Problem 46 - Thomas Calculus

Question

Derivatives of triple scalar products a. Show that if \(\mathbf{u}, \mathbf{v},\) and \(\mathbf{w}\) are differentiable vector functor functons of \(t,\) then $$ \begin{array}{r}{\frac{d}{d t}(\mathbf{u} \cdot \mathbf{v} \times \mathbf{w})=\frac{d \mathbf{u}}{d t} \cdot \mathbf{v} \times \mathbf{w}+\mathbf{u} \cdot \frac{d \mathbf{v}}{d t} \times \mathbf{w}+} \\\ {\mathbf{u} \cdot \mathbf{v} \times \frac{d \mathbf{w}}{d t}}\end{array} $$ b. Show that Equation \((7)\) is equivalent to $$ \frac{d}{d t}\left|\begin{array}{ccc}{u_{1}} & {u_{2}} & {u_{3}} \\ {v_{1}} & {v_{2}} & {v_{3}} \\ {w_{1}} & {w_{2}} & {w_{3}}\end{array}\right|=\left|\begin{array}{ccc}{\frac{d u_{1}}{d t}} & {\frac{d u_{2}}{d t}} & {\frac{d u_{3}}{d t}} \\ {v_{1}} & {v_{2}} & {v_{3}} \\\ {w_{1}} & {w_{2}} & {w_{3}}\end{array}\right| $$ $$ +\left|\begin{array}{lll}{u_{1}} & {u_{2}} & {u_{3}} \\ {\frac{d v_{1}}{d t}} & {\frac{d v_{2}}{d t}} & {\frac{d v_{3}}{d t}} \\ {w_{1}} & {w_{2}} & {w_{3}}\end{array}\right| $$ $$ +\left|\begin{array}{lll}{u_{1}} & {u_{2}} & {u_{3}} \\ {v_{1}} & {v_{2}} & {v_{3}} \\ {\frac{d w_{1}}{d t}} & {\frac{d w_{2}}{d t}} & {\frac{d w_{3}}{d t}}\end{array}\right| $$

Step-by-Step Solution

Answer

The derivative of the triple scalar product is the sum of each partial derivative. This is shown using both vector calculus and determinant operations.

Step 1: Understanding the Triple Scalar Product

The triple scalar product of vectors \( \mathbf{u}, \mathbf{v}, \) and \( \mathbf{w} \) is given by \( \mathbf{u} \cdot (\mathbf{v} \times \mathbf{w}) \). This is a scalar quantity and involves both dot product and cross product operations.

Step 2: Applying the Product Rule to Derivatives

To differentiate \( \mathbf{u} \cdot (\mathbf{v} \times \mathbf{w}) \) with respect to \( t \), we use the product rule. The product rule for scalars states that the derivative of a product is the derivative of the first function times the second plus the first times the derivative of the second, adapted here to vector calculus.

Step 3: Calculating the Derivative of Triple Product

Apply the derivative to each component:- Differentiate \( \mathbf{u} \), keep \( \mathbf{v} \times \mathbf{w} \) constant: \( \frac{d\mathbf{u}}{dt} \cdot (\mathbf{v} \times \mathbf{w}) \).- Keep \( \mathbf{u} \) constant, differentiate \( \mathbf{v} \times \mathbf{w} \). Using the product rule for cross products: \( \mathbf{u} \cdot \left( \frac{d\mathbf{v}}{dt} \times \mathbf{w} + \mathbf{v} \times \frac{d\mathbf{w}}{dt} \right) \).

Step 4: Combine the Results

Combine all derived parts into one expression: \( \frac{d}{dt}( \mathbf{u} \cdot (\mathbf{v} \times \mathbf{w}) ) = \frac{d\mathbf{u}}{dt} \cdot (\mathbf{v} \times \mathbf{w}) + \mathbf{u} \cdot (\frac{d\mathbf{v}}{dt} \times \mathbf{w}) + \mathbf{u} \cdot (\mathbf{v} \times \frac{d\mathbf{w}}{dt}) \).

Step 5: Understand the Determinant Representation

The triple scalar product can also be represented as a determinant: \( \left| \begin{array}{ccc} u_1 & u_2 & u_3 \ v_1 & v_2 & v_3 \ w_1 & w_2 & w_3 \end{array} \right| \). This geometric interpretation relates the scalar product to areas and volumes.

