Derivatives Related Rates

Derivative related rates is the normal derivatives whereas the differentiation is carried out with respect to the time function t. The Differentiation of a given function with respect to time is called related rates derivatives. The related rates derivatives are also one of the parts of calculus which deals with calculating the rate of change of function with respect to time. The following are the example problems for related rates derivatives.

Related Rates Example Problems:

Example 1:

Find the related rate derivatives for the given function.

f(t) = t 4 – 18t + 16

Solution:

The given function is

f(t) = t 4 – 18t + 16

The first derivative f ' is given by

f '(t) = 4 t 3 – 18

Example 2:

Find the related rate derivatives for the given function.

f(t) = t 5 – 6 t 3  + 11

Solution:

The given function is

f(t) = t 5 – 6 t 3  + 10

The first derivative f ' is given by

f '(t) = 5t 4 – 6(3 t 2 )

f '(t) = 5t 4 – 18 t 2

Example 3:

Find the related rate derivatives for the given function.

f(t) = t2 – 4t + 8

Solution:

The given function is

f(t) = t 2 – 4t + 8

Differentiate the above equation with respect to t.

f '(t) = 2 t  – 4

Example 4:

Find the related rate derivatives for the given function.

f(t) = t 3 – 5 t 2  + 11t

Solution:

The given function is

f(t) = t 3 – 5 t 2  + 11t

The first derivative f ' is given by

f '(t) = 3t 2 – 5(2 t  ) + 11

f '(t) = 3t 2 – 10 t + 11

Example 5:

Find the related rate derivatives for the given function.

f(t) = t4 – 3t 3 – 4t 2  + t

Solution:

The given function is

f(t) = t4 – 3t 3 – 4 t 2  + t

The first derivative f ' is given by

f '(t) = 4 t 3 – 3(3t 2 ) – 4( 2 t  ) + 1

f '(t) = 4 t 3 – 9t 2  – 8 t  + 1

Related Rates Practice Problems:

1) Find the related rate derivatives for the given function.

f(t) = t 3 – 6 t 2  + 11t

Answer: f '(t) = 3t 2 – 12 t

2) Find the related rate derivatives for the given function.

f(t) = t 2 – 6 t   + 11

Answer: f '(t) = 2t – 6

Construct Probability Table

Construct Probability table is defined as an equation or table with its probability happenings. Usually, probability table is for constant occurrence for much number of data produces results. For construct probability table function we learn about discrete random variable, if a random variable obtains only a finite or a countable number of values, it is known as discrete random variable. To construct Probability table it contains probability mass function and moments.

Classification of Construct Probability Table:

Probability Mass Function:

The statistical description of discrete probability table function p (x)  functions and satisfies the following properties:

(1.) The probability which x can take a particular value x is p (x)

That is P(x = x) = p (x) = px.

(2). P(x) is a non – negative for every real x.

(3). The amount of p (x) over all likely values of x is one. Which is Σpi = 1 here j denotes that x and pi is the probability at x = xi

Moments:

A probable value of a function of a chance variable x is used for calculating the moments. Let us see the two types of moments.

(i) Moments about the origin.
(ii) Moments about the mean that are known as central moments.

Example Problem for Construct Probability Table:

A box has 4 green and 3 black pens. Construct a probability table distribution of number of black pens in 3 draws one by one from the box. (i) With replacement

Solution:

(i) With replacement

Let x be the approximate variable of drawing number of black pens in three draws.

X will takes the values 0,1,2,3.

P (Black pen) = `3 / 7 ` = P (B)

P (Not Black pen) = `4 / 7` = P (G)

Hence P(X = 0) = P (GGG) = `4/ 7` * `4 / 7` * `4 / 7` = `64 / 343`

P(X = 1) = P (BGG) + P (GBG) + P (GGB)

=  ( `3 / 7` * `3 / 7` * `4 / 7` )  + ( `4 / 7` * `3 / 7`   * `4 / 7` ) +  ( `4 / 7` * `3 / 7`   * `3 / 7` )

= 3 * (`48 / 343` )

= `144 / 343 `

P(X = 2) = P (BBG) + P (BGB) + P (GBB)

