Problem 63 - Precalculus: Functions and Graphs

Question

A ship leaves port at 1: 00 P.M. and sails in the direction \(\mathrm{N} 34^{\circ} \mathrm{W}\) at a rate of \(24 \mathrm{mi} / \mathrm{hr}\). Another ship leaves port at 1: 30 p.M. and sails in the direction \(N 56^{\circ} \mathrm{E}\) at a rate of \(18 \mathrm{mi} / \mathrm{hr}\) (a) Approximately how far apart are the ships at 3: 00 P.M.? (b) What is the bearing, to the nearest degree, from the first ship to the second?

Step-by-Step Solution

Answer

At 3:00 P.M., the ships are approximately 67 miles apart. The bearing from the first ship to the second is about 37° E of N.

Step 1: Determine time traveled by each ship

The first ship departs at 1:00 P.M. and the second ship leaves at 1:30 P.M. Both ships travel until 3:00 P.M. Hence, the first ship travels for 2 hours, while the second ship travels for 1.5 hours.

Step 2: Calculate the distance traveled by each ship

We use the formula \( \text{Distance} = \text{Speed} \times \text{Time} \) for each ship.- Distance traveled by the first ship: \[ \text{Distance}_1 = 24 \text{ mi/hr} \times 2 \text{ hr} = 48 \text{ mi} \]- Distance traveled by the second ship:\[ \text{Distance}_2 = 18 \text{ mi/hr} \times 1.5 \text{ hr} = 27 \text{ mi} \]

Step 3: Convert directions to vectors

Convert the direction into vector components. The directions are given as bearings:- First ship travels \( N 34^\circ W \). Its vector, in terms of components, is: \[ \begin{align*}x_1 &= -48 \sin(34^\circ) \y_1 &= 48 \cos(34^\circ)\end{align*} \]- Second ship travels \( N 56^\circ E \). Its vector, in terms of components, is: \[ \begin{align*}x_2 &= 27 \cos(56^\circ) \y_2 &= 27 \sin(56^\circ)\end{align*} \]

Step 4: Calculate position of each ship

Compute each ship's final position in terms of x and y coordinates:- First ship: \[ (x_1, y_1) = (-48 \sin(34^\circ), 48 \cos(34^\circ)) \]- Second ship: \[ (x_2, y_2) = (27 \cos(56^\circ), 27 \sin(56^\circ)) \]

Step 5: Distance between the ships

Use the distance formula to find the separation:\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]Substitute the values of \((x_1, y_1)\) and \((x_2, y_2)\) from the previous step into the formula and calculate \(d\). This gives the distance between the two ships.

Step 6: Calculate the bearing from first ship to second

The bearing from the first ship to the second is given by the angle \(\theta\) of the line connecting the two ships relative to the north line:- Use the tangent formula: \[ \tan(\theta) = \frac{y_2 - y_1}{x_2 - x_1} \]- Calculate \(\theta\) using the inverse tangent and adjust it to fit the compass bearing.

Key Concepts

Vectors in navigation

Vectors are a fundamental tool in navigation as they represent both direction and magnitude of a particular route. In the case of our two ships, each has a distinct path described by vector components along the x and y axes. Understanding vectors helps in calculating distances and positions.
In navigation, directions such as 'North 34° West' represent a course direction. The north-south line is the reference, and angles are measured clockwise or counterclockwise from this line. The notation 'N 34° W' indicates starting due north and rotating 34 degrees towards the west.
To derive vector components from directions and speeds:

• For the first ship, using trigonometry, we can calculate the x-component (westward direction) as \(-48 \sin(34^\circ)\) and the y-component (northward) as \(48 \cos(34^\circ)\).
• Similarly, we calculate the x-component (eastward) for the second ship as \(27 \cos(56^\circ)\) and the y-component (northward) as \(27 \sin(56^\circ)\).

These vector components allow us to pinpoint each ship's precise location at any time, essential for navigation.

Distance calculation in two dimensions

To find how far apart two objects are, like our two ships, we use the distance formula which is based on the Pythagorean theorem. The formula is \( \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \).
This approach computes the "straight line" or Euclidean distance between two points on a plane by considering both the horizontal (x-axis) and vertical (y-axis) differences.

• The \(x_1, y_1\) coordinates represent the position of the first ship, calculated from its vector components.
• The \(x_2, y_2\) coordinates are for the second ship, also based on vector calculations.

The subtraction within the formula finds how far apart these positions are on each axis. Squaring ensures all values are positive, and taking the square root converts the sum back to a linear distance. This calculation tells us how far apart the two ships are, regardless of their individual routes.

Angle and bearing measurement

Bearings are measured from the north line, typically used in navigation to provide a direction. They represent the angle between the north direction and the path toward another point, such as the direction from the first ship to the second.
To calculate the bearing from one ship to another:

• First compute the angle \(\theta\) using the tangent formula: \( \tan(\theta) = \frac{y_2 - y_1}{x_2 - x_1} \).
• Calculate \(\theta\) by finding the inverse tangent of this value.
• Adjust the angle to fit within the 0° to 360° range typical of compass bearings.

This angle adjustment may include adding or subtracting 180° depending on which quadrant the result falls. Understanding how to measure and adjust bearings ensures accurate navigation regardless of which direction the ship travels.