Question

Find the function \(\mathbf{r}\) that satisfies the given conditions. $$\mathbf{r}^{\prime}(t)=\left\langle 1,2 t, 3 t^{2}\right\rangle ; \mathbf{r}(1)=\langle 4,3,-5\rangle$$

Short answer

Question: Determine the vector function \(\mathbf{r}(t)\) given its first derivative \(\mathbf{r}^{\prime}(t) = \langle 1, 2t, 3t^2 \rangle\) and the initial condition \(\mathbf{r}(1) = \langle 4, 3, -5 \rangle\). Answer: The vector function \(\mathbf{r}(t)\) is given by \(\mathbf{r}(t) = \langle t+3, t^2+2, t^3-6 \rangle\).

Step-by-step solution

Integrate the First Derivative to Find Vector Function r(t) Without Constants

To find the vector function \(\mathbf{r}(t)\), we need to integrate each component of the first derivative individually: $$\mathbf{r}(t) = \int \mathbf{r}^{\prime}(t) \, dt=\int \left\langle 1,2t,3t^2 \right\rangle \, dt = \left\langle \int 1 \, dt, \int 2t \, dt, \int 3t^2 \, dt \right\rangle$$ Integrate each component and introduce the constants \(C_1\), \(C_2\), and \(C_3\): $$\mathbf{r}(t) = \left\langle t+C_1, t^2+C_2, t^3+C_3 \right\rangle$$

Apply the Initial Condition to Find the Constants

The given initial condition is \(\mathbf{r}(1)=\langle 4,3,-5\rangle\). Substitute \(t=1\) into the expression for \(\mathbf{r}(t)\), set it equal to the initial condition, and solve for \(C_1\), \(C_2\), and \(C_3\): $$\langle 1+C_1, 1+C_2, 1+C_3 \rangle =\langle 4,3,-5\rangle$$ From the equation, we get the system of equations: $$\begin{cases} 1 + C_1 = 4 \\ 1 + C_2 = 3 \\ 1 + C_3 = -5 \end{cases}$$ Solving the system of equations, we find \(C_1 = 3\), \(C_2 = 2\), and \(C_3 = -6\).

Write the Final Expression for the Vector Function r(t)

Substitute the constants \(C_1\), \(C_2\), and \(C_3\) back into the expression for \(\mathbf{r}(t)\) to obtain the final expression: $$\mathbf{r}(t) = \left\langle t+3, t^2+2, t^3-6 \right\rangle$$

Key concepts

Vector Calculus

Vector calculus is an essential branch of mathematics that deals with differential and integral calculus of vector fields. It primarily focuses on vector functions, which are mathematical expressions that assign a vector to each point in a subset of space.
In our exercise, we encountered the derivative of a vector function, represented as \( \mathbf{r}'(t) = \langle 1, 2t, 3t^2 \rangle \) where \( t \) is the variable. To find the original vector function, we integrate each component separately. This concept is fundamental as it extends the idea of integrating scalar functions to vectors, allowing for operations on multi-dimensional spaces.
For students learning vector calculus, envision vector quantities as arrows with direction and magnitude, evolving as \( t \) changes. Picture the integration process as a reverse-engineering task—simplifying and understanding complex motion into basic components that describe position, velocity, and acceleration.

Initial Condition

The initial condition in calculus plays a vital role in finding the specific solution to differential equations. It's like setting a starting point or providing a snapshot of the system at a particular moment. In vector calculus, the initial condition can determine a unique position vector in a field of possibilities.
Our problem presents the initial condition \( \mathbf{r}(1) = \langle 4, 3, -5 \rangle \), effectively pinning down the function at \( t = 1 \) and allowing us to find the particular constants in the general solution. When solving similar problems, always ensure to plug in the given initial conditions precisely, as this step secures the accuracy and uniqueness of the solution.

Antiderivative

An antiderivative, in the simplest terms, is a function that

Parameterization of Curves

Parameterization is the process of expressing a curve as a vector function of a parameter, usually designated as \( t \). This powerful concept allows us to describe complex curves and surfaces within three-dimensional space using simpler, one-dimensional variables.
In our textbook problem, we utilized the parameter \( t \) to express each component of the vector function \( \mathbf{r}(t) \) as a function of time. This parameterization is particularly important in physics and engineering, as it can describe the trajectory of an object over time. To become adept at parameterization, practice by visualizing how the components of the vector function change as the parameter varies, and you'll gain a deeper understanding of spatial movement and curves.