Category: Maths Notes PPT

In calculus, a line integral is represented as an integral in which a function is to be integrated along a curve. A line integral is also known as a path integral, curvilinear integral, or curve integral. Line integrals have several applications such as in electromagnetic, line integral is used to estimate the work done on a charged particle travelling along some curve in a force field defined by a vector field. In classical mechanics, line integral is used to compute the word performed on mass m moving in a gravitational field. In this article, we will study a line integral, line integral of a vector field, line integral formulas, etc.

 

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Line Integral Definition

A line integral is an integral in which a function is integrated along some curve in the coordinate system. The function which is to be integrated can either be represented as a scalar field or vector field. We can integrate both scalar-valued function and vector-valued function along a curve. The value of the vector line integral can be evaluated by summing up all the values of the points on the vector field.

Line Integral of the Vector Field

A line integral (also known as path integral) is an integral of some function along with a curve. One can also incorporate a scalar-value function along a curve, obtaining such as the mass of wire from its density. We can also incorporate certain types of vector-valued functions along a curve. These vector-valued functions are the ones whose input and output size are similar and we usually define them as vector fields.

 

The line integral of the vector field is also interpreted as the amount of work that a force field does on a particle as it moves along a curve. 

Application of Line Integral

• Line integrals can be used to find the three-dimensional surface areas. It is an extension of simple integrals and is mostly applicable for curvy surfaces. 

• In the field of classical mechanics, line integrals are used to calculate the work done by an object of mass m, moving in a gravitational field. 

• Also, if one wants to figure out how many calories a swimmer might burn in swimming along a certain route, provided the currents in all areas can be accurately predicted. The total work that he needs to do would vary upon the strength and direction of the current. So, a line integral over his route will help to determine the total work done or calories that a swimmer will burn in swimming along the desired path.

• In electrical engineering, line integrals are used to determine the exact length of power cable needed to connect two substations that could be miles away from each other.

• Space flight engineers regularly use line integrals for long missions. While launching exploratory satellites, they consider the path of the different orbiting velocities of Earth and the planet the probe is targeted for.

• Line integrals are also used to find the velocity and trajectory of an object, predict the position of the planets, and understand electromagnetism in depth.

• They are also used in statistics to evaluate survey data and help draw out useful strategies. In chemistry, line integrals are used to determine the rate of reaction and know some necessary information regarding radioactive decay reactions.

• A line integral can also be used to calculate the mass of a wire, its moment of inertia, and the center of mass of the wire. It is also used to calculate the magnetic field around a conductor when using Ampere’s law.

• Ampere’s law states that the line integral of a magnetic field B around a closed path ‘C’ is equal to the total current flowing through the area bounded by boundary ‘C’.
  

Step-by-Step Guide to solving Line Integrals

• Identify the function f(x,y,z) in the given function and the curve ‘C’ over which the integration will take place. If it is a problem involving the work done on an object,  then f(x,y,z) represents the force on the object.

• Determine its parametric equations which are represented as x(t), y(t), z(t). The given equation would be a function of x, y, and z. So, it first needs to be translated into its parametric form x(t), y(t), and z(t).

For example, the equation of a circle is given as,
x² + y² = r².
Its parametric form would be,
x(t)= rcos(t) and y(t)= rsin(t), where t: 0→2π.

However, after determining the parametric equation, you move along the curve in the opposite direction as ‘t’ increases, the value of the line integral is multiplied by -1.

Difference between Line Integrals and Definite Integrals

Definite integral involves infinite summation of infinitesimal elements between two given boundaries called limits. It is an open integral which means the starting and ending boundaries need not be the same. It can be an integration of over a line, surface, volume, etc.

 

Line integral on the other hand is a closed integral which has a particular direction of travel in the direction of the given function. Most line integrals are definite integrals but the reverse is not necessarily true. Examples of line integrals are stated below. This will help you understand the concept more clearly.

Line Integral Examples with Solutions

The line integral example given below helps you to understand the concept clearly.

