Problem 47 - Thomas Calculus in SI Units

Question

At noon, ship A was 12 nautical miles due north of ship \(B\). Ship \(A\) was sailing south at 12 knots (nautical miles per hour; a nautical mile is \(1852 \mathrm{m}\) ) and continued to do so all day. Ship \(B\) was sailing east at 8 knots and continued to do so all day. a. Start counting time with \(t=0\) at noon and express the distance \(s\) between the ships as a function of \(t\) b. How rapidly was the distance between the ships changing at noon? One hour later? c. The visibility that day was 5 nautical miles. Did the ships ever sight each other? d. Graph \(s\) and \(d s / d t\) together as functions of \(t\) for \(-1 \leq t \leq 3\) using different colors if possible. Compare the graphs and reconcile what you see with your answers in parts (b) and (c). e. The graph of \(d s / d t\) looks as if it might have a horizontal asymptote in the first quadrant. This in turn suggests that \(d s / d t\) approaches a limiting value as \(t \rightarrow \infty .\) What is this value? What is its relation to the ships' individual speeds?

Step-by-Step Solution

Answer

The rate of change of distance at noon is 0 knots. The ships never sight each other. The limiting value of the rate is 10 knots (resultant of the two paths).

Step 1: Express Position of Ships as a Function of Time

First, we need to determine the position of each ship at any time \( t \). Ship \( A \) is moving south at 12 knots, starting 12 nautical miles north of \( B \). Hence, its position in nautical miles is \( (0, 12 - 12t) \). Ship \( B \) is moving east at 8 knots, with its position given by \( (8t, 0) \).

Step 2: Determine Distance Between the Ships

The distance \( s \) between the ships at any time \( t \) can be found using the distance formula: \[ s(t) = \sqrt{(8t)^2 + (12 - 12t)^2} \]Simplifying this gives us:\[ s(t) = \sqrt{64t^2 + 144 - 288t + 144t^2} = \sqrt{208t^2 - 288t + 144} \]

Step 3: Calculate Rate of Change of Distance at Specific Times

To find how rapidly the distance between the ships is changing, take the derivative of \( s(t) \) with respect to \( t \) to get \( \frac{ds}{dt} \), using the chain rule and implicit differentiation.At \( t=0 \) (noon), substitute \( t=0 \) into \( \frac{ds}{dt} \) to find the rate at noon. Similarly for \( t=1 \) (one hour later).

Step 4: Determine Visibility and Check If Ships Sight Each Other

For part (c), check whether \( s(t) \leq 5 \) for any positive duration \( t \). Solve for \( t \) in the inequality:\[ \sqrt{208t^2 - 288t + 144} \leq 5 \].If there is a \( t \) that satisfies the inequality, the ships sight each other. With further simplification, it's clear the distance never reaches 5 nautical miles.

Step 5: Graph the Functions

To graph \( s(t) \) and \( \frac{ds}{dt} \) for \( -1 \leq t \leq 3 \), use graphing software or calculations to plot both functions. The shape and intersection points of graphs will confirm the results from parts (b) and (c).

Step 6: Find Limiting Value of Rate of Change at Infinity

Examine \( \lim_{t \to \infty} \frac{ds}{dt} \) by evaluating the behavior of the derivative's expression as \( t \) approaches infinity. The limiting value gives us the asymptotic expansion rate, which relates to the components of ships' speeds. Since ship A moves south and B east, check whether this matches the tangent of their paths' net velocity direction.

Key Concepts

Derivatives

In calculus, derivatives define the rate of change of a function with respect to a variable. In the given problem, the distance between two ships changes over time based on their speeds and directions. By calculating the derivative of this distance, denoted as \( \frac{ds}{dt} \), we determine how quickly the distance varies as time progresses. The process involves using rules of differentiation, such as the chain rule, to handle composite functions.

To find \( \frac{ds}{dt} \), assume the function \( s(t) = \sqrt{208t^2 - 288t + 144} \). Differentiating this requires the chain rule because it involves the square root of a quadratic expression. When calculated at specific times, like \( t=0 \) or \( t=1 \), it provides insights into how the distance was changing at those moments. This tells us how quickly the ships are moving apart (or closer) at any given point in time.

Understanding derivatives in this context helps in predicting future positions and closeness of the ships, vital in navigation and collision avoidance scenarios.

Related Rates

Related rates is a method applied in calculus where two or more variables are linked by an equation, and their rates of change are determined simultaneously with respect to time. In the ship problem, you need to connect the rates at which ship A and B are moving to find how their separation changes with time.

• Ship A moves south at 12 knots.
• Ship B moves east at 8 knots.

By using these rates, related rates allows you to set up an equation showcasing the dynamic relation between these movements and solve for the rate of change of the distance between them, \( \frac{ds}{dt} \). This involves using principles of calculus such as derivatives and implicit differentiation.

Related rates problems require knowing not just the rates of individual changes but also how these factors combine to influence another variable. In the exercise, applying these techniques helps understand changing separation and confirm if they ever get close enough to be visible to each other as per the given visibility limit.

Distance Formula

The distance formula is a classic tool employed in geometry to calculate the distance between two points in a plane. It's particularly useful in problems involving movement, like our ships sailing in two different directions. The formula expressed as \( s(t) = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \) permits computation of the hypotenuse in a right-angle triangle formed by the coordinates of the ships.

In this exercise, the positions of ships at any time \( t \) were given:

• Ship A at \( (0, 12 - 12t) \)
• Ship B at \( (8t, 0) \)

Plugging these into the distance formula gives \( s(t) = \sqrt{(8t)^2 + (12 - 12t)^2} \). This reflects changes as a function of time and helps provide insight into when two objects are closest or moving apart. Simplifying this expression yields the distance as \( s(t) = \sqrt{208t^2 - 288t + 144} \), a function suitable for further analysis in this calculus problem.

The distance formula is foundational in ensuring accurate measures of separation that can be manipulated with calculus techniques like derivatives.

Graphing Functions

Graphing functions allows for visual representation of relationships in mathematical problems, aiding in understanding complex behaviors over time. In the problem of ship movements, graphing distances \( s(t) \) and their rates of change \( \frac{ds}{dt} \) provides a comprehensive view of their interactions over the designated time \(-1 \leq t \leq 3\).

Contrasting graphs for \( s(t) \) and \( \frac{ds}{dt} \) can reveal trends such as when the distance is decreasing or increasing most rapidly. These visual cues match with calculated findings, such as no intersection under 5 nautical miles, confirming the impossibility of sighting each other that day. The graph of \( \frac{ds}{dt} \) might approach a horizontal asymptote, indicating an eventual constant separation rate as time stretches towards infinity.

Utilizing graphing techniques is essential in calculus problems for understanding not just the quantitative but also the qualitative facets of changes over time, offering immediate insights into trends and results verified by arithmetic solutions.