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Mean Value Theorem for Integrals: What is It?

In this calculus guide, we are going to help you understand the mean value theorem for integrals. If you are having difficulties with it, read this!

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The mean value theorem for integrals is a crucial concept in Calculus, with many real-world applications that many of us use regularly. If you are calculating the average speed or length of something, then you might find the mean value theorem invaluable to your calculations.

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What is the Mean Value Theorem?
Why Is the Mean Value Theorem for Integers Important?
How to Use the Mean Value Theorem
Conclusion
Mean

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Ultimately, the real value of the mean value theorem lies in its ability to prove that something happened without actually seeing it. Whether it’s a speeding vehicle or tracking the flight of a particle in space, the mean value theorem provides answers for the hard-to-track movement of objects.

What is the Mean Value Theorem?

Based on the first fundamental theorem of calculus, the mean value theorem begins with the average rate of change between two points. Between those two points, it states that there is at least one point between the endpoints whose tangent is parallel to the secant of the endpoints.

A Frenchman named Cauchy proved the modern form of the theorem. One of the most prolific mathematicians of his time, Cauchy proved the mean value theorem as well as many other related theorems, one of which bears his name.

It is also possible for a function to have more than one tangent that is parallel to the secant. The derivative, or slope, of each tangent line, is always parallel to the secant in the mean value theorem.

As an addition to the mean value theorem for integers, there is the mean value theorem for integrals. This theorem allows you to find the average value of the function on at least one point for a continuous function.

Stipulations for this theorem are that it is continuous and differentiable. That means that the line acts as a traditional function, without any odd stops, gaps, drop-offs, or any other non-continuous feature.

It also must be differentiable, which means you can find the slope of a point on the function. For cube roots or the absolute value of x, you cannot find a derivative because they are either undefined or not tangential to the average rate of change. 

Floor and ceiling functions also do not have derivatives because they are not continuous functions. Thus the mean value theorem of integers does not apply to them.

Why Is the Mean Value Theorem for Integers Important?

Like many other theorems and proofs in calculus, the mean value theorem’s value depends on its use in certain situations. Since this theorem is a regular, continuous function, then it can theoretically be of use in a variety of situations. Any instance of a moving object would technically be a constant function situation.

Real-world applications for the mean value theorem are endless, and you’ve probably encountered them either directly or indirectly at some point in your life. 

One of the classic examples is that of a couple of police officers tracking your vehicle’s movement at two different points. You are then issued a ticket based on the amount of distance you covered versus the time it took you to complete that distance. 

You were not speeding at either point at which the officer clocked your speed. But, they can still use the mean value theorem to prove you did speed at least once between the two officers. 

More specifically, consider modern-day toll roads. These roads have cameras that track your license plate, instantaneously clocking your time spent on the road and where and when you exited and entered. Law enforcement could quickly begin to crack down on speeding drivers on these roads, by merely finding the average rate of change between the two points. Drivers could then blame the mean value theorem of integers as the reason for their ticket.

Further use occurs in sports, such as racing. When investigating the speeds of various racing objects, such as horses or race cars, technicians and trainers need to know the performance of horses or race cars at specific points during the race. 

Using data obtained throughout the race, individuals can determine how their horse or car was performing at certain times. For horses, this can mean altering training patterns or other variables to improve performance related to results. Race car drivers can use the data to tune equipment in various ways to better utilize the car’s speed.

The critical part of the theorem is that it can prove specific numbers. It can determine the velocity of a speeding car without direct visual evidence, or the growth, length, and myriad other instances where an object or thing changes over time.

How to Use the Mean Value Theorem

In the real-world, a continuous function could be the rate of growth of bacteria in a culture, where the number of bacteria is a function of time. You could divide the difference in the number of bacteria by time to find out how fast they multiplied.

Applying the mean value theorem to the above situation would allow you to find the exact time where the bacteria multiplied at the same rate as the average speed. This might be useful to researchers in various ways, to determine the characteristics of certain bacteria.

Another more practical situation would be to determine the average speed of a thrown baseball. The distance of the ball thrown is a function of time. Dividing the difference in the length by the time it took for the ball to get from point a to point b would tell you how fast the ball goes.

When the mean value theorem is applied, a coach could analyze at which point the ball achieved the average speed. If the speed was faster before or after the tangential point, then the coach could alter the mechanics or delivery of the player’s throw. This would make for more optimal speed with the throw reaches the batter.

Finally, let’s find the average speed of the vehicle and then at which point during the drive, the car reached a speed equal to the average rate.

  1. 1

    First, find the total distance traveled by the vehicle. To do this, check the odometer before and after driving. Calculate the difference between the two readings.

  2. 2

    Determine the amount of time spent driving the car between those two points.

  3. 3

    Divide the distance by the time. Let’s say it’s 40 mph.

  4. 4

    Now you need to find the point – or points – during which the car was traveling at 40 mph.

  5. 5

    Graph the function.

  6. 6

    The point at which the vehicle traveled 40 mph will show as the highest or lowest point on the slope connecting the beginning of the drive and the end.

  7. 7

    Using the graph, you can then find the exact time at which the car was traveling at 40 mph.

When using the mean value theorem in practical applications like vehicle speed, it is essential to note that the average rate of change is just that – an average. If your vehicle speed is 50 mph, then at some point during your drive you drove over and under 50 mph. Of course, you would hit that speed at least twice at a minimum.

