Problem 51 - 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. \begin{equation} \begin{array}{l}{\text { a. Start counting time with } t=0 \text { at noon and express the distance }} \\ \quad{\underline{\phantom{xxx}} s \text { between the ships as a function of } t \text { . }} \\ {\text { b. How rapidly was the distance between the ships changing at }} \\ \quad{\text { noon? One hour later? }} \\\ {\text { c. The visibility that day was } 5 \text { nautical miles. Did the ships }} \\ \quad{\text { ever sight each other? }} \\ {\text { d. Graph } s \text { and } d s / d t \text { together as functions of } t \text { for }-1 \leq t \leq 3} \\ \quad {\text { using different colors if possible. Compare the graphs and }} \\ \quad {\text { reconcile what you see with your answers in parts (b) and (c). }} \\ {\text { e. The graph of } d s / d t \text { looks as if it might have a horizontal asymptote }} \\ \quad {\text { in the first quadrant. This in turn suggests that } d s / d t} \\ \quad {\text { approaches a limiting value as } t \rightarrow \infty . \text { What is this value? }} \\ \quad {\text { What is its relation to the ships' individual speeds? }} \end{array} \end{equation}

Step-by-Step Solution

Answer

The ships never got within 5 nautical miles of each other.

Step 1: Define Positions as Functions of Time

The position of Ship A at time \( t \) is given by \( y_A(t) = 12 - 12t \) because it starts 12 nautical miles north of Ship B and moves south at 12 knots. For Ship B, its position is given by \( x_B(t) = 8t \) as it moves east at 8 knots. These equations give us the coordinates \((0, y_A(t))\) for Ship A and \((x_B(t), 0)\) for Ship B at any time \( t \).

Step 2: Construct the Distance Function

The distance \( s(t) \) between the two ships at time \( t \) can be determined using the Pythagorean theorem: \[ s(t) = \sqrt{(x_B(t) - 0)^2 + (0 - y_A(t))^2} = \sqrt{(8t)^2 + (12 - 12t)^2}. \] By simplifying, \( s(t) = \sqrt{64t^2 + (12 - 12t)^2} \).

Step 3: Differentiate the Distance Function

We find the rate of change of \( s(t) \) with respect to time, \( \frac{ds}{dt} \). Using chain rule, differentiate \( s(t) = \sqrt{64t^2 + (12-12t)^2} \). The derivative \( \frac{ds}{dt} \) can be computed as: \[ \frac{ds}{dt} = \frac{1}{2\sqrt{64t^2 + (12-12t)^2}} \cdot (128t - 2(12)(12-12t)(-12)). \] Simplify to find the expression of \( \frac{ds}{dt} \).

Step 4: Calculate \( \frac{ds}{dt} \) at \( t = 0 \) and \( t = 1 \)

Substitute \( t = 0 \) and \( t = 1 \) into the expression for \( \frac{ds}{dt} \) calculated in Step 3 to find how rapidly the distance between the ships is changing at noon (\( t = 0 \)) and one hour later (\( t = 1 \)).

Step 5: Determine Sight Visibility

Determine if the ships were ever within the 5 nautical miles visibility limit by solving \( s(t) < 5 \). Use the expression for \( s(t) \) derived to check if \( s(t) \) is ever less than 5 nautical miles between \( t = 0 \) and when \( s(t) \) becomes greater than 5. Calculate the range of \( t \) for which the inequality holds.

Step 6: Graph \( s(t) \) and \( \frac{ds}{dt} \)

Plot \( s(t) \) and \( \frac{ds}{dt} \) against time \( t \) from \( t = -1 \) to \( t = 3 \). Different colors should be used to differentiate the graphs of \( s(t) \) and \( \frac{ds}{dt} \). Notice where \( s(t) \) approaches or equals 5 and how \( \frac{ds}{dt} \) behaves to confirm results from previous steps.

Step 7: Identify Horizontal Asymptote

Analyze the graph of \( \frac{ds}{dt} \) in the first quadrant to predict asymptotic behavior as \( t \to \infty \). This involves looking at the constant term in the derivative's expression. Determine how the ships' speeds relate to this limiting behavior in terms of trigonometry or vector addition of their speeds.

Key Concepts

Pythagorean theorem

The Pythagorean theorem is a fundamental mathematical principle that helps us find the distance between two points. In the context of moving ships, it can be particularly handy.
To apply it, we first imagine the ships forming a right triangle with their positions as vertices. The distance between the ships as the hypotenuse of this right triangle can be expressed using the Pythagorean theorem:

• If Ship A's southward distance is represented as a vertical side, and Ship B's eastward movement as a horizontal side, we use their respective positions to construct the triangle.
• The formula is given by: \( s(t) = \sqrt{(x_B(t) - 0)^2 + (0 - y_A(t))^2} \)
• Substituting the known values: the positions of Ship A and B at time \( t \) yield \[ s(t) = \sqrt{(8t)^2 + (12 - 12t)^2}. \]

Using the Pythagorean theorem offers a straightforward way to calculate distances, helping us solve one part of the original exercise. This approach gives the exact distance between the ships at different times.

differentiation

Differentiation is a powerful tool to understand how variables change with respect to others. In physics and applied mathematics, we often use it to find rates of change.
Here, we need the derivative of the distance function \( s(t) \) to find how quickly the distance between ships A and B changes over time.
Steps involved in differentiation:

• Recognize \( s(t) = \sqrt{64t^2 + (12 - 12t)^2} \) as our function of interest.
• Apply the chain rule to derive \( \frac{ds}{dt} \), which gives us the rate of change of distance concerning time.
• This involves differentiating under the square root by treating it as \( \frac{1}{2} \) times the derivative of the expression inside divided by the square root itself.

Through differentiation, we simplify to get the derivative: \[ \frac{ds}{dt} = \frac{1}{2\sqrt{64t^2 + (12-12t)^2}} \cdot (128t - 2(12)(12-12t)(-12)). \]
This result helps us understand how fast the ships are distancing or approaching each other at different moments.

rate of change

The rate of change is all about how one variable shifts concerning another. It answers questions like how fast or slow things happen. In the scenario of the two ships, it tells us how quickly the distance between them changes over time.
The derivative \( \frac{ds}{dt} \) gives us this rate.
Here's what to consider:

• Initial rate analysis involves calculating \( \frac{ds}{dt} \) at \( t = 0 \) and \( t = 1 \). This gives the rate per hour data, indicating how much the distance is changing at those specific moments.
• At noon, substitute \( t = 0 \) into the derivative function and obtain the exact rate. Do the same for one hour later \( t = 1 \).
• These computations reveal how dynamic the scenario is, showing us if the ships are getting closer or pulling away at specified intervals.

Having this rate inline allows observers to anticipate future ship positions and ensure navigation safety.

visibility limit

A visibility limit defines how far objects can be seen on the horizon given clear conditions. For our exercise, the ships' visibility limit is set to 5 nautical miles.

• This constraint is crucial for naval operations since limited visibility could increase the likelihood of undetected approaches and potential collisions.
• Using the function \( s(t) \), we determine if the ships ever close within this critical range. Solving \( s(t) < 5 \) helps identify when such an event occurs.
• By calculating where the distance between the ships is within the visibility limit throughout their paths, safety protocols can be established, ensuring they remain apart.

Understanding this limitation is key in real-life scenarios to avoid close encounters between vessels, enabling safe navigation and operation even in restricted visibility conditions.