Problem 48 - Calculus Early Transcendentals

Question

Find an expression for $$ \frac{d}{d t}[\mathbf{u}(t) \cdot(\mathbf{v}(t) \times \mathbf{w}(t))] $$

Step-by-Step Solution

Answer

Use the product and cross product rules to differentiate the expression.

Step 1: Understand the Problem

We need to differentiate the expression \( \mathbf{u}(t) \cdot (\mathbf{v}(t) \times \mathbf{w}(t)) \) with respect to time \( t \). This involves both dot and cross products of time-dependent vectors.

Step 2: Apply Derivative Rule

We'll apply the product rule for differentiation, since the expression involves a time-dependent dot and cross product of vectors. The general rule is: \( \frac{d}{dt}[\mathbf{a}(t) \cdot \mathbf{b}(t)] = \frac{d\mathbf{a}}{dt} \cdot \mathbf{b} + \mathbf{a} \cdot \frac{d\mathbf{b}}{dt} \).

Step 3: Differentiate Cross Product

Differentiate \( \mathbf{v}(t) \times \mathbf{w}(t) \) using: \( \frac{d}{dt}[\mathbf{v}(t) \times \mathbf{w}(t)] = \frac{d\mathbf{v}}{dt} \times \mathbf{w} + \mathbf{v} \times \frac{d\mathbf{w}}{dt} \).

Step 4: Apply Derivative to Original Expression

Now, substitute the result from Step 3 into the product rule application:\[ \frac{d}{dt}[\mathbf{u}(t) \cdot (\mathbf{v}(t) \times \mathbf{w}(t))] = \frac{d\mathbf{u}}{dt} \cdot (\mathbf{v} \times \mathbf{w}) + \mathbf{u} \cdot \bigg( \frac{d\mathbf{v}}{dt} \times \mathbf{w} + \mathbf{v} \times \frac{d\mathbf{w}}{dt} \bigg) \].

Step 5: Simplify the Expression

Combine terms in Step 4 to obtain the final expression:\( \frac{d}{dt}[\mathbf{u}(t) \cdot (\mathbf{v}(t) \times \mathbf{w}(t))] = \frac{d\mathbf{u}}{dt} \cdot (\mathbf{v} \times \mathbf{w}) + \mathbf{u} \cdot (\frac{d\mathbf{v}}{dt} \times \mathbf{w}) + \mathbf{u} \cdot (\mathbf{v} \times \frac{d\mathbf{w}}{dt}) \).

Key Concepts

Derivatives of Vector Functions

Vector functions are expressions that associate a vector with each point in its domain, often dependent on a parameter like time, \( t \). When we talk about the derivative of a vector function, we're essentially looking at how this vector changes with respect to the parameter. Imagine you have a vector that points in different directions at different times—its derivative tells us how quickly and in what direction it's shifting as time progresses.

• The derivative of a vector function \( \mathbf{r}(t) \) is a vector that indicates the direction and speed of \( \mathbf{r}(t) \) as \( t \) changes.
• For a vector function \( \mathbf{a}(t) = [a_1(t), a_2(t), a_3(t)] \), its derivative \( \frac{d\mathbf{a}}{dt} \) is \( [\frac{da_1}{dt}, \frac{da_2}{dt}, \frac{da_3}{dt}] \).

Understanding these derivatives is crucial for physics and engineering where vectors often describe paths, forces, or fields. By breaking down the vector into its components, one can apply standard differentiation rules to each component independently, simplifying the process.

Product Rule for Differentiation

When dealing with vector functions, you might have to differentiate expressions involving multiple functions multiplied together. This is where the product rule comes into play. The product rule is a formula that allows us to differentiate such products efficiently. The product rule for scalar functions is familiar: \( \frac{d}{dt}(f(t)g(t)) = \frac{df}{dt}g + f\frac{dg}{dt} \). For vector functions, a similar form exists.

• If \( \mathbf{a}(t) \) and \( \mathbf{b}(t) \) are vector functions, the derivative of their dot product is: \( \frac{d}{dt}[\mathbf{a}(t) \cdot \mathbf{b}(t)] = \frac{d\mathbf{a}}{dt} \cdot \mathbf{b}(t) + \mathbf{a}(t) \cdot \frac{d\mathbf{b}}{dt} \).

The product rule helps us manage the complexity of these vector operations in calculus, ensuring that we account for changes in each function separately before bringing them together in the final expression.

Cross Product Differentiation

The cross product in vector calculus results in a vector perpendicular to the plane created by two vectors. Differentiating a cross product is somewhat more complex than differentiating scalar or dot products because it involves spatial orientation.To differentiate a cross product \( \mathbf{v}(t) \times \mathbf{w}(t) \), we use a rule similar to the product rule.

• The derivative rule for a cross product is: \( \frac{d}{dt}[\mathbf{v}(t) \times \mathbf{w}(t)] = \frac{d\mathbf{v}}{dt} \times \mathbf{w}(t) + \mathbf{v}(t) \times \frac{d\mathbf{w}}{dt} \).

This rule demonstrates that the derivative of a cross product combines the rates of change of both vectors into a new perpendicular vector. It's an extension of the product rule, adapted for the geometric nature of the cross product. By applying this rule, we maintain the orientation and magnitude relationships dictated by the original vectors, reflecting how their combined rate of change functions in a spatial context.