Problem 32 - Calculus Early Transcendentals: Pearson New International Edition

Question

Prove that \(\mid \mathbf{r}(t) \|\) is constant if and only if \(\mathbf{r}(t) \cdot \mathbf{r}^{\prime}(t)=0\).

Step-by-Step Solution

Answer

The magnitude of \( \mathbf{r}(t) \) is constant if and only if it is orthogonal to its derivative \( \mathbf{r}'(t) \).

Step 1: Understand the Problem

We are given a vector function \( \mathbf{r}(t) \) and need to prove that the magnitude (i.e., length) of this vector is constant if and only if the dot product of \( \mathbf{r}(t) \) and its derivative \( \mathbf{r}'(t) \) is zero.

Step 2: Express the Magnitude of \( \mathbf{r}(t) \)

The magnitude of a vector \( \mathbf{r}(t) \) is given by \( ||\mathbf{r}(t)|| = \sqrt{\mathbf{r}(t) \cdot \mathbf{r}(t)} \). We need to show conditions under which this magnitude remains constant.

Step 3: Differentiate and Apply Chain Rule

If \( ||\mathbf{r}(t)|| \) is constant, then its derivative with respect to \( t \) is zero: \( \frac{d}{dt} ||\mathbf{r}(t)|| = 0 \). Use the chain rule to find the derivative: \( \frac{d}{dt} \mathbf{r}(t) \cdot \mathbf{r}(t) = 2 \mathbf{r}(t) \cdot \mathbf{r}'(t) \).

Step 4: Analyze the Result

Setting the derivative to zero, we get \( 2 \mathbf{r}(t) \cdot \mathbf{r}'(t) = 0 \), which simplifies to \( \mathbf{r}(t) \cdot \mathbf{r}'(t) = 0 \). This means that for the magnitude to be constant, \( \mathbf{r}(t) \) and \( \mathbf{r}'(t) \) must be orthogonal.

Step 5: Reverse Implication

Assume \( \mathbf{r}(t) \cdot \mathbf{r}'(t) = 0 \). Then \( \frac{d}{dt} ||\mathbf{r}(t)||^2 = 2 \mathbf{r}(t) \cdot \mathbf{r}'(t) = 0 \), which implies \( ||\mathbf{r}(t)|| \) is constant. This completes the proof.

Key Concepts

Magnitude of a Vector

The magnitude of a vector represents its length. For a vector function \( \mathbf{r}(t) \), the magnitude is calculated using the formula:

• \( ||\mathbf{r}(t)|| = \sqrt{\mathbf{r}(t) \cdot \mathbf{r}(t)} \).

This formula involves taking the dot product of the vector with itself. The square root gives us the length of the vector in the vector space it inhabits.
The magnitude is a crucial concept because it provides a way to measure the size of a vector, irrespective of its direction. If the magnitude is constant, it suggests that the vector is not getting longer or shorter over time. In this context, if \( ||\mathbf{r}(t)|| \) is constant, we explore how it relates to the vector's nature and behavior over time.

Derivative of a Vector Function

The derivative of a vector function like \( \mathbf{r}(t) \) is similar to finding the derivative of a scalar function, but it applies to each component of the vector. Symbolically, this derivative is written as \( \mathbf{r}'(t) \). It provides information on how the vector \( \mathbf{r}(t) \) changes as the parameter \( t \) varies.
In the context of the problem, if the magnitude of \( \mathbf{r}(t) \) is constant, then its rate of change, as given by its derivative, must be zero. According to the steps provided, when we differentiate \( ||\mathbf{r}(t)|| \) (following chain rule), and find that it's zero, it reinforces the condition for a constant magnitude. This leads us to look further into the properties that enable such a condition.

Dot Product Properties

The dot product is a fundamental operation in vector calculus that combines two vectors to yield a scalar. For vectors \( \mathbf{a} \) and \( \mathbf{b} \), the dot product is expressed as:

• \( \mathbf{a} \cdot \mathbf{b} = ||\mathbf{a}|| \, ||\mathbf{b}|| \cos(\theta) \)

where \( \theta \) is the angle between the two vectors. If the dot product is zero, the vectors are orthogonal (perpendicular) to each other, since \( \cos(90^\circ) = 0 \).
In this exercise, we consider the dot product \( \mathbf{r}(t) \cdot \mathbf{r}'(t) \). If it equals zero, the derivative vector \( \mathbf{r}'(t) \) is orthogonal to the original vector function \( \mathbf{r}(t) \). This orthogonality is the key condition ensuring that the magnitude of \( \mathbf{r}(t) \) remains constant.

Chain Rule

The chain rule is a method in calculus for finding the derivative of a composition of functions. When dealing with vector-valued functions, the chain rule helps us understand the rate of change of a vector function's magnitude.
In the solution, we apply the chain rule to differentiate the square of the magnitude of \( \mathbf{r}(t) \), expressed as:

• \( \frac{d}{dt}(\mathbf{r}(t) \cdot \mathbf{r}(t)) = 2 \mathbf{r}(t) \cdot \mathbf{r}'(t) \)

By this differentiation, we can see explicitly how the change in \( \mathbf{r}(t) \) interacts with its own derivative. If the result is zero, the vector maintains a constant magnitude, reinforcing the connection between the vector and its rate of change. The chain rule here reveals the interconnectedness of these aspects, making it essential for understanding dynamics of vector functions.