If these sorts of applications make your eyes cross, then this quick tutorial will hopefully ease the pain. We’re going to start with some very basic concepts and work our way, gently, up to the idea of the limit, which lies at the heart of Calculus.

Galileo’s Law

We’ll begin our exploration of Calculus by investigating a simple and classic concept: the rate of a falling object over a period of time.

In the 16th century, Galileo discovered an equation that relates the distance and time it takes a solid object dropped from rest to hit the ground. The equation accounts for the effect of Earth’s gravity on the object and assumes negligible air resistance. The equation is pretty famous, and you’ve probably seen it before. It says that the height or distance from the ground in feet,(y) is equal to the fall time (t) in seconds squared and multiplied by 16.

Falling Rocks

Let’s put our formula to use. Say we have a rock that breaks loose and falls from a cliff, and it takes 20 seconds for the rock to hit the ground. Using Galileo’s law we can calculate just how high that cliff was by simply substituting t=20 into the equation.

Therefore, we know that assuming negligible air resistance, the cliff has a height of 6400 feet. Not too difficult, right?

Calculating Average Speed

Next, we want to find the rock’s average speed over an interval.

One thing we know about gravity’s pull on falling objects is that the longer the object falls the faster it’s speed. In other words, free-falling is an acceleration. This means that the speed differs over different time intervals of the fall.

For this first example, we are going to find the average speed during the first 3 seconds of the fall.

Since we want to find the speed, we know we’ll be looking for an answer in distance over time. In this case, our units will be feet per second.

And we want the average, so we’ll average both the change in height from the beginning of the interval to the end, Δy, (pronounced “delta y” where the delta symbol represents “change”) as well as the change in time, Δt.

Our interval is the first three seconds, so we can think of this as the interval [0,3] in mathematical notation. I’ll label the first time in the interval as T1 and the second time as T2. That means we’ll be calculating the height (y) for T1 and subtracting it from the height of T2 to get the change in height, Δy, over the time period.

Similarly, I’ll also be subtracting the start time from the end time in the time period to get Δt. This will provide us with the change in height divided by the change in time, which, if you think about it, is the average speed over the interval.

Filling in the formula with T1 = 0 and T2 = 3, for the interval [0,3], we get:

Finishing out the arithmetic we get 48 ft/sec is the average speed during the first 3 seconds of the fall.

How to Find Instantaneous Rate of Change

So far this has all been pretty straightforward. Now it’s time to start working our way towards the concept of the limit.

Suppose we want to know the speed of the falling rock at a specific time during the fall, say we want to know the speed at 2 seconds. (A speed at a specific point in time is called the Instantaneous Rate of Change by the way.) Using what you know from above, how could we go about calculating this? Or better yet if we don’t know how to calculate it precisely, how could we make a really good estimate?

Any ideas??

Well, we already know how to calculate the average speed over an interval. So one way to tackle this problem would be to calculate the average rate of speed over a small interval beginning at 2 seconds to see if we can make an educated guess at what the instantaneous speed at 2 seconds is.

Let’s figure out what the average speed is in the 1-second interval starting at 2 seconds. AKA let’s calculate the average speed on the time interval [2,3]. Plugging in the values to our average speed formula we get:

Which equals 80 feet/second. This may surprise you since the average speed on the interval [0,3] was 48 feet/second, yet on the interval [2,3] it’s 80 feet/second. Remember that our rock is accelerating as it falls, so the last second of the [0,3] interval will have a greater speed than the entire interval.

80 feet/second is an alright estimate, but we could do better by analyzing a smaller interval. How about the interval [2, 2.1]?

Finishing out this calculation we get 65.6 feet/second. This is an even better estimate. But why stop there? We could do an even smaller interval to find an even better estimate. How about [2, 2.01]?

Now we get 64.16 feet/second. I bet you know what I’m going to say next…why don’t we do an even smaller interval, like [2, 2.001]?

This time we get an answer of 64.016 feet/second, which is actually pretty close to our last interval which resulted in 64.16 feet/second. Interesting… let’s try one more teeny tiny interval and see what happens. This time let’s try [2, 2.0001]?

This one has an answer of 64.0016 feet/second, barely any different than our last estimate of 64.016 feet/second. Since it seems that we’ve finally stumbled upon small enough intervals where the average speed has only negligible differences, we now have enough info to truly make an educated guess.

What would you guess the instantaneous speed at 2 seconds is?

The smaller we make the interval beginning at 2 seconds, the closer our average speed gets to 64 feet/second. Therefore we can guess that the instantaneous speed at 2 seconds is 64 feet/second.

Congratulations, You Found the Limit!

This idea that we can use smaller and smaller intervals to find the speed of a falling object at a specific point in time leads us directly to the concept of limits.

In this case, we are saying that 64 feet/second is the limiting value for the speed of the rock at 2 seconds. Which is why we call it the limit. It is the value that our function is nearing in on.

Limits are truly the backbone of Calculus. It’s the concept that both separated Calculus from the other branches of mathematics and laid the foundation for derivatives and integrals. When you first start your study of limits, you’ll begin like we did today. You’ll be asked to make tables, where you’re intervals, get smaller and smaller until you can determine what value your function is going to, but as you progress you’ll leave behind the tedious task of table-making for more efficient methods.

The first shortcut you’ll learn is the algebraic approach.

Finding the Limit Algebraically

Instead of making smaller and smaller intervals manually, we could assign a variable to represent a really, really, really tiny interval. We will denote this teensy, tiny interval as h.

We can add h onto the time we wanted to calculate the instantaneous change above to create a really small interval. In other words, we’re going to evaluate the interval [2, 2 + h] where h represents a really, really, really, really small value.

This is where the Algebra comes in. Now we have to FOIL out the binomial and collect like terms. Once you do all that you’ll get:

Since every term left has an h, we can reduce everything by h leaving:

Since we were able to remove the h from the denominator, we can do something really cool now. We can make h=0, which makes it the smallest possible interval and gives us the instantaneous speed at 2 seconds.

(Note: we couldn’t do this until after we canceled out h in the denominator because we would have had a divide by zero error, and wouldn’t have been able to complete the calculation.)

