Rotational Inertia Formula: Things You Should Know

If you take a physics course, there’s a good chance you will learn about the rotational inertia formula. Click here to find out more!

If you take a physics course, there’s a good chance you will learn about the rotational inertia formula. Learn more about the formula and how knowing calculus can help you understand the formula better.

What Is The Rotational Inertia Formula?

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Whether you’re currently enrolled in a physics class or will be taking one in the near future, you are likely to learn about rotational inertia and how to use the formula. Like in calculus, some people understand physics a little easier and better than others. 

Even if you find physics to be a bit of a challenge, your knowledge and experience with calculus may actually be beneficial to working through various physics formulas. In our article, we’ll discuss how calculus can help you in physics, take a closer look at rotational inertia, and how to work through the rotational inertia formula.

What Is Rotational Inertia?

Before we take a look at its formula, it’s a good idea to take a closer look at rotational inertia. As you most likely
already know, the general definition of physics is the study of matter and energy and how they interact. Since physics is ever-changing, it is often a more complex topic than its seemingly simple definition.

Inertia describes a property of matter that remains at rest or in uniform motion unless changed by external force. Considering the definition of inertia, rotational inertia is best described as the measure of an object’s resistance to change during rotation.

Rotational inertia is also known as the moment of inertia or the second moment of mass. The second moment of mass refers to the length of the moment arm squared. The rotational inertia is the property of any object that has the ability to rotate.

Rotational inertia also has a scalar value, rather than a vector, which tells us how difficult it is to change the velocity of an object while in rotation.

Another thing that’s important to note about rotational inertia is that it is similar to mass in linear mechanics. The rotational inertia of an object greatly depends on the mass of the object, as well as the distribution of the mass in relation to the axis of rotation.

As the object moves further away from the axis of rotation, it’s harder to manipulate the rotational velocity. Why? The mass carries more momentum with it as it rotates, thanks to the increase of speed, as well as the momentum vector changing more quickly.

Before we jump into the rotational inertia formula, let’s take a look at rotational inertia in an everyday application.

Understanding Rotational Inertia In Everyday Life

It’s not uncommon to become easily overwhelmed or confused by the technical terms used in physics or calculus. Even if you consider yourself to be a “math-minded” person, it’s often more helpful to shift gears and take a look at a concept differently.

Since many concepts in physics are easier to understand in a real-world application, let’s break down rotational inertia in a “real world” sense.

Merry-Go-Round

Think of a merry-go-round (or a roundabout or carousel) on a playground; not the automated kind with animals at a carnival, but the type on a playground that children ride on. To get a merry-go-round started, someone needs to push the equipment. The harder the push, the faster it will accelerate.

A smaller merry-go-round will accelerate more quickly than a larger one, even if the same force of a push is used to start the equipment in motion. How does rotational inertia fit into this example?

Let’s say there are seven children on the merry-go-round. If the children are seated at the outer edge of the merry-go-round, it will be more challenging for the equipment to pick up speed, but if the same children move towards the center of the equipment (or the axis), it will accelerate more quickly.

The rotational inertia (or moment of inertia) is larger when the children are on the edge rather than closer to the axis; it’s important to note that the mass remains the same as well as the force used to rotate the merry-go-round.

Holding A Baseball Bat

If you have ever played softball or baseball, you may have experienced rotational inertia without even realizing. Anyone who’s ever held a bat knows that they aren’t all the same, as some are heavier than others.

Imagine only having the option to use the heaviest ball bat available. Depending on your strength and size, you might find it relatively easy or hard to use the bat. If you feel comfortable with the heavy bat, you’re likely to grip it on the handle (in the standard position) and have no problem hitting the ball with a strong force that accelerates it way into the outfield.

If you struggle with the weight of the bat, you can still hit the ball with the same force if you reposition your grip further up on the bat. Similar to what happens on the merry-go-round, repositioning your grip is decreasing the radius of the rotation, which will result in speed and force.

Something You Can Try At Home

As you begin to understand rotational inertia, you can probably think of more applications where this occurs. Here’s one more example that you can try at home.

If you have some sturdy cardboard on hand, cut out a circle with a 10 cm radius. Near the edge of the cardboard circle, write numbers 1 through 12, much like a clock face. Place a horizontal axis through its center and make sure your circle can spin freely; you may want to tack the circle to the wall using a nail or do something similar.

Place a lump of poster putty on the circle at number 3. If you don’t have poster putty on hand, you could try something like “silly putty,” but it may not stick as well. After you apply the putty to the circle, make sure that the circle is positioned so that the number 12 is at the top (like a clock).

Play around with the putty by adjusting the amount to see how much you need to make the circle spin in a full rotation. You may also try to move the putty closer to the axis to see how the speed and time are affected. While you probably won’t be keeping track of the weight of the putty, it’s a fun, hands-on experiment to try and gain a better understanding of rotational inertia.

Rotational Inertia Formula

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Now that you have a better understanding of the basics of rotational inertia, it’s time to take a look at the formula. You may see the rotational inertia formula written a little differently, but here’s how we break it down.

Rotational inertia = mass x radius^2 or I = mr

As long as you keep the formula, I = mrin mind, and what it means, you should be able to solve problems with relative ease and a little practice.

The best way to test out this formula is by jumping right into a problem. Let’s consider another real-life application using another classic piece of playground equipment: a tetherball. If you’re unfamiliar with a tetherball, it has a simple design of a pole and a ball attached to a rope.

The object is to hit the ball and wrap the rope around the pole as many times as you can before another player blocks the motion and intercepts the play. Let’s assume that the ball is 3 kg and the pole is 4 meters away. What’s the rotational inertia? Keeping the formula in mind, we are multiplying the mass by the radius squared. Since we know that the mass of the ball is 3 kg and the radius is 4 meters, our equation will look like this:

I= 3 x 4

Since 4 meters squared is 16 meters times 3 kilograms, the rotational inertia will be 16 times 3, which equals 48 kilograms times meters squared.

Even though our example is pretty basic, you should have a better understanding of rotational inertia and its formula overall. As you learn more about rotational inertia, you’re likely to have more challenging and involved questions, but practicing with simple examples can help you keep the concept and the formula fresh in your mind.

How Does Calculus Help With Understanding The Rotational Inertia Formula?

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Regardless of your experience in calculus or physics, you might be wondering how calculus can help you understand rotational inertia and its formula.

Newcomers to calculus and physics often don’t realize the strong connection to the two studies. As we already explained, physics is the study of matter and energy and how they change. Calculus analyzes things that change. Calculus allows you to take an up-close look at one part of a problem and use the math skills you already know.

You can undoubtedly succeed at physics without having knowledge of calculus, but even knowing the basic concepts of calculus may help you understand rotational inertia, its formula, and countless other formulas in physics.

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Cantor Set: Generalization, Properties & Everything You Need To Know

The Cantor set is a set of points lying on a single line segment that has a number of remarkable and deep properties. Learn more about it on this article.

