Problem 39 - College Trigonometry

Question

For the given vector \(\vec{v}\), find the magnitude \(\|\vec{v}\|\) and an angle \(\theta\) with \(0 \leq \theta<360^{\circ}\) so that \(\vec{v}=\|\vec{v}\|\langle\cos (\theta), \sin (\theta)\rangle\) (See Definition 11.8.) Round approximations to two decimal places. $$ \vec{v}=\langle-2.5,0\rangle $$

Step-by-Step Solution

Answer

Magnitude is 2.5, and the angle \(\theta\) is 180°.

Step 1: Calculate Magnitude of the Vector

To find the magnitude \(\|\vec{v}\|\), use the formula for the magnitude of a vector \(\vec{v}=\langle v_1, v_2 \rangle\): \(\|\vec{v}\| = \sqrt{v_1^2 + v_2^2}\). For \(\vec{v}=\langle -2.5, 0 \rangle\), the magnitude is: \[\|\vec{v}\| = \sqrt{(-2.5)^2 + 0^2} = \sqrt{6.25} = 2.5\]

Step 2: Express Vector Using Polar Coordinates

The task is to express \(\vec{v}\) in the form \(\|\vec{v}\|\langle \cos(\theta), \sin(\theta) \rangle\). Since \(\vec{v}=\langle -2.5, 0 \rangle\), this implies \(\cos(\theta) = -1\) and \(\sin(\theta) = 0\).

Step 3: Determine the Angle \(\theta\)

The angle \(\theta\) needs to satisfy \(\cos(\theta) = -1\) and \(\sin(\theta) = 0\). \(\cos(\theta) = -1\) occurs when \(\theta = 180^\circ\). Thus, \(\theta = 180^\circ\) is the correct angle for the given conditions.

Key Concepts

Polar Coordinates

Polar coordinates are a fascinating way of representing points in the plane using distances and angles. Instead of the usual Cartesian coordinates where you use an X and Y axis, polar coordinates use a central point called the origin, along with two parameters: the radius and the angle. The radius is the distance from the origin to the point, and the angle is measured from a reference direction, usually the positive X axis, counterclockwise.

Polar coordinates are perfect for vector description because they simplify the process of moving between the length of a vector and its orientation. In our case, the vector \( \vec{v} = \langle -2.5, 0 \rangle \) can be expressed in polar coordinates as \( \|\vec{v}\| \langle \cos(\theta), \sin(\theta) \rangle \). Here, \( \|\vec{v}\| \) is the radius, and \( \theta \) is the angle that describes the vector's direction.

Trigonometric Functions

Trigonometric functions, namely sine and cosine, are essential tools in the realm of polar coordinates. These functions relate the angles of a right triangle to the lengths of its sides and can extend to describe circular motion and oscillations.

For the vector \( \vec{v} = \langle -2.5, 0 \rangle \), the task is to find both the cosine and sine of the angle \( \theta \). This is because every point or vector plotted in polar coordinates can be described using these trigonometric functions:

• \( \cos(\theta) \) aligns with the X-coordinate (horizontal component).
• \( \sin(\theta) \) aligns with the Y-coordinate (vertical component).

In our example, \( \cos(\theta) = -1 \) and \( \sin(\theta) = 0 \). These values suggest that the vector lies entirely along the negative X-axis with no deviation in the Y direction.

Angle Determination

Determining an angle in polar coordinates is all about finding the correct orientation of the vector relative to the positive X axis. In essence, it involves figuring out the precise direction in which the vector points.

The vector \( \vec{v} = \langle -2.5, 0 \rangle \) has a cosine value of \( -1 \) and a sine value of \( 0 \). Such values are unique and direct indicators that the angle \( \theta \) is precisely \( 180^\circ \), also known as \( \pi \) radians.

• \( \theta = 180^\circ \) (or \( \pi \) radians) means the vector is pointing directly left along the negative X axis.

This makes sense because the negative value of the cosine function implies that the vector must be oriented in the opposite direction to the positive X axis.