Plugging in zero for h we get:

Our instantaneous speed at 2 seconds is exactly 64 feet/second!

That’ s a lot to take in for one lesson! The good news is that through this one application we’ve learned all about the concept of limits AND we have come dangerously close to learning about the derivative. Take some time understanding this lesson, and when you get to learn about derivatives you’ll see a lot of similarities!

Need More Math Help?

• Check out Math Hacks on YouTube for more hands-on math tutorials covering popular topics from Algebra through College Math.
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Thanks for joining me!

❤ Brett

Intro To Limits: Average Speed vs Instantaneous Rate of Change was originally published in Math Hacks on Medium, where people are continuing the conversation by highlighting and responding to this story.

You might be surprised to find that the heart of Trigonometry lies in Geometry. Understanding the relationships used to solve right triangles geometrically is fundamental to pretty much everything you do Trigonometry. So whether you’re learning this for the first time or are here for a little refresher you’ll walk away from today’s tutorial with a good grasp at how to solve right triangles.

Three Key Concepts

We’ll be using three key concepts in today’s lesson:

• Trig Functions: Sine, Cosine, and Tangent (aka SOH CAH TOA)
• The Pythagorean Theorem: a² + b² = c²
• 180 degrees in a triangle

That’s it! Master those three things and you’ve learned how to solve the majority of right triangles. (For the final scenarios, you can check out part two of this series!)

But first, a Few Questions You Might Have…

What does “Solving a Triangle” Mean?

Before we jump into solving some problems. Let’s recap some pertinent information. First of all, what do we mean by “solving a right triangle”?

Solving the triangle simply means that we want to find values for all of the missing parts. This includes any missing side lengths and angle measures.

How Do You Know You Have a Right Triangle?

First of all, everything we do today only works when you have a right triangle. How do you know you have a right triangle? Look for a little square drawn in one of the angles of the triangle. That square indicates the angle is a 90-degree angle, and therefore you have a right triangle.

Is There Any Special Triangle Notation?

Yes. Often the angles will be labeled with capital letters such as A, B, and C. The side lengths will be labeled with lower case letters: a, b, and c. The sides and angles will be arranged so that the angle measure is across from the corresponding side length. For example, angle A would be across from side a, angle B opposite leg b, and angle C across from leg c.

When using the Pythagorean Theorem, c is always the hypotenuse. And remember the hypotenuse is always the leg across from the 90-degree right angle, and it is also always the longest leg.

Example: Solve the Following Triangle

*To see this example, plus a bonus example, worked out live check out my video tutorial above*

We’re going to start off by jumping straight into an example and I’ll explain things as we go. Let’s solve this right triangle where we are given a 23-degree angle and know that the leg adjacent to the 23-degree angle is 4.5 units long.

To solve the triangle we’ll need to find the length of the other leg and the hypotenuse as well as the measure of the remaining angle.

Solve for Leg b

You can start solving in whatever order you like. I’m going to start by solving for the side length of leg b. To solve for b, we need to use one of our three trig functions: Sine, Cosine, and Tangent. Each of these functions takes an angle as its input and is equivalent to the ratio of two specific side lengths of a right triangle.

As we walk through the different relationships Sine, Cosine and Tangent are equal to remember to keep in mind that you must always work from the perspective of the angle you are inputting into the function. You’ll see what I mean as we walk through the next part.

SOH CAH TOA Explained

You’ve probably heard of the mnemonic SOH CAH TOA. It’s the most popular way to remember the relationship of the trig functions Sine (abbreviated as Sin), Cosine (abbreviated as Cos) and Tangent (abbreviated as Tan).

• SOH stands for Sin(x) = Opposite leg / Hypotenuse
• CAH stands for Cos(x) = Adjacent leg / Hypotenuse
• TOA stands for Tan(x) = Opposite leg / Adjacent leg

Where x represents the measure of the angle you are working from.

In this problem, we are working from the perspective of the 23-degree angle. To determine which trig function you want to use to solve for the missing leg b, imagine you are standing at the 23-degree angle. From that position side b is directly opposite you on the far side of the triangle. Therefore you know that the function you want to use must include the “opposite leg”. So we’ll be using either Sine or Tangent.

Next, you want to use the side length that you were given (if you weren’t given any side lengths then you can’t solve the triangle without more information). In this problem, that means that you want to use the 4.5 unit leg. That leg is next to you so refer to it as the adjacent leg. Therefore we’ll use the Tangent function since TOA stands for Tan(x) = Opposite leg / Adjacent leg.

• It’s important to note here that the hypotenuse (the leg across from the 90-degree angle) is never referred to as anything but the hypotenuse. You don’t want to accidentally misidentify it as the adjacent or opposite leg. No matter what angle you’re working from it’s always called the hypotenuse.

Now that you’ve determined which function you want to use, it’s time to set up the equation. Start by writing down Tangent with 23 degrees as it’s input.

Then set it equal to the opposite leg divided by the adjacent leg. The opposite leg is b, which we are solving for. The adjacent leg has a length of 4.5, so go ahead and write down 4.5 in its place.

Next, let’s evaluate the Tan(23) using a calculator. Before you punch it in make sure you are in degree mode (not radians). Then multiply both sides by 4.5 to isolate b. Therefore b has a length of approximately 1.91 units.

Solve for c, the Hypotenuse

Next, we can solve for c, the hypotenuse. Because we now know the lengths of sides of a right triangle you have two options for solving for c: you can use one of the SOH CAH TOA trig functions or you can use the Pythagorean Theorem. Which you choose may depend on the instruction in your assignment or maybe just your preference.

Solving Using Cosine

If you choose to solve for the hypotenuse using the trig functions, you’ll first need to determine which function to use. Again you are standing at the 23-degree angle, but this time you want to solve for the hypotenuse. So you’ll be using either Sine (SOH) or Cosine (CAH) since both include the hypotenuse in their ratios.

Next, you want to include the side length you know. At this point, you know both the opposite and the adjacent legs’ lengths. I’d recommend using the 4.5 unit adjacent leg. That way if you made an error in calculating the opposite leg, you won’t cause this problem to be incorrect as well.

