# Analytic Geometry in FE Electrical and Computer

Welcome to the Electrical and Computer Engineering world, where the prolonged encounter with Analytic Geometry is inevitable. Analytic Geometry in FE Electrical and Computer engineering is essential for solving circuits and signal processing problems.

Analytic Geometry is a branch of mathematics that studies geometric shapes using algebraic equations. Using algebraic expressions, it is a powerful tool that allows engineers and computer experts to represent and manipulate geometric objects, such as points, lines, curves, and surfaces.

In Electrical and Computer Engineering, Analytic Geometry plays a vital role in various applications such as designing circuits, analyzing waveforms, and modeling systems. By utilizing the principles of Analytic Geometry, engineers can visualize complex systems in a more intuitive way, making it easier to identify and solve problems.

This blog will explore the exciting domain of Analytic Geometry in Electrical and Computer Engineering. We will delve into various Analytic Geometry applications and provide insights on using this tool effectively to solve field problems. So, fasten your seat belts, put on your thinking caps, and let’s take a deep dive into the world of Analytic Geometry

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## Coordinate Systems

Cartesian and Polar coordinate systems are two different ways of representing points in two-dimensional space.

### Cartesian Coordinate Systems

The Cartesian coordinate system, or the rectangular coordinate system, is based on a grid of perpendicular axes that intersect at the origin point. The two axes are typically labeled x and y. Each point in the plane is represented by an ordered pair of numbers (x, y), where x represents the horizontal distance from the origin. Y represents the vertical distance from the origin. The x-axis is usually considered the horizontal axis, while the y-axis is the vertical axis.

For example, consider the point (3, 4) in the Cartesian coordinate system. This point is located 3 units to the right of the y-axis and 4 units above the x-axis.

The location of the point (3,4) in different quadrants with variations of sign convention will be given as:

Point | Quadrant | x | y | Description |

(3, 4) | 1st | 3 | 4 | The point (3, 4) lies in the first quadrant because both its x and y coordinates are positive. |

(-3, 4) | 2nd | -3 | 4 | The point (-3, 4) lies in the second quadrant because its x coordinate is negative, and its y coordinate is positive. |

(-3, -4) | 3rd | -3 | -4 | The point (-3, -4) lies in the third quadrant because its x and y coordinates are negative. |

(-4, -3) | 4th | -4 | -3 | The point (-4, -3) lies in the fourth quadrant because its x coordinate is negative, and its y coordinate is positive. |

The four quadrants in the Cartesian coordinate system are numbered counterclockwise from the positive x-axis. The first quadrant is where both x and y coordinates are positive, and the second is where x is negative. Y is positive, the third quadrant is where both x and y coordinates are harmful, and the fourth quadrant is where x is positive and y is negative.

Understanding which quadrant a point lies in is crucial in many applications, such as determining the direction of vectors or calculating trigonometric functions.

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### Polar Coordinate Systems

The Polar coordinate system, on the other hand, is based on a different set of coordinates. In this system, each point is represented by an ordered pair (r, θ), where r represents the distance of the point from the origin and θ represents the angle made by the line connecting the point to the origin concerning a reference line, usually the positive x-axis.

To represent the point (3, 4) in polar coordinates, we need to find their distance from the origin (r) and the angle it makes with the positive x-axis (θ).

We can use the Pythagorean theorem to find the distance r:

r = sqrt(3^{^}2 + 4^{^}2) = 5

So, the distance from the origin to the point (3, 4) is 5 units.

To find the angle θ, we can use trigonometric functions. The tangent of θ is the ratio of the opposite side to the adjacent side in a right triangle, which in this case is 4/3. Therefore, we can use the arctangent function (tan^{^}-1) to find θ:

θ = tan^{^}-1(4/3) ≈ 0.93 radians ≈ 53.13 degrees

So, in polar coordinates, the point (3, 4) can be represented as (5, 0.93), where 5 is the distance from the origin, and 0.93 radians (or 53.13 degrees) is the angle made with the positive x-axis.

In the context of trigonometry, finding the angle and distance of a point in polar coordinates is similar to finding the magnitude and direction of a vector, which we will discuss in the following section (Vectors can be represented using polar coordinates, where the magnitude is the length of the vector, and the direction is the angle it makes with a reference line).