Another exciting application of the mean value theorem is its use in determining the area. When the point at which the tangent line occurs is understood, draw a line from the new point parallel to the x-axis. This line is the top of your rectangle. The bottom is the x-axis. The left side is the y-axis, and the right is the endpoint of your continuous function.

Once this is complete, the area of your rectangle will be the same as the area beneath the curve of your function. One practical application of this instance is determining the exact height of a liquid in a container. If the liquid is suspended or not at rest, then calculating the mean value theorem of integers for the endpoints of the liquid will help you to determine the resting volume.

Determining amounts of liquid or the properties of a substance are just a few of the many applications of the mean value theorem. All fields of science use this theorem, and merely finding the volume of a liquid at rest is just scratching the surface. As sport becomes more science-based, the value of this theorem will only continue to increase.

Conclusion

While a fundamental calculus theorem may not change your life, it can make your life a tiny bit more manageable. Understanding the movement of an object and the properties within that movement can help you make a variety of educated conclusions. 

When working in scientific fields such as physics or biology, the use of the theorem can aid in the research of particles or microscopic organisms. In sports, you can use the theorem to develop a better understanding of fast-moving objects. On the highway, the police can issue more speeding tickets. The mean value theorem of calculus is an invaluable tool for all types of people.

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Indeterminate Forms in Calculus: What are They?

Indeterminate forms in calculus begin with algebraic functions that utilize a limit for the independent variable to find a solution. Learn more here!

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Indeterminate forms in calculus begin with algebraic functions that utilize a limit for the independent variable to find a solution. The indeterminate aspect of the form results when different rules vie to apply to the solution.

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What is an Indeterminate Form in Calculus?
Why are Indeterminate Forms in Calculus Important?
How to Use Indeterminate Forms
Conclusion
Calculus

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Since there is no clear way to determine which rule of calculus will govern the answer, the function then becomes an intermediate form. There are many indeterminate forms, but there is only a handful that commonly occurs. For instance, 0/0 and 0 to the power of 0 are examples of more common indeterminate forms.

What is an Indeterminate Form in Calculus?

Replacing algebraic combinations within a function with its limits, assuming an independent variable sometimes results in an answer such as 0/0. In this case, there are several competing rules vying for an opportunity to define this solution.

Indeterminate forms, officially coined by a student of the famed French mathematician Augustin Cauchy, have been around for as long as calculus. However, they have only been studied in the last 150 years or so. 

Substituting a limit that results in a zero, infinity, negative infinity, or any combination of these may result in an indeterminate form. When both functions approach the given limit that results in the indeterminate form, there is not enough information to determine what the behavior of the function is at that point.  

Some indeterminate forms can be solved by factoring through elimination or using L’Hopital’s rule. The functions resulting in 0/0 and infinity over negative infinity can achieve a solution through various means.

An indeterminate form is a limit that is still easy to solve. It only means that in its current form as a limit put into a function, it presents too many unknowable characteristics to form an appropriate answer properly. You can’t just solve for the quotient.

When solving for a limit, we are looking at two functions so that they make a ratio. Ratios are the most common, but not the only, way to discern an indeterminate function. Several geometric functions are also indeterminate forms, but not ratios.

Finally, while limits resulting in zero, infinity, or negative infinity are often indeterminate forms, this is not always true. Infinity, negative or positive, over zero will always result in divergence. As well, one over zero has infinite solutions and is therefore not indeterminate. We can determine a universal solution, but an indeterminate answer is one that needs more information.

Why are Indeterminate Forms in Calculus Important?

Indeterminate forms hover over the calculus no matter where you turn. When dealing with ratios such as 1/0, 0/0, or infinity in any form, you will most likely need to use a further theorem, such as L’Hopital’s, to solve for a limit. Knowing that indeterminate forms are sometimes solvable can elicit clarity in a function that previously was not available.

Delving further into the particulars of indeterminate forms allows us to utilize a variety of methods to determine the characteristics of a function at a particular limit. Not enough information determines an indeterminate form. However, sometimes, we have too much information and need to whittle down a solution to one technique.

Take the fraction 0/0. An elementary view of this fraction would tell someone that it equals one because a numerator that equals a denominator equals one. Or more practically, we could think of the fraction as just zero. Further, a zero in the denominator could indicate infinity or does not exist. Therefore, there are options for classifying this ratio, but no clear winner stands out.

So why are indeterminate forms necessary in practical, real-life situations? The answer is that they aren’t, at least not by themselves. However, their mere existence provides the ground for individuals to find ways to make indeterminate forms solvable.  

Many areas of physics and math will use a denominator of zero as a potential starting point in a variety of situations and formulas. Laws involving physics, heat, and quantum mechanics will use indeterminate forms at some point. Understanding these forms allows you to solve them using an appropriate method, whether that is L’Hopital’s method or another.

From an even more practical standpoint, there could be very general problems involving motion, velocity, and time that require the use of indeterminate forms. Questions surrounding the speed and time of a moving object will have to start at zero. Determining an infinite limit of two of these objects would result in an indeterminate form.

Again, it is critical to note  the reason for classifying an indeterminate form as “indeterminate.” It is to differentiate it from other ratios that are zero or does not exist. To alleviate confusion, possibly substituting the word “temporary” for indeterminate would clear up some of the misconceptions surrounding the use of these forms.

Understanding these forms as a transient is a better way to think of them. An indeterminate form, therefore, is just a vehicle for further computation that is up to the discretion of the user.

