Problem 3 - 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). $$\mathbf{x}=\left[\begin{array}{l}2 \\ 2\end{array}\right]$$

Step-by-Step Solution

Answer

The vector is \((2, 2)\), has a length of \(2\sqrt{2}\), and forms a \(45^\circ\) angle with the \(x_1\)-axis.

Step 1: Represent the Vector on the Plane

In this step, we plot the vector \( \mathbf{x} = \begin{bmatrix} 2 \ 2 \end{bmatrix} \) as a point and a directed line segment in the \( x_1-x_2 \) plane. The point has coordinates \((2, 2)\), and the vector originates from the origin \((0, 0)\) to \((2, 2)\).

Step 2: Calculate the Length of the Vector

The length (or magnitude) of the vector \( \mathbf{x} = \begin{bmatrix} 2 \ 2 \end{bmatrix} \) is calculated using the formula: \( \|\mathbf{x}\| = \sqrt{x_1^2 + x_2^2} \). Substituting the values, we get \( \|\mathbf{x}\| = \sqrt{2^2 + 2^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \).

Step 3: Determine the Angle with the Positive \(x_1\)-Axis

To find the angle \( \theta \) between the vector and the positive \( x_1 \)-axis, we use the arctangent function. The formula is \( \theta = \tan^{-1} \left( \frac{x_2}{x_1} \right) \). Substituting the values, we get \( \theta = \tan^{-1} \left( \frac{2}{2} \right) = \tan^{-1}(1) = 45^\circ \).

Step 4: Summary of the Results

The vector \( \mathbf{x} \) is represented by the point (2, 2) on the \( x_1-x_2 \) plane, its length is \( 2\sqrt{2} \), and it forms an angle of \( 45^\circ \) counterclockwise with the positive \( x_1 \)-axis.

Key Concepts

Vector Representation

Vectors are fundamental elements in vector calculus that represent both magnitude and direction. In the exercise, we are working with a vector \( \mathbf{x} = \begin{bmatrix} 2 \ 2 \end{bmatrix} \). This notation allows us to place the vector within a coordinate plane, specifically the \( x_1-x_2 \) plane.

Here's how you can understand vector representation better:

• The vector \( \mathbf{x} \) starts at the origin \((0,0)\) and ends at the point \((2, 2)\).
• This vector is depicted as a line segment pointing from the origin to the point \((2, 2)\), which indicates its direction.
• The coordinates of the endpoint \((2, 2)\) are determined by the vector's components, \(x_1\) and \(x_2\).

This visual representation helps us understand how vectors operate within a plane, offering insight into both their orientation and their application in navigating multi-dimensional spaces.

Vector Magnitude

The magnitude of a vector refers to its length, which is a fundamental concept in vector calculus. Calculating a vector's magnitude involves using its components. For our vector \( \mathbf{x} = \begin{bmatrix} 2 \ 2 \end{bmatrix} \), the length of the vector is determined by the formula:

\[ \|\mathbf{x}\| = \sqrt{x_1^2 + x_2^2} \]

This means:

• Substitute the respective values: \( \|\mathbf{x}\| = \sqrt{2^2 + 2^2} \).
• Simplify under the square root: \( \|\mathbf{x}\| = \sqrt{4 + 4} = \sqrt{8} \).
• Conclude with the result: \( \|\mathbf{x}\| = 2\sqrt{2} \).

This magnitude provides a sense of scale or size of the vector, irrespective of its direction. It is one of the key properties when assessing vectors and is crucial in measuring distances within vector contexts.

Angle with Axis

Understanding the angle a vector forms with an axis is crucial in vector calculus, as it relates to the vector's direction within the plane. For our exercise with vector \( \mathbf{x} = \begin{bmatrix} 2 \ 2 \end{bmatrix} \), the angle \( \theta \) with the positive \( x_1 \)-axis is determined using trigonometry.

The angle calculation involves the arctangent function:

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

Breaking it down:

• Substitute the component values: \( \theta = \tan^{-1} \left( \frac{2}{2} \right) \).
• As \( \frac{2}{2} = 1 \), we find \( \theta = \tan^{-1}(1) \).
• Finally, recognize the angle: \( \theta = 45^\circ \).

This angle shows that the vector \( \mathbf{x} \) is equally inclined to both the \( x_1 \)-axis and \( x_2 \)-axis, indicating a symmetrical orientation in the plane.