Maybe you’ve glanced ahead at your calculus syllabus and notice that there will be a section about the Cantor set. Learning about the Cantor set can be intriguing for many students, but also very challenging.

In this article, we give you a basic introduction to the set and how to construct it. By introducing you to the basics, the goal is to help you feel more comfortable and familiar with the material when it’s presented to you in class. Our information can also be helpful to refreshing your memory if you’ve already learned about the set or just need to brush up on the basics.

Your Introduction To A Cantor Set

Before we jump into constructing the set and doing some hands-on work, let’s take a look at the background. If you’re like many students of mathematics, you might assume that most of the concepts that we learn and discuss in calculus have a history that dates back to centuries ago. This can be an easy assumption as much of what we learn feels like it’s always just been here.

Compared to other concepts, Cantor sets are relatively new and the concept was first introduced by a German mathematician, Georg Cantor in 1883. Cantor sets are sometimes referred to as the “Cantor comb” or the “no middle third set.” If you are unfamiliar with these terms, they will make more sense when we talk about constructing a set.

Like many concepts in calculus and other areas of mathematics, mathematicians may define the Cantor sets a little differently. In general, Cantor sets are closed and totally disconnected. They are a perfect subset of a closed interval, which is traditionally (0,1); we will go more in-depth on this a bit later.

Cantor sets are also the result of an iterative process, or getting the desired result by repeating a cycle of operations, and it’s often referred to as an example of a fractal (or neverending pattern).

You might be wondering why Cantor’s discovery holds any value today. Some mathematicians note that Cantor’s set is a tool that was created and used to disprove certain properties. Although mathematics may seem fairly stagnant, properties and concepts are constantly challenged and sometimes even proven wrong.

Many people had limited knowledge about what sets would measure to zero. While many assumed that cubed rational numbers were some of the only that could measure to zero, Cantor’s discovery proved that there are many numbers that can measure to zero.

Constructing A Cantor Set

As we noted before, cantor sets are sometimes referred to as a “cantor comb,” once you begin to construct one on your own or watch someone construct the set, you’re likely to see why it’s been given that nickname.

At first glance, you might think the early construction of a set looks something like your bracket for Fantasy Football; in some ways it does, but it’s a process that’s infinite. You don’t need many tools to construct a set, just a pen and paper or a dry erase board.

Start by drawing a line segment of no particular length. On one end write “0” and on the other end write “1.” This line segment (0,1) is your starting point.

The next step is to take out the middle third of your line segment; this is also known as the open middle third. After you take out the middle third you are left with one end of the closed line segment is 0 to ⅓ and the other is ⅔ to 1. All the points between ⅓ and ⅔ have been removed, but ⅓ and ⅔ remain.

You should continue to remove the open middle third of each closed line segment. The next set of results should be four shorter closed line segments with each one having a length of 1/9. So, one closed line segment will be 0 to 1/9, 2/9 to ⅓, ⅔ to 7/9, and 8/9 to 1; everything else in between is gone.

As you construct your set, you continue to break it down and remove the open middle third. After your four short closed line segments, you should have eight closed line segments that each have a length of 1/27.

While this may be difficult to visualize without actually drawing it out yourself or watching someone else construct a set, you probably have a good idea where this is going. You continue to break down the set by removing the open middle third and creating smaller closed line segments. This process goes on and on until your left with what many people  call, “cantor dust.”

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What happens to all the numbers that are removed? While they don’t appear in our set anymore, if you were to put them all together they would equal 1, which would bring us back to where we started (the original line segment of 0 to 1).

Now that we’ve given you an explanation on how to construct the set, let’s take a look at how you would write in down are part of your formula.

After you draw the your line segment, you can call this set . Once you remove the first middle third, you will call this set . Your subsequent sets should be , , and so on.

It’s important to note that at each step of the set, you are doubling the number of closed line segments. has one segment and has two segments. , which is also or 4 segments. has or 8 segments.

By the time you get down to you have or 1,024
segments. While this may seem like a lot, it’s really just the beginning.

The length of your first segment, , is zero because there are no gaps in the segment. Your second set, or , has a gap length of ⅓ and has a gap length is 1/9, which results in the total length of 5/9. The total gap lengths of is 19/27.

By the time you get to , your formula for finding the total length will look something like this. Since your result is ≈ .982658 you can see that the further you get in your set, the closer you get back to the number 1 (at which the set originated).

The Structure of Cantor’s Set

We’ve given you the basics of Cantor’s set, and you will undoubtedly dive deeper into the details during your calculus course. Depending on which level of class you take, you may only touch the basics like we have here or you will spend more time playing around with numbers.

Your instructor may even suggest removing the middle fourth of the line segment (rather than the middle third) just to see how your results vary. One of the unique features of Cantor’s set is that it is relatively easy if you don’t become too overwhelmed by the formula and keep certain things in mind, such as the structure of the set.

Here are some points to consider and remember, which may help you understand the set even better:

  • The set has no interior points, which leads it to “Cantor’s dust.” Why? The length is zero, which means it contains no intervals.

  • The set is bounded, which means it lives inside of the interval of (0,1)

  • The set is compact, which means that every task might require an infinite number of steps. It can, however, be solved with a finite number of steps.

  • The set has not isolated points

At first glance, Cantor’s set may look pretty easy to tackle and depending on your previous experience with calculus; you might pick up the concepts pretty quickly.

Like all concepts in math, Cantor’s set often takes a lot of practice and patience. Ask your instructor for help as soon as you don’t understand something. Many students spend time trying to catch up to what their instructor is explaining but are too nervous or frustrated to ask for help, especially in an advanced math course.

Don’t hesitate to reach out for other additional resources. Even if your instructor is one of the best, sometimes getting another perspective can be beneficial to learning something new.

If you don’t have access to additional resources, consider forming a study group or see if there are resources at your school (such as math tutoring) that can give you the extra attention you might need to succeed in understanding Cantor’s set and calculus overall.

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Area Of A Trapezium – Different Ways To Calculate The Area

Wondering how to find the area of a trapezium? Learn and revise how to calculate the area of dimensional shapes like trapezoids.

Finding the area of various shapes like trapeziums may sound difficult in the start due to their forms. However, all you need is the right formula to calculate the area of a trapezium. The area covers a dimensional space that it occupies.

In this article, we’ll define what a trapezium is and show you different ways to find the area.

What Is A Trapezium?

A trapezium is a quadrilateral with a set of parallel sides. The quadrilateral has four parts, but if it has a pair of parallel sides, it’s a trapezoid. However, if both pairs of sides are parallel, it’s a parallelogram.

Some of a trapezium’s properties include:

  • No angles, sides, and the diagonals
    are congruent
  • Parallel bases which mean AB/CD

  • Four parts

  • Each lower base angle supplements
    the upper base angle on a similar side

What Is An Isosceles Trapezoid?

