Problem 24 - Calculus for Biology and Medicine

Question

Find the angle between \(\mathbf{x}=[-1,2]^{\prime}\) and \(\mathbf{y}=[-2,4]\).

Step-by-Step Solution

Answer

The angle between the vectors is 0 degrees.

Step 1: Dot Product Calculation

The first step to find the angle between two vectors is to calculate the dot product of these vectors. The formula for the dot product is \( \mathbf{x} \cdot \mathbf{y} = x_1y_1 + x_2y_2 \). For vectors \( \mathbf{x} = [-1,2] \) and \( \mathbf{y} = [-2,4] \): \[ \mathbf{x} \cdot \mathbf{y} = (-1)(-2) + (2)(4) = 2 + 8 = 10 \]

Step 2: Calculate Magnitudes of Vectors

Next, calculate the magnitudes (lengths) of the vectors \( \mathbf{x} \) and \( \mathbf{y} \) using the formula \( \| \mathbf{v} \| = \sqrt{v_1^2 + v_2^2} \). For \( \mathbf{x} \): \[ \| \mathbf{x} \| = \sqrt{(-1)^2 + 2^2} = \sqrt{1 + 4} = \sqrt{5} \] For \( \mathbf{y} \): \[ \| \mathbf{y} \| = \sqrt{(-2)^2 + 4^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5} \]

Step 3: Apply the Cosine Formula

Using the dot product and magnitudes, apply the cosine formula: \( \cos{\theta} = \frac{\mathbf{x} \cdot \mathbf{y}}{\| \mathbf{x} \| \| \mathbf{y} \|} \). Substitute the values obtained: \[ \cos{\theta} = \frac{10}{\sqrt{5} \times 2\sqrt{5}} = \frac{10}{10} = 1 \]

Step 4: Find the Angle

Since \( \cos{\theta} = 1 \), this means \( \theta = \cos^{-1}(1) \), which implies that \( \theta = 0 \) degrees.

Key Concepts

Dot Product

The dot product is a fundamental concept in vector calculus. It is essentially a way to multiply two vectors to get a single number, otherwise known as a scalar. This operation helps us understand the relationship between vectors, especially in terms of angles and projections.
To calculate the dot product of two vectors, each with two components, use the formula:

• \( \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 \)

In the given exercise, we have two vectors, \( \mathbf{x} = [-1, 2] \) and \( \mathbf{y} = [-2, 4] \). Applying the formula, we find

• \( \mathbf{x} \cdot \mathbf{y} = (-1)(-2) + (2)(4) = 2 + 8 = 10 \)

This tells us that the dot product of these two vectors is 10. The significance of the dot product value is not just a number; it gives insight into the angular relationship between the vectors. A dot product of zero would mean the vectors are perpendicular, but in this case, it is positive, pointing to a different relationship.

Magnitude of Vectors

The magnitude of a vector, often referred to as its length or norm, provides a measure of how long the vector is. To calculate the magnitude of a vector with components \( [v_1, v_2] \), the formula is:

• \( \| \mathbf{v} \| = \sqrt{v_1^2 + v_2^2} \)

For vector \( \mathbf{x} = [-1, 2] \), its magnitude is:

• \( \| \mathbf{x} \| = \sqrt{(-1)^2 + 2^2} = \sqrt{1 + 4} = \sqrt{5} \)

Similarly, for vector \( \mathbf{y} = [-2, 4] \):

• \( \| \mathbf{y} \| = \sqrt{(-2)^2 + 4^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5} \)

The magnitudes \( \sqrt{5} \) and \( 2\sqrt{5} \) tell us how large each vector is. The magnitude plays a crucial role in defining the geometric representation of vectors and is indispensable when determining the angle between vectors.

Cosine of Angle Between Vectors

To find the angle between two vectors using vector calculus, we use the cosine formula, which connects the dot product and magnitudes of the vectors. The formula is:

• \( \cos{\theta} = \frac{\mathbf{x} \cdot \mathbf{y}}{\| \mathbf{x} \| \| \mathbf{y} \|} \)

For the vectors \( \mathbf{x} \) and \( \mathbf{y} \), substituting the known values gives:

• \( \cos{\theta} = \frac{10}{\sqrt{5} \times 2\sqrt{5}} = \frac{10}{10} = 1 \)

This result indicates that \( \cos{\theta} = 1 \).
In trigonometry, when the cosine of an angle is 1, the angle \( \theta \) is equal to 0 degrees. This implies that the two vectors \( \mathbf{x} \) and \( \mathbf{y} \) are in the same direction. Understanding how to use these calculations helps in analyzing vector directions and angles, which is widely applicable in physics and engineering.