Problem 6 - Calculus for Biology and Medicine

Question

In Problems , represent each given vector \(\mathbf{x}=\left[\begin{array}{l}x_{1} \\ x_{2}\end{array}\right]\) 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).$$ \mathbf{x}=\left[\begin{array}{l} -1 \\ -1 \end{array}\right] $$

Step-by-Step Solution

Answer

The vector's length is \(\sqrt{2}\) and the angle is \(225^\circ\).

Step 1: Plot the Vector

Place the vector \(\mathbf{x} = \begin{bmatrix} -1 \ -1 \end{bmatrix}\) on the coordinate plane. Represent the vector starting from the origin (0,0) and ending at the point (-1,-1) in the \(x_1-x_2\) plane.

Step 2: Calculate the Length of the Vector

The length of a vector \(\mathbf{x} = \begin{bmatrix} x_1 \ x_2 \end{bmatrix}\) is given by the formula \(\|\mathbf{x}\| = \sqrt{x_1^2 + x_2^2}\). Substitute \(x_1 = -1\) and \(x_2 = -1\) to find \(\|\mathbf{x}\| = \sqrt{(-1)^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2}\).

Step 3: Determine the Angle with the Positive x1-axis

The angle \(\theta\) that the vector forms with the positive \(x_1\)-axis can be found using the arctangent function: \(\theta = \tan^{-1}\left(\frac{x_2}{x_1}\right)\). Substitute \(x_1 = -1\) and \(x_2 = -1\) to get \(\theta = \tan^{-1}\left(\frac{-1}{-1}\right) = \tan^{-1}(1) = 45^\circ\). Since the vector is in the third quadrant, add 180° to the angle: \(\theta = 45^\circ + 180^\circ = 225^\circ\).

Key Concepts

Understanding the Coordinate Plane

The coordinate plane is a two-dimensional space where we can visually represent mathematical concepts and equations. It consists of an x-axis (horizontal) and a y-axis (vertical), meeting at a point called the origin, labeled as (0,0). Every point on this plane can be represented with a pair of coordinates, usually in the form (x, y).
For vectors, this means we can plot them starting at the origin and reaching a specific point that corresponds to their coordinates. In this case, our vector \(\mathbf{x} = \begin{bmatrix} -1 \ -1 \end{bmatrix}\) translates to a line that starts at (0,0) and ends at (-1, -1).
The representation of vectors on a coordinate plane helps us to not just visualize direction and magnitude, but also to perform calculations such as determining angles and lengths. By plotting the vector, we can see exactly which quadrant it resides in, which is helpful for further analysis like angle calculations.

Calculating Vector Length

The length of a vector is also known as its magnitude. It's a measure of how long the line is from the origin to its endpoint on the coordinate plane. Mathematically, we find the length using the Pythagorean theorem.
Given a vector \(\mathbf{x} = \begin{bmatrix} x_1 \ x_2 \end{bmatrix}\), the length is calculated using the formula: \[ \| \mathbf{x} \| = \sqrt{x_1^2 + x_2^2} \] For our example, substitute \(x_1 = -1\) and \(x_2 = -1\), giving us:

• \(\| \mathbf{x} \| = \sqrt{(-1)^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2}\)

The length here is \(\sqrt{2}\), which is a crucial value because it helps us understand the scale of the vector compared to others and is essential for determining the angle it forms.

Determining the Angle of a Vector

An angle formed by a vector with the positive x-axis is vital in understanding vector direction. The standard way to find this angle is using the arctangent function of the y-coordinate divided by the x-coordinate, expressed as:

• \(\theta = \tan^{-1}\left(\frac{x_2}{x_1}\right)\)

For our vector \(\mathbf{x} = \begin{bmatrix} -1 \ -1 \end{bmatrix}\), we substitute in our values:

• \(\theta = \tan^{-1}\left(\frac{-1}{-1}\right) = \tan^{-1}(1) = 45^\circ\)

Because the vector lies in the third quadrant of the coordinate plane, we must adjust for this by adding 180° to our previous result. This accounts for the direction in which the vector points:

• \(\theta = 45^\circ + 180^\circ = 225^\circ\)

This calculation ensures we have the correct direction and orientation of the vector concerning the positive x-axis.