Problem 35 - Calculus Single Variable

Question

A policeman stands 60 feet from the edge of a straight highway while a car speeds down the highway. Using a radar gun, the policeman ascertains that, at a particular instant, the car is 100 feet away from him and the distance between himself and the car is changing at a rate of 92 feet per second. At that moment, how fast is the car traveling down the highway?

Step-by-Step Solution

Answer

The car is traveling at 115 feet per second.

Step 1: Understanding the Problem Setup

The problem involves a right triangle scenario, where the policeman is at point P, 60 feet away from the highway. The car is at point C on the highway, and at a particular instant, the distance PC is 100 feet.

Step 2: Applying the Pythagorean Theorem

Since the problem forms a right triangle, apply the Pythagorean theorem: \( d^2 = a^2 + b^2 \), where \( d = 100 \) feet, \( a = 60 \) feet, and \( b \) is the distance along the highway. Thus, \( 100^2 = 60^2 + b^2 \).

Step 3: Solving for the Unknown Distance

Solve for \( b \) using the equation from Step 2: \( 100^2 = 60^2 + b^2 \). Simplify to find \( b = \sqrt{100^2 - 60^2} = \sqrt{10000 - 3600} = \sqrt{6400} = 80 \) feet.

Step 4: Utilizing Related Rates

We need to find how fast the car is traveling, denoted as \( \frac{db}{dt} \). Derive the Pythagorean relation with respect to time \( t \): \( 2d \frac{dd}{dt} = 2a \frac{da}{dt} + 2b \frac{db}{dt} \). Since \( a \) is constant, \( \frac{da}{dt} = 0 \).

Step 5: Simplifying the Related Rates Equation

Substitute the known values into the derived equation: \( 2 \times 100 \times 92 = 2 \times 0 + 2 \times 80 \times \frac{db}{dt} \) to solve for \( \frac{db}{dt} \).

Step 6: Solving for the Car's Speed

Calculate the car’s speed using the simplified equation from Step 5: \( 200 \times 92 = 160 \times \frac{db}{dt} \). Therefore, \( \frac{db}{dt} = \frac{200 \times 92}{160} = 115 \) feet per second.

Key Concepts

Pythagorean Theorem

The Pythagorean Theorem is a fundamental concept in geometry that relates the sides of a right triangle. You might remember it from school: in any right-angled triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides. This relationship is expressed in the formula\[ c^2 = a^2 + b^2 \]where \( c \) is the hypotenuse and \( a \) and \( b \) are the other two sides.
In the context of the exercise, the policeman, car, and highway form a right triangle. Here, the distance between the policeman and the car (100 feet) is the hypotenuse \( c \), the policeman's distance from the highway (60 feet) is side \( a \), and the distance along the highway \( b \) is what we solve for using:\[ 100^2 = 60^2 + b^2 \]
This gives us in our scenario that \( b = 80 \) feet, representing the distance along the highway at that instant.

Calculus Problem Solving

Calculus is all about understanding how things change over time. In many real-world problems, conditions are not static, and calculus provides the tools to deal with this dynamic nature. The original problem involves finding the rate at which a car is moving along the highway.
To solve this, we used a calculus technique called 'related rates', which helps us establish relationships between the rates of change of different quantities. First, we solve the Pythagorean relationship for the static positions, and then we apply calculus to find how fast these distances change in relation to each other over time. We differentiate the Pythagorean relation to incorporate time and solve for unknown rates. This approach allows us to connect measurable and understandable changes in the scenario like the distances and speeds involved.

Differentiation with Respect to Time

In related rates problems, differentiation with respect to time is key. It allows us to understand how quantities change over time by differentiating equations that relate to those quantities.
For the exercise, once we determined the static distances, we differentiated the relationship\[ d^2 = a^2 + b^2 \]with respect to time \( t \), resulting in\[ 2d \frac{dd}{dt} = 2a \frac{da}{dt} + 2b \frac{db}{dt} \].
This equation links the change rates: \( \frac{dd}{dt} \) (rate of change of the hypotenuse distance between the car and policeman), \( \frac{da}{dt} \) (which is zero because it's constant), and \( \frac{db}{dt} \) (the rate of the car moving down the highway, our unknown). By substituting known values, we solve for \( \frac{db}{dt} \), revealing how fast the car travels. This process showcases how powerful differentiation with respect to time is in analyzing changing systems.