Step 6: Differentiate the Determinant

Using matrix determinant properties, the derivative of each element affects only its row. Thus, differentiating each row individually gives the expanded determinant: \( \left| \begin{array}{ccc} \frac{du_1}{dt} & \frac{du_2}{dt} & \frac{du_3}{dt} \ v_1 & v_2 & v_3 \ w_1 & w_2 & w_3 \end{array} \right| + \left| \begin{array}{ccc} u_1 & u_2 & u_3 \ \frac{dv_1}{dt} & \frac{dv_2}{dt} & \frac{dv_3}{dt} \ w_1 & w_2 & w_3 \end{array} \right| + \left| \begin{array}{ccc} u_1 & u_2 & u_3 \ v_1 & v_2 & v_3 \ \frac{dw_1}{dt} & \frac{dw_2}{dt} & \frac{dw_3}{dt} \end{array} \right| \).

Step 7: Concluding the Equivalence

The explanation and calculations show that the derivative of a triple scalar product, both in vector and determinant form, conclude equivalently as: - Each component derivative is added to the expression, confirming the expanded sum of the partial derivatives.

Key Concepts

Derivative

In calculus, the derivative represents the rate at which a quantity changes. Differentiate a function to determine how it evolves over time. Derivatives in vector calculus, like those of scalars, follow specific rules, including the product rule. In this scenario, we apply derivatives to vector functions. Each vector can change over time, and our task is to track these changes.
When dealing with the triple scalar product, which involves a dot product (\(\cdot\)) and a cross product (\(\times\)), we need to use the modified product rule. It states the derivative of a combined product breaks into parts: one part focuses on differential of one vector while holding others constant.
In mathematical terms, the derivative (\(\frac{d}{dt}(\mathbf{u} \cdot (\mathbf{v} \times \mathbf{w})) \)) becomes the sum of three terms: (\(\frac{d\mathbf{u}}{dt} \cdot (\mathbf{v} \times \mathbf{w})\)), (\(\mathbf{u} \cdot (\frac{d\mathbf{v}}{dt} \times \mathbf{w})\)), and (\(\mathbf{u} \cdot (\mathbf{v} \times \frac{d\mathbf{w}}{dt})\)). This method allows for precise change tracking in multidimensional contexts.

Triple Scalar Product

The triple scalar product integrates the operations of both dot product and cross product in vector calculus. It takes three vectors (\(\mathbf{u}, \mathbf{v}, \mathbf{w}\)) and produces a scalar. This result is essentially a volume computation.
Think of (\(\mathbf{v} \times \mathbf{w}\)) as generating a vector orthogonal to the plane containing (\(\mathbf{v}\)) and (\(\mathbf{w}\)). When dot-multiplied by (\(\mathbf{u}\)), the triple scalar product assesses how (\(\mathbf{u}\)) aligns with this perpendicular direction. This alignment provides a measure of the three-dimensional spanned space.
Mathematically, the expression is represented as: (\( \mathbf{u} \cdot (\mathbf{v} \times \mathbf{w}) \)) and it's both useful and insightful in physics and engineering, especially in contexts dealing with torque and rotational motion.

Cross Product

The cross product is a fundamental operation in vector calculus with notable features. It involves two three-dimensional vectors and results in a third vector, perpendicular to both initial vectors.
To understand, consider vectors (\(\mathbf{v}\)) and (\(\mathbf{w}\)). Their cross product, (\(\mathbf{v} \times \mathbf{w}\)), creates a vector orthogonal to the plane formed by (\(\mathbf{v}\)) and (\(\mathbf{w}\)). The direction follows the right-hand rule.
The magnitude equals the area of the parallelogram formed by (\(\mathbf{v}\)) and (\(\mathbf{w}\)). This operation is crucial in determining torque, where force impacts rotational motion around an axis.
Mathematically, the cross product is defined for vectors (\(\mathbf{v} = (v_1, v_2, v_3)\)) and (\(\mathbf{w} = (w_1, w_2, w_3)\)) as: \[ \mathbf{v} \times \mathbf{w} = (v_2w_3 - v_3w_2, v_3w_1 - v_1w_3, v_1w_2 - v_2w_1) \].

Determinants

Determinants are a key concept in linear algebra, closely connected to matrix operations. A determinant of a matrix offers insights into geometry, such as area or volume, and properties such as invertibility of matrices.
For a triple scalar product, a determinant represents the scalar outcome of vectors (\(\mathbf{u}, \mathbf{v}, \mathbf{w}\)). Arranged in a matrix, their determinant equals the volume of a parallelepiped they form.
In the context of the exercise, visualizing the triple scalar product as a determinant aids in understanding its derivative. Each row shift corresponds to a differentiated vector, leading to efficient differentiation.
The basic property is: If (\(\mathbf{M}\)) is the matrix, its determinant (\(|\mathbf{M}|\)) is: \[ \left| \begin{array}{ccc} a & b & c \ d & e & f \ g & h & i \end{array} \right| = a(ei-fh) - b(di-fg) + c(dh-eg) \].
This structural understanding is intuitive in areas such as physics and engineering, assisting in streamlined solutions of vector problems.