Problem 7 - 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). $$\mathbf{x}=\left[\begin{array}{r}-\sqrt{3} \\ -1\end{array}\right]$$

Step-by-Step Solution

Answer

The vector length is 2, and it forms an angle of \(210^\circ\) with the positive \(x_1\)-axis.

Step 1: Identifying the Vector Components

The vector given is \( \mathbf{x} = \left[ \begin{array}{r} -\sqrt{3} \ -1 \end{array} \right] \). This means that \( x_1 = -\sqrt{3} \) and \( x_2 = -1 \).

Step 2: Calculating the Length of the Vector

The length (or magnitude) of vector \( \mathbf{x} \) is calculated using the formula: \( ||\mathbf{x}|| = \sqrt{x_1^2 + x_2^2} \). Substituting the values, \( ||\mathbf{x}|| = \sqrt{(-\sqrt{3})^2 + (-1)^2} = \sqrt{3 + 1} = 2 \). Thus, the length of the vector is 2.

Step 3: Determining the Angle with the Positive x-axis

The angle \( \theta \) that the vector makes with the positive \( x_1 \)-axis can be found using the arctangent function. The formula is \( \theta = \tan^{-1}\left(\frac{x_2}{x_1}\right) \). For \( x_1 = -\sqrt{3} \) and \( x_2 = -1 \), \( \theta = \tan^{-1}\left(\frac{-1}{-\sqrt{3}}\right) = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) \). This results in \( \theta = \frac{\pi}{6} \). Since both components of the vector are negative, the vector is located in the third quadrant of the plane, meaning we need to add \( \pi \) to the angle: \( \theta = \frac{\pi}{6} + \pi = \frac{7\pi}{6} \).

Step 4: Convert to Degrees (Optional)

For applications where angles are commonly expressed in degrees, convert radians to degrees. Using \( 1 \text{ radian} = \frac{180}{\pi} \text{ degrees} \), the angle in degrees is: \( \theta = \frac{7\pi}{6} \times \frac{180}{\pi} = 210^\circ \).

Key Concepts

Magnitude of a Vector

The magnitude of a vector provides valuable information about the vector's length, regardless of its direction. To find the magnitude, you use the formula:\[||\mathbf{x}|| = \sqrt{x_1^2 + x_2^2}\]This formula is essentially a direct application of the Pythagorean theorem. Think of the vector as forming a right triangle with the axes, where the vector is the hypotenuse. By squaring the components, adding them, and taking the square root, you find the vector's magnitude or length.
In our specific case, with the vector \( \mathbf{x} = \left[ \begin{array}{r} -\sqrt{3} \ -1 \end{array} \right] \), substituting these values into the formula gives:\[||\mathbf{x}|| = \sqrt{(-\sqrt{3})^2 + (-1)^2} = \sqrt{3 + 1} = 2\]Thus, the length of our vector \( \mathbf{x} \) is 2.

Angle with x-axis

Understanding the angle between a vector and the positive x-axis is crucial in determining its direction. The angle \(\theta\) is calculated using the inverse tangent function, specifically:\[\theta = \tan^{-1}\left(\frac{x_2}{x_1}\right)\]This formula helps you identify how the vector tilts away from the positive x-axis. Typically, arctangent gives the angle, but it's essential to consider the vector's position in the coordinate plane. In scenarios where both the x and y components are negative, the vector is situated in the third quadrant.For the vector \( \mathbf{x} = \left[ \begin{array}{r} -\sqrt{3} \ -1 \end{array} \right] \), we calculate:\[\theta = \tan^{-1}\left(\frac{-1}{-\sqrt{3}}\right) = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6}\]Since the vector is in the third quadrant, we add \(\pi\) to \(\frac{\pi}{6}\), resulting in:\[\theta = \frac{7\pi}{6}\]To convert radians to degrees, remember:\[1 \text{ radian} = \frac{180}{\pi} \text{ degrees}\]Thus, our angle in degrees is:\[\theta = \frac{7\pi}{6} \times \frac{180}{\pi} = 210^\circ\]This conversion is especially useful when degrees are the preferred unit measurement.

Vector Components

A vector is made up of several components that show its direction and magnitude. In a two-dimensional plane, we have two components: an x-component and a y-component. These are the coordinates \(x_1\) and \(x_2\), respectively, representing the vector's projection on the x-axis and y-axis.For the vector \(\mathbf{x} = \left[ \begin{array}{r} -\sqrt{3} \ -1 \end{array} \right]\):

• The x-component \(x_1 = -\sqrt{3}\)
• The y-component \(x_2 = -1\)

Understanding these components is vital because they allow us to represent the direction and magnitude of vectors in algebraic form. Similar to how a location on a map needs both latitude and longitude, a vector needs both components to fully define its position and effect in a coordinate system.
These components are essential when determining how the vector interacts with forces, velocity, or any other vector-based quantities in physics and engineering.