How to Use Indeterminate Forms

In the real-world, an indeterminate form would take the form of algebraic expression with a defined limit. Let’s say the numerator was x-4, and the denominator was 2x-8. With a limit of 8, what is the value of the ratio? The ratio becomes 0/0.

Once again, 0/0 is an indeterminate form. Solving it further requires some factoring. If we cancel out the variable, we get a ratio of ⅔. Using simple algebra, we can get away from the indeterminate form. 

When simple algebra is unavailable for use in solving an indeterminate form, then L’Hopital’s rule becomes necessary. A simple search of the internet results in some handy lists of indeterminate forms.

Typically these lists are arrayed into three columns. The first column lists the indeterminate form. The second shows how to use L’Hopital’s rule to get a ratio to either 0/0 or infinity/infinity. Once the ratio becomes either of those, then they are solved either algebraically or using derivatives. 

While L’Hopital’s rule is not precisely real-world applicable on its own, it is also a vehicle for further calculation much like indeterminate forms. Without the rule and the existence of indeterminate forms, calculations for meaningful and applicable formulas in other areas would not be possible.

How do we know when an indeterminate form needs the use of L’Hopital’s rule or needs to use algebra?

  1. 1

    First, input the limit into the functions.

  2. 2

    Solve each function.

  3. 3

    If both the numerator and denominator equal zero or infinity, then you’ve probably got an indeterminate form.

  4. 4

    Attempt to factor functions without inputting the limit.

  5. 5

    If factoring results in a determinate form, then you are done. Otherwise, go on to the next step.

  6. 6

    Find the derivative of both the numerator and the denominator.

  7. 7

    Input your limit into the derivative ratio, and you’ve got your answer. You’ve just made use of L’Hopital’s rule.

When using indeterminate forms, be sure you have a proper understanding of what a derivative is and how to use them. The first step in finding the derivative of a polynomial is to bring each exponent down by one. Multiply the initial exponent by the coefficient. Multiplying allows you to find the derivatives and use L’Hopital’s formula.

It is critical to forming a thorough understanding of L’Hopital’s formula if you will be dealing with indeterminate forms. Instances where you could get end up with zero minus infinity, or infinity minus infinity all call for the use of this rule. The use of derivatives allows you to transform this indeterminate form into 0/0 or infinity over infinity. From there, you can solve using algebra. 

There are instances where L’Hopital’s rule will not work with an indeterminate form. If you have zero over infinity, finding the derivative of the polynomials will only lead to the wrong answer. Using the inverse of the infinite polynomial will then change the ratio into an L’Hopital-friendly 0/0 ratio. 

Instances, where a function equals zero to the zero power, requires the use of natural logarithms. Subtracting to infinities calls for using the laws of trigonometry and making calculations using cos, sin, and tan. Any indefinite forms that you find in the course of your calculus journey have a method for solving.

Conclusion

While an indeterminate form may not change your life, it can provide the means for further understanding of broader calculus concepts and rules. As well, indeterminate forms are primarily made up of infinity, zero, and one, which is the primary values often dealt with in calculus. Understanding their indeterminate forms is crucial.

Many subject areas will use formulas and calculations involving indeterminate forms. A thorough understanding of how to solve them, and the parts that make up their solutions such as derivatives and trigonometric functions, are fundamental.

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What Is the Tangent Line Equation?

In calculus you will come across a tangent line equation. What exactly is this equation? This article will explain everything you need to know about it.


In calculus you will inevitably come across a tangent line equation. What exactly is this equation? This article will explain everything you need to know about it.

In calculus, you learn that the slope of a curve is constantly changing when you move along a graph. This is the way it differentiates from a straight line. You can describe each point on a graph with a slope. 

A tangent line is just a straight line with a slope that traverses right from that same and precise point on a graph. When we want to find the equation for the tangent, we need to deduce how to take the derivative of the source equation we are working with.

When looking for the equation of a tangent line, you will need both a point and a slope. You will be able to identify the slope of the tangent line by deducing the value of the derivative at the place of tangency. This is where both line and point meet. 

In regards to the related pursuit of the equation of the normal, the “normal” line is defined as a line which is perpendicular to the tangent. This line will be passing through the point of tangency. Now that we have briefly gone through what a tangent line equation is, we will take a look at the essential terms and formulas which you will need to be familiar with to find the tangent equation.

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Key Terms and Formulas Defined
The Primary Method of Finding the Equation of the Tangent Line
Write Down the Function and Draw the Tangent Line
Use the Derivative For the Slope
Input the x-coordinate Into f(x)
Convert the Tangent Line Equation Into Point-Slope Form
Confirm the Tangent Line Equation
Methods to Solve Problems Related to the Tangent Line Equation
Identify the Maximum and Minimum Points
Uncover the Equation of the Normal
Vertical and Horizontal Tangent Lines

Key Terms and Formulas Defined

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Before we get to how to find the tangent line equation, we will go over the basic terms you will need to know. By having a clear understanding of these terms, you will be able to come to the correct answer in your search for the equation.

  • Tangent line – This is a straight line which is in contact with the function at a point and only at that specific point. This line is barely in contact with the function, but it does make contact and matches the curve’s slope. This line is also parallel at the point of the meeting. You can also simply call this a tangent.

  • Secant line – This is a line which is intersecting with the function. This line will be at the second point and intersects at two points on a curve. You can also just call this a secant.

  • Slope-intercept formula – This is the formula of y = mx + b, with m being the slope of a line and b being the y-intercept. You will use this formula for the line.

