Question

Canoe in a current A woman in a canoe paddles due west at \(4 \mathrm{mi} / \mathrm{hr}\) relative to the water in a current that flows northwest at \(2 \mathrm{mi} / \mathrm{hr} .\) Find the speed and direction of the canoe relative to the shore.

Short answer

Answer: The speed of the canoe relative to the shore is \(2\sqrt{10} \mathrm{mi/h}\) and its direction is \(71.6^{\circ}\) west of north.

Step-by-step solution

Define the velocity vectors

Let's denote the velocity of the canoe relative to the water as vector \(\overrightarrow{c}\) and the velocity of the water current as vector \(\overrightarrow{w}\). We also need to find the velocity of the canoe relative to the shore (\(\overrightarrow{s}\)). We are given: - The speed of the canoe relative to the water is \(4 \mathrm{mi/h}\) and the direction is due west. - The speed of the current is \(2 \mathrm{mi/h}\) and the direction is northwest. Since the problem involves directions, let's use the standard unit vectors \(\boldsymbol{i}\) (east) and \(\boldsymbol{j}\) (north) to represent the velocity vectors. \(\overrightarrow{c} = -4\boldsymbol{i}\text{ }(\text{due west})\) Since the current is flowing northwest, it has equal components in the north and west directions. So, the current vector can be represented as: \(\overrightarrow{w} = -2\boldsymbol{i} + 2\boldsymbol{j}.\) Now, we need to find the velocity of the canoe relative to the shore, vector \(\overrightarrow{s}\).

Add the velocity vectors

To find the velocity vector of the canoe relative to the shore, we add the vectors \(\overrightarrow{c}\) and \(\overrightarrow{w}\): \(\overrightarrow{s} = \overrightarrow{c}+\overrightarrow{w} = (-4\boldsymbol{i})+(-2\boldsymbol{i} + 2\boldsymbol{j})=(-6\boldsymbol{i}+2\boldsymbol{j}).\)

Compute the magnitude of the velocity vector

The magnitude of the canoe's velocity relative to the shore represents the speed. To find the magnitude of the vector \(\overrightarrow{s}\), apply the Pythagorean theorem: \(|\overrightarrow{s}| = \sqrt{(-6)^2+(2)^2} = \sqrt{36+4} = \sqrt{40} = 2\sqrt{10} \mathrm{mi/h}.\)

Find the direction

To determine the direction of the canoe's velocity relative to the shore, we need to find the angle \(\theta\) formed between the vector \(\overrightarrow{s}\) and the north direction. To find the angle, use the tangent function: \(\tan{\theta} = \frac{|\text{east component}|}{|\text{north component}|} = \frac{6}{2} = 3.\) Now, calculate the angle from the inverse tangent function: \(\theta = \tan^{-1}(3) \approx 71.6^{\circ}.\) As the east component of the velocity vector is negative and the north component is positive, the direction of the canoe relative to the shore is \(71.6^{\circ}\) west of north. The speed of the canoe relative to the shore is \(2\sqrt{10} \mathrm{mi/h}\) and its direction is \(71.6^{\circ}\) west of north.

Key concepts

Vector Addition

When understanding relative velocity, vector addition is a key concept. In the given exercise, the woman's canoeing speed and the water current speed are uniquely described by their vectors. These vectors represent magnitude (speed) and direction.

The process of vector addition combines these two velocities to find a new resultant velocity vector. This resultant vector represents the canoe's speed relative to the shore. By visually or mathematically lining up vectors head-to-tail, you derive their combined effect in space.

In the scenario provided, the vectors are defined as:

• The canoe’s vector, \(\overrightarrow{c}\), is aligned straight west due to its direction: \(-4\mathbf{i}\).
• The current’s vector, \(\overrightarrow{w}\), flows northwest with components in both west and north directions: \(-2\mathbf{i}+2\mathbf{j}\).

Combining these, the resultant vector, \(\overrightarrow{s}\), is calculated by adding corresponding components: \(-6\mathbf{i}+2\mathbf{j}\).This step determines how the combined effects of the current and paddling will influence the canoe's trajectory.

Pythagorean Theorem

To find the actual speed or magnitude of the canoe's resultant velocity vector relative to the shore, the Pythagorean theorem is used. This theorem is extremely useful in calculating the magnitude of vectors which are orthogonal—their directions are at right angles to each other.

In this exercise, you have a resultant vector \(\overrightarrow{s} = -6\mathbf{i} + 2\mathbf{j}\). To use the Pythagorean theorem, we consider the values \-6\ and \2\ as sides of a right triangle. The magnitude of the vector, \|\overrightarrow{s}|\, represents the hypotenuse.

Applying the Pythagorean theorem:

• Calculate \((-6)^2+(2)^2\) to sum the squares of the components.
• Then, take the square root: \|\overrightarrow{s}| = \sqrt{40} = 2\sqrt{10} \, \text{mi/h}\.

This step confirms the overall speed at which the canoe moves relative to the shore. It is crucial because it connects the vector's geometric properties to a tangible interpretation of speed.

Angle Calculation

Understanding the direction of the resultant velocity requires calculating its angle with respect to a known axis, such as the north. This involves determining the angle at which the canoe travels with respect to the north-south line.

You first find the tangent of the angle \(\theta\) using the ratio of the absolute values of the vector's components. Here, the east component (though negative) and the north component are used:

\(\tan{\theta} = \frac{6}{2} = 3\).

To find \(\theta\), you apply the inverse tangent function, \(\theta = \tan^{-1}(3)\). This yields an angle of \71.6^{\circ}\. However, it's significant to pay attention to signs (directions).

Since the east component is negative and north is positive, the actual direction is \71.6^{\circ}\ west of north. This angle calculation adds clarity to the vector's directional sense, guiding you from abstract vectors to real-world direction.

Tangent Function

In trigonometry, the tangent function uniquely helps in determining the angle associated with direction in vector problems. Here, it represents the relationship between the velocity components of the vector, \(\overrightarrow{s}\).

To correctly utilize the tangent, consider both the numerical values \-6\ and \2\, which are the canoe's net east and north components, respectively. The formula for tangent summarizes this relationship:

\(\tan{\theta} = \frac{6}{2} = 3\).

This quotient describes the steepness or inclination between the components. The inverse tangent, \(\tan^{-1}(3)\), directly provides the direction angle which is then interpreted as degree west of north.

Being mindful of the sign of each component is vital, as it affects the angle’s quadrant. For the canoe problem, clarity in orientation is achieved by this mathematical understanding—it's what lets us conclude the direction as \71.6^{\circ}\ west of north.