Problem 41 - Precalculus : Building Concepts and Connections

Question

Find the components of the vector in standard position that satisfy the given conditions. Magnitude \(19 ;\) direction \(34^{\circ}\)

Step-by-Step Solution

Answer

The components of the vector are approximately \(15.91, 10.58\)

Step 1: Convert Degree to Radian

First, convert the angle from degrees to radians, since radians are a more natural unit of measurement for angles in mathematics. To do this, multiply the angle by \(\frac{\pi}{180}\).

Step 2: Calculate X and Y Components

The x component of the vector is equal to the magnitude of the vector times the cosine of the angle. Therefore, the x component of the vector is \(19*\cos(34^{\circ})\). The y component of the vector is equal to the magnitude of the vector times the sine of the angle. Therefore, the y component of the vector is \(19*\sin (34^{\circ})\).

Step 3: Evaluate for X and Y components

Evaluate the expression \(19*\cos(34^{\circ})\) for the x component, and \(19*\sin(34^{\circ})\) for the y component.

Key Concepts

Magnitude and Direction of Vectors

When dealing with vectors in physics or mathematics, two fundamental characteristics must be understood: the magnitude and the direction. Think of a vector as an arrow pointing from one point to another. The magnitude is the length of this arrow, representing how much force or how far an object would move in that direction.

In our exercise, the provided vector has a magnitude of 19 units. This means if the vector represents a force, for example, it is a force of 19 units. The direction, on the other hand, tells us where the vector is pointing. It's typically measured in degrees from a reference direction (usually the positive x-axis) going counterclockwise. In the exercise, the direction is given as 34 degrees, which tells us the vector is 34 degrees above the positive x-axis.

Understanding these two qualities helps us visualize vectors and later on, calculate their components which are fundamental in vector-related calculations in both two and three-dimensional spaces.

Converting Degrees to Radians

Since trigonometric functions in the context of vectors often require angles to be in radians, it’s crucial to know how to convert degrees to radians. The conversion factor is \(\frac{\pi}{180}\) because there are \(2\pi\) radians in a full circle (360 degrees).

For our exercise, we take the given angle of 34 degrees and multiply it by \(\frac{\pi}{180}\), so \(34\times\frac{\pi}{180}\) radians. This conversion is step one in solving the exercise and it's the stepping stone for successfully calculating the vector components through trigonometric functions.

Calculating Vector Components

Knowing the magnitude and direction of a vector allows us to find its components along the x- and y-axes in a coordinate system. A vector's component along the x-axis is found by multiplying the magnitude of the vector by the cosine of the angle it makes with the x-axis. Similarly, the y-component is found by multiplying the magnitude by the sine of the angle.

In formula terms, if the magnitude is \(M\) and the direction angle is \(\theta\):

• X-component: \(M\cdot\cos(\theta)\)
• Y-component: \(M\cdot\sin(\theta)\)

For our exercise, we are calculating these components given a magnitude of 19 and a direction of 34 degrees, after converting it to radians. The result are the specific values that describe how far along the x- and y-axes the vector extends from the origin.

Trigonometric Functions in Vectors

The applications of trigonometry are vast in vector calculus, particularly sine (sin) and cosine (cos), which are used to resolve a vector into its components. The sine function relates the opposite side to the hypotenuse in a right-angled triangle, while the cosine relates the adjacent side to the hypotenuse.

When applying these to vectors, the hypotenuse would be the magnitude of the vector, and the sides become the x and y components of the vector. Therefore, trigonometric functions enable us to navigate between the magnitude-direction form of a vector and its component form.

Utilizing these functions in our initial exercise, we see that they are the bridge connecting the vector’s magnitude and direction to its x and y components, ultimately allowing us to analyse vectors in a more algebraic and less geometric way.