  • Point-slope formula – This is the formula of y – y1 = m (x-x1), which uses the point of a slope of a line, which is what x1, y1 refers to. The slope of the line is represented by m, which will get you the slope-intercept formula.

With the key terms and formulas clearly understood, you are now ready to find the equation of the tangent line. You should retrace your steps and make sure you applied the formulas correctly. Otherwise, you will get a result which deviates from the correctly attributed equation.

The Primary Method of Finding the Equation of the Tangent Line

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When we are ready to find the equation of the tangent line, we have to go through a few steps. If you take all these steps consecutively, you will find the result you are looking for. 

There is more than one way to find the tangent line equation, which means that one method may prove easier for you than another. We will go over the multiple ways to find the equation. The following is the first method.

Write Down the Function and Draw the Tangent Line

While you can be brave and forgo using a graph to illustrate the tangent line, it will make your life easier to graph it so you can see it. This is because it makes it easier to follow along and identify if everything is done correctly on the path to finding the equation. You will want to draw the function on graph paper, with the tangent line going through a set point. 

Use the Derivative For the Slope

What you will want to do next is take the first derivative (f’x), which represents the slope of the tangent line somewhere, anywhere, on f(x), as long as it is on a point. 

Input the x-coordinate Into f(x)

Take the point you are using to find the equation and find what its x-coordinate is. When you input this coordinate into f'(x), you will get the slope of the tangent line.

Convert the Tangent Line Equation Into Point-Slope Form

What you need to do now is convert the equation of the tangent line into point-slope form. The conversion would look like this: y – y1 = m(x – x1). In this equation, m represents the slope whereas x1, y1 is a point on your line. Congratulations! You have found the tangent line equation.

Confirm the Tangent Line Equation

While you can be fairly certain that you have found the equation for the tangent line, you should still confirm you got the correct output. It helps to have a graphing calculator for this to make it easier for you, although you can use paper as well. You will graph the initial function, as well as the tangent line. If confirming manually, look at the graph you made earlier and see whether there are any mistakes.

Methods to Solve Problems Related to the Tangent Line Equation

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There are a few other methods worth going over because they relate to the tangent line equation. Knowing these will help you find the extreme points on the graph, the equation of the normal, and both the vertical and horizontal lines.

Identify the Maximum and Minimum Points

With this method, the first step you will take is locating where the extreme points are on the graph. These are the maximum and minimum points, given that one is higher than any other points, whereas another is lower than any points. Remember that a tangent line will always have a slope of zero at the maximum and minimum points. 

A caveat to note is that just having a slope of 0 does not completely ensure the extreme points are the correct ones. To be confident that you found the extreme points, you should take the following steps:

  • Take the first derivative of the function, which will produce f'(x). The resulting equation will be for the tangent’s slope.

  • Solve for f'(x) = 0. This will uncover the likely maximum and minimum points.

  • Take the second derivative of the function, which will produce f”(x). What this will tell you is the speed at which the slope of the tangent is shifting.

  • For the likely maximum and minimum points that you uncovered previously, input the x-coordinate, a, into f”(x). Now you will have to check whether this is positive or negative. If it is positive, you have found the minimum at a. If it is negative, you have found the maximum. A third outcome you would get is the inflection point, when f”(a) equals zero. 

  • When you discover an extreme point at a, you will have to find f(a) to reveal the y-coordinate.

Uncover the Equation of the Normal

The “normal” to a curve at a specific point will go through that point. However, its slope is perpendicular to the tangent. When you want to find the equation of the normal, you will have to do the following:

  • Find the slope of the tangent line, which is represented as f'(x).

  • If you have the point at x = a, you will have to find the slope of the tangent at that same point.

  • You will now want to find the slope of the normal by calculating -1 / f'(a).

  • Write down the equation of the normal in the point-slope format.

Vertical and Horizontal Tangent Lines

To find out where a function has either a horizontal or vertical tangent, we will have to go through a few steps. 

  • When looking for a horizontal tangent line with a slope equating to zero, take the derivative of the function and set it as zero. 

  • Obtain and identify the x value.

  • Take the original function to deduce the y value. The result is that you now have the location of the point.

  • When looking for a vertical tangent line with an undefined slope, take the derivative of the function and set the denominator to zero. 

  • Obtain and identify the x value.

  • Take the original function to deduce the y value. The result is that you now have the location of the point.

There are two things to stay mindful of when looking for vertical and horizontal tangent lines. In the case of horizontal tangents, you will want to make sure that the denominator is not zero at either the x or y points. In the case of vertical tangents, you will want to make sure that the numerator is not zero at either the x or y points.

Wrapping Up

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Congratulations on finding the equation of the tangent line! You can now be confident that you have the methodology to find the equation of a tangent. It may seem like a complex process, but it’s simple enough once you practice it a few times. The key is to understand the key terms and formulas. Having a graph as the visual representation of the slope and tangent line makes the process easier as well. 

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Ultimate Guide On How To Calculate The Derivative Of Arccos

Read this full guide on how to calculate the derivative of arccos that is used in trigonometry and understand the importance of degrees and radians.

The derivative of arccos in trigonometry is an inverse function, and you can use numbers or symbols to find out the answer to a problem. It uses a simple formula that applies cos to each side of the equation. Some people find using a drawing of a triangle helps them figure out the solutions easier than using equations.

The derivative of Arccos is used in trigonometry. It’s an inverse function, and you can manipulate it with numbers or symbols. There are several terms you’ll need to know when working with Arccos, including radian.

Arccos means arccosine. You may also work with arcsin, or arcsine when working with trigonometry problems.

