Problem 42 - Thomas Calculus

Question

Derivatives of triple scalar products a. Show that if \(\mathbf{u}, \mathbf{v},\) and \(\mathbf{w}\) are differentiable vector functions of \(t,\) then $$\frac{d}{d t}(\mathbf{u} \cdot \mathbf{v} \times \mathbf{w})=\frac{d \mathbf{u}}{d t} \cdot \mathbf{v} \times \mathbf{w}+\mathbf{u} \cdot \frac{d \mathbf{v}}{d t} \times \mathbf{w}+\mathbf{u} \cdot \mathbf{v} \times \frac{d \mathbf{w}}{d t}$$ b. Show that $$\frac{d}{d t}\left(\mathbf{r} \cdot \frac{d \mathbf{r}}{d t} \times \frac{d^{2} \mathbf{r}}{d t^{2}}\right)=\mathbf{r} \cdot\left(\frac{d \mathbf{r}}{d t} \times \frac{d^{3} \mathbf{r}}{d t^{3}}\right)$$ Hint: Differentiate on the left and look for vectors whose products are zero.)

Step-by-Step Solution

Answer

Both results use vector calculus product rules for differentiation.

Step 1: Recognize the Triple Product Rule

To solve part (a), recognize that the expression involves a cross product inside a dot product, which is a triple scalar product. We need the product rule for differentiation in this vector context. Our task is to differentiate each vector function within the scalar product.

Step 2: Apply the Derivative to the Triple Scalar Product (a)

Apply the derivative to each term in the scalar product: \[ \frac{d}{dt} (\mathbf{u} \cdot \mathbf{v} \times \mathbf{w}) = \frac{d\mathbf{u}}{dt} \cdot (\mathbf{v} \times \mathbf{w}) + \mathbf{u} \cdot \frac{d}{dt}(\mathbf{v} \times \mathbf{w}) \]For the second term, apply the product rule to \( \mathbf{v} \times \mathbf{w} \): \[ \frac{d}{dt} (\mathbf{v} \times \mathbf{w}) = \frac{d\mathbf{v}}{dt} \times \mathbf{w} + \mathbf{v} \times \frac{d\mathbf{w}}{dt} \].

Step 3: Substitute Back and Simplify (a)

Substitute the derivative of the cross product back into the expression: \[ \frac{d}{dt} (\mathbf{u} \cdot \mathbf{v} \times \mathbf{w}) = \frac{d\mathbf{u}}{dt} \cdot (\mathbf{v} \times \mathbf{w}) + \mathbf{u} \cdot \left(\frac{d\mathbf{v}}{dt} \times \mathbf{w} + \mathbf{v} \times \frac{d\mathbf{w}}{dt}\right) \]. Distribute \(\mathbf{u}\) to get the final expression: \[ \frac{d\mathbf{u}}{dt} \cdot (\mathbf{v} \times \mathbf{w}) + \mathbf{u} \cdot \frac{d\mathbf{v}}{dt} \times \mathbf{w} + \mathbf{u} \cdot \mathbf{v} \times \frac{d\mathbf{w}}{dt} \].

Step 4: Define Triple Scalar Products for Derivatives (b)

For part (b), consider the expression \( \mathbf{r} \cdot \frac{d \mathbf{r}}{d t} \times \frac{d^2 \mathbf{r}}{d t^2} \). This is another triple scalar product involving derivatives of the same vector \( \mathbf{r} \).

Step 5: Differentiate using Product Rule (b)

Differentiate the given expression: \[ \frac{d}{dt}\left(\mathbf{r} \cdot \frac{d \mathbf{r}}{d t} \times \frac{d^2 \mathbf{r}}{d t^2}\right) = \frac{d\mathbf{r}}{dt} \cdot \left( \frac{d\mathbf{r}}{dt} \times \frac{d^2\mathbf{r}}{dt^2} \right) + \mathbf{r} \cdot \frac{d}{dt}\left(\frac{d\mathbf{r}}{dt} \times \frac{d^2\mathbf{r}}{dt^2}\right) \].

