Problem 68 - Precalculus with Limits

Question

In Exercises 67-74, find the component form of \(\mathbf{v}\) given its magnitude and the angle it makes with the positive \(x\)-axis. Sketch \(\mathbf{v}\). Magnitude - ||\(\mathbf{v}\)|| \(= 1\) Angle - \(\theta = 45^{\circ}\)

Step-by-Step Solution

Answer

The component form of the vector \(\mathbf{v}\) is \(\mathbf{v} = (\sqrt{2}/2 , \sqrt{2}/2)\).

Step 1: Determine the x-component of the vector

Use the formula \(v_x = r \cdot \cos(\theta)\). We know \(r = 1\) and \(\theta = 45^{\circ}\), but we should convert degrees to radians for appropriate usage in the cosine function. So, \(45^{\circ} = \pi / 4\) radians. Then, \(v_x = 1 \cdot \cos(\pi / 4) = \sqrt{2}/2\).

Step 2: Determine the y-component of the vector

Use the formula \(v_y = r \cdot \sin(\theta)\). We know \(r = 1\) and \(\theta = 45^{\circ}\), but we should convert degrees to radians for appropriate usage in the sin function. So, \(45^{\circ} = \pi / 4\) radians. Then, \(v_y = 1 \cdot \sin(\pi / 4) = \sqrt{2}/2\).

Step 3: Write down the component form of the vector

Now that we have computed \(v_x\) and \(v_y\), the component form of the vector: \(\mathbf{v} = (\sqrt{2}/2 , \sqrt{2}/2)\).

Step 4: Sketch \(\mathbf{v}\)

In order to sketch the vector \(\mathbf{v}\), start from the origin of the coordinate plane and draw a line to the point (\(\sqrt{2}/2 , \sqrt{2}/2\)). The line represents a vector with magnitude 1 and directed towards 45 degrees from the positive x-axis.

Key Concepts

Vector Magnitude

Vector magnitude tells us the length or size of a vector. It's like stretching a rubber band—knowing how long it is. For any vector \(\mathbf{v}\), we write its magnitude as \(||\mathbf{v}||\). In our example, the vector's magnitude is 1, meaning it has a length of 1 unit. Think of it like measuring a straight line on a piece of paper; it gives us a sense of how much space the vector covers.
To calculate the magnitude, use the Pythagorean theorem in 2D:

• \(||\mathbf{v}|| = \sqrt{v_x^2 + v_y^2}\)

With both components equal, as with our \( (\sqrt{2}/2, \sqrt{2}/2) \), this formula affirms that the magnitude is indeed 1.

Trigonometric Functions

Trigonometric functions like sine and cosine relate angles to the sides of a right triangle. They're vital for converting between angles and distances in vector problems. For a given angle \( \theta \),
the trigonometric functions help us find:

• The cosine part (\(\cos(\theta)\)) gives the horizontal distance—how much right or left.
• The sine part (\(\sin(\theta)\)) shows the vertical distance—how much up or down.

In our example, multiplying\( \cos(45^{\circ})\) and \(\sin(45^{\circ})\) by the vector's magnitude (1) gives the components \(\sqrt{2}/2\), showing the vector's direction and stretch.

Coordinate Plane

The coordinate plane is essentially a grid to locate points in two-dimensional space. Acting like a map, it helps us visualize vectors.
It consists of:

• The horizontal axis \((x\)-axis) representing left-right movements.
• The vertical axis \((y\)-axis) representing up-down movements.

Vectors are depicted as arrows starting from the origin \((0,0)\) and pointing to their coordinates. For our vector \(\mathbf{v} = (\sqrt{2}/2, \sqrt{2}/2)\), start at the origin and draw to these coordinates. This helps in understanding both direction and length visually.

Radians and Degrees

Radians and degrees are two units for measuring angles. Degrees are familiar from everyday life; a full circle is 360 degrees. Radians connect angle measures to a circle's radius, with a full circle being \(2\pi\) radians.
When dealing with trigonometric functions, using radians is often more natural mathematically. Converting between the two is key:

• To go from degrees to radians, multiply by \(\pi/180\).
• To go from radians to degrees, multiply by \(180/\pi\).

For example, \(45^{\circ}\) equals \(\pi/4\) radians, making it simple to use with cosine and sine in vector problems. This conversion bridges the gap between everyday angle measurements and mathematical calculations.