Results for the derivative of Arccos are written in radians. Solving a problem involving Arccos first requires knowledge of radians.

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An Overview Of Radians
The Difference Between Geometry And Trigonometry
Arccos X Functions

An Overview Of Radians

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Radians are used to measure angles. The angle of a single radian subtends or creates an angle. At a certain point, extremities present straight lines that join at that particular point. When two rays pass through the arc endpoints, the angle is subtended.

The radian has an angle of a single radian subtended from a unit circle’s center. A radian creates an arc length of 1. Therefore, we can determine that a full angle measures 2pi radians. The angle contains 360 degrees for every 2pi radians, which is equivalent to 57.29577951 degrees per radian or 180 degrees pi.

A right angle measures pi/2 radians. A straight angle measures pi radians. Radians let you write integrals and derivative easily, like d/(dx)six equals cosx when x is measured by radians.

The Difference Between Geometry And Trigonometry

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The three branches of mathematics are algebra, arithmetic, and geometry. Geometry is the study of sizes, properties, and shapes of spaces of three- dimensional and two-dimensional objects. Euclid, the Father of Geometry, is often mentioned in conjunction with theorems and postulates about the field.

Geometry is a combination of the Greek terms Geo (earth) and metron (measure). You will encounter solid, spherical and plane geometry in your studies.

Spherical geometry is the study of three-dimensional objects such as spherical polygons and spherical triangles. Plane geometry features lessons about two-dimensional objects such as points, curves, lines, and planes, including polygons, circles, and triangles. Solid geometry focuses on spheres, prisms, cubes, pyramids, and other three dimensional objects.

Euclidean Geometry is the study of flat surfaces, and it’s just one of the branches of geometry. Riemannian geometry, the study of curved surfaces, is another major branch of geometry.

Trigonometry is a branch of geometry. The field emerged in about 150 B.C. when a mathematician named Hipparchus used sine to make a trig table.

Trigonometry deals with triangles, their lengths, and their angles. You can also use trigonometry to study waves and oscillations. You will study the way the side lengths of right triangles relate to each other. The basic relationships between triangles and their sides are referred to as Sine, Tangent and Cosine.

If you have a right angle triangle, the longest base is referred to as the hypotenuse. The side in front of an angle is its opposite side, and the adjacent side is the side behind that angle. The relationship for a right angle triangle then is sin A, or the opposite of the hypotenuse, cos A, the adjacent side of the hypotenuse, and tan A the opposite/adjacent side.

The secondary relationships in trigonometry are:

  • Secant

  • Cotangent

  • Cosecant

Circumference

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These measurements are the respective dimensions of Sine, Cosine, and Tangent.

Spherical trigonometry that deals with triangles in 3D scenarios, and it’s used in navigation and astronomy.

Trigonometry helped ancient sailors navigate the seas. Today, trigonometry has many uses beyond math class.

Oceanographers use trigonometry to figure out the height of ocean tides or measure sea animals. Trigonometry can be used to build ships and help them navigate. This mathematical field also helps criminologists determine the trajectory of bullets and other projectiles.   

Archeologists divide their excavation sites up for work using trigonometry. Flight engineers use trigonometry to determine the best flight course for a plane. Trigonometry helps fill in the “third side” of a flight’s wind, speed, and direction equation to make sure the plane goes in the right direction.  

Example 1

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In trigonometry, the Arccos means cos to the negative first power (z) or the arc cosine of the complex number represented by (z). You will consider that this problem involves an inverse trigonometry function and write down two identities:

Cos (cos to the negative 1st power(x)) equals x, then cos to the negative 1st (power (cos(x)) equals x.

You need to decide which formula to use. You should use the first formula, due to the chain rule. The chain rule allows you to find the derivative for the term cos−1

(x) more easily than using the first formula. Simply the formula to cos (y) equals x. Now you have an implicit formula to solve the problem and determine y.

Use implicit differentiation to find out the answer for y’.  

(cos(y))0 = (x)followed by y’ (− sin(y)) = 1 and y’ equals negative 1 over sin (y), then y’ equals negative 1 over sin(cos to the negative 1st power(x)).

The formula written above is an excellent tool, but there is an even better way to write it using geometry instead of algebra. Cos has an angle and a number between negative 1 and 1, cos negative one will also give angle and have a number between negative 1 and 1.

Now define θ equals cos minus 1(x)to show the angle cos negative 1. Therefore, x equals cos (θ), according to the inverse function. You can also use cos with both sides of the function.

Drawing a triangle can also help you find the answer to this problem. Draw a triangle with points ABC and θ will be the angle less than C. Cos (θ) equals ACBC in geometry, and cos (θ) also equals x.

Consider that BC equals 1 and AC equals x. You could choose any identity in the above values, but the most obvious ones make the math easier.

In the triangle example, sin(θ) = AB over BC. Use the Pythagorean Theorem to calculate the answer for sin(cos−1(x)), which is equal to sin(θ).

Here’s the solution per the Pythagorean Theorem:

BC squared equals AB squared plus AC squared, followed by AB squared equals BC squared minus AC squared. Therefore, AB squared equals one minus x squared, and AB equals the square root of 1 minus x squared.

The equation calculated above answers the question correctly. Sin(cos negative 1 (x)) equals sin(θ) equals AB over BC equals AB equals the square root of 1 minus x squared. Complete the answer with

Example #2

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This problem demonstrates how to determine the derivative of Arccos x and Arcsin x.