So we have the hypotenuse and the adjacent leg, which means we’ll be using Cosine. Go ahead and set up your equation using 23 as the input degree measure.

The algebra is a little trickier with this one. I’d recommend using cross multiplication to help isolate c.

Alternate Method: Pythagorean Theorem

It’s really simple to solve for the hypotenuse using the Pythagorean Theorem when you already know the lengths of the two legs. The Pythagorean Theorem states that the sum of the squares of the two legs is equal to the square of the hypotenuse. So simply fill in the side lengths for a and b and solve for c.

Find the Measure of Angle A

Our last task is to solve for the missing angle A. Since we already know the degree measure of the two other angles in the triangle we can use the fact that the angles of a triangle always sum to 180 degrees to create an equation to solve for the missing angle.

Therefore the measure of Angle A is 67 degrees.

In part two of this series, we’ll learn about the inverse trig functions of Sine, Cosine, and Tangent and how to use them to solve for missing angles.

Need More Math Help?

• Check out Math Hacks on YouTube for more hands-on math tutorials covering popular topics from Algebra through College Math.
• You can find a number of interesting math topics and problems right here on Medium, just click that follow button!

Thanks for joining me!

❤ Brett

Right Triangle Trigonometry Explained was originally published in Math Hacks on Medium, where people are continuing the conversation by highlighting and responding to this story.

What are Algebraic Expressions?

The first question you might have is “What exactly are algebraic expressions?” As the name implies you have expressions involving algebra, but what does that mean?

An expression is a grouping that can include any combination of numbers, symbols, and operators (addition, subtraction, multiplication, division). The one thing that expressions don’t have are equal signs. The presence of an equal sign makes an equation, not an expression, although an expression can be part of an equation.

Since the expressions we’ll be dealing with are algebraic we know that they will include algebraic variables, making them a tad bit more abstract. You’ll often be asked to “simplify algebraic expressions”, that means you want to rewrite the expression as shortly as possible, so combine all the things you can!

Watch my video tutorials for fully worked-out examples and read on to learn all the notation and exponent rules you need to know 😁

Level 1: Simplifying Algebraic Expressions

Level 2: Challenging Algebraic Expressions

Yes, there are always rules.

Like everything you have ever done in mathematics, there are some rules you must obey. But don’t worry the rules are pretty straightforward and once you have the hang of it you’ll hardly notice them. In fact, these rules build on your prior knowledge of mathematics, so they’ll be a breeze for you!

Rule #1: When Multiplying Like Bases, Add the Exponents

Basically, when you multiply two of the same type of things together (same base), you add the exponents. This is nothing new. You’ve been doing this for a while. For example, when you multiply 2³ x 2⁵ you get 2⁸, since 3 + 5 = 8.

The same applies when you are working with variables. For instance, a³ x a⁵ = a⁸. Now you must be careful not to accidentally combine two things that are different together. You can’t simplify a³ x b⁵ because they are different types of things.

Rule #2: When Dividing Like Bases, Subtract the Exponents (or use the Canceling technique)

Whenever you stumble across two like things being divided, you have two options for handling it. They are essentially the same idea, but depending on how your brain works you’ll probably prefer one way over the other.

The first way you can handle a division of like things is to subtract the exponents. So for example, if I have a⁸/a⁵ I can handle the operation by subtracting the exponent of 5 from 8 to get an answer of a³.

Again, I must have like things. So I can’t do any simplification to the expression a⁸/b⁵ since they have different bases.

The second way you can handle a division of like things is to cancel out like terms. You’ll be using the canceling technique lots in simplifying algebraic expressions, so this is a great opportunity to learn about it early on!

Remember that a⁸ = a ∙ a ∙ a ∙ a ∙ a ∙ a ∙ a ∙ a, and a⁵ = a ∙ a ∙ a ∙ a ∙ a, and recall that a / a = 1. That means that if I’m dividing the product of eight a’s by the product of five a’s, I have five a/a’s. Those in turn divide or cancel to 1, leaving only a ∙ a ∙ a, or a³.

Remember that you can use the canceling technique whenever you have a number or variable in the numerator and denominator that will divide to 1. In the video tutorial, we’ll walk through lots of instances where you can cancel out like terms to rapidly simplify your expressions.

Rule #3: When Raising a Power to a Power, Multiply Powers

You’ll often run into the scenario where you have a value with an exponent in parenthesis and then a power on the parenthesis. It’ll look something like this:

In this case, you want to multiply the powers together. So (a³)⁵ is equivalent to a¹⁵. Let’s take a look at why this is true.

Applying a power of 5 on the outside of the parenthesis means that we have whatever is inside the parenthesis multiplied with itself five times. In mathematical terms we have:

Using Rule #1 from above, we can see how (a³)⁵ equals a¹⁵ since a³ ∙ a³ ∙ a³ ∙ a³ ∙ a³ = a¹⁵.

Rule #4: When You See Negative Exponents, Flip It

The last thing that typically shows up in simplifying algebraic expressions problems is negative exponents. Whenever you see a number or variable with a negative exponent, move it to the opposite side of the fraction bar and remove the negative symbol.

So if you have a negative exponent in the numerator, such as:

Move it to the denominator, remove the negative, and simplify from there.

If you have a negative exponent in the denominator, move it to the top, remove the negative and simplify it.

And if you don’t have a fraction bar, you can always write your expression over 1 to create a fraction and simplify from there.

Putting It All Together

Of course, the difficult part is putting all of this together to simplify more complicated algebraic expressions. For this part, I think it’s best to see real problems worked out step-by-step. Check out the following tutorials for more help with algebraic expressions of all kinds!

Need More Math Help?

• Check out Math Hacks on YouTube for more hands-on math tutorials covering popular topics from Algebra through College Math.
• You can find a number of interesting math topics and problems right here on Medium, just click that follow button!

Thanks for joining me!

❤ Brett

Simplifying Algebraic Expressions was originally published in Math Hacks on Medium, where people are continuing the conversation by highlighting and responding to this story.

In this guide, you’ll learn about two different techniques you can use to solve these complicated systems. Let’s jump in!

What is a Three-Variable System of Equations?

