Problem 88 - Precalculus

Question

There is a branch of calculus devoted to the study of vectorvalued functions; these are functions that map real numbers onto vectors. For example, \(v(t)=\langle t, 2 t\rangle\). Calculate the dot product of the vector-valued functions \(\mathbf{u}(t)=\langle\cos t, \sin t\rangle\) and \(\mathbf{v}(t)=\langle\cos t,-\sin t\rangle\).

Step-by-Step Solution

Answer

The dot product is \(\cos(2t)\).

Step 1: Understand the Problem

We are asked to find the dot product of two vector-valued functions \(\mathbf{u}(t)\) and \(\mathbf{v}(t)\). The given functions are \(\mathbf{u}(t) = \langle \cos t, \sin t \rangle\) and \(\mathbf{v}(t) = \langle \cos t, -\sin t \rangle\).

Step 2: Recall the Dot Product Formula

The dot product of two vectors \(\mathbf{a} = \langle a_1, a_2 \rangle\) and \(\mathbf{b} = \langle b_1, b_2 \rangle\) is calculated as \(\mathbf{a} \cdot \mathbf{b} = a_1 \times b_1 + a_2 \times b_2\).

Step 3: Identify the Components

For \(\mathbf{u}(t) = \langle \cos t, \sin t \rangle\), the components are \(a_1 = \cos t\) and \(a_2 = \sin t\). For \(\mathbf{v}(t) = \langle \cos t, -\sin t \rangle\), the components are \(b_1 = \cos t\) and \(b_2 = -\sin t\).

Step 4: Apply the Dot Product Formula

Substitute the identified components into the dot product formula: \[ \mathbf{u}(t) \cdot \mathbf{v}(t) = (\cos t) \times (\cos t) + (\sin t) \times (-\sin t) \]

Step 5: Simplify the Expression

Simplify the expression:\[ \cos^2 t - \sin^2 t \]This expression can be recognized as a trigonometric identity.

Step 6: Interpret the Result

The expression \(\cos^2 t - \sin^2 t\) is known as the cosine double angle identity, \(\cos(2t)\). Therefore, the dot product of the two vector-valued functions simplifies to \(\cos(2t)\).

Key Concepts

Dot Product

The dot product is a fundamental concept in vector algebra, particularly useful when dealing with vector-valued functions. Simply put, the dot product of two vectors gives a single number, called a scalar. For two vectors

• \( \mathbf{a} = \langle a_1, a_2 \rangle \)
• \( \mathbf{b} = \langle b_1, b_2 \rangle \)

the dot product is calculated by multiplying their corresponding components and summing the results:
\[ \mathbf{a} \cdot \mathbf{b} = a_1 \times b_1 + a_2 \times b_2 \]This operation combines both directional and magnitude information of vectors into a single scalar value. In the given exercise, the vectors are **time-dependent**, meaning they change as time \( t \) changes.
The calculation of the dot product \( \mathbf{u}(t) \cdot \mathbf{v}(t) \) involves substituting the given components

• \( \mathbf{u}(t) = \langle \cos t, \sin t \rangle \)
• \( \mathbf{v}(t) = \langle \cos t, -\sin t \rangle \)

into the formula, leading to \( \cos^2 t - \sin^2 t \). Recognizing patterns in this result can connect to more advanced mathematical concepts.

Trigonometric Identities

Trigonometric identities are useful mathematical tools that relate the angles and lengths of a triangle. These identities simplify complex expressions and are instrumental in calculus. In the context of the given vector-valued functions, understanding trigonometric identities is crucial. For instance, when dealing with the expression \( \cos^2 t - \sin^2 t \), recognizing it as a known identity makes solving problems much faster.
This identity, \( \cos^2 t - \sin^2 t \), simplifies using the double-angle formula for cosine:

• \( \cos(2t) = \cos^2 t - \sin^2 t \)

Such identities come handy in simplifying expressions and solving equations in calculus.
By understanding and recognizing these identities, you can transform challenging trigonometric expressions into simpler ones, ultimately making calculations more efficient.

Calculus

Calculus is a branch of mathematics that studies how things change. Within calculus, vector-valued functions allow us to deal with entities that have magnitude and direction, offering a way to model physical phenomena.
In the original exercise, we see an application of calculus in the computation of the dot product of vector-valued functions: \( \mathbf{u}(t) \) and \( \mathbf{v}(t) \). These functions depend on a parameter \( t \), often representing time, which makes them versatile in modeling moving objects.
Calculus helps us in differentiating and integrating these functions.

• **Differentiation** allows us to find rates of change, like velocity and acceleration.
• **Integration** helps in understanding accumulated quantities, such as distance or area under curves.

Working with vector-valued functions in calculus not only deepens our understanding of mathematical concepts but also enables us to tackle real-world problems involving dynamics and kinematics.