Problem 31 - Precalculus

Question

Find the vector, given its magnitude and direction angle. $$|\mathbf{u}|=9, \theta=335^{\circ}$$

Step-by-Step Solution

Answer

The vector is approximately \((8.1567, -3.8034)\).

Step 1: Understand the Problem

In this problem, you are given the magnitude (or length) of the vector, which is 9, and the direction angle, which is \(335^{\circ}\). The task is to find the vector's components (\(x\) and \(y\)).

Step 2: Use the Magnitude and Direction Angle Formula

The formulas to find the components of the vector \(\mathbf{u}\) from its magnitude and direction angle \(\theta\) are: \( u_x = |{u}| \cos \theta \) and \( u_y = |{u}| \sin \theta \). These formulas will help us to find the \(x\) (horizontal) and \(y\) (vertical) components.

Step 3: Calculate the x-component

Substitute the given values into the formula for the \(x\)-component: \( u_x = 9 \cos 335^{\circ} \). Calculate \( \cos 335^{\circ} \) using a calculator or a trigonometry table. \(\cos 335^{\circ} = \cos (360^{\circ} - 25^{\circ}) = \cos 25^{\circ} \). Approximate this value to get \(\cos 335^{\circ} \approx 0.9063\). Therefore, \( u_x \approx 9 \times 0.9063 \approx 8.1567 \).

Step 4: Calculate the y-component

Substitute the values into the formula for the \(y\)-component: \( u_y = 9 \sin 335^{\circ} \). Calculate \( \sin 335^{\circ} \) using a calculator. \(\sin 335^{\circ} = \sin (360^{\circ} - 25^{\circ}) = -\sin 25^{\circ} \). Approximate this value to get \(\sin 335^{\circ} \approx -0.4226\). Therefore, \( u_y \approx 9 \times (-0.4226) \approx -3.8034 \).

Step 5: Write Down the Vector

Now that you've calculated both components, you can write the vector as \( \mathbf{u} = (8.1567, -3.8034) \).

Key Concepts

Magnitude of a Vector

The magnitude of a vector is essentially its length or size. Think of it as the distance from the vector's starting point to its endpoint.
A vector's magnitude is always a non-negative number. It measures how 'strong' or 'big' the vector is in absolute terms, without considering its direction.
The notation for the magnitude of a vector \( \mathbf{u} \) is \( |\mathbf{u}| \). For example, in our exercise, the magnitude is given as 9. This implies that the 'strength' or length of the vector is 9 units.

• The magnitude is specifically important as it dictates the scale of the vector.
• Understanding magnitude independently of direction helps in comparing vectors based on their size alone.

Calculating or using magnitude in problems involves basic arithmetic when combined with component formulas.

Direction Angle

The direction angle of a vector is what gives the vector its orientation. In simple terms, it tells you which way the vector is pointing.
In our example exercise, the direction angle is \( 335^{\circ} \). This angle is measured in a counterclockwise direction from the positive x-axis.

• Angles in vector problems often require conversion or recalculation depending on the reference point.
• For instance, \(335^{\circ}\) can also be visualized as \(360^{\circ} - 25^{\circ}\), indicating a 25-degree move clockwise past the positive x-axis.

Understanding the direction angle is crucial as it allows for the decomposition of the vector into its components, letting you see the impact of the vector in any given direction.
It's the combination of the angle and magnitude that defines a vector's true nature.

Trigonometric Functions

Trigonometric functions play a pivotal role in breaking down vectors into components. In vector problems, you'll often encounter sine (\(\sin \)) and cosine (\(\cos \)) functions.
These functions are used to calculate how much of the vector is acting in the 'x' direction (horizontal) and the 'y' direction (vertical).

• For any angle, \( \theta \), \( \cos(\theta) \) gives the horizontal influence, while \( \sin(\theta) \) gives the vertical influence.
• These trig values are vital for resolving vectors into components using formulas like \( u_x = |\mathbf{u}| \cos \theta \) and \( u_y = |\mathbf{u}| \sin \theta \).

In the exercise, you saw cosine and sine used with \(335^{\circ}\) to find the vector's components.
The trigonometric functions are critical as they transform directional data (angles) into numerical values you can work with in equations.