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An Isosceles trapezoid is a unique trapezium with equal length legs, with the legs being the non-parallel sides. Some properties of isosceles trapezoid include:

  • Congruent diagonals

  • The features of trapezoid work by
    definition

  • The lower and upper base angles are
    congruent

What Is The Origin Of The Word Trapezium?

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Euclid wrote his geometry work about 300 B.C. in his book; he did use the word trapezion, as the name of a four-sided figure that wasn’t a rhomboid, rectangle or a rhombus. The name later got a Latin translation to now the trapezium.

Later, Proclus a geometer incorporated specific refinements to Euclid’s work, which have now become geometry basics. Proclus came up with a new meaning of a trapezium, where he referred to it as any four-sided figure with two parallel sides.

The geometer later came up with a new term, trapezoid which denotes a four-sided figure with no two parallel sides. However, in 1795, an English mathematician, Charles Hutton got the descriptions reversed. The US still follows the Hutton’s definition although the British have reverted to Proclus original meaning.

How To Calculate The Area Of A Trapezium

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For you to calculate the area of a trapezium, you need to have a formula. Find the length of the nonparallel side, the height of the trapezoid, and the length of one base. The height of a trapezoid is the diagonal line that extends from one corner to another opposite edge.

The basic formula is (B1+B2) x1/2x (H)

  • Where B1 is the upper parallel
    length of the trapezium

  • Where B2 is the lower parallel
    length of the trapezium

  • Where H is the height of the
    trapezium

Also, you can get the area of a trapezoid by dividing it into two triangles and a rectangle. At this point, you’ll need to calculate the area of both triangles and find the area of the rectangle. Sum these two to get the area of the trapezoid.

The formula for calculating the area of the triangle will be ½ (b-a) h, where b is the length minus the height, and h is the height of the triangle.

The formula for creating the area of the rectangle is ah, where a is the length of the triangle and h is the height.

For example, if you have the height of the trapezium is 8 cm, the lower height 14 cm, and its height 4 cm, you’ll have a rectangle with a length of 8 cm and a height of 4 cm.

The triangle will have 6 cm length when you subtract 14cm from 8 cm. The height of the triangle remains 4 cm.

Now to calculate the area of the trapezium, you’ll need the area of both triangles and the area of the rectangle.

The area of the triangle= (6X4)/2=12

The area of the rectangle= (8X4) =32

Add both these areas to get 44cm2 to get the area of a trapezium.

Also, you can use the previous formula of ½ (b1+b2) h ½ (14+8) x4=44cm2

What If You Have The Median?

You can find the area of a trapezoid if you have
the median. The median is the line segment that links the midpoints of the
non-parallel sides. Also, the median is the average of two parallel sides.

When you have the median and the height, you can
get the area of a trapezoid using the formula:

A=mh where h is the height and m the median.

Finding The Area Of A Trapezium Without The length Of One  Parallel Side

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In some cases, you may only have the length of the diagonal, length of the diagonal, height of the trapezoid, and one nonparallel side. You now need to calculate the area of the trapezoid. Fortunately, you can calculate the area despite having only one nonparallel side.

All you need is to find the length of the base, the height, and the distance of the given nonparallel side. For example, if you
have a trapezoid with 4 inches height, a nonparallel side of 5 inches, and the base equal to 6 inches, you need to identify the length of the other side.

If the trapezoid has 8 inches of diagonal, use the Pythagorean Theorem to find the length of the unknown side.

The formula is (a x 2) + (b x2) =(C x 2)

The letters a and b represent the two other sides, and c is the hypotenuse.

With the trapezium, you’ll have two triangles. The total of the two unknown sides of the triangles is the length of the hidden side. Use the Pythagorean formula to find the unknown sides.

The second triangle will have a length of 6 inches and the first triangle a length of 3 inches. Add both inches to get the
length of the unknown base as 9 inches.

To now, calculate the area of a trapezoid, ½ (b1 + b2) h= ½ (6+9) 4= 30 inches.

Finding the Area of a Trapezium without Parallel Sides

Sometimes you may need to find the area of a trapezium that doesn’t have parallel sides. To get started, you’ll need to divide the trapezium into two triangles.

The next step is to get the values for an angle of the side, one side that you know, and the other side indicated. You could name the sides a, b, and c. Each triangle will have all the sides above.

To calculate the areas of the triangles, use the formula

A=be/2 (Sin x a)

Put in the values of a, b, and c into the formula to calculate the area of both triangles. Add the areas for the triangles to get the area of a trapezium without parallel sides.

Calculating Missing Values

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You may need to figure out the base or the height when you have the area. Here’s how you can get the missing values.

How To Calculate The Height When You Have The Area

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While sometimes you can calculate the area using a standard formula, you may be required to figure out the height when given the area. To do this, you’ll need to transpose the conventional method.

You can use algebra to figure out the formula. For example, to get the height, you can use the formula:

H=2A/ (a+ b)

Where A is the area and a, b two bases.

To get the value of the base, you need the height, area, and two bases. Rearrange the main formula to get the base.

The formula will be:

B=2A/ (h)-a where A is the area of the trapezium and a is one side of the trapezium.

Applications Of Trapeziums

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In biology, you’ll find terms like trapeziform or trapezoidal which refer to specific organs or forms. You can also find the term in architecture as it refers to buildings, windows, symmetrical doors with a broad base and tapering to the top.

The windows on A-frame gables resemble trapezoids with the bottom sides horizontal and the parallel sides running from left to right.

When looking at Egyptian style of building, you’ll notice most of their properties have an isosceles trapezoid shape with straight sides and angular corners. The Inca has this standard style for their windows and doors.

Truss bridges have several trapezoids that connect the base of the bridge to the overhead structure. The aluminum or steel supports create adjacent trapezoids that have parallel sides, which form the bottom and top of the bridge sides.

In Geometry, you’ll find out that the crossed ladders problem is the method of finding the distance between the parallel sides of the right trapezoid when you have the distance from the vertical; base to the slanting intersection and the diagonal lengths.

Most handbags have two trapezoids as the larger sides of the purse. You’ll notice that the top and bottom are parallel, although the top edge is shorter than the one at the bottom.

Final Thoughts

Calculating the area of a trapezium isn’t complicated. All you need is to use the above formula. Alternatively, you can divide the trapezium into two triangles and a rectangle. Calculate the area of the triangles and add this to the area of the rectangle.

Also, you can use trigonometric functions like tangent, cosine, and sine to find the missing sides of either of the triangles. These functions also work when you know the angles of the trapezium.

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Slope Formula In Calculus: Understanding The Application

Learn about the slope formula and its applications in the real world. Along with math, have numerous applications in your everyday life.

The history of calculus and its many problems is almost as fascinating as its many modern applications and challenges. Gottfried Wilhelm Leibniz and Sir Isaac Newton discovered calculus in the 17th century. However, they did so independently then accused each other of stealing the other’s work. Their mathematical feud continued until the end of their respective lives.