The equation you use is d over dx (arcsin x plus arccos x) equals zero. Note that arcsin x plus arccos x equals pi over 2. You can explain this equation with the following calculations:

If arcsin x equals zero, then x equals sinθ equals cos( pi over 2 minus θ), then arccos x equals pi over 2 minus θ equals pi over 2 minus arcsin x; therefore arcsin x plus arccos x equals pi over 2.

Example #3

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The derivative you find for arccos x will be the negative of arcsin x’s derivative. This derivative stature holds for the inverse of each cofunction pair.

The same formulas apply to similar trigonometry problems. Arccot x’s derivative is the negative of arctan x’s derivative. Arcsec’s derivative is the negative of the derivative of arcsecs x.

The variable y equals arcsec x, represent tan y equals plus-minus the square root of x to the second power minus one.

Begin solving the problem by using y equals arcsec x, which shows sec y equals x. Now use Pythagorean identity b to figure the next step.  

Tan y equals plus minus the square root of sec squared y minus 1 equals plus minus the square root of x squared minus 1. Y’s derivative equals arcsin x.

The equation then is d over dx arcsin x equals one over the square root of one minus x squared. Prove this by looking at y equals arcsin x, which stands for sin y equals x. Figure the derivative of x with the following equation:

Cos y followed by dy over dx equal 1, then dy over dx equals 1 over cos y’, then dy over dx equals 1 over the square root of 1 minus x squared ‘.

An alternate theorem for this would be  d arcsin x over dx equals 1 over din sin y over dy equals 1 over cos y equals 1 over the square root of 1 minus x squared.

Arccos X Functions

The arccosine of x is the representation of the inverse cosine function of the variable x when negative 1 is less than x and x is less than 1. If the cosine of y equals x, the arccosine of the variable x is equal to x’s inverse cosine function, which equals y. The equation for this function reads arccos x equals cos to the inverse cosine x equals y.

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The equation is written arccos 1 equals cos, followed by the negative 1 symbol, which stands for inverse cosine function in this problem.

Therefore, arccos 1 equals cos negative 1 (inverse function), and 1 equals zero, and Rad equals zero degrees.

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What is a Relative Minimum in Calculus?

Learn about the relative minimum in calculus and how to find it along with when you might use the knowledge. Read further here!

It’s easy to visualize the minimum and maximum values of a function since the minimum is the lowest point on a graph while the maximum is the highest. If you’re looking at a graph with points on it, just choose the lowest one. You can find four possible values of a function on a graph including:

  • Absolute maximum
  • Relative maximum
  • Absolute minimum
  • Relative minimum

For the purpose of this article, we want to look at the relative minimum. The relative minimum won’t be the lowest point on the graph but a relative interval between the lowest and highest points. For instance, assume we have a graph with an x and y-axis. Let’s put three points five points on the chart that crisscross x as demonstrated.

The first three points to the left of y end up being -9, 11, and -6. To the right of y, we have two points at 13 and -8. In this very simple scenario, the relative minimum is -6 and the relative maximum is 11 because they fall within the domain of x. The domain might get a little tricky at times, but it’s all simple if you think of it as merely being all the variables that make the function work.

Understanding the Domain of a Function

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The domain of x could be a simple set of numbers beginning, from left to right, with negative six and ending with positive six. However, we don’t have to use the entire domain. Your function could end up confined between negative four and positive four. That’s how we end up with relative minimums and relative maximums. Other points may exist, obviously, but they don’t fall into the correct domain.

Let’s look at a simple equation like y is equal to the square root of x plus four. The domain of this equation is illustrated with the function x is greater than or equal to negative four. We can’t go any lower than negative four of the equation breaks down because the denominator of a fraction can’t be a zero and number under our square root sign is positive and must remain that way.

If you try to solve this using a number lower than negative four, it won’t work or produce an answer. Try it using -11 or even -5, and the result will always be incorrect. The domain of a function is a little more complicated than our simple scenario here, but to understand the relative minimum this definition is all you need at the moment.

My Relative Minimum is Missing

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Sometimes on a test, you may have a problem that asked you to solve the equation x is equal to x squared using only two intervals like negative one and positive two. If you graph that, the relative and absolute minimum are both stuck on zero. The negative one can’t be the relative minimum because it’s at the end of our interval.

In cases like this, the answer is always one of several multiple-choice solutions. The correct answer here would likely be none of the above. Passing a math test and math don’t always go together well, and little tricks like this get put on the test to make sure you’re paying attention not guessing the answer when you can’t solve the problem.

Revisiting Relative Minimum

If you didn’t notice, the first explanation of relative minimum is overly complicated and an almost robotic way of explaining it. You need to understand how to find the relative minimum using the most complicated methods possible if you want to excel at calculus. However, an easier way to find the relative maximum or minimum exists, if you graph everything.

Assuming your graph has a line with a starting and ending point, and more than two points, the relative minimum is the lowest part of your line where it changes direction. The opposite is true for the relative maximum. If your line starts and negative eleven and moves to a positive number then back to negative six before climbing again, the negative six if where the direction changes and the relative minimum.

Why Do I Need to Learn This?

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Well, for one, you need to know this to pass your algebra class. That seems like an excellent reason. Aside from that, many jobs require strong math skills and they pay well. If you check the Bureau of Labor Statistics website, you can look up your dream job and learn more about it. You can also use that website to look into other positions, find out what they pay, and possibly find a new career.

