Problem 6 - 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). $$\mathbf{x}=\left[\begin{array}{l}-1 \\ -1\end{array}\right] \quad$$

Step-by-Step Solution

Answer

Length is \( \sqrt{2} \); angle is \( 225^{\circ} \).

Step 1: Draw the Vector

The vector \( \mathbf{x} = \begin{bmatrix} -1 \ -1 \end{bmatrix} \) is represented in the \( x_1-x_2 \) plane. The point corresponding to this vector is \( (-1, -1) \). Starting from the origin \( (0,0) \), draw a line to the point \( (-1, -1) \).

Step 2: Calculate the Length of the Vector

To find the length of the vector \( \mathbf{x} = \begin{bmatrix} -1 \ -1 \end{bmatrix} \), use the formula for the magnitude of a vector: \( \| \mathbf{x} \| = \sqrt{x_1^2 + x_2^2} \). Substituting \( x_1 = -1 \) and \( x_2 = -1 \), we get:\[ \| \mathbf{x} \| = \sqrt{(-1)^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2} \].

Step 3: Determine the Angle with the x1 Axis

The angle \( \theta \) that the vector makes with the positive \( x_1 \)-axis can be found using the inverse tangent function: \( \theta = \tan^{-1} \left( \frac{x_2}{x_1} \right) \). For \( \mathbf{x} = \begin{bmatrix} -1 \ -1 \end{bmatrix} \), \( \theta = \tan^{-1} \left( \frac{-1}{-1} \right) = \tan^{-1}(1) \). The angle whose tangent is 1 is \( 45^{\circ} \), but since the vector is in the third quadrant (both components negative), we add \( 180^{\circ} \): \[ \theta = 225^{\circ} \].

Key Concepts

Vector Representation

Vectors are essential tools in mathematics and physics because they allow us to describe both magnitude and direction. In this exercise, we need to represent the vector \( \mathbf{x} = \begin{bmatrix} -1 \ -1 \end{bmatrix} \) in the \( x_1-x_2 \) plane. A vector in two dimensions can be represented by an arrow starting from the origin \( (0,0) \) and pointing to the coordinates specified by the vector's components.

• Identify the components: The vector \( \begin{bmatrix} -1 \ -1 \end{bmatrix} \) has components \( x_1 = -1 \) and \( x_2 = -1 \), meaning it points to \( (-1, -1) \) in the plane.
• Draw the vector: From the origin, draw a line to the point \( (-1, -1) \). This line represents your vector in the graph.

It's important to realize that vector representation in a plane provides an effective way to visualize complex problems in simpler geometric shapes, which can be particularly helpful when dealing with multiple vectors and forces.

Vector Magnitude

The vector magnitude, also known as the "length" of the vector, provides us with a scalar indicating how long the vector is, regardless of its direction. For a vector \( \mathbf{x} = \begin{bmatrix} x_1 \ x_2 \end{bmatrix} \), the magnitude is calculated using the formula:
\[ \| \mathbf{x} \| = \sqrt{x_1^2 + x_2^2} \]

In our case, the vector components are \( x_1 = -1 \) and \( x_2 = -1 \). Substituting these into the formula gives:

• Squares of the components: \( (-1)^2 = 1 \), so for both components, it results in 1.
• Sum of the squares: \( 1 + 1 = 2 \).
• Sqare root of the sum: \( \sqrt{2} \).

Thus, the magnitude \( \| \mathbf{x} \| \) of the vector in this exercise is \( \sqrt{2} \). Understanding the vector magnitude is key to many applications, such as determining speed if the vector represents velocity.

Angle Calculation

Determining the angle a vector makes with a reference axis, such as the positive \( x_1 \)-axis, allows us to understand the vector's direction in the plane. For our vector \( \mathbf{x} = \begin{bmatrix} -1 \ -1 \end{bmatrix} \), we use the inverse tangent function:
\[ \theta = \tan^{-1} \left( \frac{x_2}{x_1} \right) \]

Inserting our values where both \( x_1 \) and \( x_2 \) are -1:

• Ratio calculation: \( \frac{-1}{-1} = 1 \).
• Inverse tangent: \( \tan^{-1}(1) = 45^{\circ} \).
• Adjust for the quadrant: Since our vector lies in the third quadrant (both components are negative), we add 180° to \( 45^{\circ} \).

Thus, the angle \( \theta \) is \( 225^{\circ} \). Note that understanding which quadrant the vector lies in is crucial to adjust the base angle provided by the inverse tangent function, ensuring the correct direction is depicted.