Some historians argue that calculus was commonplace much earlier than the 17th century. Some claim it got used as early as 2000 BC. It’s true that some parts of calculus had been around for centuries before Newton’s and Leibniz’s discoveries, but they get credited with bringing all the functions, derivatives, integrals, and other terms into a group and calling it infinitesimal calculus.

For clarity, calculus is here to help us solve problems, so we use the term problems as a way to describe an equation since it’s a problem until we answer it. The challenge in solving many problems is determining which formula applies and how to write the equation. Even a small error in a formula can wildly change the outcome when you’re dealing with math.

It sounds simple enough, right? However, if you knew a house cost X dollars in 1980 and it’s value increased until it was sold again in 1990 and you need to show at what rate the value of the house increased over time, how do you write that statement in mathematical terms? It sounds mind-numbingly hard, but the solution is simple if we use the right formula.

The calculus way to finding a solution to our problem is by creating a graph and finding the slope of the line that connects the original value of the house to its value at the time it got sold. The slope of the line indicates the rate at which the house increased in value. We’re assuming that the rate of increase created a straight line for this problem.

Defining The Slope Formula In Calculus

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The slope of a line is always the same anywhere along the route. If we get technical, the slope is merely a ratio that defines the changes in a line as it goes from its starting point to its end. It’s easier to understand if you picture a graph and imagine the line starting at the bottom left corner of a graph and growing until it reaches any point along the right wall of the chart.

The changes in the line’s rise as it travels to its ending point on the right wall is the line’s slope. Let’s look at a real-world scenario like determining the slope of a roof. Picture a graph with a bottom row of evenly spaced numbers from zero to nine. The right wall of our graph has the same numbers on it except they go up vertically beginning with zero.

Our roof begins at the bottom left over the number zero and travels up until it ends at the number seven on the right wall of the graph. We know the graph is nine points across the bottom and the roof intersects the right wall at the number eight, we can write a simple formula to explain the slope which is: the slope equals eight divided by nine or 0.888 repeating.

If the numbers on our graph indicate feet, we now know the roof rises about 0.88 feet per foot of roof length. This application of the slope formula has a lot of applications in the real world. For instance, an accessible wheelchair ramp at a public facility in some states must maintain a specific ratio. The ramp must be one foot long for every inch of rising. If the ramp is four inches tall, it must be four feet long.

Our problem is simple to keep the math simple for the sake of explaining the slope formula. The math gets more complicated based on the type of slope. There are four types of slopes to contend with including:

  • Zero slope: the line is perfectly horizontal

  • Positive slope: this is when a line increases in height

  • Negative slope: this is a positive slope in reverse

  • Undefined slope: primarily used to define the slope of a vertical line because you can’t divide by zero

Real World Examples Of The Importance Of The Slope Formula

We use math every day in hundreds of ways. Without math, we couldn’t drive cars or browse the internet. If you look closely, you’ll find some form of math in everything you do from cooking to exercising. Aside from built-in math that we all take advantage of, math is essential to many careers. You just can’t exist and not use math directly or indirectly.

For most of us, we use math indirectly almost constantly. For instance, we know it takes 20 minutes to drive from home to work when traffic is slow. However, some days traffic is heavier than others so we use variables and math to determine when we should leave to ensure we make it to our job on time. Imagine the horror of baking without using math to measure ingredients.

You may use math in a more direct sense if you work in medicine, construction, or marketing. If you do work in these fields, you know that the slope formula may be one of the most critical math formulas that we take for granted. It has different names depending on the occupation, but it’s all the same when you reduce it to pure math.

The most apparent jobs that require a solid understanding of the slope formula are architects and construction workers. They use the slope formula or a variation of it when designing or building roofs, stairs, ramps, and just about every part of a building above ground. A typical house is little more than a box with a triangle nested on top of it. The slope formula is used to build the triangle on top.

The slope of the roof determines the number of materials needed to build it along with how to measure and cut each piece of material. It is much easier to construct a triangle out of wood and nails if you know how long each piece of wood needs to be to make the slope correct. In some cases, getting the slope wrong could cause the roof to collapse or perform poorly in high winds.

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The slope formula makes everything around us, in respect to construction, safer. A slope is used to make roads safer and easier to travel across. It makes stairs and accessibility ramps safer and easier to use as well. Imagine building stairs without using the slope formula. Some might stretch out into the street while others may be too steep to climb. The same principle applies to roads and bridges.

Social scientists and marketing professionals use the slope formula as well. Remember our example of a graph and determining the rise of a roof based on its run? The same principle gets used to determine things like how well a product is increasing concerning sales or how a population is performing based on specific economic indicators.

The slope of a line on a marketing graph tells a company how their product is performing. A negative slope indicates poor performance while a positive slope means the product is doing well. Determining the slope of the line on the marketing graph lets them know exactly how well the product is doing over time and may even indicate when the product’s sales are at their best.

People that work in healthcare use the slope formula in a more critical sense. For example, epidemiologists may use a graph and the slope formula to determine how rapidly a disease is progressing over time. Doctors may use it to determine how long it might take for a treatment to cure a patient. The slope of a line may tell a doctor to increase or decrease medications as well.

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It’s critically important to healthcare to know the slope of lines on a chart if we expect doctors to find cures for terminal disease and many chronic ailments that don’t have treatments. It’s all about knowing how high a line rises over a specific period. That’s all math and 100 percent slope formula.

The list of uses for the slope formula is vast. If your career is impacted by it, you recognize its importance and understand why we think it’s one of the most crucial calculus topics. Beyond the examples we touched on above, the slope formula has many other uses including:

  • Determining speed over time

  • Determining distance over time

  • Calculating stock prices over time

  • Determining weight loss over time

  • Calculating win versus loss rates for sports teams

  • Calculating pay rate increases over time

Some Final Notes

Nothing we see or do today would be possible without math. Okay, a few things might be possible, but roads and stairs would be unsafe while some buildings might fall randomly. You wouldn’t be able to read this article without math because the internet and your computer could not exist without it. Appreciate the power of it and how the slope formula impacts your life at every angle.

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Parallel Planes: How It Is Used In Calculus And Geometry?

Understand the basics of parallel planes and how it is used in calculus and geometry. Learn more of parallel postulate and other calculus formulas.

A parallel plane is a flat, two-dimensional surface. If you have two planes that don’t intersect, they’re parallel. Use calculus formulas to determine the distance between two parallel planes. You can solve some problems by finding random points on the plane, and then use an equation to figure the distance.

What Are Parallel Planes In Calculus?

You probably deal with parallel parking now and then, so you may have a vague idea of the definition of parallelism and how it might be used with figures in calculus and geometry.

You’ll encounter parallel planes in your Calculus 3 classes, and focus on equations of planes and other problems.  

