Problem 47 - University Calculus: Early Transcendentals

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 ds/dt 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 minimum distance between ships was 6 nautical miles, so they never sighted each other (visibility 5 miles).

Step 1: Define the Positions of the Ships Over Time

At time \( t = 0 \), ship \( A \) is 12 nautical miles due north of ship \( B \). Ship \( A \) is moving south at 12 knots (nautical miles per hour), so its position relative to its initial position at noon can be expressed as \( y_A(t) = 12 - 12t \). Ship \( B \) is moving east at 8 knots, so its position can be expressed as \( x_B(t) = 8t \).

Step 2: Express the Distance Between Ships as a Function of Time

We can use the Pythagorean theorem to express the distance \( s(t) \) between the ships as a function of time: \[ s(t) = \sqrt{(x_B(t))^2 + (y_A(t))^2} = \sqrt{(8t)^2 + (12 - 12t)^2}. \] Simplifying, we get: \[ s(t) = \sqrt{64t^2 + 144 - 288t + 144t^2} = \sqrt{208t^2 - 288t + 144}. \]

Step 3: Find the Rate of Change of Distance at Noon

To determine how rapidly the distance between the ships is changing at any time \( t \), we need to compute the derivative \( \frac{ds}{dt} \). First, write down \( s(t) \) again to solve for \( \frac{ds}{dt} \): \[ \frac{ds}{dt} = \frac{d}{dt} \left( \sqrt{208t^2 - 288t + 144} \right). \] Use the chain rule: \[ \frac{ds}{dt} = \frac{1}{2\sqrt{208t^2 - 288t + 144}} \times (416t - 288). \] At noon (\( t = 0 \)), \( \frac{ds}{dt}(0) \) is 12 knots (since ships are moving away), and one hour later (\( t = 1 \)), substitute \( t = 1 \) into the derivative to find \( \frac{ds}{dt}(1) = 8 \).

Step 4: Determine if Ships Came Within Visibility Range

Visibility is 5 nautical miles. We need to find the minimum value of \( s(t) \) over the interval \( t \geq 0 \). Find critical points by setting \( \frac{ds}{dt} = 0 \): \[ 416t - 288 = 0 \implies t = \frac{288}{416} = \frac{3}{4}. \] Calculate \( s(\frac{3}{4}) \approx 6 \). Since \( 6 > 5 \), the ships were never within sight of each other.

Step 5: Graph and Compare

Graph \( s(t) \) and \( \frac{ds}{dt} \) for \( -1 \leq t \leq 3 \):- Plot \( s(t) = \sqrt{208t^2 - 288t + 144} \) to examine the function's behavior and values.- Plot \( \frac{ds}{dt} = \frac{1}{2\sqrt{208t^2 - 288t + 144}}(416t - 288) \) to see how the rate of change of distance evolves over time.The graphs support the results of parts (b) and (c), confirming no interaction under 5 nautical miles and changes in velocity.

Step 6: Analyze Long-Term Behavior of Rate of Change

As \( t \rightarrow \infty \), the influence of the constant terms in \( s(t) \) diminishes and \( \frac{ds}{dt} \approx \frac{1}{2} \times \frac{416t}{\sqrt{208t^2}} = 8 \). This means \( ds/dt \) approaches a limiting value of 8 knots, equivalent to ship \( B \)'s speed, since \( A \)'s vertical speed reaches equilibrium with \( B \)'s horizontal expansion.

Key Concepts

Pythagorean Theorem in Related Rates

The Pythagorean Theorem is a fundamental principle in geometry, used to relate the lengths of the sides of a right triangle. In related rates problems, especially involving moving objects, this theorem helps express distances as a function of time.
In the context of our problem, the positions of Ships \( A \) and \( B \) form the legs of a right triangle at any given time \( t \). The distance \( s(t) \), the hypotenuse, between the two ships can be calculated using the Pythagorean Theorem. We have \( y_A(t) = 12 - 12t \) for Ship \( A \'s \) southward movement, and \( x_B(t) = 8t \) for Ship \( B \'s \) eastward movement. Thus, the distance \( s(t) \) is given by

• \( s(t) = \sqrt{(x_B(t))^2 + (y_A(t))^2} \)
• \( = \sqrt{(8t)^2 + (12 - 12t)^2} \)
• \( = \sqrt{208t^2 - 288t + 144} \)

Using these equations, you can find how the distance between Ships \( A \) and \( B \) changes as a function of time, a perfect example of applying the Pythagorean Theorem to a dynamic situation.

The Concept of Derivatives in Motion

Derivatives are essential in calculus for measuring how a quantity changes over time. In problems like the one involving ships, derivatives help determine how the distance changes between two moving objects - essentially the rate of change of that distance.
To find the rate of change of distance \( \frac{ds}{dt} \) between the ships, we use the derivative of \( s(t) = \sqrt{208t^2 - 288t + 144} \). Applying the chain rule, we differentiate:

• First, identify the function inside the square root, \( u(t) = 208t^2 - 288t + 144 \)
• The derivative is \( \frac{ds}{dt} = \frac{1}{2\sqrt{u(t)}} \times \frac{du}{dt} \)
• Where \( \frac{du}{dt} = 416t - 288 \)

This gives us a clear picture of how quickly the separation \( s(t) \) is increasing or decreasing over time. Calculating at specific values of \( t \), such as noon \( t = 0 \) or one hour later \( t = 1 \), can show us the real-time rate at which the ships' separation changes.

Function of Time in Distance and Motion

In many physics and calculus problems, quantities vary over time, and it's crucial to express those as functions of time to understand their behavior. In this problem, the distance \( s \) between the ships is a function of time: \( s(t) = \sqrt{208t^2 - 288t + 144} \). This function lets us analyze how the distance changes as time progresses.
Understanding a function of time involves:

• Recognizing how variables \( x_B(t) \) for Ship \( B \) and \( y_A(t) \) for Ship \( A \) change with time \( t \)
• Substituting these expressions into a relationship such as the Pythagorean Theorem to find \( s(t) \)

Being able to model real-world scenarios, such as ship movements, in terms of functions of time is key to making predictions and understanding dynamics in various fields like physics and engineering.

Understanding Rate of Change in Related Rates Problems

Rate of change is a crucial concept in calculus used to describe how a dependent variable changes with respect to an independent variable. In the context of related rates problems, we're often interested in how quickly one quantity is changing with respect to time.
For our ships' distance problem, calculating \( \frac{ds}{dt} \) helps in knowing how fast the distance between the ships is increasing or decreasing. This involves:

• Differentiate the function \( s(t) \) with respect to time \( t \)
• At time \( t = 0 \), \( \frac{ds}{dt} = 12 \) knots, meaning the ships are moving apart rapidly at noon
• One hour later \( t = 1 \), \( \frac{ds}{dt} = 8 \) knots, indicating the rate of separation has adjusted

This concept helps not only in theoretical scenarios but in understanding everyday situations where rates play a role, such as speed in driving or growth in populations.