Question

Prove the property. In each case, assume that \(\mathbf{r}, \mathbf{u},\) and \(\mathbf{v}\) are differentiable vector-valued functions of \(t,\) \(f\) is a differentiable real-valued function of \(t,\) and \(c\) is a scalar. $$ D_{t}[\mathbf{r}(f(t))]=\mathbf{r}^{\prime}(f(t)) f^{\prime}(t) $$

Short answer

Given exercise is proven. Derivative of a vector-valued function composed with a real-valued function is indeed the derivative of the vector-valued function evaluated at the real-valued function times the derivative of the real-valued function.

Step-by-step solution

Understanding the Chain Rule

Recall the chain rule, which states that the derivative of a composition of functions is the derivative of the outer function evaluated at the inner function times the derivative of the inner function. In this case, the vector-valued function \( \mathbf{r}(t) \) is the outer function and the real-valued function \( f(t) \) is the inner function.

Differentiate

We know that \( D_{t}[\mathbf{r}(f(t))] \) means the derivative of the vector-valued function \( \mathbf{r}(t) \), evaluated at \( f(t) \). Apply the chain rule to find this derivative: the derivative of \( \mathbf{r} \) at \( f(t) \) times the derivative of \( f(t) \). This gives us \( \mathbf{r}^{\prime}(f(t)) f^{\prime}(t) \).

Statement Verification

The calculus performed demonstrates that \( D_{t}[\mathbf{r}(f(t))]=\mathbf{r}^{\prime}(f(t)) f^{\prime}(t) \). This successfully proves the given property using the principle of Chain Rule.

Key concepts

Vector-Valued Functions Differentiation

When dealing with vector-valued functions, it's essential to understand how to differentiate them in the context of vector calculus. A vector-valued function assigns a vector to each point in its domain, which in many cases is a subset of real numbers. The differentiation of these functions extends the concept of the derivative from real-valued functions to functions that take vectors as inputs or outputs.

For example, consider the function \( \mathbf{r}(t) \), which maps a real number \( t \) to a vector in a space. To find the derivative of \( \mathbf{r}(t) \) with respect to \( t \)—written as \( \mathbf{r}'(t) \) or \(\frac{\text{d}\mathbf{r}}{\text{d}t} \)—we take the derivative of each of the function's component functions individually, with regards to \( t \).

This means if \( \mathbf{r}(t) = \langle r_1(t), r_2(t), ..., r_n(t) \rangle \) in an \( n \) dimensional space, then \( \mathbf{r}'(t) = \langle r_1'(t), r_2'(t), ..., r_n'(t) \rangle \). This concept is crucial for understanding the dynamics of physical systems, among other applications, where the position, velocity, and acceleration of objects are vector-valued functions of time.

Application of the Chain Rule

The Chain Rule is a fundamental tool in calculus, enabling us to differentiate a function of a function. Its application is not limited to real scalar functions, but also extends to vector-valued functions. When we deal with the composition of a vector-valued function \( \mathbf{r}(t) \) and a real-valued function \( f(t) \), we're interested in the rate of change of \( \mathbf{r} \) with respect to \( t \) when \( t \) itself changes.

The Chain Rule tells us that to differentiate \( \mathbf{r}(f(t)) \) with respect to \( t \), we need to multiply the derivative of \( \mathbf{r} \) with respect to its argument—here, that's \( f(t) \), which we represent as \( \mathbf{r}'(f(t)) \)—by the derivative of \( f \) with respect to \( t \)—denoted as \( f'(t) \). In formula terms, the Chain Rule for this composition is written as \[ D_{t}[\mathbf{r}(f(t))] = \mathbf{r}'(f(t))f'(t) \].

Understanding this concept is crucial when analyzing complex systems where one quantity depends on another through some function, as often found in physics and engineering contexts.

Differentiable Functions

Differentiability is a property of a function that allows it to be linearly approximated at any point in its domain. For a real-valued function \( f(t) \), being differentiable at a point \( t \) means the function's graph has a tangent at that point, and therefore, the derivative \( f'(t) \) exists.

A vector-valued function \( \mathbf{r}(t) \) is said to be differentiable if each of its component functions is differentiable. In the exercise, \( \mathbf{r} \) and \( f \) are assumed to be differentiable, which means that near any point \( t \) we can find linear approximations that closely reflect the behavior of \( \mathbf{r} \) and \( f \) around \( t \). This condition being met is essential for the Chain Rule to be applicable, as non-differentiable functions can have discontinuities or ‘sharp turns’ at points, invalidating linear approximations.

Differentiability is thus a cornerstone concept in calculus because it ensures smooth transitions and predictable behavior—paving the way for the application of various calculus tools, such as optimization and integration, to real-world problems.