Question

Let \(\mathbf{r}=\mathbf{r}(t)\) be a vector whose length is always 1 (it may vary in direction). Prove that either \(\mathbf{r}\) is a constant vector or \(d \mathbf{r} / d t\) is perpendicular to \(\mathbf{r} .\) Hint: Differentiate \(\mathbf{r} \cdot \mathbf{r}.\)

Short answer

\( \textbf{r} \) is constant or \( \frac{d\textbf{r}}{dt} \) is perpendicular to \( \textbf{r} \).

Step-by-step solution

Express the magnitude of the vector

Since \(\textbf{r}\) has a constant magnitude of 1, we can write \(\textbf{r} \cdot \textbf{r} = 1\).

Differentiate both sides with respect to time

Differentiate \(\textbf{r} \cdot \textbf{r} = 1\) with respect to \(t\). This gives \( \frac{d}{dt}(\textbf{r} \cdot \textbf{r}) = \frac{d}{dt}(1) \).

Apply the product rule

By using the product rule, we have \( \frac{d}{dt}(\textbf{r} \cdot \textbf{r}) = \textbf{r} \cdot \frac{d\textbf{r}}{dt} + \frac{d\textbf{r}}{dt} \cdot \textbf{r} \).

Simplify the differentiation result

Since \( \frac{d}{dt}(1) = 0 \), the result from Step 3 becomes \( 2 \textbf{r} \cdot \frac{d\textbf{r}}{dt} = 0 \).

Analyze the resulting equation

The equation \( 2 \textbf{r} \cdot \frac{d\textbf{r}}{dt} = 0 \) implies that \( \textbf{r} \cdot \frac{d\textbf{r}}{dt} = 0 \). This means that the derivative \( \frac{d\textbf{r}}{dt} \) is either zero or orthogonal to \( \textbf{r} \).

Conclude the proof

Hence, \( \textbf{r} \) is either a constant vector (if \( \frac{d\textbf{r}}{dt} = 0 \)) or \( \frac{d\textbf{r}}{dt} \) is perpendicular to \( \textbf{r} \).

Key concepts

Exploring Constant Magnitude in Vectors

The concept of a vector having constant magnitude is fundamental in vector calculus. A vector \(\textbf{r} \(t\)\) has a constant magnitude when its length does not change over time. To express this mathematically, we say that \(\textbf{r} \cdot \textbf{r} = 1\). The dot product \( \textbf{r} \cdot \textbf{r} \) essentially computes the square of the length of the vector. Since the length is constant and given as 1, the dot product is 1.

Understanding Vector Differentiation

Vector differentiation involves taking the derivative of a vector with respect to time or any other variable. For our vector \( \textbf{r}\(t\) \), its time derivative is expressed as \( \frac{d \textbf{r}}{dt} \). Differentiation helps in analyzing how the vector changes over time. When we differentiate \( \textbf{r} \cdot \textbf{r} = 1 \), we apply the product rule, which gives us \( \textbf{r} \cdot \frac{d\textbf{r}}{dt} + \frac{d\textbf{r}}{dt} \cdot \textbf{r} \). Here, using the given problem, since 1 differentiated is 0, we simplify to obtain \( \textbf{r} \cdot \frac{d \textbf{r}}{dt} = 0 \). This suggests that either the vector does not change over time or the direction of change is perpendicular to the original vector.

Orthogonality in Vector Calculus

Orthogonality means that two vectors are perpendicular to each other. In the context of our exercise, the derivative \( \frac{d \textbf{r}}{dt} \) is either zero or orthogonal to \(\textbf{r} \). Mathematically, if the dot product of two vectors is zero, \( \textbf{r} \cdot \frac{d\textbf{r}}{dt} = 0 \), then the vectors are orthogonal. When vectors are orthogonal, they form a 90-degree angle. Therefore, if \( \frac{d\textbf{r}}{dt} \) is not zero, it must be perpendicular to \(\textbf{r} \), signifying no change in magnitude, only a change in direction.