Problem 4 - Calculus for Biology and Medicine

Question

Represent each given vector \(\mathrm{x}=\left[\begin{array}{l}x_{1} \\\ x_{2}\end{array}\right]\) in the \(x_{1}-x_{2}\) plane, and determine its length and the angle that it forms with the positive \(x_{1}\) -axis (measured counterclockwise). $$\mathbf{x}=\left[\begin{array}{r}-2 \\ 0\end{array}\right]$$

Step-by-Step Solution

Answer

The vector has a length of 2 and forms a 180° angle with the positive x1-axis.

Step 1: Plot the Vector

Let's start by plotting the vector \( \mathbf{x} = \begin{bmatrix} -2 \ 0 \end{bmatrix} \) on the \(x_1 - x_2\) plane. This vector has its tail at the origin (0,0) and its head at the point (-2, 0), on the negative \(x_1\)-axis.

Step 2: Determine the Length of the Vector

The length (or magnitude) of a vector \( \mathbf{x} = \begin{bmatrix} x_1 \ x_2 \end{bmatrix} \) is given by \( \| \mathbf{x} \| = \sqrt{x_1^2 + x_2^2} \). For the vector \( \mathbf{x} = \begin{bmatrix} -2 \ 0 \end{bmatrix} \), the length is \( \| \mathbf{x} \| = \sqrt{(-2)^2 + 0^2} = \sqrt{4} = 2 \).

Step 3: Calculate the Angle with the Positive x1-Axis

To find the angle \( \theta \) the vector forms with the positive \(x_1\)-axis, we use the arctangent function: \( \theta = \tan^{-1}\left(\frac{x_2}{x_1}\right) \). Since \( x_2 = 0 \) and \( x_1 = -2 \), \( \theta = \tan^{-1}(0) = 0 \). However, because the vector points directly left along the \(x_1\)-axis, the angle counterclockwise from the positive \(x_1\)-axis is \(180^\circ\) or \(\pi\) radians.

Key Concepts

Vector Representation

Vectors are mathematical objects that have both a direction and a magnitude. In the plane, they are often represented as arrows starting from one point and ending at another. When we talk about vector representation, we refer to how these vectors can be shown in a specific coordinate system like the Cartesian plane. A vector \[ \mathbf{v} = \begin{bmatrix} x_1 \ x_2 \end{bmatrix} \]can be visualized as an arrow originating from the origin (0, 0) to the point (\(x_1, x_2\)). This means that the vector has a starting point at (0,0) and points towards the coordinates given by \(x_1\) and \(x_2\).
- The first element, \(x_1\), is the horizontal movement.- The second element, \(x_2\), is the vertical movement.
In our exercise, the vector \(\mathbf{x} = \begin{bmatrix} -2 \ 0 \end{bmatrix}\) starts at the origin and points directly left to the coordinate (-2, 0). Hence, it's entirely along the negative horizontal axis with no vertical displacement.

Magnitude of a Vector

Understanding the magnitude of a vector is crucial, as it tells us the length of the vector. This is similar to calculating the straight-line distance between two points. For a vector \[ \mathbf{v} = \begin{bmatrix} x_1 \ x_2 \end{bmatrix} \]its magnitude is given by the formula:\[ \| \mathbf{v} \| = \sqrt{x_1^2 + x_2^2}\]
This formula is an application of the Pythagorean theorem, where \(x_1\) and \(x_2\) form the two sides of a right-angled triangle, and the magnitude is the hypotenuse.
For our specific vector \[ \mathbf{x} = \begin{bmatrix} -2 \ 0 \end{bmatrix} \],we calculate the magnitude as follows:
1. Square each component: \((-2)^2 = 4\), and \(0^2 = 0\).2. Sum these squares: \(4 + 0 = 4\).3. Take the square root of the result: \( \sqrt{4} = 2 \).
Thus, the length of our vector is 2 units.

Angle Calculation

Calculating the angle a vector makes with the positive horizontal axis involves understanding the vector's orientation in the plane. The angle, typically denoted as \(\theta\), can be calculated using the inverse tangent (arctangent) function when the vector has non-zero components. The function is:\[ \theta = \tan^{-1}\left(\frac{x_2}{x_1}\right)\]
- \(x_2\) is the vertical component.- \(x_1\) is the horizontal component.
In the case where a vector has a horizontal component along the axis and a zero vertical component, as our vector \(\mathbf{x} = \begin{bmatrix} -2 \ 0 \end{bmatrix}\), the calculation results in\[ \theta = \tan^{-1}(0) = 0\].However, this calculation only holds when the vector points in the positive direction. Our vector instead points left, or in the negative direction. Consequently, we need to consider the full counterclockwise angle from the positive \(x_1\)-axis, which is
\(180^\circ\) or \(\pi\) radians.