= (`3 / 7` * `3 / 7` * `4 / 7` ) + (`3 / 7` * `4 / 7` * `3 / 7` ) + (`4 / 7` * `3 / 7` * `3 / 7` )

= 3 * ( `3 / 7` ) * ( `3 / 7` ) * ( `4 / 7` )

= 3 * (`36 / 343` )

= `108 / 343`

P(X = 3) = P (BBB) = ( `3 / 7` ) * ( `3 / 7` ) * ( `3 / 7` )

= `27 / 343`

Therefore the wanted probability table distribution is

x	0	1	2	3
p (x =x)	`64 / 43`	`144 / 343`	`108 / 343`	`27 / 343`

Example 2:

Find out the probability mass function, and the collective distribution function for obtaining ‘3’s while two dice are thrown.

Solution:

2 dice are thrown. Let x be the approximate variable of obtaining number of ‘3’s. Hence x can take the values 0, 1, 2.

P (no ‘3’) = P (X = 0) =` 25 / 36`

P (one ‘3’) = P (X = 1) = `10 / 36`

P (two ‘3’s) = P (X = 2) = `1 / 36`

Obtained Probability mass function is

x	0	1	2
P (x = x)	`25 / 36`	`10 / 36`	`1 / 36`

Limit Point of a Sequence

In this article, we will discuss the limit point of a sequence. A set of numbers said to be a limit point of a sequence. It has two types of sequences.

1. Arithmetic sequence and

2. Geometric sequence.

Arithmetic sequence means that, the sequence of a numbers such that the difference between two consecutive members of the sequence is a constant. Geometric sequence means that, the sequence of a numbers such that the ratio between two consecutive members of the sequence is a constant. The limit point of a sequence formulas and example problems are given below.

Formulas and Example Problems for Limit Point of a Sequence

Sequences formulas are given below.

Formula for arithmetic sequence:

nth term of the sequence : an = a1 + (n - 1)d

Series of the sequence: sn = `(n(a_1 + a_n))/2 `

Formula for geometric sequence:

nth term of the sequence: an = a1 * rn-1

Series of the sequence: sn = `(a_1(1-r^n))/(1 - r)`


Example problem 1:

Find the 11th term of the given series 11, 12, 13, 14, 15,......

Solution:

First term of the series, a1 = 11

Difference of two consecutive terms, d = 12 - 11 = 1

n = 11

The formula to find the nth term of an arithmetic series, `a_n = a_1 + (n-1)d`

So, the 11th term of the series  11, 12, 13, 14, 15,... = 11 + (11 - 1) 1

= 11 + 10 * 1

= 11 + 10

After simplify this, we get

= 21

So, the 11th term of the sequence 11, 12, 13, 14, 15,... is 21.

More Example Problems for Limit Point of a Sequence

Example problem 2:

Find out the 5th term of a geometric sequence if a1 = 70 and the common ratio (C.R) r = 2

Solution:

Use the formula `a_n = a_1 * r^(n-1)` that gives the nth term to find `a_5` as follows

`a_5 = a_1 * r^(5-1)`

= 70 * (2)4

= 70 * 16

After simplify this, we get

= 1120.

The 5th term of a geometric sequence is 1120.

The above examples are helpful to study of limit point of a sequence.

Parts of Quadrilateral

Quadrilateral is a type of polygon with four sides and four vertices's or corners. The quadrilaterals are whichever convex or concave. The entire convex quadrilateral covers the plane by continual revolving around the midpoints of their ends.

Parts of Quadrilateral

• Sides
• Vertices's
• Interior angles
• Diagonals
• Adjacent sides
• Opposite angles
• Opposite sides
• Consecutive angles

Parts of Quadrilateral-side

The quadrilateral is of many types, special cases are tangent quadrilaterals and cyclic quadrilateral. The sides of the tangent quadrilateral are opposite and have equal length, in cyclic quadrilateral the product of the opposite sides are the same, are called a harmonic quadrilateral.

Vertices's

Vertex is the two side’s meets at the end point. They are four vertices's on a quadrilateral. The consecutive vertices's are same at the endpoints.