1. Find the line integral of

[_{c}int (1+x^2y)ds]

Where C is considered as an ellipse

 

r(t) = (2cos t) + (3sin t)              

 

for 0 ≤ t ≤ 2π

 

Solution:

 

We calculate, 

 

ds =[sqrt{(-2sint)^2+(3cost)^2dt}]

    = [sqrt{4sin^2t + 9cost^2t}]

 

We have the integral

 

[_{0}int^{2pi} (1+(2cost)^2(3sint))] [sqrt{4sin^2t + 9cost^2t}] 

 

Hence, we get line integral = 15.87.

2. Evaluate [{_c}int 4x^3 ds]

 

where ‘C’ is the line segment from (1,2) to (-2,-1).

 

Solution:

 

Here is the parameterization of the curve 

 

r(t) = (1-t) (1, 2) + t (-2,-1)

 

( 1-3t, 2 – 3t)

 

For, 0≤ t ≤ 1. 

 

Note: we are changing the direction of the curve and this will also change the parameterization of the curve, so we can ensure that we start/end at the proper point.

Here, you can see the line integral

 

[{_c}int4x^3ds]=[{_0}int^{1} 4(1-3t)^3] [sqrt{9+9dt}] (substituted the parametric form)

= [12sqrt{2(-1/12)(1-3t)^4 {_0}int{^1}}] (integrated the term (1-3t)³)

 

= [12sqrt{2(-1/12)[1-3(1)]^4 -[1-3(0)]^4}] (applying the limits)

 

= [12sqrt{2(-1/12)[16-1]}]

 

= [12sqrt{2(-5/4)}]

= [-15sqrt{2}]

 

= -21.213

Quiz Time

1. The line integral is used to calculate

1. Force

2. Length

3. Area

4. Volume

 

2. The integral form of potential and field relation is given by the line integral.

1. True

2. False

 

3. The value of c∫dI along a circle of radius 2 units is

A. Zero

B. 2π

C. 4π

D. 8π

Tips to understand Line Integrals:

Understanding line integrals takes time. It isn’t like you read this post and have fully understood line integrals. Read and reread the above article and make your own revision notes. 

 

Also, go through solved examples and understand the concept applied at each step. Once students understand the fundamentals, then it becomes easier to solve problems of line integrals. 

 

Practice often. Every problem of line integral is different and requires consistent practice to understand the chapter well. The more you practice different kinds of problems, the easier it would be to write during the exams.

Linear regression is used to predict the relationship between two variables by applying a linear equation to observed data. There are two types of variable, one variable is called an independent variable, and the other is a dependent variable. Linear regression is commonly used for predictive analysis. The main idea of regression is to examine two things. First, does a set of predictor variables do a good job in predicting an outcome (dependent) variable? The second thing is which variables are significant predictors of the outcome variable? In this article, we will discuss the concept of the Linear Regression Equation, formula and Properties of Linear Regression.  

 

Examples of Linear Regression

The weight of the person is linearly related to their height. So, this shows a linear relationship between the height and weight of the person. According to this, as we increase the height, the weight of the person will also increase. It is not necessary that one variable is dependent on others, or one causes the other, but there is some critical relationship between the two variables. In such cases, we use a scatter plot to simplify the strength of the relationship between the variables. If there is no relation or linking between the variables then the scatter plot does not indicate any increasing or decreasing pattern. In such cases, the linear regression design is not beneficial to the given data.

 

Linear Regression Equation

The measure of the relationship between two variables is shown by the correlation coefficient. The range of the coefficient lies between -1 to +1. This coefficient shows the strength of the association of the observed data between two variables.

 

Linear Regression Equation is given below:

 

Y=a+bX

 

where X is the independent variable and it is plotted along the x-axis

 

Y is the dependent variable and it is plotted along the y-axis

 

Here, the slope of the line is b, and a is the intercept (the value of y when x = 0).

