates your car’s velocity. 3. **Dependence on Distance:** Examine whether the conclusion in part (b) depends on the cop’s car being 10 feet from the road, rather than some other positive distance. #### Part (a): Diagram and Relationships **Diagram Description:** - Draw a right triangle where the cop is at one vertex 10 feet away from the road (adjacent side). - Your car is a point moving along the opposite side of the triangle. - The hypotenuse \( s \) represents the distance from the cop to your car. - The base \( x \) represents the distance along the road from the point closest to the cop to your car. **Mathematical Relationships:** Using the Pythagorean theorem in the right triangle: \[ s^2 = x^2 + 10^2 \] Differentiate both sides with respect to time \( t \): \[ 2s \frac{ds}{dt} = 2x \frac{dx}{dt} \] Simplify: \[ \frac{dx}{dt} = \frac{s}{x} \frac{ds}{
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