The first question you may have is what exactly is this three-variable system? You may remember from two-variable systems of equations, the equations each represent a line on an XY-coordinate plane, and the solution is the (x,y) intersection point for the two lines. In other words, the solution is the value or values for x and y that hold true for both equations.

Three variable systems of equations aren’t so different. The solution still represents the values for x, y, and z (or whatever variables your equations are using) that when plugged into each equation holds true.

Visually, a three-variable equation is represented in 3-dimensional space as a plane in the xyz-coordinate axis. Don’t worry, you seldom will be asked to graph a set of three variable systems in 3D space. It’s probably sufficient for you to understand that the solution to your three-variable system of equations is the values that will make all three equations true and represents the intersection of three planes in 3D space.

Three Variables, Three Equations

In general, you’ll be given three equations to solve a three-variable system of equations. This is similar to how you need two equations to solve a standard system of linear equations. In some cases, you may be able to solve a three-variable system of equations with only two equations, but it isn’t as common.

Two Methods

Just like when you were solving two-variable systems of equations you had multiple methods to choose from, you have a couple of options here as well. The two main methods are called the Backsolving Method and Gaussian Elimination/Row Reduction Method (sometimes called Gauss-Jordan Elimination). The main difference is whether you want or need to solve the system using equations or matrices.

If you are in second-year Algebra or PreCalculus, most likely you’ll be using the Backsolving Method. If you are in a Linear Algebra or computer programming course, you’ll probably be asked to solve using Gaussian Elimination with matrices.

The Backsolving Method

The Backsolving Method is reminiscent of the Elimination Method from our standard systems of equations arsenal. In fact, you’ll be using the Elimination method multiple times in the course of the backsolving method.

The basic idea behind the backsolving method is to take your 3 three-variable equations and make 2 two-variable equations so that you can solve them using the traditional substitution or elimination methods. Once you find one variable’s solution, you can then back solve to find the solutions for the other two variables.

It’s a bit tedious, but a fairly simple process once you get the hang of it. In the following video tutorial, I demonstrate how to choose two sets of two equations to perform the elimination method on, and then back solve to solve for all three variables.

The key to completing the backsolving method successfully is to eliminate the same variable from your two sets of three-variable equations. This ensures that the two resulting two-variable equations are capable of being solved using the substitution or elimination methods. Again, this is easier in practice so check out the tutorial below to see it in action :)

Gaussian Elimination

If you’re familiar with matrices, Gaussian Elimination is a wonderful way to solve three-variable systems of equations as well as systems with more variables and more equations. There is quite a bit of set up work involved in order to use this method, but the payoff is once you have it set up the system is straightforward to solve.

The basic idea is to translate your system of equations into a matrix using placeholder zeroes and ones where needed. Once you have the matrix set up, you can begin manipulating it by using basic properties of matrices. The process is like a game where you add, subtract, multiply, and divide the rows in your matrix in order to make zeroes and ones in specific locations.

As you work through your matrix, you’ll be working to “row-echelon” and/or “reduced row echelon” form. Row echelon is when you have a diagonal of ones going from the upper left to the second-to-the-right bottom row and under all the ones you have zeroes.

Reduced row echelon form is when in addition to having the diagonal of ones and zeroes beneath, you also have zeroes above the ones. Reduced row echelon form is ideal because you can easily read off the solution to your system of equations without any additional work.

To learn how to solve a 3x3 system of equations using Gaussian Elimination, check out the video tutorial below. Once again it’s easier in practice than it sounds in writing!

For more help with Gaussian Elimination check out this tutorial with step-by-step instructions.

Infinitely Many & No Solution Explained

Just like with two variable systems of equations you can have infinitely many solutions or no solution. Since we’re now working with three planes instead of two lines, we have more scenarios that can result in infinite or no solution.

Infinitely Many Solutions

There are multiple ways you could wind up with infinite solutions in a three-variable system of equations. The most obvious is that the three planes could be the same so that a solution to one equation will be the solution to all of the equations. Another possibility is that all of the equations are different but they intersect in a line, and we know from our standard system of equations that a linear solution set is infinitely many solutions. Lastly, we could have two equations that are the same plane that intersect the third plane forming a line.

No Solution

There are a few different scenarios that leave you with no solution when dealing with three variable systems. The most common is that you have three parallel planes that never intersect and therefore have no intersection points. Another possibility is you have two parallel planes and one intersecting plane or three planes that intersect but not at the same location.

In all of the scenarios, you are unable to produce a point or set of points that represent the intersection of all three planes. Fortunately, if you are using Gaussian Elimination no solution results are really easy to detect because you will stumble across a row that translates into a false statement, such as 0=1 or some similar contradiction. This is your indicator that the system has no solution, and remember if it has no solution the system is inconsistent.

Dependent, Independent, & Inconsistent Systems

Any system you have will fall into one of three categories: dependent, independent, or inconsistent.

Dependent System

A dependent system has infinitely many solutions. Recall from above that there are multiple ways your system could produce an infinite number of solutions (all three planes are the same plane or they intersect to form a line).

Independent System

If a dependent system has infinitely many solutions, then an independent system has a single solution. For an intersection of three planes, you’ll end up with an (x, y, z) coordinate as your solution, or in other words a value for each variable in your system.

Inconsistent System

A system is inconsistent if there is no solution.

The Opposite of Inconsistent must be Consistent

And finally, if it’s not inconsistent it must be consistent, which means that it has at least one solution. A system is consistent if it has a solution — it can be one solution (i.e a coordinate point answer) or infinitely many solutions, so long as there is a solution!

Need More Math Help?🙏

• Check out Math Hacks on YouTube for more hands-on math tutorials covering popular topics from Algebra through Advanced Math.
• You can find a number of interesting math topics and problems right here on Medium, just click that follow button!

Thanks for joining me!

❤ Brett

How to Solve 3 Variable Systems of Equations: Beginner’s Guide was originally published in Math Hacks on Medium, where people are continuing the conversation by highlighting and responding to this story.

Here are three reasons why RTD became the standard from Algebra to Calculus.