 

Linear Regression Formula

As we know, linear regression shows the linear relationship between two variables. The equation of linear regression is similar to that of the slope formula.  We have learned this formula before in earlier classes such as a linear equation in two variables. Linear Regression Formula  is given by the equation

 

Y= a + bX

We will find the value of a and b by using the below formula

a= [frac{left ( sum_{Y}^{} right )left ( sum_{X^{2}}^{} right )-left ( sum_{X}^{} right )left ( sum_{XY}^{} right )}{nleft ( sum_{x^{2}}^{} right )-left ( sum_{x}^{} right )^{2}}]

 b= [frac{nleft ( sum_{XY}^{} right )-left ( sum_{X}^{} right )left ( sum_{Y}^{} right )}{nleft ( sum_{x^{2}}^{} right )-left ( sum_{x}^{} right )^{2}}]

Simple Linear Regression

Simple linear regression is the most straight forward case having a single scalar predictor variable x and a single scalar response variable y. The equation for this regression is given as y=a+bx

 

The expansion to multiple and vector-valued predictor variables is known as multiple linear regression. It is also known as multivariable linear regression. The equation for this regression is given as Y = a+bX. Almost all real-world regression patterns include multiple predictors. The basic explanations of linear regression are often explained in terms of multiple regression. Note that, in these cases, the dependent variable y is yet a scalar.

 

Least Square Regression Line or Linear Regression Line

The most popular method to fit a regression line in the XY plot is found by using least-squares. This process is used to determine the best-fitting line for the given data by reducing the sum of the squares of the vertical deviations from each data point to the line. If a point rests on the fitted line accurately, then the value of its perpendicular deviation is 0. It is 0 because the variations are first squared, then added, so their positive and negative values will not be cancelled. Linear regression determines the straight line, known as the least-squares regression line or LSRL. Suppose Y is a dependent variable and X is an independent variable, then the population regression line is given by the equation;

Y= B₀+B₁X

Where

B₀ is a constant

B₁ is the regression coefficient

When a random sample of observations is given, then the regression line is expressed as;

ŷ = b₀+b₁x

where b₀ is a constant

b₁ is the regression coefficient, 

x is the independent variable, 

ŷ is known as the predicted value of the dependent variable.

 

Properties of Linear Regression

For the regression line where the regression parameters b₀ and b₁are defined, the following properties are applicable:

• The regression line reduces the sum of squared differences between observed values and predicted values.

• The regression line passes through the mean of X and Y variable values.

• The regression constant b₀ is equal to the y-intercept of the linear regression.

• The regression coefficient b₁ is the slope of the regression line. Its value is equal to the average change in the dependent variable (Y) for a unit change in the independent variable (X)

Regression Coefficient

The regression coefficient is given by the equation :

Y= B₀+B₁X

Where

B₀ is a constant

B₁ is the regression coefficient

Given below is the formula to find the value of the regression coefficient.

B₁=b₁ = ∑[(x_i-x)(y_i-y)]/∑[(x_i-x)²]

Where x_i and y_i are the observed data sets.

And x and y are the mean value.

Importance of Regression Line 

A regression line is used to describe the behaviour of a set of data, a logical approach that helps us study and analyze the relationship between two different continuous variables. Which is then enacted in machine learning models, mathematical analysis, statistics field, forecasting sectors, and other such quantitative applications. Looking at the financial sector, where financial analysts use linear regression to predict stock prices and commodity prices and perform various stock valuations for different securities. Several well-renowned companies make use of linear regressions for the purpose of predicting sales, inventories, etc. 

Key Ideas of Linear Regression 

• Correlation explains the interrelation between variables within the data.

• Variance is the degree of the spread of the data.

• Standard deviation is the dispersion of mean from a data set by studying the variance’s square root.

• Residual (error term) is the actual value found within the dataset minus the expected value that is predicted in linear regression. 

Important Properties of Regression Line 

• Regression coefficient values remain the same because the shifting of origin takes place because of the change of scale. The property says that if the variables x and y are changed to u and v respectively u= (x-a)/p v=(y-c) /q, Here p and q are the constants.Byz =q/p*bvu Bxy=p/q*buv.

• If there are two lines of regression and both the lines intersect at a selected point (x’, y’). The variables x and y are considered. According to the property, the intersection of the two regression lines is (x`, y`), which is the solution of the equations for both the variables x and y. 

