Question

Prove the property. In each case, assume that \(\mathbf{r}, \mathbf{u},\) and \(\mathbf{v}\) are differentiable vector-valued functions of \(t,\) \(f\) is a differentiable real-valued function of \(t,\) and \(c\) is a scalar. $$ \begin{array}{l} D_{t}\\{\mathbf{r}(t) \cdot[\mathbf{u}(t) \times \mathbf{v}(t)]\\}=\mathbf{r}^{\prime}(t) \cdot[\mathbf{u}(t) \times \mathbf{v}(t)]+ \\ \mathbf{r}(t) \cdot\left[\mathbf{u}^{\prime}(t) \times \mathbf{v}(t)\right]+\mathbf{r}(t) \cdot\left[\mathbf{u}(t) \times \mathbf{v}^{\prime}(t)\right] \end{array} $$

Short answer

The property is verified through using the product rule of differentiation for vector-valued functions, and applying it to both the dot product and the cross product. Given that \( \mathbf{r}, \mathbf{u}, \mathbf{v} \) are vector functions of \( t \) and \( f \) is a real-valued function of \( t \), we can differentiate the scalar triple product using the product rule for differentiation to achieve the required result.

Step-by-step solution

Apply the product rule

The differentiation of a product of two functions is given by \(f(x)g(x) = f'(x)g(x) + f(x)g'(x)\). Similarly, for scalar triple product, use the product rule: \(D_{t}\{ \mathbf{r}(t) \cdot [\mathbf{u}(t) \times \mathbf{v}(t)] \} = D_{t}\{\mathbf{r}(t)\} \cdot [\mathbf{u}(t) \times \mathbf{v}(t)] + \mathbf{r}(t) \cdot D_{t}\{ [\mathbf{u}(t) \times \mathbf{v}(t)] \}\).

Apply the product rule to cross product

Next, apply the product rule to the cross product of \(\mathbf{u}(t)\) and \(\mathbf{v}(t)\) in the second part from step 1: \(\mathbf{r}(t) \cdot D_{t}\{ [\mathbf{u}(t) \times \mathbf{v}(t)] \} = \mathbf{r}(t) \cdot [D_{t}\{\mathbf{u}(t)\} \times \mathbf{v}(t) + \mathbf{u}(t) \times D_{t}\{\mathbf{v}(t)\}\].\)

Put it together

Combine them to get our final result, just substituting \(\mathbf{r}'(t)\), \(\mathbf{u}'(t)\), and \(\mathbf{v}'(t)\) for \(D_{t}\{\mathbf{r}(t)\}\), \(D_{t}\{\mathbf{u}(t)\}\), \(D_{t}\{\mathbf{v}(t)\}\) respectively. Thus, we get: \(D_{t}\{\mathbf{r}(t) \cdot[\mathbf{u}(t) \times \mathbf{v}(t)]\} =\mathbf{r}^{\prime}(t) \cdot[\mathbf{u}(t) \times \mathbf{v}(t)] + \mathbf{r}(t) \cdot[\mathbf{u}^{\prime}(t) \times \mathbf{v}(t)] + \mathbf{r}(t) \cdot[\mathbf{u}(t) \times \mathbf{v}^{\prime}(t)]\).

Key concepts

Scalar Triple Product

The scalar triple product is an operation that involves three vectors and results in a scalar value. It is defined as the dot product of one vector with the cross product of the other two vectors. Mathematically, it can be represented as \( \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) \), where \( \mathbf{a} \) , \( \mathbf{b} \) , and \( \mathbf{c} \) are vectors.

The geometric interpretation of the scalar triple product is the volume of the parallelepiped formed by the vectors. If the product is zero, it implies that the vectors are coplanar.

In the context of the given exercise, the property described involves differentiating the scalar triple product of three vector-valued functions. This process requires the application of the product rule for differentiation, which will be further discussed in the following sections.

Product Rule Differentiation

The product rule is a fundamental concept in calculus used when differentiating the product of two functions. The rule states that if you have two differentiable functions, \( f(t) \) and \( g(t) \) , the derivative of their product \( f(t)g(t) \) is given by \( f'(t)g(t) + f(t)g'(t) \).

This rule is essential when dealing with vector-valued functions or the scalar triple product as in the exercise. When applied to vector calculus, it allows us to find the derivative of the dot product or cross product of vector functions, ensuring that each component of the vectors is appropriately accounted for during differentiation.

The step-by-step solution of the exercise demonstrates the application of the product rule to the differentiation of a scalar triple product. It is important to consider each component of the vectors and apply the rule accordingly to obtain the correct derivative.

Cross Product Rule

The cross product rule in vector calculus refers to the product rule as applied to the cross product of two vector-valued functions. Just as with scalar functions, when finding the derivative of the cross product of two vectors, the product rule is applied. For vectors \( \mathbf{u}(t) \) and \( \mathbf{v}(t) \), the derivative of their cross product is given by \( \mathbf{u}'(t) \times \mathbf{v}(t) + \mathbf{u}(t) \times \mathbf{v}'(t) \).

This rule is particularly noteworthy in the provided exercise, as it is used to differentiate the scalar triple product. The cross product creates a new vector that is perpendicular to both \( \mathbf{u}(t) \) and \( \mathbf{v}(t) \), and its magnitude relates to the area of the parallelogram they span. When the derivative is involved, the cross product rule ensures that this geometric interpretation is preserved and the rate of change with respect to \( t \) is correctly calculated.

By understanding and applying the cross product rule, as well as the product rule for differentiation, one can solve complex vector calculus problems such as the one illustrated in the exercise.