Question

Determine the general form of \(A(t)\) by constructing antiderivatives as needed and imposing any given constraints.$$ A^{\prime}(t)=\left[\begin{array}{c} -1 \\ 2 t \end{array}\right] $$

Short answer

Question: Given the vector function A'(t) = [-1, 2t], find the general form of A(t). Answer: The general form of A(t) is A(t) = [-t + C_1, t^2 + C_2], where C_1 and C_2 are constants of integration.

Step-by-step solution

Antiderivative of individual components

First, we need to determine the antiderivative of each component of the vector function A'(t). The first component is -1, which has the antiderivative: $$ \int{-1}~dt = -t + C_1 $$ where C_1 is the constant of integration. The second component is 2t, which has the antiderivative: $$ \int{2t}~dt = t^2 + C_2 $$ where C_2 is the constant of integration.

Reassemble into a general form of A(t)

Now that we have found the antiderivative of each component, we can reassemble them into the general form of the vector function A(t): $$ A(t)= \left[\begin{array}{c} -t + C_1 \\\ t^2 + C_2 \end{array}\right] $$ This is the general form of A(t) using the provided derivative and no constraints have been given.

Key concepts

Antiderivatives

Antiderivatives, also known as indefinite integrals, are the reverse operation of differentiation. They help us determine the original function when given its derivative. If you start with a function's derivative and you want to find the original function, you compute its antiderivative.
Given a derivative function, say \( f'(x) \), the antiderivative is denoted by \( F(x) \) or \( \int f'(x) \, dx \). For example, if \( f'(x) = 2x \), then an antiderivative \( F(x) \) could be \( x^2 + C \), where \( C \) is a constant of integration. This process is applicable to each component of a vector function, such as in the problem given.

• The antiderivative of a constant \(-1\) is \(-t + C_1\).
• The antiderivative of \(2t\) is \(t^2 + C_2\).

Finding these antiderivatives allows us to reconstruct the original function from its derivative, step by step. They are the foundation for solving many problems involving differential equations.

Vector Functions

A vector function is a function where each input is associated with a vector output. These are commonly represented as combinations of functions that represent different dimensions. For example, a vector function in two dimensions might look like \( \textbf{A}(t) = \begin{pmatrix} x(t) \ y(t) \end{pmatrix} \), where \( x(t) \) and \( y(t) \) are real-valued functions that define the vector's components.
In this problem, we work with a vector function \( A(t) \) that is constructed from its derivative \( A'(t) \). This involves finding the antiderivatives of each component function:

• The first component, \(-1\), integrates to \(-t + C_1\).
• The second component, \(2t\), integrates to \(t^2 + C_2\).

The vector function \( A(t) \) is then the combination of these two antiderivatives as \( \begin{pmatrix} -t + C_1 \ t^2 + C_2 \end{pmatrix} \). This technique of working with vectors is crucial in fields like physics and engineering, where many quantities are best described as vectors.

Constants of Integration

When you find an antiderivative, it includes an arbitrary constant known as the constant of integration. This constant is essential because differentiating a constant gives zero, meaning that any antiderivative can differ by a constant. For a given derivative, there's not just one antiderivative, but rather, a family of functions differentiated only by different values of this constant.
Consider the antiderivative of a derivative function \( f'(x) \). The general antiderivative could be expressed as \( \int f'(x) \ dx = F(x) + C \). Each value of \( C \) represents a different member of the family of original functions.
In vector functions, each component function will have its own constant of integration, often denoted differently (such as \( C_1, C_2 \), etc.) to avoid confusion. In the case of \( A(t) \):

• \( -t + C_1 \) comes from the component \(-1\).
• \( t^2 + C_2 \) comes from the component \(2t\).

Understanding how to handle these constants is key in solving differential equations, as they can often be determined by initial conditions or constraints provided in a particular problem.