Step 6: Apply the Vector Product Rule (b)

Use the vector product rule on the derivative of the cross product: \[ \frac{d}{dt}\left(\frac{d\mathbf{r}}{dt} \times \frac{d^2\mathbf{r}}{dt^2}\right) = \frac{d^2\mathbf{r}}{dt^2} \times \frac{d^2\mathbf{r}}{dt^2} + \frac{d\mathbf{r}}{dt} \times \frac{d^3\mathbf{r}}{dt^3} \]. Note that any vector crossed with itself is zero, thus simplifying to \[ \mathbf{r} \cdot \left( \frac{d\mathbf{r}}{dt} \times \frac{d^3\mathbf{r}}{dt^3} \right) \].

Step 7: Conclude on Simplification Zero Products (b)

The first term \( \frac{d\mathbf{r}}{dt} \cdot \left( \frac{d\mathbf{r}}{dt} \times \frac{d^2\mathbf{r}}{dt^2} \right) \) is zero because a vector dotted with its cross product with another vector is zero. This leads to the simplified result: \[ \mathbf{r} \cdot \left( \frac{d\mathbf{r}}{dt} \times \frac{d^3\mathbf{r}}{dt^3} \right) \].

Key Concepts

Vector Calculus

Vector calculus is a branch of mathematics that deals with vector fields and functions. In this exercise, we explore the differentiation of a triple scalar product, which involves vectors in three-dimensional space. Understanding vector calculus is essential because it provides the groundwork for physics and engineering concepts that deal with forces, acceleration, and more.

In a triple scalar product, you combine three vectors using a dot and cross product. For example, in the expression \( \mathbf{u} \cdot (\mathbf{v} \times \mathbf{w}) \), \( \mathbf{v} \times \mathbf{w} \) forms a new vector that is perpendicular to both \( \mathbf{v} \) and \( \mathbf{w} \). Then, \( \mathbf{u} \cdot (\mathbf{v} \times \mathbf{w}) \) results in a scalar value, which is the volume of the parallelepiped formed by the vectors.

Key operations in vector calculus include:

• Dot Product: Measures the extent to which two vectors point in the same direction.
• Cross Product: Gives a vector that is perpendicular to the plane containing the two input vectors.

These operations are crucial in manipulating and understanding how different vectors interact in space.

Product Rule in Calculus

The product rule is a fundamental concept in calculus that allows us to find the derivative of a product of two functions. When applied to vector functions, it becomes essential for dealing with complexities of vector differentiation as seen in this exercise.

For the usual product of two functions \( f(t) \) and \( g(t) \), the rule states:
\[ \frac{d}{dt}[f(t)g(t)] = f'(t)g(t) + f(t)g'(t) \]
In the context of the triple scalar product from the exercise, the product rule extends to handle vector functions:

• Differentiate each component of the scalar triple product separately.
• Reassemble the derivative by utilizing the results from differentiating individual parts.

This approach is particularly useful because:

• Vectors have both magnitude and direction, thus requiring careful differentiation of each aspect.
• Understanding this helps with complex problems in dynamics and electromagnetic theory where terms are often expressed as products of multiple functions.

Cross Product Differentiation

Cross product differentiation involves understanding how to handle derivatives of vector cross products. It's an extended application of the product rule to vectors and this concept is pivotal for operations involving vector fields. In the given exercise, the differentiation of cross products is a critical step.

When differentiating \( \mathbf{v} \times \mathbf{w} \), use: \[ \frac{d}{dt}(\mathbf{v} \times \mathbf{w}) = \frac{d\mathbf{v}}{dt} \times \mathbf{w} + \mathbf{v} \times \frac{d\mathbf{w}}{dt} \]

• First term represents the rate of change of \( \mathbf{v} \) and how it affects the cross product.
• Second term accounts for changes in \( \mathbf{w} \).

This step-by-step method ensures that each element of the cross product is accurately differentiated, maintaining the vector properties throughout the calculation.

Important considerations include:

• The order of vectors in a cross product affects the sign of the result (anti-commutative property).
• A vector crossed with itself is zero, simplifying calculations where such terms arise.

Understanding these principles is vital not only in calculus but in fields like physics where motion, rotation, and other vector behaviors are analyzed.