Problem 23 - Calculus Early Transcendentals: Pearson New International Edition

Question

At \(7: 00 \mathrm{~A} . \mathrm{M}\). one ship was 60 miles due east from a second ship. If the first ship sailed west at 20 miles per hour and the second ship sailed southeast at 30 miles per hour, when were they closest together?

Step-by-Step Solution

Answer

The ships are closest together at approximately 8:12 AM.

Step 1: Define the Problem

We need to determine when two ships are closest to each other given their speeds and directions. The first ship starts 60 miles east of the second ship, sailing west at 20 mph. The second ship sails southeast at 30 mph.

Step 2: Establish the Coordinate System

Let's place the second ship's starting point at the origin (0,0) of a coordinate plane. The first ship starts at (60,0) since it is 60 miles due east.

Step 3: Determine Positions Over Time

At time \(t\) hours after 7:00 AM, the first ship's position is \((60 - 20t, 0)\) as it moves west. The second ship moves southeast with diagonally equal components, resulting in a position \((30t/\sqrt{2}, -30t/\sqrt{2})\).

Step 4: Calculate the Distance Between Ships

The distance \(d(t)\) between the ships at time \(t\) is given by:\[d(t) = \sqrt{\left((60 - 20t) - \frac{30t}{\sqrt{2}}\right)^2 + \left(0 + \frac{30t}{\sqrt{2}}\right)^2}\]

Step 5: Differentiate the Distance Function

Differentiate \(d(t)^2\) with respect to \(t\) to find the minimum distance. It simplifies the differentiation process since the distance and its square will reach extrema simultaneously.

Step 6: Set the Derivative to Zero

By setting \(\frac{d}{dt}(d(t)^2) = 0\), solve for \(t\) to determine the critical points. Simplify and solve for \(t\) where the derivative equals zero.

Step 7: Solve for the Critical Time

Solving the equation from the previous step leads us to determine the value of \(t\). This value is approximately \(t \approx 1.2\) hours after 7:00 AM.

Step 8: Determine the Closest Time

Add \(t = 1.2\) hours to 7:00 AM to find the time when the ships are closest together, which is around 8:12 AM.

Key Concepts

Coordinate System

In Calculus Problem Solving, establishing a coordinate system is crucial. The coordinate system helps us define the locations or trajectories of objects in space. This system works similarly to a map, allowing for precise tracking of positions over time.

• In our exercise, the second ship's starting point was strategically chosen as the origin at (0,0). This simplifies calculations.
• The first ship starts at point (60,0) as it is 60 miles due east, following the problem's initial conditions.

Using coordinates allows us to depict movements on a 2D plane, helping visualize and calculate complex motion paths. This setup provides clarity in determining how each ship's position evolves with time and their ensuing proximity to one another.

Distance Function

Understanding how a distance function works is vital for finding how close two objects are, such as the ships in this problem. The distance function provides a formula for the separation between two entities in a coordinate system.

• Our goal is to express how distance changes over time as each ship travels.
• The general distance formula between two points, \( (x_1, y_1) \) and \( (x_2, y_2) \), is given by: \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
• For these ships, the distance \(d(t)\) changes as time \(t\) increases because both ships move continuously.

Once set, the distance function lets us explore how the separation evolves, necessary for determining when they are nearest.

Derivative

The derivative in this context helps pinpoint the critical points where changes in distance occur most significantly. Derivatives give us rates of change, which are integral in finding optima, or points of closest approach, in this task.

• We differentiate the square of the distance function \(d(t)^2\) rather than \(d(t)\) itself because squaring avoids handling square root functions, simplifying calculations.
• The derivatives help identify when the rate of change in distance between two objects - the ships - is zero, suggesting a point of minimum or maximum distance.

By understanding derivatives, we shift from finding static distances to examining how they change and pinpointing minimum distance intervals.

Critical Points

Critical points represent moments in time where certain behaviors of a function, like minimum distance here, occur. They are found by setting derivatives to zero. In our problem, they indicate times when the ships are closest or farthest from each other.

• Finding critical points involves solving the condition \(\frac{d}{dt}(d(t)^2) = 0\), where the rate of change of the squared distance comes to a halt.
• These points provide insight into when two paths converge most closely, crucial for concluding the exact encounter moments.

Once identified, any critical point needs assessing to confirm it signifies a minimum - solving for optimum scenarios leads to solutions like determining closeness around 8:12 AM in this exercise.