• You will understand that the correlation coefficient between the two variables x and y is the geometric mean of both the coefficients. Also, the sign over the values of correlation coefficients will be the common sign of both the coefficients. So, if according to the property regression coefficients are byx= (b) and bxy= (b’) then the correlation coefficient is r=+-sqrt (byx + bxy) which is why in some cases, both the values of coefficients are negative value and r is also negative. If both the values of coefficients are positive then r is going to be positive.

• The regression constant (a₀) is equal to the y-intercept of the regression line and also  a₀ and a₁ are the regression parameters.

 

Regression Line Formula: 

A linear regression line equation is written as-
Y = a + bX

where X is plotted on the x-axis and Y is plotted on the y-axis. X is an independent variable and Y is the dependent variable. Here, b is the slope of the line and a is the intercept, i.e. value of y when x=0. 

Multiple Regression Line Formula: y= a +b₁x₁ +b₂x₂ + b₃x₃ +…+ b_tx_t + u 

Assumptions made in Linear Regression

• The dependent/target variable is continuous.

• There isn’t any relationship between the independent variables.

• There should be a linear relationship between the dependent and explanatory variables.

• Residuals should follow a normal distribution.

• Residuals should have constant variance.

• Residuals should be independently distributed/no autocorrelation.

Solved Examples

1. Find a linear regression equation for the following two sets of data:

Sol: To find the linear regression equation we need to find the value of Σx, Σy, Σx

2

2

and Σxy 

Construct the table and find the value

The formula of the linear equation is y=a+bx. Using the formula we will find the value of a and b

a= [frac{left ( sum_{Y}^{} right )left ( sum_{X^{2}}^{} right )-left ( sum_{X}^{} right )left ( sum_{XY}^{} right )}{nleft ( sum_{x^{2}}^{} right )-left ( sum_{x}^{} right )^{2}}]

Now put the values in the equation

[a=frac{25times 120-20times 144}{4times 120-400}]

a= [frac{120}{80}]

a=1.5

b= [frac{nleft ( sum_{XY}^{} right )-left ( sum_{X}^{} right )left ( sum_{Y}^{} right )}{nleft ( sum_{x^{2}}^{} right )-left ( sum_{x}^{} right )^{2}}]

Put the values in the equation

[b=frac{4times 144-20times 25}{4times 120-400}]

b=[frac{76}{80}]

b=0.95

Hence we got the value of a = 1.5 and b = 0.95

The linear equation is given by

Y = a + bx

Now put the value of a and b in the equation

Hence equation of linear regression is y = 1.5 + 0.95x

Mathematics is the abstract study of different topics such as quantity, number theory, structure (algebra). It is evolved from counting, measuring, and describing the shape of objects. The word Mathematics originates from the Greek word Mathema.

There are many branches of Mathematics such as algebra, geometry, trigonometry, calculus, probability, and statistics.

Logarithmic differentiation is a part of calculus. The technique is used in cases where it is easier to differentiate the logarithm of a function rather than the function itself. In some cases, it is easier to differentiate the logarithm of a given function than to differentiate it from the function itself.

It is a famous concept, and it applies to the majority of the non-zero functions. The only condition is that the non-zero functions should be differentiable in the future. Differentiation and integration are the two main concepts of calculus. Differentiation is used to study the small change of a quantity. Integration is different from differentiation. It is used to add small and discrete data and cannot be added singularly.

Derivative of Logarithm

When a logarithmic function is represented as:

The derivative of a logarithmic function is given by:

1. Here, x is called as the function argument.

2. b is the logarithm base.

3. ln b is the natural logarithm of b.

We can differentiate log in this way.

The derivative of ln(x) is 1/x

This is the way of differentiating ln. The derivative of ln(x) is a well-known derivative.

