Problem 31 - Contemporary Precalculus

Question

Find the component form of the vector \(v\) whose magnitude and direction angle \(\theta\) are given. $$\|\mathbf{v}\|=6, \theta=40^{\circ}$$

Step-by-Step Solution

Answer

Answer: The component form of the vector is approximately ⟨4.56, 3.85⟩.

Step 1: Write down the given information

The given information is the magnitude of the vector \(\|v\|=6\) and its direction angle \(\theta=40^{\circ}\).

Step 2: Convert the angle from degrees to radians

In order to use the formulas for finding the components of the vector, we need the angle in radians. To convert the given angle to radians, use the following formula: $$\text{Radians} = \text{Degrees} \times \frac{\pi}{180}$$ So, we have: $$\theta = 40^{\circ} \times \frac{\pi}{180} = \frac{2\pi}{9} \text{ radians}$$

Step 3: Calculate the x-component of the vector

To find the x-component of the vector, use the formula: $$v_x = \|v\| \cos{\theta}$$ Plug in the values for the magnitude and direction angle: $$v_x = 6 \cos{\left(\frac{2\pi}{9}\right)}$$ Compute the value: $$v_x \approx 4.56$$

Step 4: Calculate the y-component of the vector

To find the y-component of the vector, use the formula: $$v_y = \|v\| \sin{\theta}$$ Plug in the values for the magnitude and direction angle: $$v_y = 6 \sin{\left(\frac{2\pi}{9}\right)}$$ Compute the value: $$v_y \approx 3.85$$

Step 5: Write the vector in component form

Now that we have the x and y-components of the vector, we can write the vector in component form as: $$\mathbf{v} = \langle 4.56, 3.85 \rangle$$

Key Concepts

Vector Magnitude

Understanding the concept of vector magnitude is fundamental when dealing with vectors in any dimension. The magnitude of a vector evaluates how long the vector is, regardless of its direction. You can think of it as the vector's length. In mathematics, when we say a vector \(|\mathbf{v}|\), we are often referring to its magnitude. For a vector defined in two-dimensional space, you can calculate the magnitude using the Pythagorean theorem if you know its components, using the formula: \\[ |\mathbf{v}| = \sqrt{v_x^2 + v_y^2} \]In our problem, the magnitude is given as 6. This means the distance from the origin to the vector's point is 6 units. This is another way to understand the strength or the size of the vector, independent of where it's pointing to.

Direction Angle

The direction angle of a vector gives us information about the direction in which the vector is pointed. It is usually measured from the positive x-axis toward the vector, using degrees or radians.If you imagine standing at the origin of a coordinate plane, the direction angle is like a compass direction pointing you to your vector's orientation. In our specific example, the given direction angle \(\theta\) is \(40^{\circ}\). This indicates that the vector is pointed \(40^{\circ}\) away from the rightward horizontal, in a counterclockwise direction. Calculating the components of the vector later allows us to understand precisely in which direction it's headed.

Radians Conversion

Working with angles in radians can often be necessary, especially in trigonometry and calculus, as many mathematical functions and formulas assume radian measure.To convert angles from degrees into radians, the conversion factor \(\frac{\pi}{180}\) is used. This stems from the fact that a full angle of 360° corresponds to \(2\pi\) radians.For the problem at hand, converting \(40^{\circ}\) into radians, we calculate as: \[ \theta_{\text{radians}} = 40^{\circ} \times \frac{\pi}{180} = \frac{2\pi}{9} \]Now, with the angle in radians, you can effectively use trigonometric functions to find the components of the vector.

Trigonometric Components

To express a vector in its component form, we use its magnitude and direction angle with trigonometric functions to find the x and y components.These components are calculated as:

• X-Component: \(v_x = |\mathbf{v}| \cos(\theta)\)
• Y-Component: \(v_y = |\mathbf{v}| \sin(\theta)\)

Substituting in the values from the problem, with \(|\mathbf{v}| = 6\) and \(\theta = \frac{2\pi}{9}\), we find:

• \(v_x = 6 \cos\left(\frac{2\pi}{9}\right) \approx 4.56\)
• \(v_y = 6 \sin\left(\frac{2\pi}{9}\right) \approx 3.85\)

Thus, the vector in component form is \(\langle 4.56, 3.85 \rangle\), which gives a complete picture of the vector's direction and magnitude.