Here’s a look at planes in calculus, and how parallelism relates to them. We’ll also look at parallel postulates, and how parallel lines and planes are used in geometry and calculus.

What Is A Parallel Plane?

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In calculus or geometry, a plane is a two-dimensional, flat surface. Two non-intersecting planes are parallel. You can find three parallel planes in cubes. The planes on opposite sides of the cube are parallel to each other.

Parallel lines are mentioned much more than planes that are parallel. They are the lines in a plane that don’t meet. A plane and a line, or two planes in a 3D Euclidean space are parallel if they don’t share a point.

Parallelism is used in Euclidean and affine geometry. Hyperbolic geometry may have lines with analogous features that fall under parallelism’s properties.  

Skew lines are two lines in a 3D space that don’t meet in a common plane.

The Parallel Postulate 

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Euclid’s parallel postulate says that for every straight line and a point that’s not on it, there’s only one straight line passing through that point that doesn’t intersect the first line, regardless of how far the line or point are extended.

This mathematical postulate corresponds with Euclid’s Fifth Postulate. Euclid didn’t use this postulate until Proposition 29 of the Elements.

Many experts didn’t believe the Fifth Postulate was accurate but considered it a theorem taken from Euclid’s first four postulates. The term absolute geometry refers to geometry that is based on Euclid’s first four postulates.  

Many parallel postulate proofs have been written and discussed by the mathematical community over the centuries.  

The dissertation of G. S. Klügel in 1763 called Euclid’s parallel postulate a necessary tool to prove mathematical results. The postulate wasn’t intuitive, Klügel wrote, but it was helpful.

An axiom proposed by 17th Century mathematician John Wallis stated that a triangle could be changed to be larger or smaller with any distortion of its angles or proportions.

Lobachevsky and Janos Bolyai, in two separate 1823 studies, concluded that you could create non-Euclidean geometry that didn’t adhere to the parallel postulate.

The parallel postulate refers to modern day Euclidean geometry. Change the phrase so only one straight line passes or no line which passes exists, or two lines or more pass, you will be describing elliptic (no line passing) or hyperbolic geometry (two or more lines passing).

The following theorems and axioms are equal to the parallel postulate:

  • Playfair’s Axiom,

  • Triangle Postulate,

  • Equidistance Postulate,

  • Pythagorean Theorem

  • Proclus’ Axiom.

One of Hilbert’s parallel axioms is also equivalent to the parallel postulate.

Find The Distance Between Two Parallel Planes 

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Our planes are IIand II2.  These planes are parallel if II1’s normal (n= abc1) is the scalar multiple k of II2’s normal n_2 = (ka_2, kb_2, kc_2)

Example #1

Example #2

Example #3

Let’s look at the planes II2:4x+8y+12z+6=0 and II1:2x+4y+6z+1=0. Get the norms of these planes to arrive at n1=(2,4,6) and n2=(4,8,12) You’ll get 2n1=n2 and finally II1∥II2.

Find an arbitrary point on II1 or II2. After choosing your arbitrary point on one of the planes, use the equation for the other plane in the formula for the distance between a plane and point. You’ll now determine the distance between the two planes.  

The planes II2x+3y+4z−3=0 and II2:−4x−6y−8z+8=0 are the basis of our second problem. Find a random point on the first plane, for example, 0,0, and four over 3, and then use formula D (distance) equals ax0 plus by0 plus cz0 plus d over the square root of a squared plus b squared plus c squared.  

Now figure out the distance between the two planes using this formula.

D equals 4(0) plus negative 6(0) plus negative 8(3/4) plus 8 over the square root of negative 4 to the second power plus negative 6 to the second power plus negative 8 to the second power, followed by D equals negative 6 plus 8 over the square root of 16 plus 36 plus 64, then D equals 2 over the square root of 116.

Practicing With Parallel Planes

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Remember to practice equations involving parallel planes several times, using problems other than those assigned by your teacher. There are plenty of practice questions in textbooks and online. You should also join a study group or contact a tutor if you need more practice with geometry or calculus.  

Logical Learning In Calculus And Geometry

The logical or mathematical style of learning works for any subject, but it works best for algebra, geometry, and calculus. Logical learning enables you to recognize patterns easily and draw connections between content that may seem unrelated. You know how to group information do you arrive at a correct conclusion.    

A person with a propensity for logical learning remembers the basics of geometry, algebra, and calculus without referring to a textbook. You can perform moderately difficult calculations in your head.

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You use a system to work through problems, and apply it to all types of equations and mathematical questions. Setting budgets and numerical goalposts helps you make progress when you solve complicated equations. A scientific thought process allows you to support your arguments with statistics and facts.

You should identify and point out flaws in other people’s’ logic, and work out strategies for all types of projects. You may play video games that involve detective work and strategizing simulated war plans. If you have a logical thinking style, you may like science, computer programming or law as well as mathematics.

You look for the logical way to solve a calculus or geometry problem. You strive to understand all the details behind why you perform certain steps in solving an equation. You aren’t satisfied with merely memorizing formulas. Explore the logical steps you use to apply to a problem, and keep them in mind when you tackle a new equation.   

Don’t overanalyze when you work on a difficult problem. Work with the formula and prior knowledge of similar equations. If you don’t get the answer right at first, try again, but avoid developing “analysis paralysis” when figuring out what you did wrong.

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Substitution Method In Calculus: Integration, Derivatives, And More

Read this guide to learn how to use substitution method in calculus including derivatives, definite integral, trigonometric functions, and more.

Tired of solving messy expressions when working on integrals in calculus? Learn to use the substitution method and help clean them up.

What Is The Substitution Method In Calculus?

If you’ve studied calculus for any length of time, you’ve encountered integration and likely looked for different ways to work out an integral. The substitution method for integrals is nothing more than another way to find it, but bear in mind that it only works when the integral is written in a relatively simplified form, namely the product of a function and its derivative.

Using u-substitution makes it easier to read and simplify composite functions, which can otherwise start to look messy and intimidating.

This method is also called u-substitution. The idea behind doing u-substitution is to rewrite the integral into a form that easily fits with the known rules of integrals.  We’ll look at these rules, go over the basics of what integrals are to provide a refresher, then go into how to do the substitution method. You can only integrate by using the known rules.

What Is Integration?

Integration in calculus is the idea of taking a generalized area underneath the curve of a function. It is impossible to get an exact area, but you can approximate it with an integral, which is the combination of different measurements of the area under a given curve. On a Cartesian graph, you would have to use rectangular or trapezoidal sections, but that isn’t accurate.

Of course, taking a higher number of thinner slices of x gives you a more accurate measurement of the area under the curve. In this case, x is the horizontal distance between the dividing lines you use to measure. The integral is what happens when x approaches zero. You can’t use a zero value for x, because there is no change in x to measure.

All of this said, there are some rules to integration you need to be aware of. We’ll run down a brief list of the most common ones.  The expression ‘dx’ indicates that the change in x between points is infinitely small without being zero.


