Problem 13 - Calculus for Biology and Medicine

Question

Suppose a vector \(\left[\begin{array}{l}x_{1} \\ x_{2}\end{array}\right]\) has length 3 and is \(15^{\circ}\) clockwise from the positive \(x_{1}\) -axis. Find \(x_{1}\) and \(x_{2}\).

Step-by-Step Solution

Answer

The vector components are approximately \(x_1 = 2.898\) and \(x_2 = 0.776\).

Step 1: Understand the given problem

We have a vector \( \left[\begin{array}{l} x_{1} \ x_{2} \end{array}\right] \) whose length is 3 and makes an angle of \(15^{\circ}\) with the positive \(x_{1}\)-axis. We need to find the components \(x_1\) and \(x_2\) of the vector.

Step 2: Use trigonometric definitions

The components of a vector can be found using trigonometry. For a vector \([x_1, x_2]\) at an angle \( \theta \) with the horizontal axis, \(x_1 = r \cdot \cos \theta\) and \(x_2 = r \cdot \sin \theta\), where \(r\) is the magnitude or length of the vector.

Step 3: Substitute the known values

Substitute the known values into the trigonometric formulas. We have \(r = 3\) and \(\theta = 15^{\circ}\) clockwise. Thus, the formulas become:\[x_1 = 3 \cdot \cos(15^{\circ})\]\[x_2 = 3 \cdot \sin(15^{\circ})\]

Step 4: Calculate \(x_1\) using cosine

Using the formula \(x_1 = 3 \cdot \cos(15^{\circ})\), calculate the value. Cosine of \(15^{\circ}\) is approximately 0.9659. Therefore, \[x_1 = 3 \cdot 0.9659 = 2.8977\]

Step 5: Calculate \(x_2\) using sine

Using the formula \(x_2 = 3 \cdot \sin(15^{\circ})\), calculate the value. Sine of \(15^{\circ}\) is approximately 0.2588. Therefore, \[x_2 = 3 \cdot 0.2588 = 0.7764\]

Step 6: Final answer

Thus, the components of the vector are \(x_1 = 2.8977\) and \(x_2 = 0.7764\).

Key Concepts

Vector Components

Understanding vector components is crucial when dealing with trigonometric vectors. Imagine you have a vector pointing in a specified direction, which can be described using these components. A vector in a 2D plane can be represented by two perpendicular components: one along the horizontal axis (often the x-axis) and another along the vertical axis (typically the y-axis). These components are like the vector's x and y parts.

• The horizontal component is often denoted as \(x_1\) in mathematical problems. It represents how far the vector stretches along the x-axis.
• The vertical component is typically represented as \(x_2\). This shows the vector's extent along the y-axis.

Identifying these components is a fundamental step in vector analysis, allowing you to understand and express the vector in a coordinate system.

Trigonometry

Trigonometry is a powerful mathematical tool used to explore relationships between the angles and sides of triangles. In the context of vectors, trigonometry helps to find the vector's components using angles. For a vector making an angle \(\theta\) with the x-axis and having a magnitude \(r\), you can employ trigonometric functions to resolve it into its components.

• Cosine function: This function is used to find the horizontal component of the vector. The formula is \(x_1 = r \cdot \cos \theta\).
• Sine function: This one helps in finding the vertical component. The formula is \(x_2 = r \cdot \sin \theta\).

By substituting the known angle and vector magnitude into these formulas, you can accurately determine the measurements for each component. This method simplifies complex vector diagrams into easy-to-understand components.

Vector Magnitude

The magnitude of a vector can be thought of as its length or size. It tells you how strong or powerful the vector is. In a 2D space, the magnitude is often the starting point for calculations involving vectors, especially when dealing with trigonometric vectors.

• Magnitude is indicated by the symbol \(r\) or sometimes noted as \(|\vec{v}|\), representing the total length of the vector in its space.
• Knowing the magnitude is vital since it provides the scale for component calculations via trigonometry. In our current example, the magnitude is given as 3 units.

Understanding vector magnitude is essential because it forms the base for determining the direction and components of the vector using trigonometric formulas. With these, the vector can be broken down into as many smaller vectors as necessary for calculations.