## Vectors

A vector is a physical quantity with magnitude and direction (unlike scalers that only contain magnitude), such as Force, Torque, and Weight. It is represented by an arrow pointing from its tail to its head, where the length of the arrow represents the magnitude, and the direction of the arrow represents the direction.

To represent the point (3, 4) as a vector, we need to subtract the origin (0, 0) from the point (3, 4) to get the vector components:

V = (3-0, 4-0) = (3, 4)

This means that the vector V has a length of 5 units (which we found in the previous answer) and points in the direction of the point (3, 4) from the origin.

To find the unit vector in the direction of V, we need to divide the vector by its magnitude:

Unit vector of V = uV = V / |V| = (3/5, 4/5) or ⅗ i + ⅘ j

So, the unit vector in the direction of the vector V is (3/5, 4/5).

The vector’s magnitude is the vector’s length and is denoted by |V|. In this case, the magnitude of vector V is 5 units, the same as the distance from the origin to the point (3, 4) we found earlier in the polar coordinate representation.

The magnitude and direction of a vector are essential concepts in physics, engineering, and many other fields. They describe and analyze physical quantities such as force, velocity, and acceleration.

In vector form, the point (3, 4) can be represented as a displacement vector from the origin (0, 0) to the point (3, 4). This can be done by subtracting the origin’s position vector from the point’s position vector (3, 4).

The point (3, 4) position vector can be written as (3i + 4j), where i and j are unit vectors in the x and y directions, respectively. The “i” vector points in the positive x direction, and the “j” vector points in the positive y direction.

Similarly, the position vector of the origin can be written as (0i + 0j) or simply 0.

Therefore, the displacement vector from the origin to the point (3, 4) can be written as:

(3i + 4j) – 0 = 3i + 4j

So, the point (3, 4) can be represented in vector form as the displacement vector 3i + 4j.

In physics and engineering, vector notation is commonly used to represent physical quantities such as forces and velocities. Vector addition and subtraction can combine or resolve forces or velocities, and dot and cross products can calculate work, energy, and torque. Vector notation is a powerful tool for analyzing and understanding physical phenomena.

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### Work & Force (Scalar & Vector Products)

Now consider the example of two concepts in physics, work, and torque, to uncover the difference between scalar and vector products. Work and torque are two crucial concepts in physics and engineering. Both are represented as vectors and related to force and distance.

Work is defined as the scalar product of the force and the distance over which it acts. It measures the energy transferred to or from an object by force, denoted by the symbol W.

Torque, also known as moment of force, is a vector product of force and the moment arm, which is the perpendicular distance from the axis of rotation to the line of action of the force. It measures the rotational force applied to an object and is denoted by the symbol T.

Given the force F = 4i+3j and the distance vector d = 3i+4j, we can calculate the work and torque as follows:

#### Work (W)

The work done by a force F over a distance d is given by the scalar product of the force and distance vectors, which is:

W = F.d = |F||d|cos(θ)

where θ is the angle between the force and distance vectors.

Substituting the given values, we get:

W = (4i+3j).(3i+4j) = 12 + 12 = 24 (where i.i=1, j.j=1, i.j=0, and j.i=0)

The magnitude of work done is 24 units.

#### Torque (T)

The torque vector T is given by the vector product of the force F and the moment arm d, which is:

T = F x d

The direction of the torque vector is perpendicular to both the force and moment arm vectors, and follows the right-hand rule. The magnitude of the torque is given by:

|T| = |F| |d|sin(θ),

where θ is the angle between the force and moment arm vectors.

Substituting the given values, we get:

T = (4i+3j) x (3i+4j) = 0k + 0k + (16K – 9K) = 7K (where ixi=0, jxj=0, ixj=K, and jxi=-K)

So, the torque vector is 7K (in the direction of z-axis that is perpendicular to the plane of x and y), and its magnitude is:

|T| = |7K| = 7 = 49 units

Why we are finding the magnitude of the torque, unlike work? Yes, you got it right. The work is a scalar product, and torque is the vector product. Unlike work, you must find its magnitude to find its scalar quantity (magnitude).

Alternatively, you can also find work and torque by finding the magnitudes of F and d and cos(θ) & sin(θ) followed by putting the values in equations:

|T| = |F||d|sin(θ)

&

W = F.d = |F| |d|cos(θ)

## Straight Lines

A line is a straight one-dimensional object that extends infinitely in both directions. It is the collection of collinear points or the shortest distance between two points. It can be represented by an equation of y = mx + b in the two-dimensional Cartesian coordinate system, where m is the slope of the line and b is the y-intercept.