Following are some of the examples of logarithmic derivatives:

1. Find the Value of dy/dx if,y=ex4

Solution: Given the function y=ex4

Taking the natural logarithm of both the sides we get,

ln y = ln ex4

ln y = x 4 ln e

ln y = x4

Now, differentiating both the sides w.r.t we get,

1ydydx = 4×3

⇒dy dx =y.4×3

⇒dydx =ex4×4×3

2. Find the Value of dy dx if y = 2x{cos x}.

Solution: Given the function y = 2x{cos x}

Taking logarithm of both the sides, we get

log y = log(2x{cos x})

⇒log y=log2+log x cosx(As log(mn)=logm+logn)

⇒logy=log2+cosx×logx(As logmn=nlogm)

Now, differentiating both the sides w.r.t by using the chain rule we get,

1ydydx=cosx–(sinx)(logx)

This is a way used for differentiating logarithmic functions.

y “(x)=(logax)′=1 xlna. 

If a=e, we obtain the natural logarithm the derivative of which is expressed by the formula (lnx)′=1x, 

where the number M is equal to M=log10e≈0.43429.

We derived the formula (logax)′=1xlna from first principles using the derivative’s limit definition.

This is the way of deriving logarithms.

Deriving log functions becomes possible because of the use of exponents.

Following are some of the log derivative rules:

Common Functions    

Function Derivative

• Constant   c       0

• Line  x       1

         ax     a

• Square      x2     2x

• Square Root       √x     (½)x-½

• Exponential        ex     ex

         ax     ln(a) ax

         loga(x)       1 / (x ln(a))

         cos(x)        −sin(x)

         tan(x)        sec2(x)

         cos-1(x)     −1/√(1−x2)

         tan-1(x)     1/(1+x2)

Rules Function

• Derivative

• Multiplication by constant  cf      cf’

• Power Rule         xn     nxn−1

• Sum Rule  f + g f’ + g’

• Difference Rule  f – g  f’ − g’

• Product Rule      fg     f g’ + f’ g

• Quotient Rule    f/g    (f’ g − g’ f )/g2

• Reciprocal Rule  1/f    −f’/f2                 

• Chain Rule

(as “Composition of Functions”)   f º g  (f’ º g) × g’

Chain Rule (using ’ )    f(g(x))        f’(g(x))g’(x)

Chain Rule (using  ddx )        dy dx =  dy du du dx

1. These are logarithmic differentiation rules.

The logarithmic function with base a (a>0, a≠1) and exponential function with the same base form a pair of mutually inverse functions; the log function’s derivative is also found using the inverse function theorem. (logax)′=f′(x)=1φ′(y)=1(ay)′=1aylna=1alogaxlna=1xlna.

The differentiation of natural log ln(x) is 1 divided by x.

2. Logarithmic differentiation steps are as follows:-

A natural log is supposed to be taken on both sides.

Use the property of the log of the product.

Differentiate on both sides. For every term on the right side of the equation, a chain rule should be used.

The last step is to multiply both sides by f(x).

Following are the logarithm derivative rules we always need to follow:-

The slope of a constant value (for example 3) is always 0.

The slope of a line like 2x is 2, or 3x is 3, etc. 

One can use logarithmic differentiation when applied to functions raised to the power of variables or functions. 

Logarithmic differentiation relies on the chain rule as well as the properties of the logarithm.

Differentiation in y

The derivative of function y = f(x) of a variable x is the measure of the rate at which the value y of the function changes concerning the change of the variable x. The derivative of ln y is 1/ (derivative of f = e[^x]) = 1/e[^x].

Understanding the Principle of Mathematical Induction

Proof by Induction will help you understand the meaning of mathematical induction. It consists of –

1) The basis or base case proves that statement for n = 0 without assuming knowledge of other cases. 

2) The 2nd case or the inductive step proves if the statement holds for any given case n = k, it must also hold for the next case n = k + 1.

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Application of Mathematical Induction in Real Life – ‘The Domino Effect.’ 

If you queue a thousand dominoes and want to let them all fall by allowing the first domino to fall, how would you queue it?

The best idea is to queue it such that:

1. When the very first domino topples, it will lean against the second domino and make it fall.

2. Ensure that each domino will hit the domino next to it and that each hit makes a domino fall.

3. If conditions (1) and (2) are satisfied, then all the dominoes will fall, proving the principle of mathematical Induction.

