Problem 44 - Precalculus with Limits

Question

In Exercises 37-54, a point in rectangular coordinates is given. Convert the point to polar coordinates. \(\left(0, 5\right)\)

Step-by-Step Solution

Answer

The polar coordinates for the given rectangular coordinates (0,5) are (5, \(\frac{\pi}{2}\)).

Step 1: Calculate the magnitude of the vector \(r\)

First, calculate the magnitude of the vector using the formula \(r = \sqrt{x^{2} + y^{2}}\), where x and y are the rectangular coordinates. Here, x=0 and y=5. Hence, \(r = \sqrt{0^{2} + 5^{2}} = 5\).

Step 2: Calculate the angle \(\theta\)

Second, calculate the angle that the vector makes with the positive x-axis. This is done by finding the arctangent of y divided by x using the formula \(\theta = arctan\left(\frac{y}{x}\right)\). Here, since the denominator x=0, the angle \(\theta\) is not defined for the arctangent. When y is positive and x is 0, the angle \(\theta\) is taken as \(\frac{\pi}{2}\). Hence, the angle \(\theta = \frac{\pi}{2}\).

Step 3: Write the polar coordinates

Finally, the polar coordinates are given by the ordered pair (r, \(\theta\)). Therefore, the polar coordinates are (5, \(\frac{\pi}{2}\))

Key Concepts

Rectangular to Polar Conversion

When dealing with coordinates in a plane, two common systems are the rectangular (or Cartesian) coordinate system and the polar coordinate system. Understanding how to convert between these two systems is fundamental in various fields of mathematics and physics, so let's dive into the conversion process.

Converting a point from rectangular coordinates \( (x, y) \) where 'x' represents the horizontal axis and 'y' the vertical axis, to polar coordinates \( (r, \theta) \) requires two main steps. First, we find the magnitude \( r \) of the point from the origin (0, 0), which in simple terms, is the distance of the point from the origin. Second, we find the angle \( \theta \) that the line connecting the origin to our point makes with the positive x-axis.

To put this into practice, consider a point \( (0, 5) \) in rectangular coordinates. Using the Pythagorean theorem, we calculate \( r \) as the distance from the origin to the point, which turns out to be 5. The angle \( \theta \) in this case requires a special consideration; since the x-coordinate is zero, we cannot use our standard \( arctan \) calculation as it would be undefined. However, we know that the point lies directly above the origin on the y-axis, which corresponds to an angle of \( \frac{\pi}{2} \) or 90 degrees.

Understanding this process is important for students as it enhances their ability to visualize and solve problems in different coordinate systems, which is a skill that can be applied to many areas of study.

Magnitude of Vector

The concept of a vector's magnitude is a key element in physics and engineering, referring to the 'length' or 'size' of the vector. Whether the vector represents a force, velocity, or position in space, determining its magnitude is vital for solving many types of problems.

The magnitude of a vector given in rectangular coordinates \( (x, y) \) can be found using the formula \( r = \sqrt{x^{2} + y^{2}} \) – a direct application of the Pythagorean theorem. In our exercise, we calculated the magnitude of the vector for the point \( (0, 5) \) to be 5. This result makes intuitive sense: since the point lies on the y-axis, its distance from the origin is equal to the length of the segment along the y-axis.

Significance in Various Contexts

The magnitude is important in fields such as navigation, where it might represent the direct distance to a destination, or in engineering, where it might quantify the strength of a force. In our daily lives, the concept can be seen in action as simple as measuring how far a ball has rolled or how strong a gust of wind is. Educators emphasize on understanding the extensive use of vector magnitudes across different subjects and applications, reinforcing the idea with practical examples to help students easily grasp the concept.

Arctangent Calculation

The arctangent function, usually written as \( arctan \) or \( tan^{-1} \), plays a crucial role in trigonometry and is used to find the angle whose tangent is a given number. The arctangent is the inverse operation of the tangent, which effectively means that it helps us to find angles when we know the ratio of two sides of a right triangle.

For a point in rectangular coordinates \( (x, y) \) where \( x eq 0 \) and \( y eq 0 \), the angle \( \theta \) with the positive x-axis can be found using the formula \( \theta = arctan\left(\frac{y}{x}\right) \). However, this formula has limitations; it is undefined when \( x = 0 \) because division by zero is not possible. In our exercise, we encountered this very issue when trying to calculate \( \theta \) for the point \( (0, 5) \) since \( x = 0 \) made the \( arctan \) calculation inapplicable.

The Quadrant System

To overcome this, we refer to the quadrant system of the Cartesian coordinate plane. A point with a positive y and zero x resides in the first quadrant, indicating an angle of \( \frac{\pi}{2} \) or 90 degrees straight up from the x-axis. It's important for students to memorize the key angles and their corresponding positions on the coordinate plane to quickly identify the angle without the need for complex calculations.

Arctangent values are often used in various scientific and engineering disciplines, such as when determining the direction of a vector or when programming the trajectory of an object in simulations. Mastery of this concept will greatly aid students in their future studies and professional pursuits in mathematical or technical fields.