Problem 37 - Advanced Engineering Mathematics

Question

Find a vector function \(\mathbf{r}\) that satisfies the indicated conditions. $$ \mathbf{r}^{\prime}(t)=6 \mathbf{i}+6 t \mathbf{j}+3 t^{2} \mathbf{k} ; \mathbf{r}(0)=\mathbf{i}-2 \mathbf{j}+\mathbf{k} $$

Step-by-Step Solution

Answer

\( \mathbf{r}(t) = (6t + 1) \mathbf{i} + (3t^2 - 2) \mathbf{j} + (t^3 + 1) \mathbf{k} \)

Step 1: Integrate the Vector

To find the vector function \(\mathbf{r}(t)\), start by integrating the vector derivative \(\mathbf{r}'(t) = 6 \mathbf{i} + 6t \mathbf{j} + 3t^2 \mathbf{k}\) with respect to \(t\). Integrate each component separately. The integration of \(6\ \mathbf{i}\) is \(6t + C_1\), for \(6t\ \mathbf{j}\) is \(3t^2 + C_2\), and for \(3t^2\ \mathbf{k}\) is \(t^3 + C_3\).

Step 2: Construct the General Solution

Utilize the integrations from Step 1 to form the general vector function: \(\mathbf{r}(t) = (6t + C_1)\ \mathbf{i} + (3t^2 + C_2)\ \mathbf{j} + (t^3 + C_3)\ \mathbf{k}\). Each constant \(C_1, C_2, C_3\) is a constant of integration that must be determined.

Step 3: Apply Initial Condition to Determine Constants

Use the given initial condition \(\mathbf{r}(0)=\mathbf{i}-2 \mathbf{j} + \mathbf{k}\) to find the constants \(C_1, C_2,\) and \(C_3\).Substituting \(t = 0\) into the general solution gives: \(\mathbf{r}(0) = (C_1)\ \mathbf{i} + (C_2)\ \mathbf{j} + (C_3)\ \mathbf{k} = \mathbf{i} - 2\mathbf{j} + \mathbf{k}\). Equating components, we get \(C_1 = 1\), \(C_2 = -2\), and \(C_3 = 1\).

Step 4: Write Final Vector Function

Substitute the values of \(C_1, C_2,\) and \(C_3\) back into the general solution. Thus, the vector function \(\mathbf{r}(t)\) is \(\mathbf{r}(t) = (6t + 1)\ \mathbf{i} + (3t^2 - 2)\ \mathbf{j} + (t^3 + 1)\ \mathbf{k}\).

Key Concepts

Vector Integration

Vector integration is a crucial aspect of vector calculus, often encountered in physics and engineering problems.
It involves finding a vector function from its derivative or gradient. In our problem, we begin with the derivative of a vector function, \(\mathbf{r}'(t) = 6 \mathbf{i} + 6t \mathbf{j} + 3t^2 \mathbf{k}\), and aim to retrieve the original function \(\mathbf{r}(t)\).
Integration is performed component-wise:

• The \(\mathbf{i}\)-component: Integrate \(6\) with respect to \(t\). This gives \(6t + C_1\), where \(C_1\) is a constant of integration.
• For the \(\mathbf{j}\)-component: Integrate \(6t\), resulting in \(3t^2 + C_2\).
• Finally, the \(\mathbf{k}\)-component: Integrate \(3t^2\), yielding \(t^3 + C_3\).

These integrations form our vector function's general solution: \(\mathbf{r}(t) = (6t + C_1) \mathbf{i} + (3t^2 + C_2) \mathbf{j} + (t^3 + C_3) \mathbf{k}\).
Each component allows us to trace back the effects of the vector field described by \(\mathbf{r}'(t)\).

Initial Value Problem

The problem is essentially an initial value problem (IVP) wherein we seek a specific solution that both satisfies the differential equation and passes through a given point in space.
This process adds another layer of solving the vector integration problem by 'pinning down' the arbitrary constants introduced during integration.An IVP will provide a particular point on the trajectory described by \(\mathbf{r}(t)\) which, in this exercise, is given as \(\mathbf{r}(0) = \mathbf{i} - 2\mathbf{j} + \mathbf{k}\).
To solve an IVP, proceed by substituting the initial condition into the general solution:

• Set each initial condition component equal to its respective initial vector component.
• By substituting \(t = 0\) into the general solution, equate \((C_1, C_2, C_3)\) with \(\mathbf{i}, -2\mathbf{j}, \mathbf{k}\).

These steps give us the constants \(C_1 = 1\), \(C_2 = -2\), \(C_3 = 1\). These constants transform our general solution into the specific solution satisfying both the vector derivative and the initial condition.

Vector Functions

Vector functions are a pivotal part of vector calculus and vector integration.
They map real numbers to vectors in multi-dimensional spaces and are especially useful for describing linear motion in physics.
A vector function \(\mathbf{r}(t)\) generally has components that are individual functions of \(t\):

• The \(\mathbf{i}\), \(\mathbf{j}\), and \(\mathbf{k}\) components all describe the direction and magnitude of the vector in relation to the vector field.

Using these vector functions, one can trace the path of a moving object in three-dimensional space or solve complex engineering problems.
In our exercise, the vector function reveals the spatial path through which the described entity moves over time:\(\mathbf{r}(t) = (6t + 1) \mathbf{i} + (3t^2 - 2) \mathbf{j} + (t^3 + 1) \mathbf{k}\).
This function can describe motion, forces, fields, or any application where direction and magnitude vary with respect to time or another parameter. Understanding these movements in real-world terms provides powerful insights applicable across various scientific fields.