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Mathematical Induction Example 

Here is a mathematical induction question:

Problem 1: Family Tree 

Imagine that your great, great, … grandfather Jim had two children. Call this generation one, so that generation one contains 2 offsprings of Jim Each of these children has two children, so generation two will have 4 offsprings. Each of the four offspring has 2 children, so, at 3rd generation, we have eight offspring. If this sequence continues generation after generation, prove that at n generation, there will be 2n offsprings?

Solution:Base Case: P(1) asserts that generation 1 has 21 offspring, which is true since we are told that Jim had two children.

Inductive Step: We assume for an arbitrary integer, say k, P(k) is true. That means, that generation k has 2k offspring. We have to show that k + 1 generation has 2k + 1 offspring.

Generation k + 1 has 2 x 2k offsprings [Generation k has 2k assumed to be true]

Generation k + 1 has 21 x 2k offsprings

Generation k + 1 has 2k + 1 offsprings

Thus, P(k + 1) is true because P(k) is also true. 

Hence P(n) is true for all natural numbers.

In Mathematics, the maxima and minima (the plural of maximum and minimum respectively) of a given function are collectively known as the extrema ( the plural term of extremum). The two terms maxima and minima are the smallest and largest value of the function, either within a given range, or the entire domain. Pierre de Fermat was one of the renowned Mathematicians to introduce a general technique, adequality, for determining the maxima and minima of a function.

In set theory, the maximum and minimum of a given set are considered as the greatest and least elements of the set respectively whereas the set of real numbers has no maximum and minimum value. We can find maxima and minima using the first derivative test, and second derivative test. In the article, we will discuss how to find maxima and minima using the first derivative test.

First Derivative Test

Let us consider f real-valued function, and a,b is an interval on which function f is defined and differentiable. Further, if c is considered as the critical point of f in a,b, then

• If f’(x) > 0 ( greater than 0) for all x < c and f’(x) < 0 ( lesser than 0) for all x > C, then f (c) will be considered as the maximum value of function f in the interval a,b.

• If f’(x) < 0 ( lesser than 0) for all x > c and f’(x) > 0 ( greater than 0) for all x < C, then f (c) will be considered as the minimum value of function f in the interval a,b.

In simple words, we can say that a point is determined as the maximum of a function if the function increases before and decreases after it whereas a point is considered as the minimum if the function decreases before and increases after it.

Method to Find Whether a Critical point is Maximum, Minimum, Or Neither Using First Derivative Test

The first derivative test is a method to determine whether a critical point is maximum, minimum or neither. 

First Derivative Test: for a Given Critical Point

• If the derivative is negative on the left side of the critical point and positive on the right side of the critical point, then the critical point is considered as a minimum.

• If the derivative is positive on the left side of the critical point and negative on the right side of the critical point, then the critical point is considered as maximum.

• In any other situation, the critical point will neither be maximum nor minimum.

Example

Let us consider f(x)= 6x – x² 

The derivative of f(x)= 6x – x²  is f'(x)= 6 – 2x 

The function f has a critical point at x = 3, as 3 is the solution of 6 – 2x = 0

To determine whether the critical point i.e x = 3 is maximum, minimum, or neither, observe where f is increasing or decreasing.

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The diagram above shows x = 3 is the maximum.

How to Find Maxima and Minima Using First Derivative Test

The first derivative test is used to determine whether a function is increasing or decreasing on its domain, and to identify its local maxima or minima. The first derivative test is considered as the slope of the line tangent to the graph at a given point. When the slope is positive, the graph is increasing whereas when the slope is negative, the graph is decreasing. When the slope is 0, the point is considered as a critical point and it can be a local maximum or minimum. Given a differentiable function, the first derivative test can be applied to determine any local maxima or minima of the given function through the steps given below.

Step 1: Differentiate the given function.

Step 2: Set the derivative equivalent to 0 and solve the equation to determine any critical points.

Step 3: Test the values before and after the critical points to find whether the function that is given is increasing (positive derivative) or decreasing (negative derivative) around the point. 

Then Observe the Following Points

• If the first derivative changes from positive to negative at the given point, then the point is determined as a local maximum.

• If the first derivative changes from negative to positive at the given point, then the point is determined as a local minimum.

