Problem 63 - Precalculus Mathematics for Calculus

Question

Velocity of a Boat A straight river flows east at a speed of \(10 \mathrm{mi} / \mathrm{h}\). A boater starts at the south shore of the river and heads in a direction \(60^{\circ}\) from the shore (see the figure). The motorboat has a speed of \(20 \mathrm{mi} / \mathrm{h}\) relative to the water. (a) Express the velocity of the river as a vector in component form. (b) Express the velocity of the motorboat relative to the water as a vector in component form. (c) Find the true velocity of the motorboat. (d) Find the true speed and direction of the motorboat. GRAPH CANT COPY

Step-by-Step Solution

Answer

The true speed is \(10\sqrt{7}\) mi/h, direction \(\tan^{-1}(\frac{\sqrt{3}}{2})\) degrees north of east.

Step 1: Define the River's Velocity Vector

The river flows east at a speed of 10 mi/h. In vector terms, this is represented as \( \vec{v}_r = \langle 10, 0 \rangle \), where the x-component represents the east-west direction and the y-component represents the north-south direction. Thus, the river's velocity vector is \( \vec{v}_r = \langle 10, 0 \rangle \).

Step 2: Define the Motorboat's Velocity Relative to the Water

The motorboat is traveling at a speed of 20 mi/h relative to the water and at an angle of \(60^{\circ}\) from the shore (which is assumed to be the x-axis, or the east-west direction). To express this as a vector, we use polar to Cartesian conversion: \( \vec{v}_b = 20 \cdot \langle \cos(60^{\circ}), \sin(60^{\circ}) \rangle = \langle 20 \cdot \frac{1}{2}, 20 \cdot \frac{\sqrt{3}}{2} \rangle = \langle 10, 10\sqrt{3} \rangle \). Thus, the velocity vector of the motorboat relative to the water is \( \vec{v}_b = \langle 10, 10\sqrt{3} \rangle \).

Step 3: Calculate the True Velocity of the Motorboat

To find the true velocity of the motorboat, we need to add the river's velocity vector and the motorboat's velocity vector relative to the water: \( \vec{v}_{true} = \vec{v}_r + \vec{v}_b = \langle 10, 0 \rangle + \langle 10, 10\sqrt{3} \rangle = \langle 20, 10\sqrt{3} \rangle \). This vector \( \vec{v}_{true} = \langle 20, 10\sqrt{3} \rangle \) represents the true velocity of the motorboat.

Step 4: Find the True Speed and Direction of the Motorboat

The true speed of the motorboat is given by the magnitude of the true velocity vector \( \vec{v}_{true} = \langle 20, 10\sqrt{3} \rangle \). The magnitude is calculated as \( \| \vec{v}_{true} \| = \sqrt{20^2 + (10\sqrt{3})^2} = \sqrt{400 + 300} = \sqrt{700} = 10\sqrt{7} \). For direction, we calculate the angle \( \theta \) from the east (x-axis) using \( \tan \theta = \frac{\text{y-component}}{\text{x-component}} = \frac{10\sqrt{3}}{20} = \frac{\sqrt{3}}{2} \). Therefore, \( \theta = \tan^{-1}\left(\frac{\sqrt{3}}{2}\right) \). Thus, the true speed is \( 10\sqrt{7} \) mi/h, and the direction is \( \tan^{-1}(\frac{\sqrt{3}}{2}) \) degrees north of east.

Key Concepts

Velocity Vectors

Velocity vectors are a crucial concept in vector calculus and provide valuable information about an object's motion. In this context, the velocity vector tells us the direction and speed at which an object is moving.

When we express velocity as a vector, it typically involves breaking down the movement into its component parts: a horizontal (x-axis) and vertical (y-axis) direction. This representation offers a clear way to analyze and resolve forces acting in different directions.

• The length of the vector (or its magnitude) indicates the speed of the object, whereas the direction of the vector shows us which way the object is moving.
• In our example, the velocity vector of the river is expressed as \( \vec{v}_r = \langle 10, 0 \rangle \), indicating movement to the east at a speed of 10 mi/h.
• Similarly, the motorboat's velocity relative to the water is \( \vec{v}_b = \langle 10, 10\sqrt{3} \rangle \), demonstrating movement both eastwards and northwards, given the angle provided.

Understanding these vectors allows us to calculate true movement paths by adding them together, a process crucial in navigational tasks.

Polar to Cartesian Conversion

Converting polar coordinates into Cartesian coordinates is a valuable technique in mathematics, particularly when dealing with vectors and directional problems. In polar coordinates, a point is defined by a distance from the origin and an angle from the positive x-axis.

To convert these into Cartesian coordinates, which are more practical for calculations, we rely on trigonometric functions:

• The x-component of the vector is found using the cosine function: \( x = r \cdot \cos(\theta) \).
• The y-component is found using the sine function: \( y = r \cdot \sin(\theta) \).

In our problem, the motorboat travels at 20 mi/h at an angle of \(60^{\circ}\). Thus, applying these conversions:

• The x-component becomes \( 20 \cdot \cos(60^{\circ}) = 10 \) mi/h.
• The y-component becomes \( 20 \cdot \sin(60^{\circ}) = 10\sqrt{3} \) mi/h.

This process allows us to express directional speeds in an easily interpretable format, facilitating the combination with other vectors, like the river's flow.

Trigonometric Functions

Trigonometric functions, including sine, cosine, and tangent, play a vital role in vector calculus and help in resolving vectors into their components. These functions relate the angles of a triangle to the lengths of its sides, providing a way to move between angle-based and distance-based measurements.

In our exercise, we utilized these functions to convert the motorboat's polar coordinates into Cartesian components, which entails:

• Using \( \cos(\theta) \) to determine the horizontal part of a vector. This helps us figure out how much movement occurs parallel to the x-axis (eastward).
• Applying \( \sin(\theta) \) to calculate the vertical component, giving us the vector part moving parallel to the y-axis (northward).
• For determining direction, the \( \tan(\theta) \) function is used, where \( \theta = \tan^{-1}\left(\frac{\text{y-component}}{\text{x-component}}\right) \). It calculates the angle direction of the resultant vector based on its component ratios.

Mastery of these functions is essential in solving problems involving angles and distances, enabling effective transition between polar and Cartesian forms and enhancing our ability to work with real-world vector applications.