Problem 34 - Calculus for Biology and Medicine

Question

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

Step-by-Step Solution

Answer

The result of \( 2 \mathbf{v} - \mathbf{w} \) is \( \begin{bmatrix} -3 \\ -2 \end{bmatrix} \).

Step 1: Compute Scalar Multiplication

First, multiply vector \( \mathbf{v} \) by the scalar \( 2 \). Given \( \mathbf{v} = \begin{bmatrix} -1 \ -2 \end{bmatrix} \), you get \[ 2 \mathbf{v} = 2 \times \begin{bmatrix} -1 \ -2 \end{bmatrix} = \begin{bmatrix} 2 \times (-1) \ 2 \times (-2) \end{bmatrix} = \begin{bmatrix} -2 \ -4 \end{bmatrix} \].

Step 2: Subtract Vectors

Next, subtract vector \( \mathbf{w} \) from the result of the scalar multiplication. We have \( \mathbf{w} = \begin{bmatrix} 1 \ -2 \end{bmatrix} \), so \[ 2 \mathbf{v} - \mathbf{w} = \begin{bmatrix} -2 \ -4 \end{bmatrix} - \begin{bmatrix} 1 \ -2 \end{bmatrix} = \begin{bmatrix} -2 - 1 \ -4 - (-2) \end{bmatrix} = \begin{bmatrix} -3 \ -2 \end{bmatrix} \].

Step 3: Graphical Illustration

To illustrate \( 2 \mathbf{v} - \mathbf{w} \), plot the vectors on a Cartesian coordinate system. Begin with the origin and draw vector \( \mathbf{v} \) scaled by 2, ending at \( (-2, -4) \). From \( (-2, -4) \), draw the vector \( -\mathbf{w} \), which is the opposite of \( \mathbf{w} \), ending at \( (-3, -2) \). The resulting vector from the origin to the point \( (-3, -2) \) represents \( 2 \mathbf{v} - \mathbf{w} \).

Key Concepts

Scalar Multiplication

Scalar multiplication is a fundamental concept in vector calculus that involves multiplying a vector by a scalar, which is simply a constant number. This operation stretches or shrinks the vector, but doesn't change its direction unless the scalar is negative. For example, consider the vector \( \mathbf{v} = \begin{bmatrix} -1 \ -2 \end{bmatrix} \). When we multiply \( \mathbf{v} \) by the scalar 2, we perform the following operation on each component of the vector:

• Multiply the first component: \( 2 \times (-1) = -2 \)
• Multiply the second component: \( 2 \times (-2) = -4 \)

This results in a new vector \( 2\mathbf{v} = \begin{bmatrix} -2 \ -4 \end{bmatrix} \), which is twice as long as the original vector \( \mathbf{v} \) but points in the same direction. If the scalar had been negative, the vector would have pointed in the opposite direction.

Vector Subtraction

Vector subtraction involves finding the difference between two vectors. This is done by subtracting each component of one vector from the corresponding component of another vector. Essentially, you can think of vector subtraction as adding a negative vector.Let's look at the subtraction of vectors \( \mathbf{w} \) from \( 2\mathbf{v} \) in this context:

• Given the vectors \( 2\mathbf{v} = \begin{bmatrix} -2 \ -4 \end{bmatrix} \) and \( \mathbf{w} = \begin{bmatrix} 1 \ -2 \end{bmatrix} \), perform:
  • \( -2 - 1 = -3 \)
  • \( -4 - (-2) = -2 \)
• The resulting vector is \( \begin{bmatrix} -3 \ -2 \end{bmatrix} \).

In essence, vector subtraction combines elements of both addition and multiplication, allowing you to manipulate and arrange vectors geometrically with relative ease. This operation is crucial in fields like physics and engineering, where you often need to compute net forces or other resultant vectors.

Graphical Vector Representation

Graphically representing vectors involves plotting them on a Cartesian coordinate system. This visualization helps to understand the geometric implications of operations like addition, subtraction, and scalar multiplication. To represent the result of \( 2 \mathbf{v} - \mathbf{w} \) graphically:

• Plot the vector \( 2\mathbf{v} = \begin{bmatrix} -2 \ -4 \end{bmatrix} \). Start from the origin \((0, 0)\) and draw a line to the point \((-2, -4)\).
• From point \((-2, -4)\), subtract vector \( \mathbf{w} \) by drawing its opposite \( -\mathbf{w} = \begin{bmatrix} -1 \ 2 \end{bmatrix} \).
• This lands you at the point \((-3, -2)\).

The resultant vector \( \begin{bmatrix} -3 \ -2 \end{bmatrix} \) can be drawn from the origin to this final point \((-3, -2)\). Viewing vectors in this manner provides clear insight into how these entities interact in space, making topics like vector addition and resultant force visual accessible and intuitive.