The Commonsense Reason

The standard reason why you need to RTD is perfectly practical. As you’ve most likely discovered, in mathematics you can often write solutions in multiple different ways and forms. All of these variations are cool, but for practical purposes, they make life more difficult for those grading your papers.

Defining and requiring a standard form for answers saves your teacher the time-consuming headache of having to verify that your solution is equivalent to the answer key, or even worse, accidentally marking your answer incorrect!

Just like reducing a fraction to its simplest form, RTD is the protocol for simplifying fractions with roots in the denominator.

The Reasonable Reason

A commonly defined nomenclature makes sense and all, but still leaves us with the question: why have we decided that having a root in the numerator is okay, but having a root in the denominator is not??

Why is (2 √3) / 3 the simpler form of 2 / √3 ?

The reason is that if we need to add or subtract fractions with radicals, it’s easier to compute if there are whole numbers in the denominator instead of irrational numbers. For example, it’s easier to add (2√3/3) + (( 3−√2)/7) than the non-rationalized version: (2/√3) +(1 / (3 + √2)).

To add the first set of fractions together all we need to do is make a common denominator of 21 and then add like terms from the numerators. It’s not nearly as clear what the common denominator is of the second set of fractions.

To solve the second problem you would most likely rationalize the denominator first and then make the common denominator of 21 before adding the fractions together.

So RTD provides a common form for comparing and grading solutions as well as makes it easier for us to perform further computations by hand when needed.

The Historical Reason

At this point you may be thinking, “Why not just leave the roots in the denominator and rationalize them if and when I need to add or subtract the fractions?”

And yes, that’s a valid point. That’s why perhaps the best answer to why we are so often required to RTD is the historical one. The answer that takes us back to life before computers and calculators were ubiquitous. Back when people had to routinely do division… by hand! *gasp!*

Let’s take a look at our two equivalent fractions from above:

What would happen if we actually wanted to divide these fractions?

Let’s begin by dividing 2/√3 by hand.

Before calculators, you would begin by using an algorithm to find an approximation of √3 by hand which is 1.73205081… After approximating √3, you would then set up your long division.

If this looks complicated to you, that’s because it is. Having an irrational number as the divisor is practically unheard of.

In order to perform this division, you would need to decide how many decimal places you want to keep around, round to that decimal place, then you would need to multiply both the dividend and divisor by a large enough power of ten to make your divisor a whole number. After doing all that you could then proceed with the messy task of dividing those two unwieldy numbers.

Now let’s take a look at how we’d go about dividing the equivalent, rationalized fraction: (2√3)/3.

Without a calculator, you’d begin by computing the approximation for √3 like above, which is 1.73205081… Next, you would multiply 1.73205081… by 2 to get 3.46410162… Then simply set up your division.

This is much easier to divide by hand! We don’t have to do any prep work, we can jump directly into the division problem.

And most importantly, HOW to RTD :)

Now that you understand why it’s so important to RTD, you may want to practice up on it!

This first tutorial explains how to rationalize the denominator with a standard square root, then moves on to demonstrate how to handle the trickier scenario of rationalizing the denominator when you have the sum or difference of a whole number and a square root in the denominator. To do this you’ll be introduced to the conjugate.

In this second tutorial, you’ll learn how to rationalize more difficult denominators with cube roots and fourth roots. We’ll also talk about how you can use the alternate fractional exponent notation to simplify the process.

Need More Math Help?🙏

• Check out Math Hacks on YouTube for more hands-on math tutorials covering popular topics from Algebra, Trigonometry, Precalculus, Calculus, & Advanced Math.
• You can find a number of interesting math topics and problems right here on Medium, just click that follow button!

Thanks for joining me!

❤ Brett

Why Do We Rationalize the Denominator?? was originally published in Math Hacks on Medium, where people are continuing the conversation by highlighting and responding to this story.

In our modern world, it may seem like graphing functions by hand is...well…a bit archaic. When you can easily enter any function you wish into Desmos or Wolfram Alpha from the comfort of your phone, why should you learn to graph functions longhand? Especially the more complicated functions like the ones we encounter in trigonometry?

Well besides the obvious likelihood that you are reading this because in some Trigonometry, Precalculus, or Calculus course your instructor is requiring you to graph sans technology, I’d like to present the argument that true understanding can only be gained through *doing*. The fact remains that you can’t learn math by standing on the sidelines. Math IS a participatory sport. You can read about math, listen to lectures, watch other people do math but if you don’t actually grab a pencil and paper and struggle through the problems, you will never learn mathematics.

In a day and age where we so easily misconstrue talking about doing things as actually doing things, math remains one of those disciplines that can’t be faked. If you are struggling to progress in math ask yourself if you are actually PRACTICING mathematics or are you thinking — or talking — about doing math instead?

Notice here how we say “practicing mathematics”, just like you practice an instrument or a sport, you are a beginner seeking improvement no matter what level of math you are currently at. Yes, you are going to get problems wrong. Yes, you are going to get confused. Yes, you are going to have questions and need help. That’s just part of the practice. 🙂

Alright, enough with the real talk. Let’s jump into sketching some trigonometric graphs!

What is a Parent Function?

You’ve probably heard the term Parent Function with relation to graphing. Parent functions are the OGs of functions. They are the unaltered forms of your equations. The archetypes. For example, the equation y=x is the parent linear function; the equation y=2x+1 is still a linear function but it is not the parent function. It is an altered form of the parent where we have changed the slope and y-intercept.

This idea of parent functions applies to all types of functions. In Trigonometry our main parent functions are: y=sin x, y=cos x, y=tan x, y=csc x, y=sec x, and y=cot x.

These are the functions you will be primarily working with in trigonometry (although as you progress you might learn about other interesting functions such the inverse or hyperbolic trig functions). So these are the functions we’ll be learning how to graph today!

The MVPs: Sine, Cosine, and Tangent

These three functions are definitely the all-stars of trig. You probably remember these functions from geometry when you were first introduced to right triangle trigonometry. Although the graphs of Sine, Cosine, and Tangent may seem completely different from what you learned in geometry, we are really just experiencing a new picture of the same geometric concept.

Sine, cosine, and tangent still represent SOH CAH TOA, but in trigonmetry, we learn about some new and creative ways we can represent this information.

