Problem 32 - Calculus Early Transcendentals: Pearson New International Edition

Question

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

Step-by-Step Solution

Answer

\( \|\mathbf{r}(t)\| \) is constant if and only if \( \mathbf{r}(t) \cdot \mathbf{r}'(t) = 0 \).

Step 1: Understand the Problem

We are asked to prove that the magnitude (norm) of a vector function \( \mathbf{r}(t) \) is constant if and only if its dot product with its derivative \( \mathbf{r}'(t) \) is zero. This is an important problem in vector calculus related to motion and trajectory.

Step 2: Express the Norm Function

The norm of a vector \( \mathbf{r}(t) \) is given by \( \|\mathbf{r}(t)\| = \sqrt{\mathbf{r}(t) \cdot \mathbf{r}(t)} \). We first express its square: \( \|\mathbf{r}(t)\|^2 = \mathbf{r}(t) \cdot \mathbf{r}(t) \).

Step 3: Differentiate the Square of the Norm

To determine when \( \|\mathbf{r}(t)\| \) is constant, differentiate \( \|\mathbf{r}(t)\|^2 \) with respect to \( t \): \[ \frac{d}{dt} (\mathbf{r}(t) \cdot \mathbf{r}(t)) = 2 \mathbf{r}(t) \cdot \mathbf{r}'(t). \]

Step 4: Use the Derivative Condition for a Constant Function

For \( \|\mathbf{r}(t)\| \) to be constant, its derivative must be zero. Hence, \[ 2 \mathbf{r}(t) \cdot \mathbf{r}'(t) = 0. \] Simplifying, we obtain \( \mathbf{r}(t) \cdot \mathbf{r}'(t) = 0. \)

Step 5: Prove the 'if' and 'only if' parts

**Forward (if):** If \( \|\mathbf{r}(t)\| \) is constant, then from Step 4, \( \mathbf{r}(t) \cdot \mathbf{r}'(t) = 0 \).**Backward (only if):** If \( \mathbf{r}(t) \cdot \mathbf{r}'(t) = 0 \), differentiating, \( \frac{d}{dt} (\mathbf{r}(t) \cdot \mathbf{r}(t)) = 2 \mathbf{r}(t) \cdot \mathbf{r}'(t) = 0 \), meaning \( \|\mathbf{r}(t)\| \) is constant.

Key Concepts

Constant Magnitude of Vector Function

In vector calculus, a vector function has a constant magnitude when the length of the vector remains unchanged over time as it evolves in space. Consider the vector function \( \mathbf{r}(t) \). Its magnitude or norm is expressed as \( \|\mathbf{r}(t)\| = \sqrt{\mathbf{r}(t) \cdot \mathbf{r}(t)} \).

To understand when the magnitude is constant, think about a vector moving through space without changing its length. This means the rate of change of its norm should be zero over time.

A constant magnitude implies a derivative equal to zero because the norm's square, \( \|\mathbf{r}(t)\|^2 \), when differentiated with respect to \( t \), leads to \( 2 \mathbf{r}(t) \cdot \mathbf{r}'(t) \). For this product to be zero, the terms must either separately be zero or perpendicular. Here, \( \mathbf{r}(t) \cdot \mathbf{r}'(t) = 0 \), which indicates that the vector \( \mathbf{r}(t) \) is orthogonal to its derivative.

Dot Product

The dot product (or scalar product) is a fundamental operation in vector algebra. For two vectors \( \mathbf{a} \) and \( \mathbf{b} \), the dot product is represented as \( \mathbf{a} \cdot \mathbf{b} = \|\mathbf{a}\| \|\mathbf{b}\| \cos(\theta) \), where \( \theta \) is the angle between the vectors.

In simpler terms, the dot product measures how much one vector extends in the direction of another. It produces a scalar value that is zero when the vectors are perpendicular (i.e., \( \cos(90^) = 0 \)) and is maximal when the vectors are parallel.

• If \( \mathbf{a} \cdot \mathbf{b} > 0 \), vectors point in a similar direction.
• If \( \mathbf{a} \cdot \mathbf{b} < 0 \), vectors point in opposite directions.
• If \( \mathbf{a} \cdot \mathbf{b} = 0 \), vectors are orthogonal.

In our exercise, the expression \( \mathbf{r}(t) \cdot \mathbf{r}'(t) = 0 \) uses the zero result of a dot product, proving that the derivative vector is perpendicular to the initial vector \( \mathbf{r}(t) \), thereby ensuring constant magnitude.

Derivative and Constant Functions

The derivative is a core concept in calculus, representing the rate at which a function changes. For a function to be constant, its derivative must equal zero everywhere in its domain since a change of zero implies no variation over time.

When we apply differentiation to a vector function, such as \( \mathbf{r}(t) \), finding its derivative \( \mathbf{r}'(t) \) gives us insight into how the vector's direction and magnitude change over 'time'. The derivative reflects the instantaneous direction and magnitude of motion.

• When the derivative of a scalar or squared norm is zero, the respective function is constant.
• If \( \mathbf{r}(t) \cdot \mathbf{r}'(t) = 0 \), the magnitude \( \|\mathbf{r}(t)\| \) doesn't change, confirming it as a constant function.

In essence, connecting the concept of differentiating \( \mathbf{r}(t) \cdot \mathbf{r}(t) \) reveals that when the output is zero, as in our exercise, the magnitude doesn't change, and hence remains constant over time. This relationship forms a crucial link in understanding dynamic systems.