Problem 1 - Technical Mathematics with Calculus

Question

Given the magnitude of each vector and the angle \(\theta\) that it makes with the \(x\) axis, find the \(x\) and \(y\) components. $$\text { Magnitude }=4.93 \quad \theta=48.3^{\circ}$$

Step-by-Step Solution

Answer

The x-component of the vector is approximately 3.34 and the y-component is approximately 3.65.

Step 1: Understand the Components of a Vector

The vector components along the x and y axes can be found using trigonometry. The x-component of a vector is the magnitude of the vector times the cosine of the angle it makes with the x-axis, and the y-component is the magnitude times the sine of the angle. The components can be represented as: x-component = magnitude × cos(θ) and y-component = magnitude × sin(θ).

Step 2: Calculate the x-component

Use the cosine function to find the x-component of the vector. The magnitude of the vector is given as 4.93 and the angle θ is given as 48.3 degrees. The x-component (A_x) is calculated as A_x = 4.93 × cos(48.3°).

Step 3: Calculate the y-component

Use the sine function to find the y-component of the vector. Again, the magnitude of the vector is 4.93 and the angle θ is 48.3 degrees. The y-component (A_y) is calculated as A_y = 4.93 × sin(48.3°).

Key Concepts

Trigonometry in Vectors

When we deal with vectors in mathematics and physics, trigonometry is the bridge that connects vector quantities to their components along the Cartesian axes. Imagine breaking down the movement of an arrow flying towards a target into two separate motions - one horizontal and one vertical. Similarly, a vector can be dissected into its horizontal (x-component) and vertical (y-component) parts using trigonometric functions.

The angle the vector makes with the horizontal axis (often the x-axis) is key to finding these components. The cosine of this angle gives us the ratio of the vector's horizontal reach relative to its total length, which is why the x-component is the vector's magnitude multiplied by the cosine of the angle. The y-component, on the other hand, is the vector's reach into the vertical, given by the magnitude times the sine of the angle because sine relates to the vertical side in a right-angled triangle.

Vector Magnitude and Direction

A vector's magnitude and direction are its defining features. The magnitude is a measure of how strong or long the vector is, such as the force behind a push or the distance of a walk in a straight line. You can picture this as the length of an arrow drawn on paper. The direction, meanwhile, is the way the arrow points and is indicated by an angle, often measured from the positive x-axis going counter-clockwise.

To completely describe a vector's journey, you need to specify both how far it intends to travel (its magnitude) and the path it takes (its direction). These traits let us convert between a vector's algebraic form (with magnitude and direction) and its component form (with x and y values), which is extremely handy for doing calculations in physics and engineering, as it aligns the vector with the familiar coordinate system.

Calculating Vector Components

Pinpointing the x and y components of a vector allows us to navigate the vector with precision in the two-dimensional plane of Cartesian coordinates. It's like having the exact address of a location rather than just knowing it's somewhere in the northwest.

Here’s a handy method to find these components: first, sketch the vector with its tail at the origin and its tip pointing in the direction according to its angle. Then, draw a right triangle using the vector as the hypotenuse. The base of the triangle will represent the x-component, and the height will be the y-component. By using the magnitude of the vector as the hypotenuse's length, simple trigonometry tells us:

• The x-component equals the magnitude times the cosine of the angle (horizontal stretch).
• The y-component equals the magnitude times the sine of the angle (vertical stretch).

With the initial example, given the angle of 48.3°, the calculations would be straightforward: multiply the magnitude, which is 4.93, by the cosine and sine of 48.3° to find the corresponding components.