• If the first derivative does not change at the given point, then the given point will neither be considered as a local maximum or minimum.

Maxima and Minima Using First Derivative Test Example

Find the critical points and any local maxima or minima of a given function f(x)=1/4x⁴ -8x

Here are the steps:

1.  The first step is to differentiate f(x)= [ frac{1}{4x^{4} – 8x} ]

f’ (x)= [ left [ frac{1}{4x^{4} – 8x} right ]’ ]= 1/4 ( 4x³ – 8) = x³ – 8

2. The second step is to find the value of x 

Let us equate, x³ – 8 = 0

x³ = 8

Hence, the value of x = 2

This implies that, x³ – 8 , has a critical point at x = 2

3. The third step is to test the points around critical points at  x = 1 and x = 3.

For x = 1, f’ (x)= 1³ – 8 = 1 – 8 = -7

For x = 3, f’ (x)= 3³ – 8 = 27 – 8 = 19

At f’ (1)and f’ (3), the text point around the critical points changes from negative to positive, this implies a negative slope on the graph of f (x)before the critical point and positive slope on the graph of f'(x)after the critical point (i.e. from left to right). Hence, the critical point x = 2 is a local minimum and can be seen in the graph of f (x)as shown below.

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Points to Remember

• Maxima and minima in calculus are calculated by using the concept of derivatives.  

• The concept of the derivatives gives out the information regarding the gradient or slope of the function as the points get located where the gradient is zero.

• These points are known as turning points or stationary points. 

• These points correspond to the largest or smallest values of the function.

• Maxima and Minima are the most common concepts in differential calculus. 

• A branch
  of Mathematics known as Calculus of Variations deals with the maxima and the minima of the functional. 

• The calculus of variations is concerned with the variations in the functional during which a  small change in the function leads to the change in the functional value.

• The first variation is defined as the linear part of the change in the functional. 

• The second part of the variation is known in the quadratic part. Functional is expressed as the definite integrals that involve the functions and their derivatives.

• The functions that maximize or minimize the functional are to be found using the Euler i.e, Lagrange of the calculus of variations. 

• The two Latin words, ‘maxima’ and ‘minima’ mean the maximum and minimum value of a function respectively. 

• The maxima and minima are collectively known as the “Extrema”. 

• The Critical point of a differential function of a complex or real variable is any value in its domain where its derivative is 0. 

• It can be inferred that every local extremum is a critical point, however, every critical point does not have to be a local extremum. 

• If there is a function that is continuous, it must have maxima and minima or local extrema. 

• It implies that all such functions will have critical points. 

• If the given function is monotonic, the maximum and minimum values lie at the endpoints of the domain of the definition of that function.

• Maxima and minima are, therefore, very important concepts in the calculus of variations, which helps to find the extreme values of a function. 

• One can use the two values and where they occur for a function using the first derivative method or the second derivative method.

• If f(x) is a continuous function in its domain, then one maximum or one minimum should lie between equal values of f(x).

• Maxima and minima occur alternately which is, between two minima then there is one maximum and vice versa. 

• If f(x) tends to be infinity as x tends to a or b and f(x) = 0 only for one value x, that is, c between a and b , then f(c) is the minimum and of the least value.  If f(x) tends to minus infinity as x tends to a or b , then f(c) is of the maximum and of the highest value. 

Median Definition

So let’s say for example in a tough exam, you and your friends performed very averagely. However, this one brainiac scored 100%.  Now when you calculate the average marks of the class, it’s higher due to his score. But, this is not a fair representation of reality. Here is where the concept of Median will help you out.

Often the median is a better representative of a standard group member. If you take all the values in a list and arrange them in increasing order, the median will be the number located at the center. The median is a quality that belongs to any member of the group. Based on the value distribution, the mean may not be particularly close to the quality of any group member. The mean is also subjected to skewing, as few as, one value significantly different from the rest of the group can change the mean dramatically. Without the skew factor introduced by outliers, the median gives you a central group member. If you have a normal distribution, a typical member of the population will be the median value.

What is the Median?

