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Vector-valued functions extend the concept of regular functions by mapping a variable, typically denoted as \( t \), to a vector instead of a single value. In mathematics, a vector is an entity that has both a magnitude and a direction. This is represented in the form of a sum of vectors along standard coordinate directions, like \( \mathbf{i} \), \( \mathbf{j} \), and \( \mathbf{k} \) which correspond to the \( x \)-, \( y \)-, and \( z \)-axes respectively.
For the given function \( \mathbf{r}(t) = t^2 \mathbf{i} + t e^{-2t} \mathbf{j} - 5 e^{-4t} \mathbf{k} \), each coordinate function — \( x(t) = t^2 \), \( y(t) = t e^{-2t} \), and \( z(t) = -5 e^{-4t} \) — maps to a specific component of the vector.

• \( \mathbf{i} \) represents the horizontal or \( x \)-direction.
• \( \mathbf{j} \) captures the vertical or \( y \)-direction.
• \( \mathbf{k} \) refers to the depth or \( z \)-direction.

These functions are crucial because they allow us to model dynamic systems in multiple dimensions, such as the trajectory of a moving object in space. Each component can behave quite differently, yet together they describe a single entity — the vector \( \mathbf{r}(t) \).

The process of finding the derivative of a vector-valued function involves deriving each of its component functions independently. The derivative of a vector-valued function \( \mathbf{r}(t) = \langle x(t), y(t), z(t) \rangle \) is determined by computing the derivatives \( x'(t) \), \( y'(t) \), and \( z'(t) \).
For matrix-vector functions, we use the standard rules of differentiation such as the power rule, product rule, and chain rule. Let's break down the task using the function \( \mathbf{r}(t) = t^2 \mathbf{i} + t e^{-2t} \mathbf{j} - 5 e^{-4t} \mathbf{k} \):

• For \( x(t) = t^2 \), use the power rule to find \( x'(t) = 2t \).
• For \( y(t) = t e^{-2t} \), apply the product rule \( \frac{d}{dt}(uv) = u'v + uv' \), resulting in \( y'(t) = e^{-2t} - 2te^{-2t} \).
• For \( z(t) = -5e^{-4t} \), differentiate to get \( z'(t) = 20e^{-4t} \).

Once each component's derivative is computed, they are combined into a new vector. This new vector describes how the original vector-valued function changes at instant \( t \).

Component differentiation deals with the individual differentiation of each coordinate of a vector-valued function. Each component is a function of the parameter \( t \), and differentiation is done with respect to \( t \):
The ease of understanding comes from handling one-dimensional problems for each dimension and then combining them. Here's how it works for the provided example:

• The \( x \)-component, \( x(t) = t^2 \), differentiates to \( 2t \) using the simple power rule.
• The \( y \)-component, \( y(t) = t e^{-2t} \), requires the product rule to result in \( e^{-2t} - 2te^{-2t} \). Here, each part (\( u = t \) and \( v = e^{-2t} \)) has its own derivative: \( u' = 1 \) and \( v' = -2e^{-2t} \).
• The \( z \)-component derivative, \( z(t) = -5e^{-4t} \), illustrates the direct application of the chain rule, resulting in \( 20e^{-4t} \).

By breaking down the vector-valued function into its coordinate functions, we simplify the complex procedure into manageable steps. Once calculated, these differentiated components inform us about the rate of change in each dimension, contributing to a full picture of motion or variation in the entire vector-valued function.