Essentially what we have done here is scaled our special right triangles so that their hypotenuse is 1 unit in length, organized them around the origin of a xy-coordinate plane, and traded out the concept of degrees for radians (which is just angles measured in respect to a circular rotation).

When we lay out all these triangles around the origin, we can connect the outermost points to form a circle that has a radius of 1 unit. That circle is the Unit Circle and one of the most important topics in trig!

Now the points the lie on the circumference of the circle can be translated into the graphs we’ll be exploring today.

• According to SOH CAH TOA, we see that the x-coordinate relates to the cosine value (output) at the given angle rotation (input).
• The y-coordinate relates to the sine value (output) at the given angle rotation (input).
• And lastly tangent can be defined as cosine divided by sine, so the quotient y/x is the tangent value (output) at the given rotation (input).

Now we can graph a picture of this information that we gleaned from the Unit Circle, by creating a graph where the x-axis represents the rotation in radians, and the y-axis represents the value of our sine, cosine, or tangent functions at the given angle.

As I’m writing this, I’m realizing that this is a complicated set of relationships to understand, so I’m going to put together a separate tutorial to explain how these concepts all relate, but until then bear with me!

In the following video tutorial, I demonstrate how to graph the Sine, Cosine, and Tangent functions using a simple pattern and their periodic nature (ie repetitive) and then I relate it back to the Unit Circle, so you can see where those points come from 🙂

The Second String: Cosecant, Secant, Cotangent

The other three functions you will encounter in basic trigonometry are Cosecant, Secant, and Cotangent. These functions are often referred to as the reciprocals of Sine, Cosine, and Tangent because they are defined by the reciprocal (ie flipped upside down) ratios of SOH CAH TOA.

We can use these reciprocal identities to help us easily graph Csc, Sec, and Cot based on our knowledge of Sin, Cos, and Tan. Check it out ⬇

Advanced Topic: Trig Graph Transformations

So now that we have learned about the six standard parent functions of trigonometry, we are ready to take a look at how we can manipulate the parent functions to create and sketch interesting graphs by hand — which is one of the main reasons why we need to know how to graph the parent functions by hand!

Once you have a solid sense of how to graph your parent functions, you can apply some basic mathematical principles to yield the graphs for any variation of your parent functions 🙌

In the following tutorial, I explain how we can alter a parent function to produce x and y-axis reflections, horizontal and vertical stretches, compressions and shifts. Skip to 14 minutes in to see three examples of graphing transformed Tangent, Cosine, and Sine graphs.

Need More Math Help?🙏

• Check out Math Hacks on YouTube for more hands-on math tutorials covering popular topics from Algebra, Trigonometry, Precalculus, and Combinatorics with new videos added weekly.
• You can find a number of interesting math topics and problems right here on Medium, just click that follow button!

Thanks for joining me!

❤ Brett

How to Graph Sine, Cosine, Tangent by Hand ✍ was originally published in Math Hacks on Medium, where people are continuing the conversation by highlighting and responding to this story.

For example from the factored form, you can easily identify solutions. You get an idea of the quantity and type of solutions an equation has. You can find its roots/x-intercepts, which makes it easier to graph. You can even tell a bit about the behavior of the graph through multiplicity. All of these are wonderful concepts, but before we get to them it’s important to master the art of factoring.❤

There are a lot of different factoring techniques. From the standard here’s how to factor a quadratic to the less obvious techniques of completing the square and polynomial long division, a lot of what you learn in algebra revolves around manipulating equations, so that’s what we are focusing on today!

The Basics: How to Factor Quadratics

In this tutorial, I show you the very basics of factoring a quadratic. You will learn how to choose your factors and check your solution.

It’s important to note that in these examples, I’m not working with the full equation (notice there are no equal signs). This is a common first step as you are learning how to factor. We can set any of these polynomials equal to zero and then we have an equation. With that equation, we can do the same factoring process, and take it a step further to find the solution (if it exists) by solving for the variable, and then we can graph the quadratic if we wish to do so.

NOTE: For factoring quadratics where there is a coefficient on the x-squared term, check out the “acb method” under tricky quadratics towards the end of this post!

The Basics: How to Factor out the Greatest Common Factor (GCF)

Factoring out the greatest common factor is a handy little technique that you can use whenever there is a factor (number, variable, or both) common to ALL terms in your polynomial.

The interesting thing about this technique is that as you move on to more difficult polynomials and equations, you’ll find that often this step can be used in conjunction with other types of factoring to get your polynomial down to the very simplest form.

Formulas: How to Factor the Difference of Perfect Squares

Some types of factoring fall nice and neatly into a handful of formulas. These formulas can act as guides for factoring certain special types of polynomials. Unfortunately, it might take a bit of memorization to get these ones down.

In the video tutorial, you’ll learn about a few different formulas. These include:

• Difference of Perfect Squares Formula

• Perfect Square Trinomials Formula (positive & negative versions)

Cool Tricks: How to Factor by Grouping

Factor by grouping is a cool technique that you can use in a couple of interesting ways. You can use it to factor down polynomials with four terms, like the examples in the video, by first factoring out a GCF from two pairs of terms.

If after factoring out the GCFs you are left with two identical binomials in parenthesis, then that means your polynomial can be factored by grouping, and you can go ahead and finish the factorization.

The other interesting way you can use factor by grouping is by taking a normal trinomial quadratic and splitting the middle x-term into two terms so that they still sum to the original middle term. This will give you 4 terms and the opportunity to apply factor by grouping. I don’t cover this approach in the tutorial because it isn’t the traditional use of factor by grouping, but I do know that some people really love solving quadratics that way.

Tricky Quadratics: How to Factor Quadratics with Leading Coefficient greater than 1 | The ACB Method

In this tutorial, we take a look at one of the more difficult types of quadratics to factor: when we stumble upon a quadratic with a value on the x-squared term.

What you’ll notice when you try to factor these types of polynomials is that the extra coefficient creates a lot more possibilities of both factors and where to place those factors. Now you can still factor these types of problems using a bit of guess and check alongside the traditional quadratic factoring method, but you’ll notice that it can get pretty tricky.