To find its slope or gradient, we can rearrange the equation into the slope-intercept form, y = mx + b, where m is the slope and b is the y-intercept.

3x + 2y – 7 = 0

2y = -3x + 7

y = (-3/2)x + (7/2)

Comparing this equation with the slope-intercept form y = mx + b, we see that the slope of the line is -3/2.

Alternatively, you can also find its gradient by a formula:

m=-a/b

m=-3/2

Whereby a and b are the coefficients of x and y from the equation of a line.

Conversely, if you are given the slope of a line as m = 3/4 and its y-intercept 2, we can use the slope-intercept form to write its equation as

y = (3/4)x + 2

Rearranging the digits to simplify the equation, you will get:

4y-3x-8=0

This means that the line passes through the point (0,2) on the y-axis and has a slope of 3/4.

If a line passes through the origin (0,0), its y-intercept is 0, and its equation can be written as y = mx, where m is the slope. There is no need for a constant term because the line passes through the origin.

The slope is a measure of how steep a line is. It is defined as the change in y divided by the change in x between any two points on the line. In the slope-intercept form, the slope is the coefficient of x.

The y-intercept is the point where the line intersects the y-axis. In the slope-intercept form, it is the constant term b.

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## Curves and Surfaces

**Parabola:** A parabola is a U-shaped curve represented by y = ax^{2} + bx + c. The vertex of a parabola is where it changes direction. It can either open upward or downward depending on the sign of the coefficient a.

**Hyperbola:** A hyperbola is a curve formed by the intersection of a plane with two cones, and it can be represented by the equation x^{2}/a^{2}-y^{2}/b^{2} = 1 or y^{2}/a^{2} -x^{2}/b^{2} = 1. It has two branches that extend infinitely in opposite directions.

**Ellipse:** An ellipse is a curve formed by the intersection of a plane with a cone, and it can be represented by the equation x^{2}/a^{2}+y^{2}/b^{2} = 1 or y^{2}/a^{2}+x^{2}/b^{2} = 1. It has two axes – a central axis and a minor axis, and its shape depends on the ratio of a to b.

**Circle:** A circle is a curve formed by all the points equidistant from a fixed point called the center. It can be represented by the equation x^{2}+y^{2} = r^{2}, where r is the circle’s radius, and the center is at the origin.

Calculus (Differentiation and Integration) plays a crucial role in finding the gradients of curves and surfaces, such as the gradient of a tangent to a circle. Calculus is also used to find the area under curves, the volume of solids, and the optimization of functions.

### Finding the Gradient of a Tangent to a Circle

The gradient of a tangent to a circle at any given point can be found using calculus. The equation of a circle with radius r and center at the origin is x^{2}+y^{2} = r^{2}. To find the gradient of the tangent at any point (x,y), we first need to find the derivative of the equation concerning x:

2x + 2y(dy/dx) = 0

dy/dx = -x/y

This gives us the gradient of the tangent at any point (x,y) on the circle.

### Finding the Area of a Circle

The area of a circle can be found using the formula A=**π**r2, where A is the area and r is the radius. You can also do this by integration. To find the area of a circle from the limit angle (theta) = 0 to 90 degrees, we first need to find the area of the quarter-circle by integrating the equation x^{^}2 + y^{^}2 = r^{^}2 concerning x from 0 to r:

This gives us the area of the quarter circle. To find the area of the circle from the limit angle (theta) = 0 to 90 degrees, we simply multiply the area of the quarter-circle by 4 or integrate it with a limit of x from 0 to r or for angle 0 degrees to 90 degrees (as integration is defined as the area under the curve).

## Transformations

Translation, rotation, and scaling are all transformations that can be applied to geometric shapes in a 2D plane. Let’s apply these transformations to the given triangle with coordinates (0,0), (0,2), and (1,1).

**Translation **– A translation involves moving an object in a specific direction by a certain amount. In this case, we will translate the triangle by adding 2 to the x-coordinate of each vertex, effectively moving it 2 units to the right.

