Problem 44 - Precalculus Mathematics for Calculus

Question

Two boats leave the same port at the same time. One travels at a speed of \(30 \mathrm{mi} / \mathrm{h}\) in the direction \(\mathrm{N} 50^{\circ} \mathrm{E}\) and the other travels at a speed of \(26 \mathrm{mi} / \mathrm{h}\) in a direction \(\mathrm{S} 70^{\circ} \mathrm{E}\) (see the figure). How far apart are the two boats after one hour?

Step-by-Step Solution

Answer

The boats are approximately 48.59 miles apart after one hour.

Step 1: Determine Position Vectors

After one hour, each boat will have traveled a distance equal to its speed. Therefore, the first boat travels 30 miles, and the second boat travels 26 miles. We represent each direction using vectors. For the first boat heading north 50° east, the position vector is given by \((30 \cos(50°), 30 \sin(50°))\). The second boat heading south 70° east has a position vector of \((26 \cos(70°), -26 \sin(70°))\).

Step 2: Calculate the Vectors

Calculate the components of each boat's position vector: - For the first boat: \(x_1 = 30 \cos(50°) \approx 19.28\) and \(y_1 = 30 \sin(50°) \approx 22.98\). - For the second boat: \(x_2 = 26 \cos(70°) \approx 8.89\) and \(y_2 = -26 \sin(70°) \approx -24.47\). These components represent the eastward and northward (or southward) distances traveled by the boats.

Step 3: Find the Displacement Vector

The displacement vector between the two boats is found by subtracting the second boat's position vector from the first. So, the displacement vector is \((x_1 - x_2, y_1 - y_2)\), which calculates to approximately \((19.28 - 8.89, 22.98 - (-24.47))\), giving \((10.39, 47.45)\).

Step 4: Calculate the Distance

The distance between the two boats after one hour can be found by calculating the magnitude of the displacement vector: \[\text{Distance} = \sqrt{(10.39)^2 + (47.45)^2}\]Computing this, we have: \[\text{Distance} \approx \sqrt{107.9721 + 2252.3025} \approx \sqrt{2360.2746} \approx 48.59 \text{ miles}\].

Step 5: Conclusion

The two boats are approximately 48.59 miles apart after one hour.

Key Concepts

Position Vectors

Position vectors help us represent the location of an object using directional components. When considering a journey, such as that of a boat, a position vector shows where the boat is after traveling in a specific direction and distance.

Given a start at the origin, the position vector can be expressed as \( (x, y) \), where \( x \) is the distance moved eastward and \( y \) is the distance moved northward. For directional travel, we use angles relative to a cardinal direction (such as north or east) to break the movement into these components using trigonometric functions.

• To calculate \( x \), use cosine: \( x = \text{speed} \times \cos(\theta) \).
• To calculate \( y \), use sine: \( y = \text{speed} \times \sin(\theta) \).

In our example, after an hour, the position vector of each boat became evident by using the traveled distances and their respective angles. Each boat's movement forms part of the vector.

Displacement Vector

The displacement vector effectively shows the change in position from the starting point to another position. Here, we use it to find the difference in positioning between two objects, such as two boats. It answers the question: how far apart are the two locations?

To compute the displacement vector, subtract the position vector of one object from the other. Using components, this becomes a simple calculation.

• \( \text{Displacement vector} = (x_1 - x_2, y_1 - y_2) \)

In our scenario, the displacement vector tells us how much the two boats have separated from each other in space. It's a new vector laying out in straight terms the distance and direction from one boat to the other. Understanding this concept is key in determining how spread out objects are based on their individual paths.

Magnitude of a Vector

The magnitude of a vector represents its length or size. For vectors in two-dimensional space, this is effectively the direct distance between the original and final point in space. If you imagine a line drawn from the start of the vector to its end, the magnitude is the length of that line.

To compute a vector's magnitude: use the Pythagorean theorem concept. The magnitude is the square root of the sum of the squares of its components. For a vector \( (x, y) \), the magnitude is given by: \[\text{Magnitude} = \sqrt{x^2 + y^2}\]

For our boats, you take the displacement vector's components to calculate the straight-line distance stretching from one boat to another. It’s like figuring out the hypotenuse of a right triangle formed by the length and height represented by the vector's components.

Trigonometric Functions

Trigonometry comes in handy when working with vectors as it allows us to resolve a vector into its horizontal and vertical components. Sin and Cos come into play here as these trigonometric functions relate the angles in a triangle to ratios of the sides.

Here is how you use them in vector calculations:

• Cosine function: \( \cos(\theta) \) gives the ratio of the adjacent side (horizontal component in vector terms) to the hypotenuse (the vector's total length or distance).
• Sine function: \( \sin(\theta) \) gives the ratio of the opposite side (vertical component in vector terms) to the hypotenuse.

In our example, with each boat's heading being a specific angle from a cardinal direction, cosine and sine help break down the journey's angle into these useful components. Whether moving north-east or south-east, using trigonometry is essential to correctly find where each vector places an object along the defined axes.