Problem 34 - Calculus for Biology and Medicine

Question

Let $$ \mathbf{u}=\left[\begin{array}{l} 3 \\ 4 \end{array}\right], \quad \mathbf{v}=\left[\begin{array}{r} 1 \\ -2 \end{array}\right], \quad \text { and } \quad \mathbf{w}=\left[\begin{array}{l} -1 \\ -2 \end{array}\right] $$ Compute \(2 \mathbf{v}-\mathbf{w}\) and illustrate the result graphically.

Step-by-Step Solution

Answer

The result of \(2\mathbf{v} - \mathbf{w}\) is \([3, -2]^T\).

Step 1: Compute the Vector Scalar Multiplication

First, compute scalar multiplication of vector \( \mathbf{v} \) by 2, that is \( 2\mathbf{v} \). Vector \( \mathbf{v} = \begin{bmatrix} 1 \ -2 \end{bmatrix} \). \[ 2\mathbf{v} = 2 \times \begin{bmatrix} 1 \ -2 \end{bmatrix} = \begin{bmatrix} 2 \times 1 \ 2 \times (-2) \end{bmatrix} = \begin{bmatrix} 2 \ -4 \end{bmatrix} \]

Step 2: Subtract Vectors

Now we need to subtract \( \mathbf{w} \) from \( 2\mathbf{v} \). Vector \( \mathbf{w} = \begin{bmatrix} -1 \ -2 \end{bmatrix} \). To perform the subtraction, calculate: \[ 2\mathbf{v} - \mathbf{w} = \begin{bmatrix} 2 \ -4 \end{bmatrix} - \begin{bmatrix} -1 \ -2 \end{bmatrix} = \begin{bmatrix} 2 - (-1) \ -4 - (-2) \end{bmatrix} \] Simplifying the elements gives: \[ 2\mathbf{v} - \mathbf{w} = \begin{bmatrix} 2 + 1 \ -4 + 2 \end{bmatrix} = \begin{bmatrix} 3 \ -2 \end{bmatrix} \]

Step 3: Illustrate Graphically

Plot the vector \( \mathbf{v} = \begin{bmatrix} 1 \ -2 \end{bmatrix} \) scaled by 2 to see \( \begin{bmatrix} 2 \ -4 \end{bmatrix} \), the vector \( \mathbf{w} = \begin{bmatrix} -1 \ -2 \end{bmatrix} \), and the resultant vector \( \begin{bmatrix} 3 \ -2 \end{bmatrix} \) on a graph. The green arrow indicates \( 2\mathbf{v} \), the red arrow shows \( \mathbf{w} \), and the blue arrow is the result \( 2\mathbf{v} - \mathbf{w} \). This visual representation helps verify the calculation by showing each component's direction and magnitude change in a 2D plane.

Key Concepts

Scalar Multiplication

Scalar multiplication in vectors is akin to stretching or compressing the vector along its direction. You multiply every component of the vector by the scalar value. For instance, if you have a vector \( \mathbf{v} = \begin{bmatrix} 1 \ -2 \end{bmatrix} \) and you multiply it by a scalar 2, every component of the vector is multiplied by this scalar. This transforms \( \mathbf{v} \) into \( 2\mathbf{v} = 2 \times \begin{bmatrix} 1 \ -2 \end{bmatrix} = \begin{bmatrix} 2 \ -4 \end{bmatrix} \).

The new vector \( \begin{bmatrix} 2 \ -4 \end{bmatrix} \) is twice as long as the original vector \( \mathbf{v} \). It also retains the direction of \( \mathbf{v} \), but with every component scaled by 2. This is critical when thinking about vector applications, such as physics, where scalar multiplication can simulate changing forces or velocities.

Vector Subtraction

Vector subtraction involves finding the difference between two vectors by subtracting their corresponding components. This operation allows you to see how much one vector "lacks" compared to another in terms of both direction and magnitude. Suppose you have vectors \( \mathbf{a} = \begin{bmatrix} 2 \ -4 \end{bmatrix} \) and \( \mathbf{w} = \begin{bmatrix} -1 \ -2 \end{bmatrix} \). The subtraction \( \mathbf{a} - \mathbf{w} \) is performed element-wise:


\[ \mathbf{a} - \mathbf{w} = \begin{bmatrix} 2 - (-1) \ -4 - (-2) \end{bmatrix} = \begin{bmatrix} 3 \ -2 \end{bmatrix} \]
Each component of \( \mathbf{w} \) is subtracted from the corresponding component of \( \mathbf{a} \). The result is a vector that represents the end point of a vector starting at \( \mathbf{w} \) and ending at \( \mathbf{a} \). This results in a new vector \( \begin{bmatrix} 3 \ -2 \end{bmatrix} \), which indicates the net movement from \( \mathbf{w} \) to \( \mathbf{a} \).

Graphical Illustration of Vectors

A graphical representation of vectors helps to visualize operations like scalar multiplication and subtraction. Here’s how you can illustrate the vectors given in the problem. Start with a graph on a 2D plane:

• Plot the original vector \( \mathbf{v} = \begin{bmatrix} 1 \ -2 \end{bmatrix} \) and then scale it to become \( 2\mathbf{v} = \begin{bmatrix} 2 \ -4 \end{bmatrix} \) by drawing an arrow from the origin to the point (2, -4). This vector is longer than \( \mathbf{v} \), showing the effect of scalar multiplication.
• Next, plot the vector \( \mathbf{w} = \begin{bmatrix} -1 \ -2 \end{bmatrix} \) by drawing an arrow from the origin to (-1, -2). The position of this vector helps understand the subtraction.
• The final significant step is plotting the resultant vector \( \mathbf{r} = \begin{bmatrix} 3 \ -2 \end{bmatrix} \). This vector is plotted by drawing an arrow from the origin to (3, -2).

This visual comparison supports our calculations and clarifies how vector operations affect both direction and magnitude. It's a helpful way to visually confirm that the arithmetic operations have been executed correctly.