Fortunately, we have a tried and true method for tackling these types of problems called the ACB Method (or CAB method). This method is the secret to factoring these complicated quadratics. Using this technique takes away all the guesswork, and gives you a step-by-step process you can follow every time you encounter these types of quadratics 🎉

Advanced Techniques: Complete the Square

This is not technically a factoring technique, but I do think it is so closely related to factoring that I wanted to include it in this guide.

Completing the square is a method you can use to rewrite a quadratic from standard form (i.e. y=ax²+bx+c) to vertex form (i.e. y=(x-h)² + k) so technically you are not rewriting your equation in a fully factored form since you’ll have that constant (k) hanging out on the outside of your binomial squared. BUT you are converting your equation into a form where you can easily solve for the x-intercepts and graph, so it’s very similar to the desired goal for factoring.

If you want to explore this topic further and see it in action, check out the tutorial below!

Advanced Techniques: Polynomial Long Division

This is another technique that isn’t explicitly factoring, but it can be used to rewrite a polynomial in factored form. Polynomial long division is the process used to divide a polynomial by a smaller polynomial, most typically a binomial.

In this type of Algebra 2 problem, one could use the Rational Zero Theorem to identify potential solutions, and then use a process such as polynomial long division to test and divide out the solutions. Typically this process is done multiple times until you have identified all of the solutions.

Now if you think about dividing a value (the dividend) by another value (the divisor), the resulting value (the quotient) is really a factor of the dividend along with the divisor.

So you can use division to factor a polynomial. To learn how to perform polynomial long division, check out this tutorial.

Whew, that’s a lot!

Wow, we covered a lot of material today! If you are new to factoring, I suggest focusing on one topic at a time and make sure you get lots of practice with different problems before moving on to the next topic.

Need More Math Help?🙏

• Check out Math Hacks on YouTube for more hands-on math tutorials covering popular topics from Algebra, Trigonometry, Precalculus, and Combinatorics with new videos added weekly.
• You can find a number of interesting math topics and problems right here on Medium, just click that follow button!

Thanks for joining me!

— Brett

Starter Guide to Factoring Quadratics & Polynomials was originally published in Math Hacks on Medium, where people are continuing the conversation by highlighting and responding to this story.

(as easy as pie 🥧)

Happy Pi Day friends 🎉 It’s finally the one day of the year where math gets some love instead of the shade it gets the other 364 days! Feeling all the warm fuzzies about math makes March 14th one of my most favorite days of the year!

In celebration of Pi Day, I’m going to teach you a little pi memory trick that, for me, made memorizing the first 100 digits of pi easier than baking my Pi Day Pie 😂 I wish I were kidding but it’s true. It took me just over a day to memorize 100 digits of pi, and 4 days to bake my beautiful pi day pie! I guess some of us are better at the whiteboard than in the kitchen!

Whether you’re memorizing pi for an epic competition, a personal challenge, or just to honor our beloved mathematics here’s how to do it →

Mnemonic Major System to the rescue!

It’s no secret that memorizing strings of numbers is beyond challenging. That’s why the easiest way to memorize numbers is to use the Mnemonic Major System also known as the Phonetic Number System.

This is a classic technique where we assign a different consonant sound to each digit 0 through 9. These sound associations allow us to create silly words and phrases that correspond directly to the string of numbers we need to remember.

The handy thing about this trick is that it can be used to memorize any string of numbers you wish. Some people even get so good at it that they use it to memorize numbers directly on the spot! Talk about life goals!! 🤓

A Classic Pi Recital

Ready to see the mnemonic major system in action?? Grab a slice of pie, sit back, and watch me recite 100 digits of Pi from memory 🎉

Skip to 4 minutes in to get the skinny on the tricks I used to memorize the mnemonic system and the full explanation of how I implemented the number-sound associations to create silly, yet memorable phrases that correspond to the first 100 digits of pi.

Happy Pi Day!!

❤ STAY CONNECTED ❤

for updates & math inspiration follow Brett on social media:

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How to Memorize 100 Digits of Pi (as easy as pie 🥧) was originally published in Math Hacks on Medium, where people are continuing the conversation by highlighting and responding to this story.

Confession: I love linear algebra. Okay, maybe that’s not much of a confession, but I do love it! In a large part, because linear algebra doesn’t feel like the rest of mathematics, it feels like a puzzle.

Now that may sound crazy, but hear me out.

This new land of matrices and vectors may look and feel intimidating, and for good reason: new notation, new rules, new properties. It’s all a bit different. But maybe viewing it as a puzzle will help you jump over that new notation hurdle.

Ready to give it a try?

Where to start?

I believe the best place to start with linear algebra is with solving systems of equations because it’s something you’ve probably learned already. Remember the substitution and elimination methods?? Do those ring a bell? If not, you can refresh your memory here. 🙂

Today we are going to take a new approach to solve systems of equations using a method called Gaussian Elimination.

What is Gaussian Elimination??

Gaussian elimination is a method where we translate our equations into a matrix and use the matrix to solve the system (i.e. find the solutions for each variable that make all the equations true).

I’m going to walk you step-by-step through a simplified example today (the numbers work out nicely so there are far fewer steps!). If you would like a more in-depth example, check out my youtube tutorial on Gaussian Elimination below ⬇

The Set Up

Just like starting a card game by shuffling and dealing, the start of our game of Gaussian Elimination begins by translating our equations into a matrix.

Here’s the system we are going to solve:

The very first thing you need to recognize is that there is a lot of hidden information in our system. In red I’ll add in the placeholder zeros and ones.

Next, we are going to separate out all of the important numeric information from the extraneous symbols. Now highlighted in red, you’ll find all of the important information we’ll be working with:

We need to rewrite the numeric values above as a matrix. We’ll drop the letters, equal signs, and addition symbols (but not the minus symbols!) and simply write the numbers in the exact order and rows they appear above.

After we do that we’ll add in big bracket symbols to group them together. This is our matrix:

The Rules ✅

Like any game, there are a few rules we have to follow:

1. You can swap any two rows
2. You can multiply or divide any row by a value
3. You can add or subtract any two rows together

*Note: you can combine these rules in any one move.

