Problem 50 - Precalculus Mathematics for Calculus

Question

Find the magnitude and direction (in degrees) of the vector. $$\mathbf{v}=\langle 40,9\rangle$$

Step-by-Step Solution

Answer

The vector's magnitude is 41 and its direction is approximately 12.7 degrees.

Step 1: Understand the Components

The vector \( \mathbf{v} \) is given in component form as \( \mathbf{v} = \langle 40, 9 \rangle \), which means it has an x-component of 40 and a y-component of 9. To find the magnitude and direction, we'll analyze these components.

Step 2: Calculate the Magnitude

The magnitude of vector \( \mathbf{v} \) can be found using the formula for the magnitude of a vector, which is \( \| \mathbf{v} \| = \sqrt{x^2 + y^2} \). Substituting the given components, we have:\[ \| \mathbf{v} \| = \sqrt{40^2 + 9^2} \] Calculate the squares and the square root.\( 40^2 = 1600 \) and \( 9^2 = 81 \). The sum is \( 1600 + 81 = 1681 \).The square root is \( \sqrt{1681} = 41 \).Therefore, the magnitude of the vector is 41.

Step 3: Calculate the Direction

The direction of the vector \( \mathbf{v} \) is calculated using the tangent function. The angle \( \theta \) that the vector makes with the positive x-axis is given by:\[ \theta = \tan^{-1} \left( \frac{y}{x} \right) = \tan^{-1} \left( \frac{9}{40} \right) \]Calculate \( \tan^{-1} \left( \frac{9}{40} \right) \) using a calculator:\( \theta \approx 12.7^{\circ} \).Thus, the direction of the vector is approximately \( 12.7^{\circ} \) from the positive x-axis.

Key Concepts

Vector Components

Vectors are entities known by their magnitude and direction, and are represented through components along coordinate axes. When a vector is in 2D, it can be broken down into an x-component and a y-component. In our example, the vector \( \mathbf{v} = \langle 40, 9 \rangle \), which means:

• The x-component (horizontal direction) is 40.
• The y-component (vertical direction) is 9.

This representation of vectors is called component form. It allows us to easily apply mathematical operations to the vector, such as calculating its magnitude and direction. By isolating the effects in two perpendicular directions, the math becomes simpler and more intuitive,
making vector components a major tool in analysis of vector-related problems.

Pythagorean Theorem

The Pythagorean theorem is fundamental in geometry. It's also crucial when dealing with vectors analytically. When you have the components of a vector, you can think of them as two sides of a right triangle. Here, the x-component and y-component serve as perpendicular sides.

• For our vector \( \mathbf{v} = \langle 40, 9 \rangle \), the magnitude (or length) of the vector acts like the hypotenuse of this triangle.
• Using the theorem, the magnitude can be calculated as \( \| \mathbf{v} \| = \sqrt{40^2 + 9^2} \).

Inserting the values and solving gives us \( \sqrt{1681} = 41 \).

This shows how powerful the Pythagorean theorem is for converting vector components to a single scalar value, representing the vector's magnitude.

Inverse Tangent Function

Decision-makers often need to know not just how powerful, but also in which direction forces (or other vector quantities) act. The inverse tangent function, or \( \tan^{-1} \), helps in determining the angle of a vector relative to the positive x-axis by using its components.

• In the context of our vector, \( \mathbf{v} = \langle 40, 9 \rangle \), the direction (angle \( \theta \)) is calculated with \( \theta = \tan^{-1} \left( \frac{9}{40} \right) \).
• This ratio is the slope of the line formed by the vector from the origin.

Approximating this using a calculator yields an angle of \( 12.7^{\circ} \).

The inverse tangent function is thus indispensable when finding the direction of a vector from its components, converting between algebraic and geometric representations.