Problem 30 - Advanced Engineering Mathematics

Question

Find the indicated derivative. Assume that all vector functions are differentiable. $$ \frac{d}{d t}\left[\mathbf{r}_{1}(t) \times\left(\mathbf{r}_{2}(t) \times \mathbf{r}_{3}(t)\right)\right] $$

Step-by-Step Solution

Answer

Use the vector triple product identity and differentiate using the product rule.

Step 1: Understand the Vector Triple Product

The expression given involves a vector triple product \( \mathbf{r}_{1}(t) \times (\mathbf{r}_{2}(t) \times \mathbf{r}_{3}(t)) \). A vector triple product follows a specific expansion rule, often referred to as the vector triple product identity, which states \( \mathbf{a} \times (\mathbf{b} \times \mathbf{c}) = (\mathbf{a} \cdot \mathbf{c}) \mathbf{b} - (\mathbf{a} \cdot \mathbf{b}) \mathbf{c} \). This identity will help in differentiating the expression.

Step 2: Apply the Vector Triple Product Identity

Using the vector triple product identity, transform the expression: \( \mathbf{r}_{1}(t) \times (\mathbf{r}_{2}(t) \times \mathbf{r}_{3}(t)) = (\mathbf{r}_{1}(t) \cdot \mathbf{r}_{3}(t)) \mathbf{r}_{2}(t) - (\mathbf{r}_{1}(t) \cdot \mathbf{r}_{2}(t)) \mathbf{r}_{3}(t) \). Now you have a linear combination of vector functions which will be easier to differentiate.

Step 3: Differentiate the Linear Combination

Differentiate the expression obtained in Step 2 with respect to \( t \): \[ \frac{d}{dt} \left[ (\mathbf{r}_{1}(t) \cdot \mathbf{r}_{3}(t)) \mathbf{r}_{2}(t) - (\mathbf{r}_{1}(t) \cdot \mathbf{r}_{2}(t)) \mathbf{r}_{3}(t) \right]. \] Use the product rule for each term: \[ \frac{d}{dt}[(\mathbf{r}_{1} \cdot \mathbf{r}_{3}) \mathbf{r}_{2}] = \frac{d}{dt}(\mathbf{r}_{1} \cdot \mathbf{r}_{3}) \mathbf{r}_{2} + (\mathbf{r}_{1} \cdot \mathbf{r}_{3}) \frac{d}{dt} \mathbf{r}_{2}, \] \[ \frac{d}{dt}[ - (\mathbf{r}_{1} \cdot \mathbf{r}_{2}) \mathbf{r}_{3}] = -\frac{d}{dt}(\mathbf{r}_{1} \cdot \mathbf{r}_{2}) \mathbf{r}_{3} - (\mathbf{r}_{1} \cdot \mathbf{r}_{2}) \frac{d}{dt} \mathbf{r}_{3}. \]

Step 4: Compute the Dot Product Derivatives

Calculate the derivatives of the dot products: \[ \frac{d}{dt}(\mathbf{r}_{1} \cdot \mathbf{r}_{3}) = \frac{d}{dt}\mathbf{r}_{1} \cdot \mathbf{r}_{3} + \mathbf{r}_{1} \cdot \frac{d}{dt}\mathbf{r}_{3}, \] \[ \frac{d}{dt}(\mathbf{r}_{1} \cdot \mathbf{r}_{2}) = \frac{d}{dt}\mathbf{r}_{1} \cdot \mathbf{r}_{2} + \mathbf{r}_{1} \cdot \frac{d}{dt}\mathbf{r}_{2}. \] Substitute these into the expressions from Step 3.

Step 5: Substitute and Simplify

Substitute the dot product derivatives into the expression from Step 3 and simplify: \[ \frac{d}{dt}[\mathbf{r}_{1}(t) \times (\mathbf{r}_{2}(t) \times \mathbf{r}_{3}(t))] = \left(\frac{d}{dt}\mathbf{r}_{1} \cdot \mathbf{r}_{3} + \mathbf{r}_{1} \cdot \frac{d}{dt}\mathbf{r}_{3}\right) \mathbf{r}_{2} + (\mathbf{r}_{1} \cdot \mathbf{r}_{3}) \frac{d}{dt} \mathbf{r}_{2} - \left(\frac{d}{dt}\mathbf{r}_{1} \cdot \mathbf{r}_{2} + \mathbf{r}_{1} \cdot \frac{d}{dt}\mathbf{r}_{2}\right) \mathbf{r}_{3} - (\mathbf{r}_{1} \cdot \mathbf{r}_{2}) \frac{d}{dt} \mathbf{r}_{3}. \]

Key Concepts

Vector Triple Product

Dealing with the vector triple product might seem tricky at first, but it becomes straightforward when you know the right identity to use. The vector triple product involves three vectors, and the expression typically looks like \( \mathbf{a} \times (\mathbf{b} \times \mathbf{c}) \). This isn't the same as just multiplying straight across. Instead, it has a special expansion rule, known as the vector triple product identity.

• This identity states: \( \mathbf{a} \times (\mathbf{b} \times \mathbf{c}) = (\mathbf{a} \cdot \mathbf{c}) \mathbf{b} - (\mathbf{a} \cdot \mathbf{b}) \mathbf{c} \).
• What this means is that instead of dealing with three vectors in a product, you can reduce it to two terms. Each term consists of one vector and a dot product between two others.

This simplification is incredibly useful for differentiation, as the problem turns into a linear combination of vector functions. This transformation makes it much easier to apply calculus rules.

Derivative of Vector Functions

Understanding how to take the derivative of vector functions is key in vector calculus. A vector function takes a scalar variable (like time \( t \)) and returns a vector. When differentiating vector functions, it's important to do it component-wise.

• Each part of the vector function is treated independently under differentiation.
• If \( \mathbf{r}(t) \) is a vector function, its derivative \( \frac{d}{dt} \mathbf{r}(t) \) is found by taking the derivative of each component.
• The derivative gives you a new vector that describes how the original vector changes over time.

To address the original problem, we split the vector expressions using the triple product identity into separate terms. Each term then involves finding the derivative of dot products or vector functions, both of which follow well-established differentiation rules.

Product Rule in Calculus

The product rule is a cornerstone of calculus used when differentiating products of two functions. For scalar functions, if \( u(t) \) and \( v(t) \) are functions of \( t \), the product rule states:

• \( \frac{d}{dt}[u(t) v(t)] = u'(t) v(t) + u(t) v'(t) \).
• Here, \( u'(t) \) is the derivative of \( u \), and \( v'(t) \) is the derivative of \( v \).

This rule can also extend to vector functions. When you have vector functions like in the given problem, each component of the vectors is differentiated separately using the product rule.
For example, when differentiating \((\mathbf{r}_{1}(t) \cdot \mathbf{r}_{3}(t)) \mathbf{r}_{2}(t)\), the derivative uses the product rule on the scalar dot product \(\mathbf{r}_{1}(t) \cdot \mathbf{r}_{3}(t)\) and the vector \(\mathbf{r}_{2}(t)\).It's crucial to respect the order of operations and the nature of dot products when using the product rule with vectors.