However, many of the highest paying jobs require strong math skills. Understanding how to find the relative minimum or maximum is essential in several positions revolving around economics and marketing. However, the terms used in each industry may differ, and it’s unlikely your boss will ever email a spreadsheet and ask you to find the relative minimum of anything on it.

You may be surprised to learn that some jobs you thought might be easy once you got the right degree actually require a high level of math knowledge. For instance:

  • Robotics engineers use a lot of math
  • Pilots that fly jets or commercial airliners must have a good grasp on math
  • Game designers need trigonometry to make things move and algebra to make the blow-up
  • Animators fall into the same category as game designers
  • Sports announcers use much more math than you can imagine

Many countries require pilots to take numerical reasoning tests, and they don’t allow calculators. Take a practice test and see how well you do on it. While this isn’t a direct instance of you needing to derive the relative minimum of a function, it’s an excellent example of how vital understanding math might be to your academic and future careers.

While they may call it by another name, several highly sought and well-paying jobs use calculus heavily. Let’s look at a couple of those with a little more detail starting with a mathematician job. They don’t just teach math, and many get employed at places like NASA. Check out NASA’s basic education requirements for some of their more important and fun jobs.

A cryptographer is another excellent example of an almost a pure math-based job. They work everywhere these days from private companies that develop privacy products to governments that need to secure things for transmission. They also work at places like the NSA where they try to secure items and decipher elements that may be encrypted.

Economists work with market data, statistics, and models to figure out current and future economic trends. Highs and lows on a graph might be their entire workload sometimes. Many of them work for various government agencies, but economist jobs in the private sector may become more popular in the future.

People that work with investments like analysts and bankers need a lot of math to do their jobs well. Seeing highs and lows on a graph might be the only way they get to make a decision. Their decisions might cost someone a lot of money if they get the math wrong or don’t understand the graph.

Accountants probably use math as often as a rocket designer. Math is practically all they do. They do things like figure out payroll for companies and prepare taxes for people. A mistake on their end could mean your paycheck is wrong or you end up paying the government back more than you owe. The opposite effects are possible as well.

Geodesists tell us things like how far it is to celestial bodies like the sun or distant galaxies. Sometimes these objects move, and the distance may fluctuate which results in maximum and minimum distance measurements. They have to factor in many things mathematically like gravity and mass to get an accurate measurement.

Mathematics modelers do several things including creating complex computer simulations and investigating number theories. If we need to know whether or not a comet will hit us, these people develop models based on data from various sources, including geodesists, to determine the chances of a comet smacking into the planet. That’s high on our list of essential jobs using math.

Many of the higher paying versions of these jobs require a doctorate to compete in the job market.

Some Final Notes

Granted, out definitions and problems lean toward the simpler side of finding the relative minimum, but you should understand how it works from a broad viewpoint now. If you need more help, you can find plenty of videos online that explain how to arrive at the relative minimum and how it may apply to other pre-calculus and calculus problems.

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How To Calculate Hyperbolic Derivatives

Hyperbolic derivatives may sound scary, but they are easy. In fact, their properties are among the most elegantly simple in all of mathematics.

A hyperbolic derivative is a derivate of one of the hyperbolic functions, which are functions that utilize the exponential function (ex) to simplify otherwise complex calculations. Common uses for hyperbolic functions include representing the length of arcs such as those formed by the cables of a suspension bridge, or the shape of the Gateway Arch in St. Louis, Missouri.

As derivatives calculate the rate of change of a function (such as deriving a function for a car’s acceleration given a function of its velocity over a variable of time), calculating a hyperbolic derivative means calculating the rate of change of a hyperbolic function. This definition is especially useful for understanding hyperbolic integrals, which are helpful in arc length calculations and other applications of mathematics and engineering.

If some of these terms are unfamiliar (or outright intimidating), don’t worry! We’ll give a quick review of some of the basics before diving into the actual derivations.

Quick Navigation

A Review ​Of Trigonometry
A Review ​Of ​The Exponential Function (ex)
The Hyperbolic Functions: Combining Trigonometry ​And ​The Exponential Function
A Review ​Of Derivatives
Calculating Hyperbolic Derivatives
Conclusion

A Review ​Of Trigonometry

Hyperbolic functions are closely related to trigonometry, as the functions that they describe relate to a hyperbola in the same way that trigonometric functions relate to a unit circle. But what exactly is this relationship, or even the “unit circle” for that matter? Is there a “unit hyperbola?”

Trigonometry helps describe the “unit circle” represented by function x2+y2=1. On a two-dimensional graph, this circle has its center at the origin (point (0,0)) with a radius of one unit. Trigonometry is the study of the dimensions within this circle, where special functions provide the height and width of the circle’s radius at a certain angle. The “certain angle” part is especially important.

This concept can be challenging to visualize internally, but consider an easy example: Suppose that the unit circle’s “certain angle” starts on the x-axis at zero degrees. In other words, if we position a line within the unit circle directly on the x-axis, it will be at zero degrees relative to the rest of the circle. Rotating it about the origin towards the positive y-axis will increase this angle in the positive direction while turning it towards the negative y-axis will decrease this angle in the negative direction.

If we leave our imaginary line on the x-axis, what are the “components” of its height and width? Well, at zero degrees (or the x-axis), its width is its length. So, if our imaginary line has the same radius as the unit circle (one unit), then we could say that the “width” component of our circle is also one unit.

However, what about our height component? Where do we start measuring? We measure the height component of our line from the x-axis, just as we measured its width from the y-axis. So, since our line is lying directly on the x-axis, its height component must be zero.

