Question

Let \(r=r(t)\) 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 \(\boldsymbol{r} .\) Hint: Differentiate \(\mathbf{r} \cdot \mathbf{r}\).

Short answer

Either \(\boldsymbol{r}\) is a constant vector or \(\frac{d \boldsymbol{r}}{d t}\) is perpendicular to \(\boldsymbol{r}\).

Step-by-step solution

Understand the given condition

Given that the vector \(\boldsymbol{r}\) has a length of 1, the magnitude of \(\boldsymbol{r}\) can be represented as \(\boldsymbol{r} \cdot \boldsymbol{r} = 1\). This means \(\boldsymbol{r}\) is a unit vector.

Differentiate the dot product

Differentiate both sides of the equation \(\boldsymbol{r} \cdot \boldsymbol{r} = 1\) with respect to time \(t\). Using the product rule, we have: \(\frac{d}{dt} (\boldsymbol{r} \cdot \boldsymbol{r}) = \frac{d}{dt} 1\).

Apply the product rule

The product rule states that \(\frac{d}{dt} (\boldsymbol{r} \cdot \boldsymbol{r}) = \boldsymbol{r} \cdot \frac{d \boldsymbol{r}}{d t} + \frac{d \boldsymbol{r}}{d t} \cdot \boldsymbol{r}\). Since \(1\) is a constant, \(\frac{d}{dt} 1 = 0\). Thus, \(\boldsymbol{r} \cdot \frac{d \boldsymbol{r}}{d t} + \frac{d \boldsymbol{r}}{d t} \cdot \boldsymbol{r} = 0\).

Simplify the expression

Since the dot product is commutative, \( \boldsymbol{r} \cdot \frac{d \boldsymbol{r}}{d t} = \frac{d \boldsymbol{r}}{d t} \cdot \boldsymbol{r} \). Therefore, the equation simplifies to \(( 2 \boldsymbol{r} \cdot \frac{d \boldsymbol{r}}{d t} ) = 0\).

Conclude the proof

For the above expression to hold true, \( \boldsymbol{r} \cdot \frac{d \boldsymbol{r}}{d t} = 0 \) must be satisfied. This means that \(\frac{d \boldsymbol{r}}{d t}\) is either zero (indicating that \(\boldsymbol{r}\) is a constant vector) or perpendicular to \(\boldsymbol{r}\).

Key concepts

Unit Vectors

A unit vector is a vector with a magnitude (length) of exactly 1. Unit vectors are important because they help simplify vector calculations and provide direction without altering the magnitude of other vectors. The notation for a unit vector is usually a bold lowercase letter, such as \(\boldsymbol{r}\).

For example, if \(\boldsymbol{r}\) is a unit vector, it means that: \[\boldsymbol{r} \cdot \boldsymbol{r} = 1\] How does this relate to vectors? The dot product of a vector with itself gives the square of its magnitude. Since \(\boldsymbol{r}\) is a unit vector, we get 1.

If you need to find a unit vector from another vector \(\boldsymbol{a}\), you just divide that vector by its magnitude: \[\boldsymbol{a}_{unit} = \frac{\boldsymbol{a}}{\|\boldsymbol{a}\|}\] which normalizes \(\boldsymbol{a}\) to have a length of 1. Understanding unit vectors is a foundational step in vector calculus and many other fields of physics and engineering.

Differentiation in Vector Calculus

Differentiation in vector calculus is like differentiation in regular calculus but applied to vectors. When you differentiate a vector function, you differentiate each of its components with respect to a variable, usually time (t).

In the given exercise, the vector \(\boldsymbol{r}\) is differentiated with respect to time. The product rule is used, which states that for two functions \(u(t)\) and \(v(t)\): \[\frac{d}{dt}[u(t) \cdot v(t)] = \frac{du}{dt} \cdot v(t) + u(t) \cdot \frac{dv}{dt}\] Applying this rule to \(\boldsymbol{r}\) gives: \[\frac{d}{dt}(\boldsymbol{r} \cdot \boldsymbol{r}) = \frac{d\boldsymbol{r}}{dt} \cdot \boldsymbol{r} + \boldsymbol{r} \cdot \frac{d\boldsymbol{r}}{dt}\]

This is a key step in proving the original statement. Differentiation of unit vectors is crucial for understanding change over time in multiple directions, modeling physical phenomena like velocity and acceleration, and forming the basis of more advanced concepts in vector calculus.

Dot Product and Orthogonality

The dot product (or scalar product) measures how much two vectors point in the same direction. It is calculated as: \[\boldsymbol{a} \cdot \boldsymbol{b} = |\boldsymbol{a}| |\boldsymbol{b}| \cos \theta\] Here, \(\theta\) is the angle between the two vectors.\br\br> - If \(\cos \theta = 0\), meaning \(\theta = 90^\circ\) or the vectors are perpendicular, their dot product will be 0. This plays a critical role in the exercise where we found \(\boldsymbol{r} \cdot \frac{d \boldsymbol{r}}{d t} = 0\) . This means that the rate of change of \(\boldsymbol{r}\) over time is perpendicular to \(\boldsymbol{r}\) itself, unless \(\frac{d \boldsymbol{r}}{d t} = 0\) (i.e., \(\boldsymbol{r}\) is constant).\br\br> - If the dot product is positive, the vectors are pointing in a similar direction. \br> - If the dot product is negative, they point in opposite directions.\br\br>Understanding the dot product helps in many applications, such as projecting one vector onto another, finding angles between vectors, and solving geometric problems.