Problem 83 - Essential Calculus Early Transcendentals

Question

If $$\mathbf{u}(t)=\mathbf{r}(t) \cdot\left[\mathbf{r}^{\prime}(t) \times \mathbf{r}^{\prime \prime}(t)\right],$$ show that $$ \mathbf{u}^{\prime}(t)=\mathbf{r}(t) \cdot\left[\mathbf{r}^{\prime}(t) \times \mathbf{r}^{\prime \prime \prime}(t)\right] $$

Step-by-Step Solution

Answer

Differentiating using the product and cross product rules shows \( \mathbf{u}^{\prime}(t) = \mathbf{r}(t) \cdot \left[ \mathbf{r}^{\prime}(t) \times \mathbf{r}^{\prime \prime \prime}(t) \right] \).

Step 1: Apply the product rule to differentiate

First, we need to differentiate \( \mathbf{u}(t) = \mathbf{r}(t) \cdot \left[ \mathbf{r}^{\prime}(t) \times \mathbf{r}^{\prime \prime}(t) \right] \) with respect to \( t \). Apply the product rule for differentiation, which states that if \( f(t) = g(t) \cdot h(t) \), then \( f^{\prime}(t) = g^{\prime}(t) \cdot h(t) + g(t) \cdot h^{\prime}(t) \). Here, \( g(t) = \mathbf{r}(t) \) and \( h(t) = \mathbf{r}^{\prime}(t) \times \mathbf{r}^{\prime \prime}(t) \).

Step 2: Differentiate \( \mathbf{r}(t) \)

The derivative of \( \mathbf{r}(t) \) with respect to \( t \) is simply \( \mathbf{r}^{\prime}(t) \). This means the first term from the product rule is \( \mathbf{r}^{\prime}(t) \cdot \left[ \mathbf{r}^{\prime}(t) \times \mathbf{r}^{\prime \prime}(t) \right] \).

Step 3: Differentiate the cross product \( \left[ \mathbf{r}^{\prime}(t) \times \mathbf{r}^{\prime \prime}(t) \right] \)

The derivative of the cross product \( \mathbf{r}^{\prime}(t) \times \mathbf{r}^{\prime \prime}(t) \) is obtained by applying the product rule for cross products: \( \left( \mathbf{a} \times \mathbf{b} \right)^{\prime} = \mathbf{a}^{\prime} \times \mathbf{b} + \mathbf{a} \times \mathbf{b}^{\prime} \). So, we have:\[ \left( \mathbf{r}^{\prime}(t) \times \mathbf{r}^{\prime \prime}(t) \right)^{\prime} = \mathbf{r}^{\prime \prime}(t) \times \mathbf{r}^{\prime \prime}(t) + \mathbf{r}^{\prime}(t) \times \mathbf{r}^{\prime \prime \prime}(t) \].Note that \( \mathbf{r}^{\prime \prime}(t) \times \mathbf{r}^{\prime \prime}(t) = \mathbf{0} \), so it simplifies to \( \mathbf{r}^{\prime}(t) \times \mathbf{r}^{\prime \prime \prime}(t) \).

Step 4: Combine the results

Now substitute back in the product rule. The original differentiation gave:\[ \mathbf{u}^{\prime}(t) = \mathbf{r}^{\prime}(t) \cdot \left[ \mathbf{r}^{\prime}(t) \times \mathbf{r}^{\prime \prime}(t) \right] + \mathbf{r}(t) \cdot \left[ \mathbf{r}^{\prime}(t) \times \mathbf{r}^{\prime \prime \prime}(t) \right] \]Upon simplifying:- The first term \( \mathbf{r}^{\prime}(t) \cdot \left[ \mathbf{r}^{\prime}(t) \times \mathbf{r}^{\prime \prime}(t) \right] = 0 \) because the dot product of any vector with the cross product of itself and another vector is zero.- Therefore, we have \( \mathbf{u}^{\prime}(t) = \mathbf{r}(t) \cdot \left[ \mathbf{r}^{\prime}(t) \times \mathbf{r}^{\prime \prime \prime}(t) \right] \).