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Known Rules Of Integration

  • Power Rule: ∫f(xn dx) = [(xn+1)/n+1] + C

  • Multiplication: ∫cf(x) dx = c∫f(x)dx)

  • Sum: ∫(f + g) dx = ∫f dx + ∫g dx

  • Difference: ∫(f – g) dx = ∫f dx – ∫g dx

These rules let you calculate the integral of almost any function. There are other rules of integration, but for now we’ll focus on the simplest and most useful.


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Uses Of The Integral

Taking the integral gives you the closest possible measurement of the area under a function’s graph. You can take definite or indefinite versions. Taking a definite integral provides bounds on the two values of x, while indefinite ones focus on the whole function from negative to positive infinity.

Most practical applications will focus on definite integrals, which have their two values denoted as subscript and superscript next to the integral sign like this: ∫ab f(x) dx. In other words, you calculate the problem using the value of b, then the value of a, and use the difference.


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Derivatives

Since we know about integrals, we also need to look at derivatives because the two types of functions work together in calculus.

A derivative is nothing more than the rate of change of any function at a given point. Knowing these derivatives and their associated rules allows you to easily solve calculus problems when using the substitution method.

Here’s a list of some common derivative rules:

  • Power Rule: xn = nxn-1

  • Multiplication by Constant: cx = d/dx cx

  • Addition: f + g = f d/dx + g d/dx

  • Subtraction: f – g = f d/dx – g d/dx

  • Product: fg = f d/dx g + g d/dx + f

  • Quotient = f/g = (f d/dx g – g d/dx f)/g2


Substitution Rule

Now we’ll talk about the substitution rule. Using the u-substitution rule makes it easier to read and work with composite functions, i.e. (f(g(x)) by putting the variable u in place of the inner function, or g(x). You then multiply this by the derivative of u, also called du. Then, you place your original function in place of u and the derivatives, which gives you your answer.

Here’s how the method looks in an example problem.

We’ll start with ∫(x + 2)3 dx. The derivative of any line notated by m = ax + by + c is the variable a, which is the slope. Therefore, the integral could be rewritten as ∫(x + 2)3 * 1 dx. Now, you simply place u into the integral in place of the parenthetical expression like so: ∫(u)3 du.

Per the power rule, you can see that ∫u3 du = u4/4 + C. C is the constant of integration. In an indefinite integral, you can plug virtually any value into the function, and it will remain valid. Essentially, all you’re doing is shifting the graph of the function up or down, but not changing the actual graph. Also, the derivative of the constant is 0, so it has no real effect.

Getting back to the integral, from this step, you simply plug the original function back in where you substituted u. Completing this step yields (x + 2)4/4 + C, where again, C is the constant of integration if there was one.

Looking at another integral using the substitution method in calculus, let’s take ∫(2x + 3)4 dx. This one looks a little trickier at first, but it’s essentially the same concept. Substitute u into the parenthesis, making ∫(u)4 2 du.

Because you added the 2 into the expression, you need to do something to cancel it out, making the expression 1/2∫(u)4 2 du. Multiplying the expression by ½ effectively does this.

After this you’ll need to take the integral of the function. Do this using the power rule just like before. In turn, this yields you with ½ u5/5 + C. Putting the original function back into the equation, you get (2x + 3)5/10 as the integral.

These are just two of the simplest examples. Remember that you can only use the substitution method when the integral can be expressed in the form of ∫(f(g)x)) g d/dx, or ∫f(u) du.


Definite Integrals

We’ve talked about indefinite integrals up to this point, so let’s move on to definite ones. As we mentioned, a definite integral is solved in the same way as an indefinite one, except you’re putting in actual values for x and taking the difference between the two results.

Let’s go back to our original example of ∫(x + 2)3. This time, we’ll set bounds at 1 and 3. This puts the integral at ∫13(x + 2)3.

Now, we just have to rewrite our original integral: u4/4. Now, you’ll plug the original function back in as (x + 2), but this time, you’ll be putting in 3 for x: (3 + 2)4, or 54/4. 156.25 is the result. Then you’ll need to calculate the same for 1 as the value of x. (1 + 2)4 or 34/4. This result is 81.

Now, you just subtract 81 from 156.25, yielding 75.25 as the definite integral of the function.


Trigonometric Functions

You can also take the integrals of trigonometric functions in calculus. Sine, cosine, and tangent, are all possible to do. You can also get secant, cosecant, and cotangent, but aside from the secant, you don’t use the others much.

  • ∫cos x dx = sin x + C

  • ∫sin x dx = -cos x + C

  • ∫tan x dx = -ln|cosx| + C

The tangent integral can also be expressed as ∫(sin x/cos x) dx, just as a regular tangent operation can be expressed as sin/cos. A good example is if you try to take the tangent of a 90-degree angle: the sine of 90 is 1, and the cosine is 0; the problem becomes immediately obvious. Likewise, the tangent of 0 is 0.


Drawbacks Of The Substitution Method

The substitution method in calculus is an excellent method in most cases, but it’s easy to get wrong for beginners. One of the biggest problems beginners have with this method is not substituting u for the whole expression. The variable u is meant to be the whole inner function f(x).

Another big problem is when beginners forget to add a constant to the outside of the integral, so it can be expressed in the proper form. For example, if you have 3x inside the integral, you’ll want to multiply the whole expression by ⅓ so the answer isn’t changed.


Final Thoughts

If you learn the substitution method early in your studies of calculus, you should have fewer problems when solving integrals and derivatives than you would otherwise. Some of the functions can look intimidating, but simplifying and expressing them in terms of u and the derivative of u cleans up the problem and makes it easier to solve.


Arc Length Formula in Calculus: The Complete and In-depth Guide

If you want to know about arc length formula then this guide is for you. In this section we will discuss how to find the arc length of a parametric curve.

Figuring out the length of an arc on a graph works out differently than it would if you were trying to find the length of a segment of a circle. Where the length of a segment of a circle can be figured out with some simple knowledge of geometry (or trigonometry), finding the arc length of a function is a little more complicated. You need to use calculus for that.

Finding the length of an arc works almost the same way as finding the length between line segments. In a way, you could argue that’s what you’re doing, except taken to an infinitely precise scale because the resolution of the curve is infinite as far as you can tell. There are no jagged edges or line segments.

Using Line Segments to Approximate Arc Length

 

To start, assume you have a given function with no interruptions and uniform smoothness. To get a basic idea of how long the arc was, you would start by separating the function into line segments placed at equal values of x. For example, you might plot the arc segments at x = 2, x = 4, x = 6, and so on.

You then use the Pythagorean Theorem to calculate the distances between the two points. However, you’re connecting two points directly rather than following a curve. This means you may be missing length or some extra length was added if the section of the graph happens to be a concave curve. What do you do?

