Problem 27

Question

In Problems, find the indicated derivative. Assume that all vector functions are differentiable. $$ \frac{d}{d t}\left[\mathbf{r}(t) \times \mathbf{r}^{\prime}(t)\right] $$

Step-by-Step Solution

Verified

Answer

The derivative is $ \mathbf{r}(t) \times \mathbf{r}''(t) $.

1Step 1: Identify the Product Rule for Cross Products

To find the derivative of the cross product $ \mathbf{r}(t) \times \mathbf{r}'(t) $, we need to apply the product rule specific to cross products. The product rule for the derivative of a cross product $ \mathbf{u}(t) \times \mathbf{v}(t) $ is given by $ \mathbf{u}'(t) \times \mathbf{v}(t) + \mathbf{u}(t) \times \mathbf{v}'(t) $.

2Step 2: Apply the Product Rule

Identify the functions in the cross product: $ \mathbf{u}(t) = \mathbf{r}(t) $ and $ \mathbf{v}(t) = \mathbf{r}'(t) $. Find their derivatives: $ \mathbf{u}'(t) = \mathbf{r}'(t) $ and $ \mathbf{v}'(t) = \mathbf{r}''(t) $. Then, the derivative is $ \mathbf{r}'(t) \times \mathbf{r}'(t) + \mathbf{r}(t) \times \mathbf{r}''(t) $.

3Step 3: Simplify the Expression

Notice that $ \mathbf{r}'(t) \times \mathbf{r}'(t) = \mathbf{0} $ because the cross product of any vector with itself is zero. Thus, the expression simplifies to $ \mathbf{r}(t) \times \mathbf{r}''(t) $.

Key Concepts

Differentiable Vector FunctionsCross Product DerivativeProduct Rule for Cross ProductsDifferentiation in Vector Calculus

Differentiable Vector Functions

In vector calculus, a vector function describes vectors related to a parameter, commonly time $ t $. A differentiable vector function implies that this function has a derivative at each point of its domain. This means we can compute the derivative with respect to $ t $, which gives us another vector function. Differential calculus for vector functions extends the idea of differentiating scalar functions to vector-valued functions. Just as with scalar calculus, the derivative of a vector function $ \mathbf{r}(t) $ is expressed as $ \mathbf{r}'(t) $, which provides the rate of change at any point $ t $ of the vector's components. These derivatives reveal important information about the behavior and characteristics of the vector-valued functions. Understanding the differentiability of vector functions allows us to apply advanced operations like the product rule, and is essential in physics and engineering where vectors describe quantities like velocity and force.

Cross Product Derivative

The concept of cross product, or vector product, involves constructing a vector that is perpendicular to two given vectors. The cross product of two vectors $ \mathbf{a} $ and $ \mathbf{b} $ is denoted $ \mathbf{a} \times \mathbf{b} $. When taking the derivative of a cross product, we explore how this new vector behaves as both involved vectors vary with a parameter, such as time. The differentiation of a cross product occurs by using specific rules due to its unique properties. A key property is that the cross product of any vector with itself equals zero. This concept was crucial in simplifying the solution from the original exercise. Understanding the derivative of the cross product helps in analyzing the rotational behavior of vectors, which has practical applications in physics and computer graphics.

Product Rule for Cross Products

The product rule is a fundamental tool in calculus for differentiating products of two functions. For cross products in vector calculus, the rule adapts to suit this unique operation. When applying the derivative of the cross product $ \mathbf{u}(t) \times \mathbf{v}(t) $, we turn to the adapted product rule:

Differentiate $ \mathbf{u}(t) $ leaving $ \mathbf{v}(t) $ constant: $ \mathbf{u}'(t) \times \mathbf{v}(t) $
Add $ \mathbf{u}(t) $ and differentiate $ \mathbf{v}(t) $: $ \mathbf{u}(t) \times \mathbf{v}'(t) $

This results in:\[\frac{d}{dt}(\mathbf{u}(t) \times \mathbf{v}(t)) = \mathbf{u}'(t) \times \mathbf{v}(t) + \mathbf{u}(t) \times \mathbf{v}'(t)\]This rule enables us to differentiate complex expressions involving cross products, which is necessary for comprehending how vector functions interact and change over time. The result enhances our capability to work with vector fields and understand physical phenomena like motion.

Differentiation in Vector Calculus

Differentiation in vector calculus extends the principles of calculus to vector-valued functions. Unlike scalar functions, these incorporate directionality in addition to magnitude. Derivatives of vector functions show how vectors change, both in size and orientation, with respect to a variable such as time. Vector calculus is a pivotal mathematical tool in fields like electromagnetism, fluid dynamics, and mathematical physics. Practically, it helps in determining key physical quantities like velocity, acceleration, and the net change in momentum. A significant challenge in this differentiation is handling operations like the dot product and cross product. Understanding how to differentiate these operations aids in tackling complex problems in physics and engineering, such as rotational dynamics and field theories.

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