Problem 76 - Precalculus

Question

Convert \((-\sqrt{3}, 1)\) to polar coordinates. Solution: Label \(x\) and \(y . \quad x=-\sqrt{3}, y=1\) Find \(r\) \(r=\sqrt{x^{2}+y^{2}}=\sqrt{3+1}=\sqrt{4}=2\) Find \(\theta . \quad \tan \theta=\frac{1}{-\sqrt{3}}=-\frac{1}{\sqrt{3}}\) \(\theta=\tan ^{-1}\left(-\frac{1}{\sqrt{3}}\right)=-\frac{\pi}{4}\) Write the point in polar coordinates. \(\left(2,-\frac{\pi}{4}\right)\) This is incorrect. What mistake was made?

Step-by-Step Solution

Answer

The mistake was in calculating \(\theta\) without considering the quadrant. The right angle is \(\theta = \frac{3\pi}{4}\), resulting in \((2, \frac{3\pi}{4})\).

Step 1: Identify Cartesian Coordinates

Start by labeling the given Cartesian coordinates:\(x = -\sqrt{3}\) and \(y = 1\).

Step 2: Calculate the Radius r

The radius \(r\) in polar coordinates is calculated using the formula:\[ r = \sqrt{x^2 + y^2} \]Substitute the values of \(x\) and \(y\):\[ r = \sqrt{(-\sqrt{3})^2 + 1^2} = \sqrt{3 + 1} = \sqrt{4} = 2 \]

Step 3: Calculate the Angle \(\theta\) with Correct Signs

To find \(\theta\), use \(\tan \theta = \frac{y}{x} = \frac{1}{-\sqrt{3}} = -\frac{1}{\sqrt{3}}\).The inverse tangent of \(-\frac{1}{\sqrt{3}}\) typically gives us \(\theta = -\frac{\pi}{4}\). However, since the point is in the second quadrant (where \(x\) is negative and \(y\) is positive), \(\theta\) should be corrected to \(\theta = \pi - \frac{\pi}{4} = \frac{3\pi}{4}\).

Step 4: Write the Correct Polar Coordinates

The correct polar coordinates are thus:\[ (r, \theta) = \left(2, \frac{3\pi}{4}\right) \].

Key Concepts

Cartesian Coordinates

When working with graphs, it's important to understand Cartesian coordinates. These consist of two values, typically

• \( x \): representing the horizontal position
• \( y \): representing the vertical position

These coordinates help us to describe a point on a plane. For example, the Cartesian coordinates \((-\sqrt{3}, 1)\) tell us:
- The point is \(-\sqrt{3}\) units on the left from the origin along the x-axis while
The point is 1 unit above the origin along the y-axis.
Understanding where every point lies in each quadrant helps in transitioning to polar coordinates, especially when determining angle values.

Radius

In polar coordinates, we describe a point in terms of a radius and an angle. Calculate the radius by determining the distance of the point from the origin. Use the formula:
\[r = \sqrt{x^2 + y^2}\]
This is just an application of the Pythagorean theorem, which helps find the hypotenuse of a right triangle.
Substituting the values of the given Cartesian coordinates, we found:

• \( r = \sqrt{(-\sqrt{3})^2 + 1^2} \)
• \( r = \sqrt{3 + 1} = \sqrt{4} = 2 \)

The radius, \( r \), is always a positive value representing the distance of the point from the origin.

Angle Conversion

Once you know the radius, the next step is to calculate the angle, \( \theta \), which shows the direction from the origin. We use the tangent function to find the angle:

• \( \tan \theta = \frac{y}{x} \)

For the given point, \[ tan \theta = \frac{1}{-\sqrt{3}} = -\frac{1}{\sqrt{3}} \]
Applying inverse tangent gives \( \theta = \tan^{-1}\left(-\frac{1}{\sqrt{3}}\right) = -\frac{\pi}{4} \) initially.
But, it is important to note the quadrant of the point,
since the location affects the true angle. Here, the \( x \) is negative and \( y \) is positive, so our point lies in the second quadrant.
Adjust the angle by using:
\[ \theta = \pi - \frac{\pi}{4} = \frac{3\pi}{4}\]
This adjustment ensures the angle correctly reflects the point's position.

Trigonometric Functions

Trigonometric functions play a key role in converting between Cartesian and polar coordinates. Mainly,

• Sine (\( \sin \theta \)) and
• Cosine (\( \cos \theta \))

are used to relate angles and sides in a right triangle.
In our conversion, tangent (\( \tan \theta = \frac{\sin \theta}{\cos \theta} \)) helps us find the angle associated with a point.
By manipulating these functions,
one can derive any angle when given \( x \) and \( y \).
This understanding supports the process once you grasp the relationships between these trigonometric values and their corresponding angles.