In simplest terms, the Median is the middlemost value of a given data set. It is the value that separates the higher half of the data set, from the lower half. It may be said to be the “middle value” of any given population. To calculate the Median the data should be arranged in an ascending or descending order, the middlemost number from the so arranged data is the Median.

In order to calculate the Median of an odd number of terms, we have to arrange the data in an ascending or descending order. The middlemost term is the Median. And to calculate the Median of an even number of terms, we arrange the data in an ascending or descending order and take an average of the two middle terms.

• So, if the odd number is n, then M = ([frac{n+1}{(2)})^{th}] term 

• And if the even number is n, then [M = frac{(frac{n}{2} + 1)^{th}~term + (frac{n}{2})^{th}~term}{2}]

Relationship between Mean, Median, and Mode

Karl Pearson explains the relationship between mean median and mode with the help of his Formula as:

(Mean – Median) = [frac{1}{(3)}] (Mean – Mode)

3(Mean – Median) = (Mean – Mode)

Mode = Mean – 3 (Mean – Median)

Mode = 3 Median – 2 Mean

Therefore, the equation given above can be used if any of the two values are given and you need to find the third value.

Median of Grouped Data

Since the data is divided into class intervals, we cannot just pick the middle value anymore. Hence, we have to follow a few steps which are listed below:

Step 1: Firstly, we have to prepare a table with 3 columns.

Step 2: Class Intervals will be written in column 1. 

Step 3:  Corresponding frequencies denoted by fi will be written in column 2.

Step 4: Calculated Cumulative Frequency denoted by cf will be written in column 3.

Step 5: In this step, we have to find the total of fi denoted by N, and calculate N/2.

Step 6: here, we have to locate the Cumulative Frequency which is greater than or equal to N/2. Note down its corresponding Median Class as well.

Step 7: We have to use the formula given below to calculate the Median.

M = L + ([frac{n}{(2)}] – cf)[frac{h}{(f)}]

Where, 

L = Lower limit of Median Class

n = Total frequency

cf =The cumulative frequency of a class preceding the Median Class

f = The frequency of the Median class

h = The width of the Median Class 

Median from an Ogive Curve

We have already seen in the Cumulative Frequency topic that the Median of data can also be calculated. This can be done by plotting the Less than frequency curve and More than the frequency curve. The point at which they intersect and the corresponding value on the X-axis would be the Median of the given data.

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Solved Example 

Question 1: Find the Median Class for the Following Table:

Solution: First we have to find the total number of frequencies in order to find the median class.

So, here frequency (f) = 20

The formula to obtain the Median Class is [frac{Total~Number~of~Frequencies + 1}{2}]

Thus, Median Class = [frac{20 + 1}{2}] 

= 10.5

Therefore, the Median class of 10.5 lies between the intervals 10.5 – 13.5

Question 2: Find the median of  3, 13, 7, 5, 21, 23, 39, 23, 40, 23, 14, 12, 56, 23, 29.

Solution: Firstly we have to organize the data in an increasing order  3, 5, 7, 12, 13, 14, 21, 23, 23, 23, 23, 29, 39, 40, 56

Secondly, we have to get the middle number so, if there are 15 numbers, the middle number would be the eighth number i.e., 23. So our median is 23. 

NOTE: In case we get a pair of middle numbers then we have to add those two numbers and then divide to get the median.

In Mathematics, the Median is the middlemost value of a given data set. It is the value that separates the data into two halves. The upper part of the median is the higher half, and the lower part is the lower half.

Lesson on Median

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Learn Relationship Mean, Mode, and Median with

has explained the relationship formula given by Karl Pearson. Maths experts at have
included it in the provided notes to assist students in finding a third value if two values are available in the problems.

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Find Median with

To find the median of data, you need to remember the formula for both odd and even numbers of terms in the data set. Also, to calculate the median of group data, you have to follow a lengthy procedure. has provided a solved example of group data to give students a clear understanding of the topic. While finding the median in grouped data, students will come across some other terms, for instance, frequency, cumulative frequency. Students can find solved questions regarding the topic of Median on the website or the learning app that is available on the play store to download for free.