Problem 42 - Calculus Early Transcendentals

Question

Car A is traveling north on Highway 16 and car \(B\) is traveling west on Highway \(83 .\) Each car is approaching the intersection of these highways. At a certain moment, car \(A\) is 0.3 \(\mathrm{km}\) from the intersection and traveling at 90 \(\mathrm{km} / \mathrm{h}\) while car \(\mathrm{B}\) is 0.4 \(\mathrm{km}\) from the intersection and traveling at 80 \(\mathrm{km} / \mathrm{h} .\) How fast is the distance between the cars changing at that moment?

Step-by-Step Solution

Answer

The distance between cars A and B is decreasing at 236 km/h.

Step 1: Understand the Scenario

Car A is moving towards the north and is 0.3 km away from the intersection, traveling at a speed of 90 km/h. Car B is moving towards the west and is 0.4 km away from the intersection, traveling at a speed of 80 km/h. We need to determine how fast the distance between the two cars is changing at that moment.

Step 2: Set Up the Right Triangle

At any given moment, the positions of Cars A and B, together with the intersection, form a right triangle. The legs of the triangle are the distances each car is from the intersection: let \( x \) be the distance of Car A from the intersection and \( y \) be the distance of Car B from the intersection. The hypotenuse \( z \) is the distance between the two cars.

Step 3: Apply the Pythagorean Theorem

Using the Pythagorean theorem, we can relate the sides of the triangle: \( x^2 + y^2 = z^2 \). Differentiate both sides with respect to time \( t \) to find the rate of change of \( z \): \( 2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 2z \frac{dz}{dt} \).

Step 4: Insert Known Values

Substitute the given values into the equation: \( x = 0.3 \text{ km}, \frac{dx}{dt} = -90 \text{ km/h }, y = 0.4 \text{ km}, \frac{dy}{dt} = -80 \text{ km/h} \). Calculate \( z \) using \( z^2 = 0.3^2 + 0.4^2 \). Thus, \( z = 0.5 \text{ km} \).

Step 5: Solve for \( \frac{dz}{dt} \)

Substitute all the values into the differentiated equation: \( 2(0.3)(-90) + 2(0.4)(-80) = 2(0.5) \frac{dz}{dt} \). Simplify and solve for \( \frac{dz}{dt} \): \(-54 - 64 = z \frac{dz}{dt} \). Thus, \(-118 = 0.5 \frac{dz}{dt} \), so \( \frac{dz}{dt} = -236 \text{ km/h} \).

Step 6: Interpret the Result

The negative sign indicates that the distance between the cars is decreasing. Therefore, the distance between the cars is changing at a rate of 236 km/h.

Key Concepts

Right Triangle

When we think of a right triangle, we picture one angle measuring 90 degrees, making it resemble the letter 'L'. In the context of the cars approaching the intersection, this triangle is formed by the paths of the cars making up the two legs and the distance between the cars representing the hypotenuse.

• The leg corresponding to Car A's path is 0.3 km, pointing north.
• The other leg belongs to Car B's path, which is 0.4 km and extends west.

The interaction occurs at the intersection, characterizing a meeting point where these legs connect the triangle. Visualizing this setup as a right triangle simplifies the process of computing the changing distance between the cars. Using such a geometric setup is foundational in solving related rates problems.

Pythagorean Theorem

The Pythagorean theorem is a fundamental principle in geometry that connects the lengths of the sides of a right triangle. For our triangle, with legs referenced as \( x \) and \( y \) and the hypotenuse as \( z \), the theorem is given by the equation:\[x^2 + y^2 = z^2\]For the exercise with the cars:

• \( x = 0.3 \text{ km} \)
• \( y = 0.4 \text{ km} \)
• These squared values add up to \( z^2 \), where \( z \) turns out to be 0.5 km.

This theory provides a reliable way to calculate the numerical hypotenuse \( z \), crucial for determining how fast the separation between cars is altering. Utilizing the theorem, we connect geometry with algebra and find real-world applications such as in this traffic problem.

Differentiation

Differentiation is the key mathematical tool that allows us to find how things are changing at any particular moment. In this case, it helps to determine how the distance between the cars (hypotenuse \( z \)) changes as the cars move towards the intersection.We start by differentiating the Pythagorean equation \( x^2 + y^2 = z^2 \) to relate changes in distance:\[2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 2z \frac{dz}{dt}\]Let's break it down:- \( \frac{dx}{dt} \) is the rate at which Car A changes distance, given as -90 km/h.- \( \frac{dy}{dt} \) signifies the same for Car B, given as -80 km/h.Substituting these values along with the earlier calculated \( z \), the equation becomes solvable. We align the constant rates of \( x \) and \( y \) with the changing distance \( z \) to find \( \frac{dz}{dt} = -236 \text{ km/h} \). The negative indicates the two cars are getting closer, a common interpretation in differentiation where a negative rate signifies a decrease.