You could take more sections by placing the line segment points at x =1, x =2, x=4 and so on. This would more closely approximate the curve of the graph. However, it still wouldn’t completely match. It doesn’t matter how small you make the sections: you cannot exactly match the section length with normal math.

This is where the calculus comes into play. You have to take derivatives and make use of integral functions to get use the arc length formula in calculus. We’ll give you a refresher of the definitions of derivatives and integrals.

What is a Derivative?

 

The derivative of any function is nothing more than the slope. There are several rules and common derivative functions that you can follow based on the function. For example, if you have a constant, such as x = 6, the derivative is 0 because there is no slope. For a function f(x) – ax where a is the slope, the derivative is the variable a.

If you have x2, the derivative with respect to x is 2x. In other words, at any given point for x on the function, as the limit approaches zero for a given value of delta-x, its derivative will be 2x.

These are just a few examples of using a derivative, and you’ll need to know or have access to them when figuring out arc lengths.

Integrals

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An integral is one of the two fundamental operations in calculus, the other being differentiation. When we were talking about line segments as portions of the arc length, those are a crude form of differentiation. An integral is the act of taking discrete segments and combining them into one. As with derivatives, integrals have their own rules and basic functions.

Common Rules for Integration

 

The integral of ax dx, where dx is simply another way of saying ‘as delta-x approaches zero’, is ax + C, where C is the constant of integration. You don’t necessarily need to know the constant of integration, because it can be any number when you’re dealing with an indefinite integral.

An indefinite integral does not set upper or lower bounds for x. In practical uses, you aren’t going to see a lot of functions that have negative values for x or y, so you’ll just be taking the area under the curve in the positive x-y quadrant.

When you’re trying to find the arc length formula, you’ll need to use definite integrals. A definite integral sets boundaries on x. To find a definite integral, i.e., the area under a curve, you find the integral using the larger value of x, then the smaller value of x, and take the difference of the two numbers.

Arc Length Formula

 

The formula for arc length is ∫ab √1+(f’(x))2 dx. When you see the statement f’(x), it just means the derivative of f(x). In the integral, a and b are the two bounds of the arc segment. Therefore, all you would do is take the derivative of whatever the function is, plug it into the appropriate slot, and substitute the two values of x.

For example, let’s assume you have  ∫14 √1+(f’(x))2 dx. Let’s also assume that f(x) is 2×2 + 2. In this case, it doesn’t matter that there’s a + 2 at the end. All were looking for is the derivative, which is the same no matter where the function actually lies on the graph. In this case, the derivative of 2×2 is 4x. At any point on the graph, this will be the derivative.

Now, you can plug 4x into the formula: ∫14 √1+(4x)2 dx.

Following that, you need to substitute the first value for x, which is 1. ∫14 √1+(4(1))2 dx. Remember the order of operations when working with these numbers: you’d first multiply the 1 by 4 to get 4, then square it to get 16. 16 + 1 is 17, so the integral of this function is the square root of 17.

Next, use the same formula and plug 4 into x. ∫14 √1+4(4))2 dx. In this case, you get the square root of 257. Because these to square roots do not simplify any further, you have to express the arc length for this function and these bounds as √257 – √17. Putting the two numbers beneath the same radical will be incorrect; √9-4 is not the same as √9 – √4.

More Complex Functions

If you have a more complex function, like a cubic function or a higher degree polynomial, it isn’t going to be as easy to take derivatives and integrals. This is where the substitution method can come into play. By replacing the whole function with the variable of u and taking the appropriate derivative, it’s easier to work with the function than it otherwise would be.

Just remember that if you use the substitution method, you have to reverse the substitution by putting the original functions back into the integral where they belong. Otherwise, you will likely get an incorrect answer.

Are There Exceptions to the Arc Length Formula?

Illustration

 

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The arc length formula works only for functions that have no breaks or asymptotes.  For example, if you have a function with an asymptote at x = 4 on either side, you can’t use the arc length formula across the two sections. Similarly, if there is a limit, you cannot calculate length across that limit.

Finding the length of an arc is functionally not that different from finding the length of a diagonal line segment on a graph, although you have to use more complicated math to get the proper result. Always verify your function before starting the problem.

Things to Keep in Mind

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If you have a calculus workbook or cheat sheet with lists of rules for derivatives and integration, keep it handy or memorize it. If you work enough of these problems, then the rules will start to commit themselves to memory naturally, but at first, it’s a good idea to keep notes nearby. This makes it less likely that you’ll use the wrong derivative when you’re solving problems.

Remember to study the following:

  •  

    Integrals

  •  

    Derivatives

  •  

    Substitution Rule

Practical Applications of Finding Arc Length

If you’re given a function and you need to find the arc length, it can come in handy for a variety of situations. One of the most prominent is engineering. If you have power lines a certain length apart, and you know their bending from gravity can be expressed by a given function, you can then figure out how much wire needs to go from one pole to the next.

If you overshoot in your guess, you have a low-hanging power line, and if you underestimate, the line won’t reach at all, or it will have too little slack to be safe. Therefore, you need as close to the exact length as possible. You may also have to factor in wind resistance.

Final Thoughts

When you have the basics of derivatives and integration firmly in your mind, it becomes easy to figure out the arc length formula in calculus. Even for more complex functions, you can still use basic rules and the substitution method if you have issues. The only problem with the substitution method is that it works only on indefinite integrals.

Therefore, you’ll need to disregard the bounds of your functions when you’re using the substitution method.

Enhance Your Calculus Capabilities with These Brain Boosting Tips

Improve your problem-solving skills and calculus capabilities with our 11-best brain-boosting tips. Get an A in calculus with these simple tips.

Calculus has the reputation of being one of the most challenging subjects in school, even when compared with other advanced math classes. That’s because calculus is usually the first exposure students get to a version of math that requires more than just memorization to succeed.

Many calculus classes have specific pre-requisites, such as an understanding of advanced algebra, geometry, and trigonometry. After all, it’s essentially another language. However, you’ll need more abstract qualities as well.

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A well-developed imagination will make it much easier to understand calculus. Since calculus works with physical, real-world concepts, the ability to visualize these things in your mind is crucial to your ability to understand and solve calculus problems.

You also need to learn how to attack problems from different angles. There are specific rules in calculus that you’ll have to memorize, but not all calculus problems will follow a pre-set formula. You need to learn how to solve complicated problems without following a specific set of actions you’ve memorized.

Overall, you’ll need advanced problem-solving skills if you want to succeed in calculus. To help you learn or build upon these skills, we’ll teach you to enhance your calculus capabilities with these brain-boosting tips.

General Brain-Boosting Tips

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superBrain Boosting Secrets

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Even before you start your first day of calculus, you can work on improving your ability to learn and process advanced concepts. And it’s easier than you may think.

Dance Like Nobody’s Watching.

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It may be an overwrought cliché, but this advice not only works for improving your mood, but it can also develop your ability to solve problems.

