Problem 28 - Thomas Calculus

Question

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

Step-by-Step Solution

Answer

Both expressions verify the derivatives as given in the problem statement.

Step 1: Understand the Expression

We start by examining the triple scalar product: \( u \cdot (v \times w) \). Here, \( u, v, \) and \( w \) are vectors and functions of time \( t \). Our task is to find its derivative with respect to \( t \).

Step 2: Derivative of a Triple Scalar Product

When differentiating a product of vectors, use the product rule. Apply this rule to \( u \cdot (v \times w) \), by considering one vector fixed and differentiating the others one by one.

Step 3: Apply Product Rule to Vector Derivatives

Using the product rule for vectors, we differentiate each component: 1. Consider \( u \) as the derivative and \( v \times w \) unchanged. 2. Consider \( v \) as the derivative and \( u \) and \( w \) unchanged.3. Consider \( w \) as the derivative and \( u \) and \( v \) unchanged.

Step 4: Write the Full Derivative

Combine all components from Step 3, ensuring proper notation:\[ \frac{d}{dt}(u \cdot (v \times w)) = \frac{du}{dt} \cdot (v \times w) + u \cdot (\frac{dv}{dt} \times w) + u \cdot (v \times \frac{dw}{dt}) \]

Step 5: Simplify the Expression [Part a Solution]

Write down and verify the final expression obtained: The expression is correct as per the problem statement.

Step 6: Analyze the Second Expression

Next, consider the expression: \( r \cdot (\frac{dr}{dt} \times \frac{d^2r}{dt^2}) \). Apply the same principle to find its derivative.

Step 7: Apply Product Rule to Second Expression

Differentiating the expression using the product and chain rule:1. Focus on differentiating \( r \), keeping other parts constant.2. Analyze the derivatives of \( \frac{dr}{dt} \) and \( \frac{d^2r}{dt^2} \).

Step 8: Identify Vanishing Terms

Use the hint: look for vectors whose products yield zero. This includes the terms where vectors are orthogonal or one term's complete derivative leads to zero contribution.

Step 9: Verify and Conclude [Part b Solution]

Evaluate the expression: \[ \frac{d}{dt}(r \cdot (\frac{dr}{dt} \times \frac{d^2r}{dt^2})) = r \cdot (\frac{dr}{dt} \times \frac{d^3r}{dt^3}) \]Ensure no terms are left unjustified and verify if they align with expected mathematical results.

Key Concepts

Triple Scalar Product

The triple scalar product is a fascinating concept in vector calculus. It is denoted as \( \mathbf{u} \cdot (\mathbf{v} \times \mathbf{w}) \), involving three vectors \( \mathbf{u}, \mathbf{v}, \) and \( \mathbf{w} \). This expression computes the volume of a parallelepiped formed by the three vectors.
The operation within the parentheses, \( \mathbf{v} \times \mathbf{w} \), is a cross product that results in a vector perpendicular to both \( \mathbf{v} \) and \( \mathbf{w} \). This new vector is then dotted with \( \mathbf{u} \) to produce a scalar value.
The outcome is highly sensitive to the order of operations. Reversing the vector order in cross or dot products will alter the result, possibly changing its sign. The triple scalar product is also linked with the determinant of a matrix formed by the three vectors' components, simplifying volume calculations in 3D space.

Vector Calculus

Vector calculus extends calculus concepts to vector fields, crucial for understanding physical systems like electromagnetism or fluid flow. It involves operations across vector fields, including differentiation and integration.
Some pivotal operations in vector calculus include the gradient, divergence, and curl. The gradient points in the direction of greatest increase of a scalar function, while divergence measures the volumetric source of a vector field, and curl assesses its rotational motion.
When applying calculus operations to vectors, we need specific rules. For example, differentiation of vector functions is handled component-wise, often leveraging rules like the chain and product rule for differentiation. These concepts underpin many physical phenomena, providing a robust framework to describe changes in the multidimensional fields.

Product Rule for Differentiation

The product rule is integral for differentiating products of functions. In calculus, when dealing with simple functions \( f(x) \cdot g(x) \), the rule states that the derivative is \( \frac{d}{dx}[f(x) \cdot g(x)] = f'(x)g(x) + f(x)g'(x) \).
When translating this concept to vectors, especially in contexts like the triple scalar product, it ensures each component is considered separately. This means fixing one vector while differentiating others in turn. When solving the problem of differentiating \( \mathbf{u} \cdot (\mathbf{v} \times \mathbf{w}) \), apply the product rule component-wise:

• Differentiate \( \mathbf{u} \) while holding \( \mathbf{v} \times \mathbf{w} \) constant.
• Differentiate \( \mathbf{v} \) while holding \( \mathbf{u} \) and \( \mathbf{w} \) constant.
• Lastly, differentiate \( \mathbf{w} \) while holding \( \mathbf{u} \) and \( \mathbf{v} \) constant.

By systematically applying the product rule, we ensure accurate computation of derivatives involving multiple vector functions, foundational in advanced calculus and physics.