Key Concepts

Product Rule

The product rule is a fundamental tool in calculus for differentiating products of functions. It is especially useful when handling complex functions composed of multiple terms multiplied together. Simply put, if you have two differentiable functions, say \( g(t) \) and \( h(t) \), the product rule tells you how to find the derivative of their product. The product rule formula goes like this:

• If \( f(t) = g(t) \cdot h(t) \), then the derivative, \( f^{\prime}(t) \), is given by:
• \( f^{\prime}(t) = g^{\prime}(t) \cdot h(t) + g(t) \cdot h^{\prime}(t) \).

In our exercise scenario, this means breaking down the function \( \mathbf{u}(t) = \mathbf{r}(t) \cdot \left[ \mathbf{r}^{\prime}(t) \times \mathbf{r}^{\prime\prime}(t) \right] \) into two parts. Here, \( g(t) = \mathbf{r}(t) \) and \( h(t) = \mathbf{r}^{\prime}(t) \times \mathbf{r}^{\prime\prime}(t) \). Therefore, by applying the product rule, we can systematically consider the contributions from both \( \mathbf{r}(t) \) and the cross product \( \mathbf{r}^{\prime}(t) \times \mathbf{r}^{\prime\prime}(t) \) as they change with time.

Cross Product

The cross product is a mathematical operation that takes two vectors in three-dimensional space and returns a new vector that is orthogonal (perpendicular) to the plane formed by the original vectors. This operation is symbolically represented as \( \mathbf{a} \times \mathbf{b} \), where \( \mathbf{a} \) and \( \mathbf{b} \) are the original vectors.The cross product has the following properties:

• Orthogonality: The resulting vector is perpendicular to both \( \mathbf{a} \) and \( \mathbf{b} \).
• Magnitude: The magnitude of the cross product vector is equal to the area of the parallelogram that the original vectors span, calculated as \( \|\mathbf{a}\| \times \|\mathbf{b}\| \times \sin(\theta) \), where \( \theta \) is the angle between \( \mathbf{a} \) and \( \mathbf{b} \).

In the context of our original problem, the cross product \( \mathbf{r}^{\prime}(t) \times \mathbf{r}^{\prime\prime}(t) \) involves differentiating a cross product. To do this, we extend the product rule to handle cross product terms. Importantly, one aspect highlighted in the solution is that when a vector is crossed with itself (e.g., \( \mathbf{r}^{\prime\prime}(t) \times \mathbf{r}^{\prime\prime}(t) \)), the result is always the zero vector.

Differentiation

Differentiation is the process of finding the derivative, which represents an instantaneous rate of change. In calculus, this is expressed as the derivative of a function with respect to a variable. This concept is crucial for analyzing motion and change in various systems.For a function \( y = f(x) \), the derivative is symbolically represented as \( \frac{dy}{dx} \). In vector calculus, we deal with vector functions such as \( \mathbf{r}(t) \), which depict trajectories or movements in space.
Within the exercise, differentiation hinges on two main parts: differentiating \( \mathbf{r}(t) \) and the cross product term.

• When differentiating the vector \( \mathbf{r}(t) \), the derivative is simply \( \mathbf{r}^{\prime}(t) \).
• For the cross product \( \mathbf{r}^{\prime}(t) \times \mathbf{r}^{\prime\prime}(t) \), differentiation involves individually differentiating each vector function concerned in the product, as outlined by the product (and chain) rules discussed earlier.

By systematically applying these rules, we unravel the changes brought on by \( \mathbf{r}(t) \) alongside those brought on by its derivatives.

Dot Product

The dot product, also known as the scalar product, is another important operation in vector calculus. It combines two vectors to yield a scalar quantity, which is useful for understanding geometric and physical properties like projections and angles between vectors.The dot product is defined as:

• \( \mathbf{a} \cdot \mathbf{b} = \|\mathbf{a}\| \|\mathbf{b}\| \cos(\theta) \)

Here, \( \|\mathbf{a}\| \) and \( \|\mathbf{b}\| \) are the magnitudes of vectors \( \mathbf{a} \) and \( \mathbf{b} \), and \( \theta \) is the angle between them. Unlike the cross product, the result is not a vector but a single number.
Within the given problem, the dot product \( \mathbf{r}(t) \cdot \left[ \mathbf{r}^{\prime}(t) \times \mathbf{r}^{\prime\prime}(t) \right] \) is used to combine the vector \( \mathbf{r}(t) \) with the cross product of its derivatives. A key insight from the solution is recognizing scenarios where the dot product results in zero, such as when a vector is involved in a cross product with itself before taking a dot product.