Problem 41 - Thomas Calculus

Question

Let \(\mathbf{r}\) be a differentiable vector function of \(t .\) Show that if \(\mathbf{r} \cdot(d \mathbf{r} / d t)=0\) for all \(t,\) then \(|\mathbf{r}|\) is constant.

Step-by-Step Solution

Answer

If \( \mathbf{r} \cdot \left( \frac{d \mathbf{r}}{dt} \right) = 0 \), then \( |\mathbf{r}| \) is constant due to the constancy of \( |\mathbf{r}|^2 \).

Step 1: Understanding the Given Condition

We need to show that if the dot product \( \mathbf{r} \cdot \left( \frac{d \mathbf{r}}{dt} \right) = 0 \) for all \( t \), then the magnitude \( |\mathbf{r}| \) is constant. Here, \( \mathbf{r} \cdot \left( \frac{d \mathbf{r}}{dt} \right) = 0 \) indicates that \( \mathbf{r} \) is orthogonal to its derivative \( \frac{d \mathbf{r}}{dt} \).

Step 2: Relating to Magnitude

To relate the given condition to the magnitude of \( \mathbf{r} \), recall that the magnitude squared of a vector \( \mathbf{r} = \langle x, y, z \rangle \) is \( |\mathbf{r}|^2 = x^2 + y^2 + z^2 \).

Step 3: Derivative of Magnitude Squared

Differentiate \( |\mathbf{r}|^2 \) with respect to \( t \):\[ \frac{d}{dt}(|\mathbf{r}|^2) = \frac{d}{dt}(\mathbf{r} \cdot \mathbf{r}) = \mathbf{r} \cdot \frac{d \mathbf{r}}{dt} + \frac{d \mathbf{r}}{dt} \cdot \mathbf{r} \]

Step 4: Simplifying Using the Given Condition

Since the dot product \( \mathbf{r} \cdot \frac{d \mathbf{r}}{dt} \) is zero for all \( t \), we have:\[ \frac{d}{dt}(|\mathbf{r}|^2) = 0 + 0 = 0 \]

Step 5: Conclusion from the Derivative

Since \( \frac{d}{dt}(|\mathbf{r}|^2) = 0 \), this implies \( |\mathbf{r}|^2 \) is constant, which means \( |\mathbf{r}| \) is also constant as the square root of a constant is a constant.

Key Concepts

Dot Product

The dot product is an essential concept when dealing with vector calculus. It provides a way to calculate the interaction or projection between two vectors. Mathematically, the dot product between two vectors \( \mathbf{a} = \langle a_1, a_2, a_3 \rangle \) and \( \mathbf{b} = \langle b_1, b_2, b_3 \rangle \) is given by:\[\mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3\]This computation results in a scalar value.

• One of the critical properties of the dot product is its relationship to the angle \( \theta \) between the vectors: \( \mathbf{a} \cdot \mathbf{b} = |\mathbf{a}| \cdot |\mathbf{b}| \cdot \cos \theta \).
• This relationship means the dot product is zero if the vectors are at right angles to each other (i.e., orthogonal), since \( \cos 90^\circ = 0 \).
• The formula also expresses how one vector projects onto another, providing a measure of alignment or directionality between them.

Understanding the dot product is crucial for exploring vector-related problems, like determining the orthogonality of functions, which we'll discuss next.

Orthogonal Vectors

Orthogonal vectors are vectors that, when dotted together, yield a zero result. This characteristic stems from them being perpendicular to each other in the applicable vector space.

• To visually think about this, imagine two vectors forming the sides of a right angle.
• This relation is not just limited to visible geometric understanding but extends to more abstract spaces, too, such as function spaces or vector fields.
• In mathematical terms, if \( \mathbf{a} \cdot \mathbf{b} = 0 \), vectors \( \mathbf{a} \) and \( \mathbf{b} \) are orthogonal.

In differential vector functions, like in our initial exercise, the orthogonality between a vector function \( \mathbf{r} \) and its derivative \( \frac{d\mathbf{r}}{dt} \) implies a specific constant behavior of the vector’s magnitude over time, which leads us to our next concept.

Constant Magnitude

The condition of constant magnitude in a vector function means the size or length stays the same, irrespective of direction changes.

• To understand the magnitude of a vector \( \mathbf{r} = \langle x, y, z \rangle \), consider its calculation: \( |\mathbf{r}| = \sqrt{x^2 + y^2 + z^2} \).
• In the scenario given by the exercise, if \( \mathbf{r} \cdot (d\mathbf{r}/dt) = 0 \), then \( \frac{d}{dt}(|\mathbf{r}|^2) = 0 \).
• This formula highlights that any change over time in this squared magnitude would be zero, indicating that \( |\mathbf{r}| \) does not truly change as \( t \) advances.

When dealing with physics or engineering problems, this concept can signify stability or conservation, such as energy conservation in a system, or uniform motion in kinematic analysis.