Studies have found that dancing can improve your brain’s neural processing abilities. Structured dancing, such as tap and ballet, helps with convergent thinking—your ability to find a single, absolute answer to a problem.

On the other hand, more improvised styles of dance, like hip-hop or EDM dancing, improves divergent thinking. That means you’ll have an easier time finding multiple possible answers for a complex problem.


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Since calculus requires both types of thinking, you should consider learning multiple styles of dance. Many schools offer dance classes as an alternative to physical education, or as an elective. If you don’t relish the idea of dancing in front of other people, you can also use Youtube to find dance tutorials and practice in the privacy of your home.

Play Games.

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BrainBrain Games – Sudoku

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Over the last decade or two, there has been a rise in the popularity of logic and puzzle games like sudoku. Electronic games designed to work out your brain can be found on nearly any video game platform as well.

These types of games are not just a fun way to engage your mind when you’re bored; they’re also a great way to improve your problem-solving skills. Many logic or puzzle games require you to work a problem from multiple sides and have numerous ways of finding the solution.

For many problems, both in games and in calculus, you’ll need to approach the problem backward, not forward. By practicing this strategy with fun games in your free time, you’ll find that you have an easier time working through your calculus problems as well.

Stay Healthy.

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Obviously, exercise and a proper diet will benefit your body; however, science also proves that a healthy body leads to a healthy mind. People who regularly work out and eat wholesome, healthy food, have more energy, which helps their brain work faster and harder.

Researchers at Harvard found that regular cardio exercise can actually increase the size of your hippocampus, which is the part of the brain responsible for verbal memory and learning. It doesn’t take much, either—simply walking at a brisk pace for about two hours per week is enough to boost your mental function significantly.

A similar Harvard study also discovered a link between unhealthy food and memory problems. Scientists studied the effects of a diet high in LDL (“bad” cholesterol) and saturated fats on memory and found that participants performed much worse on tests of thinking and memory than people with a healthier diet.

Since diet and exercise have positive effects on both your body and your mind, they’re likely to enhance your calculus capabilities as well. They can help relieve stress for keeping focused on your work.

Sleep Well.

We’ve all experienced the adverse effects of sleeplessness—foggy brains, irritability, and lack of energy, to name a few.  Not getting a good night’s sleep can have a catastrophic impact on your ability to learn, as well as your problem-solving skills.

REM sleep, in particular, is vital to your ability to form connections between unrelated ideas. That means your brain works while you’re sleeping to process everything you’ve learned and allowed you to see the bigger context of different subjects and ideas.

REM sleep also enhances creative processing, your brain’s ability to form new ideas and ways of thinking about concepts you’ve learned in the past.

Advanced Problem-Solving Tips

While the above tips will prepare your brain to learn calculus, you may still need some extra help when it comes time to work through calculus exercises. If you find yourself struggling with a problem, try some of these tips.

Mind Maps

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A mind map is essentially a sketch of the problem and all the potential solutions. You can use this technique with any kind of question. Think of it as another way to brainstorm. By drawing out the issue, your mind may be stimulated to generate more ideas.

Basically, you start with your problem in the center of the page. As you brainstorm possible solutions, add branches off the center containing these answers. You can divide these main branches into sub-branches, as you plug in your potential answer to the problem and work through it. If one solution doesn’t work, you move on to the next one and continue the process until one of them works.

Music and Exercise

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We’ve already discussed how exercise can help stimulate your brain in general. However, the addition of music can further push your mind to sort through everything you’ve learned and organized it in ways that may lead to finding a solution to your problem.

Idea Journals

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Many studies have shown that the act of writing types of cement not only new ideas into your brain, but also generates new connections and creative ideas. The purpose of an idea journal is to provide a place for you to write out your problem and any subsequent ideas you generate over time.

As we’ll discuss in a little bit, sometimes your brain will come up with solutions to a problem when you least expect it, while you’re showering or watching television or hanging out with friends. Keeping an idea journal with you at all times allows you to write out these solutions as soon as they pop into your head.

Talk it Out

For many people, the act of talking through a problem helps their mind sort through ideas and generate new solutions. Whether you’re talking to yourself, a friend, or a classmate, merely verbalizing the problem may lead you to blurt out the answer without realizing you knew it all along.

One common strategy is to talk through the problem as if you were teaching the subject to someone else. By explaining the concepts involved in the question (to yourself or others), you could unlock the answers already hidden in your mind.

Create a Sketch.

As we mentioned earlier, most calculus problems can be represented physically—for instance, the shape of a curving slope, or the speed of a moving object. Sketching a visual representation of the problem could help you understand exactly what it is you’re trying to solve.

How
How to Sketch: A Beginner’s Guide to Sketching Techniques, Including…

  • Yanconsky, Liron (Author)
  • English (Publication Language)
  • 360 Pages – 12/18/2014 (Publication Date) – CreateSpace Independent Publishing Platform (Publisher)
Idea
Photo credit to Harvard Business Review

Focus on Something Else.

You’ve probably heard the story of Archimedes, who found the answer to how to measure the volume of a king’s crown while taking a bath. While the anecdote may or may not be true, the idea of the brain finding solutions to problems while you’re focusing on something else is solid.

By working on something else, particularly something that doesn’t require too much mental strain, you increase the psychological distance between yourself and your problem. That allows your brain to think more abstractly and creatively on the issue at hand while your attention is focused somewhere else.

Get Back to Basics.

If you’re still having trouble coming up with a solution, try going back to the beginning. Focus on the parts of the problem that you already understand, and work through as much of it as you can. In the process of rehashing familiar concepts, you may trick your brain into forming the complete solution.

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Photo credit to Pinterest

Final Notes

You don’t need to be a math genius to succeed at calculus. Even more than mathematical ability, calculus requires creativity, adaptability, and excellent problem-solving skills. If you work hard, take care of yourself, and use our guide to enhance your calculus capabilities with these brain-boosting tips, there should be nothing stopping you from earning that A.

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Calculus-Help now 26.2% more fun!

By popular demand, this week’s slate of updates heralds the return of music to Calculus-Help.com. Now available on the main links bar at the top of every page, you’ll find a link to “Fun.” Follow this link to listen to, or download, the (oddly) ever popular Quadratic Formula song. You may also want to settle down to the soothing lyrics of The Day Before Notebooks Are Due Blues or The A.P. Calculus Theme Song. It’s not even close to the holiday season, but does that mean you can’t go and enjoy some holiday carols involving Calculus? (Hint: no.)

If mathematical music is not your thing, that’s okay–it’s not most people’s thing either. The new page o’ fun also ushers in the return of the Interactive Cheat Sheet, a flash-based application that presents the most important calculus formulas in one location.

In honor of music’s return to Calculus-Help.com, keep your eyes fixed on this very page, because a contest is looming. Perhaps you have a calculus song burning in your very soul. You just may get the chance to unburden said soul for prizes!

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