New coordinates after translation: (2,0), (2,2), (3,1)

**Rotation **– A rotation involves rotating an object around a point by a certain angle. In this case, we will rotate the triangle 90 degrees counterclockwise around the origin (0,0).

New coordinates after rotation: (0,0), (-2,0), (-1,1)

**Scaling **– Scaling involves enlarging or shrinking an object by a specific factor in both the x and y directions. In this case, we will scale the triangle by a factor of 2 in both directions, effectively doubling its size.

New coordinates after scaling: (0,0), (0,4), (2,2)

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## 3D Graphics and Computer Vision

The concepts of geometry, such as points, lines, planes, curves, and transformations, play a significant role in 3D graphics and computer vision. These concepts provide a foundation for representing and manipulating objects in 3D space.

In 3D graphics, the concepts of points, lines, and planes represent objects and surfaces in a 3D environment. Points are used to define the vertices of 3D objects, while lines and planes are used to create edges and surfaces, respectively. Transformations, such as translation, rotation, and scaling, manipulate objects’ position, orientation, and size in the 3D environment.

Computer vision also heavily relies on geometric concepts. For example, in image processing, points and lines represent features such as corners and edges, essential for object recognition and tracking. In 3D reconstruction, points and planes represent the geometric structure of objects and scenes from multiple images or video frames.

The concepts of curves and surfaces, such as parabolas, hyperbolas, ellipses, and circles, are also crucial in 3D graphics and computer vision. These shapes create smooth surfaces and curves, critical for creating realistic 3D models of objects and scenes.

Calculus is essential in finding the gradients and slopes of curves and surfaces, which are necessary for many applications, such as computer graphics, computer vision, and machine learning. The gradient of a curve or surface represents its direction of maximum change, which helps optimize various algorithms and applications, such as image segmentation and feature detection.

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## A Guide to Pass the Analytic Geometry FE Exam

The Fundamentals of Engineering (FE) Exam can be daunting for aspiring engineers, with the Analytic Geometry section often being a source of apprehension. Unlike sections with defined content areas, the National Council of Examiners for Engineering and Surveying (NCEES) focuses on assessing fundamental understanding. While a magic formula for success may not exist, we can use strategies to excel in this section.

The key lies in establishing a solid foundation in core geometric concepts. Lines, planes, circles, and 3D shapes are the building blocks for solving problems. Ensure you fully grasp their equations, properties, and how they interact. Utilize high school geometry textbooks or online resources to refresh your knowledge. This strong foundation will serve as the bedrock for your Analytic Geometry mastery.

The FE Exam heavily utilizes formulas. However, more than memorization is required. Understanding the logic behind each formula is crucial. How does it relate to the underlying geometric principles? By grasping the “why” behind the formula, you can effectively apply and manipulate it to solve diverse problems. Numerous resources offer targeted practice problems designed for the FE Exam, allowing you to hone your formula manipulation skills.

Think of Analytic Geometry problems as puzzles. You’ll apply formulas and geometric principles to find missing pieces or solve for specific properties. Actively tackling a variety of practice problems is critical. Be bold in different problem types to develop a well-rounded understanding. As you practice, your problem-solving skills will become sharper, allowing you to approach exam questions confidently.

The FE Exam is timed, and the Analytic Geometry section is no exception. Familiarize yourself with the time constraints and practice solving problems within the allotted time frame. Mastering time management is crucial. By strategically budgeting your time, you can ensure enough time to answer all the questions and avoid feeling rushed.

Investing in high-quality FE Exam review materials dedicated to Analytic Geometry is beneficial. These resources often provide practice problems, explanations, and strategies tailored to the exam format. Utilize online resources, practice exams, and textbooks explicitly designed for the FE Exam to solidify your understanding and practice problem-solving skills.

### Conclusion

Analytic Geometry in FE Electrical and Computer is the most crucial concept for FE electrical and computer engineering aspirants, equipping them with the necessary knowledge and skills to represent, analyze, and manipulate geometrical objects and curves.

The concepts of points, lines, planes, curves, and transformations play a critical role in many applications, including computer graphics, computer vision, and machine learning. Understanding the principles of analytic geometry and their applications is essential for success in these fields.

Whether you are designing a circuit, creating a computer game, or analyzing medical images, the principles of analytic geometry will always be helpful. With its broad range of applications, mastering analytic geometry is a valuable skill for any aspiring engineer or computer scientist.