Problem 11 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

There's a 10 -mile-per-hour wind, but a bicyclist calculates that its component in his direction of motion is only 9 miles per hour. What is the angle between the velocity vector of the wind and that of the bicyclist?

Step-by-Step Solution

Answer

The angle between the wind's velocity vector and the bicyclist's velocity vector is 25.84 degrees.

Step 1: Identify Given Information

The magnitude of the wind's velocity, \(v_w\), is given as 10 mph. The component of the wind's velocity in the direction of the bicyclist's velocity (\(v_b\)) is given as 9 mph.

Step 2: Apply Cosine Rule

The angle between the vectors can be given by the equation \(\cos(\theta) = \frac{{v_b}}{{v_w}}\)

Step 3: Substitute Known Values

Substitute \(v_w = 10\) mph and \(v_b = 9\) mph into the equation: \(\cos(\theta) = \frac{{9}}{{10}} = 0.9\)

Step 4: Calculate Angle

To find the angle \(\theta\), use the inverse cosine function (also known as arccos), \(\theta = \arccos(0.9)\).

Step 5: Convert to Degrees

The output of the arccos function will be in radians; convert it to degrees as the standard unit we use in these cases. The conversion is achieved via the relationship \(1 \text{ radian} = \frac{180}{\pi} \text{ degrees}\)

Key Concepts

Vector Components

When we talk about vectors, such as the wind in the exercise, they are not just arrows pointing in a direction; they have both magnitude and direction. The magnitude is the length of the vector. When a vector is decomposed into its vector components, it means that we are breaking it down into parts that align with the axes of the system we're using - often the x and y axes in a two-dimensional plane.

This breakdown is incredibly useful in physics as it allows us to analyze the effects of forces, velocity, or any other vector quantity in separate dimensions. For instance, the exercise showed that the wind has a component in the direction of the bicyclist's travel. This is essentially projecting the wind's velocity vector onto the direction of the bicyclist's vector. Think of it like a shadow that the wind vector casts on the cyclist's path. Mathematically, we could use sine and cosine functions to determine the size of these 'shadows', or components.

Cosine Rule

In the world of triangle geometry, the cosine rule is a hero. It relates the lengths of the sides of a triangle to the cosine of one of its angles. The formula states that for any triangle with sides of lengths 'a', 'b', 'c', and an angle 'C' opposite to the side 'c', the relationship can be expressed as: \[c^2 = a^2 + b^2 - 2ab\cos(C)\]

For our purposes, when two vectors form a triangle, we can use the cosine rule to find the angle between them, as is seen in this bicyclist's problem. All we need is the magnitude of the vectors and the magnitude of their resultant vector, or in cases like our exercise, the component of one vector along the other. This formula may look daunting, but once you understand that it's about the interaction of a triangle's sides and angles, it becomes a tool you can't live without in trigonometry and physics.

Inverse Trigonometric Functions

Once you've gotten cosy with regular trigonometric functions like sine, cosine, and tangent, you'll sooner or later meet their inverse counterparts. Inverse trigonometric functions, such as arcsine, arccosine (also written as \(\arccos\)), and arctangent, are used to find the angles when the value of the trigonometric function is known.

In simple terms, they answer the question, 'What angle would give me this sine, cosine, or tangent value?' In the case of the bicyclist problem, the value of \(\cos(\theta)\) was given, and using the inverse cosine function \(\arccos\) allowed us to find the actual angle \(\theta\). Remember, this process is essentially going backwards from the cosine ratio to the angle, unlike the standard trig functions that go from the angle to the ratio. As the solution illustrated, we must always be mindful of the units and convert radians to degrees if necessary, for more understandable results.