Don’t worry if these don’t quite make sense yet. You’ll see how they operate when we work through our example.

How to Win 🎉

You win when your matrix looks like this:

Where the # symbols represent any numbers, and the rest of the matrix has zero’s in each position except for the diagonal ones. This is called Reduced-Row Echelon Form.

Let’s Play!

Start with the matrix you made during the setup step. Now we are going to use the rules to get this matrix across the finish line!

Move 1: Swap Row One and Row Two

There are lots of different ways to get this matrix into the winning form, but I think the easiest way to start is by swapping the first and second row. This way we obtain a 1 in the first position of the first row and a 0 in the first position of the second row.

Move 2: Add -2 times Row Two to Row Three

One thing we are allowed to do is use multiple rules together in one step. In this move, we will take -2 times row two and add those products to row three. This will leave row two unchanged but will help us obtain a zero in third row second position.

Move 3: Divide Row Three by -3

Next, we will simply divide row three by -3 so that we can obtain a 1 in the third-row third position.

Move 4: Add -2 times Row Three to Row Two

Now that we have all zeros in the bottom left-hand corner surrounded by a diagonal of 1, we are ready to start working on obtaining zeros in the positions above the 1’s.

👉 But before we do that, I want to make a little side note: our matrix is currently in row echelon form. In this form, you may translate your matrix back into a set of equations, if you wish, and will easily be able to solve for x, y, and z. Today we are working on getting our equation into reduced row echelon form, meaning that we want to obtain zeros in the positions above the 1’s. Often times in linear algebra you will be asked to work all the way to reduced row echelon form as it is the easiest form to read the answer from.

Okay, back to our math. In this next move, we will take -2 times the third row and add it to the second row so that we can obtain a 0 in the third position of row 2.

Move 5: Add Row Three to Row One

Next, we will simply add row three to row one since -1 + 1 = 0, that will help us attain a 0 in the third position of the first row.

Move 6: Add -2 times Row Two to Row One

Lastly, we will add -2 times row two to the first row in order to obtain a zero in the second position of the first row.

Congratulations, YOU WIN!!🎉

All we have left to do is read off the answer! To do this simply translate your matrix back into a set of equations, and you’ll see that you have discovered the solutions for x, y, and z.

Dropping all the zero terms, you get:

This means that the intersection point in 3-dimensional space where the three planes intersect is at (-1,2,-3).

Remember you can always check your solution by plugging the values back into the original equations.

Not into Matrices??

That’s okay, you can also solve 3x3 systems with good ole algebra 😉

Need More Math Help?🙏

• Check out Math Hacks on YouTube for more hands-on math tutorials covering popular topics from Algebra, Trigonometry, Precalculus, and Combinatorics with new videos added every week.
• If you need more help with Gaussian Elimination, I cover a completely different and slightly more challenging problem in the video tutorial linked at the top of this post.
• You can find a number of interesting math topics and problems right here on Medium, just click that follow button!

Thanks for joining me!

— Brett

❤ STAY CONNECTED ❤

for updates & math inspiration follow Brett on social media:

Instagram | Facebook | Twitter

The Game of Gaussian Elimination: An Introduction to Linear Algebra was originally published in Math Hacks on Medium, where people are continuing the conversation by highlighting and responding to this story.

If you’re studying advanced statistics or combinatorics, you will surely run across these, dare I say, exciting (!) factorials (yes, I’m a math nerd, and I’ll take any opportunity I get to make a pun however weak 😜).

But seriously, this mathematical notation can be confusing for beginners especially as we move from standard numeric problems to more difficult variable expressions. Don’t you worry though, this quick factorial tutorial (cheesy, I know 😊) will get you cruising in no time.

What is a Factorial?!

A factorial, denoted by an exclamation point (!), is an operation applied to a non-negative integer (i.e.the numbers 0, 1, 2, 3, etc.) that is executed by taking the product of all the positive integers less than or equal to the number stopping at 1.

So for example, if I want to know what 4! equals, I simply multiply all the positive integers together that are less than or equal to 4, like so:

You find factorials all over combinatorics because that’s where they originated. The factorial was created as a way to express the number of arrangements of a group of items, which of course we find by using, in its most basic form, the multiplication rule of counting.

The factorial is sort of the unofficial operation of the multiplication rule of counting.

Why Zero Factorial Equals 1

This is where it gets tricky because if we only think of factorials in the context of which they’re usually defined, i.e. the “product of all the positive integers less than or equal to the number” then figuring out 0! is like hitting a brick wall.

Most people will tell you that 0! is defined as 1, and if you ask why they just say “because it’s defined as one”.

Yeah, it’s pretty frustrating. It’s the mathematical equivalent to asking your parents why you have to follow some arbitrary rule they made up and being told, “because I said so.”

Although that might be an accepted parenting technique, it’s a lousy way to learn mathematics. So what’s the deal? How did we decide that zero factorial equals one?

An Intuitive Understanding

Remember how we said that the factorial originated from the mathematical operation of finding the number of permutations or arrangements of a set? (Note: not the permutations of a smaller set from a larger set, but just the arrangements of a given set.)

Zero factorial can be thought of as the number of arrangements of zero elements in a set, aka the empty set {}. (If you’ve ever studied sets, perhaps in basic statistics or discrete mathematics, you’re probably familiar with the concept of empty set. It’s literally the set of nothing.)

Now if I asked you how many arrangements there are of one thing you’d answer 1 because there is only one way to arrange one thing. The same idea follows here. Our “one thing” is the empty set, and the number of arrangements of the empty set is one. That’s it. That’s why 0! equals 1.

How to Simplify Factorial Expressions

Now that we have the basics behind us it’s time for the heart of our factorial tutorial: simplifying.

The best way to see mathematics is in action. So here I am to walk you through the six problems shown in the image at the top of this post! We’ll start nice and easy, then move into the combinations formula with numbers then variables, and end with some trickier factorial expressions.

❤ STAY CONNECTED ❤

Don’t miss a beat, or a math problem!

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Simplifying Factorials: The Easy Way was originally published in Math Hacks on Medium, where people are continuing the conversation by highlighting and responding to this story.