Problem 39 - Thomas Calculus

Question

Distance between two ships 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 2000 yd) 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

Ships moved further apart; visibility was insufficient to sight each other. The rate of change in distance settled at 8 knots.

Step 1: Understanding Initial Positions and Directions

Initially, at noon (\(t = 0\)), ship \(A\) is 12 nautical miles north of ship \(B\). Ship \(A\) is sailing southward at a speed of 12 knots. Simultaneously, ship \(B\) is sailing eastward at a speed of 8 knots.

Step 2: Defining Positions Over Time

Given the initial positions, let \(y_A\) and \(x_B\) represent the position coordinates of ships \(A\) and \(B\) at any time \(t\). At \(t = 0\), \(y_A = 12\) and \(x_B = 0\). Thus, \(y_A(t) = 12 - 12t\) and \(x_B(t) = 8t\), as ship \(A\) moves 12 miles minus 12 times \(t\) miles south, and ship \(B\) moves 8 miles \(t\) miles east.

Step 3: Deriving the Distance Function

The distance \(s(t)\) between the two ships at any time \(t\) can be found using the formula for the distance between two points: \s(t) = \sqrt{(x_B(t))^2 + (y_A(t))^2} = \sqrt{(8t)^2 + (12 - 12t)^2}.\Simplifying, \s(t) = \sqrt{64t^2 + 144 - 288t + 144t^2} = \sqrt{208t^2 - 288t + 144}.\

Step 4: Calculating the Rate of Change at Specific Times

To find how rapidly the distance between the ships is changing, we compute \(\frac{ds}{dt}\) using the chain rule:\frac{ds}{dt} = \frac{1}{2\sqrt{208t^2 - 288t + 144}} (416t - 288).\Evaluating at \(t = 0\): \frac{ds}{dt}\bigg|_{t=0} = \frac{1}{2\cdot12}(-288) = -12.\At \(t = 1\): \frac{ds}{dt}\bigg|_{t=1} = \frac{128}{28} \approx 4.57.\

Step 5: Checking Visibility

To determine if the ships ever sighted each other, we need to check if the distance \(s(t)\) ever becomes less than or equal to 5 nautical miles. Solving \ \sqrt{208t^2 - 288t + 144} \leq 5\ analytically shows it never holds true. Thus, the ships could not sight each other.

Step 6: Graphing Functions

Graph \(s(t)\) and \(\frac{ds}{dt}\) from \(t = -1\) to \(t = 3\) using graphing software. The function \(s(t)\) starts at 12 nautical miles and generally increases, while \(\frac{ds}{dt}\) moves from negative to positive, indicating the rate of expansion between the ships.

Step 7: Horizontal Asymptote Exploration

To explore the asymptotic behavior of \(\frac{ds}{dt}\) as \(t \to \infty\), consider the expression \frac{ds}{dt} = \frac{1}{2\sqrt{208t^2}}(416t)\, which simplifies to approaching 8 knots. This limiting value aligns with the eastward speed of ship \(B\).

Key Concepts

Distance Formula

When two moving objects cover different paths, the distance formula helps us calculate how far apart they are at any given moment. In our exercise, ship A moving south and ship B moving east requires us to find the distance as they navigate through the sea. The distance between the two ships can be determined by using the Pythagorean theorem. In simple terms, imagine that the paths of the ships form a right triangle:

• The leg in the north-south direction is traveled by ship A.
• The leg in the east-west direction is traveled by ship B.

Using these paths, the distance formula is applied to find the hypotenuse, giving us the separation between the ships over time: \[s(t) = \sqrt{(x_B(t))^2 + (y_A(t))^2}\] Where:

• \(x_B(t) = 8t\) is the eastward distance travelled by ship B.
• \(y_A(t) = 12 - 12t\) is the southward distance travelled by ship A.

Rate of Change

The rate of change is a crucial concept when we need to know how fast something is varying over time. In our case, it's all about how quickly the distance between the two ships is increasing or decreasing as they move. The rate of change of distance between the ships is represented by \[\frac{ds}{dt} \] and it tells us how rapidly their separation changes. By applying differentiation from calculus, we can derive this rate from our distance function. Initially, at time \(t = 0\), \[\frac{ds}{dt} = -12\] indicating the ships are getting closer. One hour later, at \(t = 1\), the computation reveals:\[\frac{ds}{dt} \approx 4.57\],demonstrating that they are moving apart. This change in sign and value shows how their relative speeds affect their growing distance.

Differentiation

Differentiation is the mathematical process we use to find the rate of change of a function. In our problem, it helps us determine \(\frac{ds}{dt}\), the rate at which the distance between the ships changes with respect to time. We employ the chain rule, a technique that simplifies differentiation for composite functions:\[\frac{ds}{dt} = \frac{1}{2\sqrt{208t^2 - 288t + 144}} (416t - 288)\]This formula might look complex, but here's the breakdown:

• The term \( \frac{1}{2\sqrt{208t^2 - 288t + 144}} \) represents a part of the derivative concerning the square root function.
• The expression \((416t - 288)\) arises from differentiating the inside function of the square root.

Using differentiation, one can check when the rate of change is zero, negative, or positive, which provides insight into how the physical distance behaves as ships move.

Graphing Functions

Graphing provides a visual interpretation of mathematical functions, making complex relationships easier to understand. By graphing the distance function \(s(t)\) and the rate of change \(\frac{ds}{dt}\) side by side, you can visually analyze their behaviors over time.

• The graph of \(s(t)\) shows how the distance starts from 12 nautical miles, influenced by ships' speeds and directions.
• Meanwhile, \(\frac{ds}{dt}\) reveals fluctuations in the rate of distance change, starting negative and turning positive as they distance themselves. This shift aligns with the ships' physical movements, validating our calculations with a graphical signature.

Additional insights like horizontal asymptotes of \(\frac{ds}{dt}\) as \(t\) approaches infinity suggest limiting behaviors akin to ship B’s speed. Observing these functions on a graph elucidates trends and confirms computations, making complex problems easier to digest visually.