As we rotate our imaginary line inside the unit circle, the height and width components of our imaginary line will change with the angle of rotation. Laying the line directly on the positive y-axis, for example, will have height and width components of one and zero units, respectively. But what about points in between?

Thankfully, the height and width components of our line are conveniently described using basic trigonometric functions. Here, the height component is the function sin⁡(x) (pronounced “sine of x”), and the width component is the function cos⁡(x) (pronounced “cosine of x”). Using zero as a value for “x” in both of these equations will yield our original width and height of one and zero, respectively.

We’ll come back to this subject later once we explore their similarities with the hyperbolic functions. Until then, we also have to review the all-important exponential function.

A Review ​Of ​The Exponential Function (ex)

The exponential function is one of the single most essential functions in all of mathematics. We won’t describe its full capabilities and applications here, but for now, understand that it adequately describes the growth of just about anything in the natural world: plant growth, radioactive decay and numerous other phenomena are all accurately quantified using the exponential function.

The exponential function is represented by y=ex, where e=2.71828…. On a graph, this forms a hyperbola that grows at an exponential rate. e, also known as “Euler’s number,” just-so-happens to be a near-universal ratio of growth.

This function is (hopefully) interesting and all, but how does it relate to hyperbolic functions?

The Hyperbolic Functions: Combining Trigonometry ​And ​The Exponential Function

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Hyperbolic functions describe the hyperbola of the exponential function in a similar way that trigonometric functions describe the unit circle. The names of the basic “height” and “width” functions are remarkably similar to those from trigonometry!

Where trigonometric functions include sin x and cos⁡(x), hyperbolic functions include sinh(x) (pronounced “cinch of x”) and koshx (pronounced “kosh of x” or “kosh” rhymes with “gosh”). These are defined using the exponential function such that:

sinh x =ex-e-x2   and   x =ex+e-x2

Note the subtle difference between the functions: The numerator of the sinh(x) function is ex-e-x, while that of the koshx function is ex+e-x.

The names and functions of the hyperbolic functions aren’t their only similarities to trigonometry; other hyperbolic functions and identities are, in fact, remarkably similar to their counterparts in trigonometry.

One example of this similarity is the “double angle” identity, which in trigonometry is sin 2x =2sin x cos x . What happens if we work backward and replace sin x and cos⁡(x) with their hyperbolic cousins? Something interesting happens:

2sinh x x =2ex-e-x2ex+e-x2=e2x-e-2×2=sinh 2x

The identity is the same as it is in trigonometry! Other trigonometric parallels are also possible but are somewhat beyond the scope of this guide.

For now, we’ve defined the two most basic hyperbolic functions— sinh(x) and koshx. At this point, you’ve probably determined that finding a hyperbolic derivative involves finding the derivative of one of these (or one of the other) functions.

Before actually calculating hyperbolic derivatives, however, we’ll have a quick review on regular derivatives. This review will be especially useful, as the exponential function has some very unique properties when it comes to derivation.

A Review ​Of Derivatives

Again, a derivative of a function is its rate of change with a variable. Returning to the car analogy, we might be able to describe a car’s distance traveled as a function of time. In other words, we could have a function that says our car moves a distance of “x” over time “t” such that x=50t.

If we assign “x” to have its unit in meters and “t” to have its unit in seconds, we could say that our car will cover a distance of 50 meters in one second (which, by the way, is very fast—about 112 miles per hour!).

We might be curious to know, then, how fast the car is going. Intuition tells us that if the car travels 50 meters in one second, it must have a speed of 50 meters per second. Just by going through this mental exercise, you’ve calculated a derivative!

The exact theory of derivation is well beyond the scope of this guide. For now, remember that a derivative of a function describes that functions rate of change.

When solving a derivative algebraically, one of the most powerful rules to remember is the “power rule.” Here, “power” refers to whatever power a function’s variable is raised to. For example, the function y=x2 has the variable “x” raised to a power of two.

What would be the derivative of such a function? Or, in other words, what would be its rate of change as we change the value of “x?” The power rule offers a convenient solution, stating that the derivative of xn=n*xn-1. So, in the case of y=x2, dydx=2x (here, dydx means “the derivative of function ‘y’ by variable ‘x’”).

What if we try that on our car’s position equation from earlier? A quick calculation shows that x’=1*50t1-1=50, which is exactly what our intuition told us.

This rule works for most algebraic equations, but some functions have special derivation rules. One of these is the exponential function from earlier, whose derivative is – wait for it – itself! Formally stated, this relationship takes the form dydxenx=nenx. Take special care to remember any constant in front of the variable “x!”

As you might be able to imagine, this relationship is especially helpful for calculating hyperbolic derivatives—which we can now finally do.

Calculating Hyperbolic Derivatives

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Now that we have a solid background, we can move forward to deriving our essential hyperbolic functions. With the knowledge from the previous sections, the derivations should be reasonably straightforward.

For calculating the derivative of sinh x , we derivate its value of ex-e-x2. Using the rules described in the previous section, this yields an exciting result: ex+e-x2. This value looks a lot like x because it is! Similarly, deriving x will produce the value of sinh x .

Conclusion

It appears that the derivatives of the two essential hyperbolic functions sinh x and x are, in fact, each other. Remembering the parallels between hyperbolic and trigonometric identities, one can easily derive hyperbolic functions such as tanh x , where tanh x =sinh⁡(x)kosh(x) just as tan x =sin⁡(x)cos(x). Can you find its derivative?

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