Problem 29 - Advanced Engineering Mathematics

Question

In Problems, find the indicated derivative. Assume that all vector functions are differentiable. $$ \frac{d}{d t}\left[\mathbf{r}(t) \cdot\left(\mathbf{r}^{\prime}(t) \times \mathbf{r}^{\prime \prime}(t)\right)\right] $$

Step-by-Step Solution

Answer

\( \mathbf{r}(t) \cdot (\mathbf{r}'(t) \times \mathbf{r}'''(t)) \)

Step 1: Understand the Problem

We are given an expression involving vector functions and their derivatives. The task is to find the derivative with respect to time of the scalar triple product \( \mathbf{r}(t) \cdot (\mathbf{r}'(t) \times \mathbf{r}''(t)) \).

Step 2: Apply Vector Derivative Rules

We apply the derivative product rule for vectors, also known as the Leibniz rule for vectors. The derivative of a scalar product \( u \cdot (v \times w) \) is given by:\[\frac{d}{dt} [u \cdot (v \times w)] = u' \cdot (v \times w) + u \cdot (v' \times w) + u \cdot (v \times w')\]where \( u, v, \) and \( w \) are vector functions. In our case, \( u = \mathbf{r}(t), v = \mathbf{r}'(t), \) and \( w = \mathbf{r}''(t) \).

Step 3: Calculate Individual Components

Substitute the vectors in our equation. Our derivative becomes:\[\frac{d}{dt} [\mathbf{r}(t) \cdot (\mathbf{r}'(t) \times \mathbf{r}''(t))] = \mathbf{r}'(t) \cdot (\mathbf{r}'(t) \times \mathbf{r}''(t)) + \mathbf{r}(t) \cdot (\mathbf{r}''(t) \times \mathbf{r}''(t)) + \mathbf{r}(t) \cdot (\mathbf{r}'(t) \times \mathbf{r}'''(t))\]

Step 4: Simplify Using Properties

Using vector product properties, we know that the cross product of any vector with itself is zero: \( \mathbf{v} \times \mathbf{v} = \mathbf{0} \). Thus, the middle term becomes zero. Additionally, \( \mathbf{r}'(t) \cdot (\mathbf{r}'(t) \times \mathbf{r}''(t)) \) also results in zero since the dot product of a vector is zero when taken with a vector perpendicular to it, which is true for all cross products. Therefore, we are left with:\[ \mathbf{r}(t) \cdot (\mathbf{r}'(t) \times \mathbf{r}'''(t))\]

Step 5: Finalize and Conclude

Thus, the derivative of the given expression is actually the simplification to a new scalar triple product involving the original function, its first and third derivative:\[ \mathbf{r}(t) \cdot (\mathbf{r}'(t) \times \mathbf{r}'''(t))\]

Key Concepts

Derivative of Vector Functions

In vector calculus, the derivative of vector functions plays a crucial role in understanding the rate of change of vector quantities. Just like derivatives in regular calculus measure how a scalar function changes, derivatives of vector functions measure the change of vector fields.
When you consider a vector function, say \( \mathbf{r}(t) = \langle x(t), y(t), z(t) \rangle \), its derivative indicates how its components change with time. The derivative \( \mathbf{r}'(t) \) is also a vector consisting of the derivatives of the individual components: \( \mathbf{r}'(t) = \langle x'(t), y'(t), z'(t) \rangle \).
Understanding this concept is fundamental when dealing with vector fields in physics and engineering, as it helps us describe motions, forces, and other physical phenomena that depend on time and space.

Scalar Triple Product

The scalar triple product is a specific type of operation involving three vectors, providing valuable results in geometrical and physical applications. It is defined as \( \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) \), where \( \mathbf{a}, \mathbf{b}, \) and \( \mathbf{c} \) are vectors. This operation results in a scalar value.
The geometric interpretation of the scalar triple product is crucial: it equals the volume of the parallelepiped formed by the vectors. If this volume is zero, the vectors are coplanar, which means they lie in the same plane.

• It helps find the orientation of a set of vectors.
• A positive result indicates a right-hand orientation, while a negative result indicates left-hand orientation.

This property is usually used in problems involving torque, angular momentum, and more.

Vector Derivative Rules

Derivative rules for vector calculus extend concepts from scalar calculus to handle vectors. The Leibniz rule for vector functions, similar to the product rule, is essential for evaluating derivatives involving multiple vector products.
In our context, the rule states:

• For a scalar product, \( \frac{d}{dt} [u \cdot (v \times w)] = u' \cdot (v \times w) + u \cdot (v' \times w) + u \cdot (v \times w') \).

Here, \( u, v, \) and \( w \) are vector functions, and their derivatives are obtained by applying basic differentiation rules individually to each function.
These rules allow the handling of more complex vector operations and ensure correct differentiation operations, which are particularly useful in dynamics, electromagnetism, and fluid mechanics, where vector products frequently appear.