Problem 16 - Advanced Engineering Mathematics

Question

Verify that the vector \(\mathbf{X}\) is a solution of the given system. $$ \mathbf{X}^{\prime}=\left(\begin{array}{rrr} 1 & 0 & 1 \\ 1 & 1 & 0 \\ -2 & 0 & -1 \end{array}\right) \mathbf{x} ; \quad \mathbf{X}=\left(\begin{array}{c} \sin t \\ -\frac{1}{2} \sin t-\frac{1}{2} \cos t \\ -\sin t+\cos t \end{array}\right) $$

Step-by-Step Solution

Answer

\( \mathbf{X} \) is a solution of the system, as verified.

Step 1: Differentiate the vector function

Differentiate the components of the vector \( \mathbf{X} \) with respect to \( t \). This will give us \( \mathbf{X}' \).- \( \frac{d}{dt}(\sin t) = \cos t \)- \( \frac{d}{dt}(-\frac{1}{2} \sin t - \frac{1}{2} \cos t) = -\frac{1}{2}\cos t + \frac{1}{2} \sin t \)- \( \frac{d}{dt}(-\sin t + \cos t) = -\cos t - \sin t \) Thus, \( \mathbf{X}' = \begin{pmatrix} \cos t \ -\frac{1}{2} \cos t + \frac{1}{2} \sin t \ -\cos t - \sin t \end{pmatrix} \).

Step 2: Multiply the matrix and vector

To verify the solution, we will multiply the matrix\[\begin{pmatrix}1 & 0 & 1 \1 & 1 & 0 \-2 & 0 & -1\end{pmatrix}\]with the vector \( \mathbf{X} \) to check if it equals \( \mathbf{X}' \):\[ \begin{pmatrix} 1 & 0 & 1 \ 1 & 1 & 0 \ -2 & 0 & -1 \end{pmatrix} \begin{pmatrix} \sin t \ -\frac{1}{2} \sin t - \frac{1}{2} \cos t \ -\sin t + \cos t \end{pmatrix} \]Calculate each component of the resulting vector.

Step 3: Validate the calculation of each entry

The result of the matrix-vector multiplication will be a vector:The first component:\( 1\times\sin t + 0\times(-\frac{1}{2} \sin t - \frac{1}{2} \cos t) + 1\times(-\sin t + \cos t) = \cos t \).The second component:\( 1\times\sin t + 1\times(-\frac{1}{2} \sin t - \frac{1}{2} \cos t) + 0\times(-\sin t + \cos t) = -\frac{1}{2} \cos t + \frac{1}{2} \sin t \).The third component:\(-2\times\sin t + 0\times(-\frac{1}{2} \sin t - \frac{1}{2} \cos t) - 1\times(-\sin t + \cos t) = -\cos t - \sin t \).Thus, the resulting vector from the multiplication is \( \begin{pmatrix} \cos t \ -\frac{1}{2} \cos t + \frac{1}{2} \sin t \ -\cos t - \sin t \end{pmatrix} \).

Step 4: Compare the results

Compare the vector obtained from the differential \( \mathbf{X}' \) to the result of the matrix multiplication:\[ \begin{pmatrix} \cos t \ -\frac{1}{2} \cos t + \frac{1}{2} \sin t \ -\cos t - \sin t \end{pmatrix} \]Both vectors match exactly, verifying that \( \mathbf{X} \) is indeed a solution of the given system.

Key Concepts

Vector Functions

A vector function is an insightful concept in mathematics, particularly in calculus and differential equations. Simply put, a vector function associates a vector to each point in time or space. For example, the function \[ \mathbf{X}(t) = \begin{pmatrix} \sin t \ -\frac{1}{2} \sin t - \frac{1}{2} \cos t \ -\sin t + \cos t \end{pmatrix} \]associates a vector to every value of \(t\), making it easier to consider changes in all three components simultaneously.

• Each component of the vector can be a function of one or more variables.
• In our example, each component is a trigonometric function of \( t \).

Vector functions become particularly useful in physics and engineering, where they can represent a point's position in space over time. Vectors can be differentiated and integrated to better understand their behavior, which is often necessary when dealing with physical systems like particle motion.

Matrix Multiplication

Matrix multiplication is a crucial operation in mathematics, especially when dealing with systems of linear equations and transformations. It involves taking two matrices and producing another matrix by multiplying rows by columns. Let's see how this works in our example.We have a matrix \[\begin{pmatrix}1 & 0 & 1 \1 & 1 & 0 \-2 & 0 & -1\end{pmatrix}\]When multiplied by the vector function \[\mathbf{X}(t) = \begin{pmatrix} \sin t \-\frac{1}{2} \sin t - \frac{1}{2} \cos t \-\sin t + \cos t \end{pmatrix}\]

• We calculate the elements of the resulting vector by multiplying elements from the matrix rows by the vector components and summing them.
• This process allows us to verify if the vector function is a solution to the system of differential equations in question.

Understanding matrix multiplication is fundamental in solving systems of linear equations, performing transformations, and even in computer graphics where objects are rotated and translated.

Differential Equations Verification

Verifying differential equations involves checking if a given function satisfies the equation. For our exercise, the function \( \mathbf{X}(t) \) needed verification as a solution to the system. This involved satisfying the equation \( \mathbf{X}' = A \mathbf{X} \), where \( A \) is a matrix.
The verification steps include:

• First, differentiating each part of the vector function \( \mathbf{X}(t) \) to find \( \mathbf{X}'(t) \).
• Second, performing matrix multiplication with \( A \) and \( \mathbf{X}(t) \).
• Finally, comparing \( \mathbf{X}'(t) \) to the result of matrix multiplication.

Consistency between \( \mathbf{X}'(t) \) and the matrix result confirms that \( \mathbf{X}(t) \) is indeed a solution. This process helps ensure that the derived expressions truly fulfill the dynamic behavior described by the differential equation.

Vector Calculus

Vector calculus extends calculus into multiple dimensions, using vector functions, derivatives, and integrals. It plays a key role in describing and solving physical problems, such as field theories and fluid dynamics.In our example, vector calculus skills are critical:

• We used differentiation to find the rate of change of the vector function \( \mathbf{X}(t) \).
• This differentiation allowed us to assess how the components of the vector changed with respect to \( t \).

Understanding vector calculus can also enrich comprehension of line integrals and flux, essential in physics and engineering. It empowers learners to delve into advanced areas of study, where vectors are used to describe three-dimensional phenomena in space and time.