Diagonal

The Line segment that joins two vertices's in opposite. In Parallelogram the diagonal bisect each other. In Rectangle the diagonals are congruent. In kite, one diagonal is perpendicular bisector of other, and the diagonal bisects a pair of opposite angles. In Rhombus, the diagonals bisect the angles and are perpendicular bisector of each other; the diagonals divide the rhombus into four congruent right triangles.

Interior angle of quadrilateral:

Interior angle of quadrilateral  is the set of all points in its plane which lie in between both the arms. The addition of the interior angles of a non-crossed quadrilateral is 360.  In a crossed quadrilateral, the addition of the one side of  interior angles equals the addition of the interior angles on the other side.

Adjacent sides

Consecutive sides or adjacent sides have a common endpoint. In kite two dissimilar pairs of adjacent sides are similar.

Opposite angles

Opposite angles are angles whose vertices's are not successive.

Opposite sides:

Opposite sides of a quadrilateral are sides that do not have a general endpoint. In parallelogram and in rectangle the opposite sides are parallel and  congruent.

Consecutive angles

In parallelogram, any pair of consecutive angles are supplementary.In Rhombus two sides are congruent

Parallelogram with Four Equal Sides

In geometry, a parallelogram is a quadrilateral with the two pairs of parallel sides. In Euclidean Geometry, the opposites or facing sides of a parallelograms are of equal length and the opposite angles of a parallelogram are of equal measure. The congruences of opposite sides and opposite angles are a direct consequence of the Euclidean Parallel Postulate and neither condition can be proven without appealing to the Euclidean Parallel Postulate or one of its equivalent formulations. The three-dimensional counterparts of a parallelogram is a parallelepiped.(Source.Wikipedia)

Parallelogram with Equal Sides are Square and Rhombus:

Parallelograms:

Area of parallelograms A = b *  h   sq. units



where h is the perpendicular height of the parallelogram.

b is the base length of the parallelogram.

Examples for Parallelogram with Equal Sides:

Example 1:

Find the area of the rhombus whose base is 150cm and the perpendicular height will be equal to 150cm.

Given:

H=150cm

B=150cm

Solution:

Area = b*h

= 150*150

= 22500cm2

Example 2:

Find the Area of a Parallelogram with a base of 12 centimeters and a height of 15 centimeters.

Solution:

A=b*h

A= (12 cm) · (15 cm)

A= 180 cm2

Example 3:

Find the area of a parallelogram with a base of 8 inches and a height of 14 inches.

Solution:

A=b*h

A= (8 in) · (14 in)

A= 112 in2

Example 4:

The area of a parallelogram is 30 square centimeters and the base is 60 centimeters. Find the height.

Solution:

A=b*h

30 cm2 = (60 cm) · h

30 cm2 ÷ (60 cm) = h

h= 0.5 cm

Example 5:

The area of a parallelogram is 100 square centimeters and the base is 50 centimeters. Find the height.

Solution:

A=b*h

100 cm2 = (50 cm) · h

100 cm2 ÷ (50 cm) = h

h= 2 cm.

These are the examples on parallelogram with four equal sides.

Fundamental Theorem of Integral Calculus

In this section we are going to discuss about the fundamental theorem of integral calculus concept. The development of short form the fundamental theorem of integral calculus corresponding to over screening the significant concept and problem in using calculus theorems are referred as review calculus. This article helps to improve the knowledge for using fundamental theorem of integral calculus problem and below the problems are helping toll for the exam. Fundamental theorem of integral calculus problem solutions also shows below. The fundamental theorem of integral calculus handled the differentiation, integration and inverse operations are process here now.

Important of Fundamental Theorem of Integral Calculus:-

Let` f(x)` is a continuous function on the closed interval `[a, b]` .

Let the area function `A(x)` be defined by `A(x) = int_a^xf(x)dx ` for` xgt=a`

Then `A'(x) = f(x)` for all `X in [a,b]`

Let` f(x)` be a continuous function defined on an interval `[a,b]` .

`If intf(x)dx = F(x) then int_a^a f(x)dx = [F(x)]^b_a`

`=F(b) - F (a)` is called the definite integral or `f (x) ` among the limits` a` and `b` .

This declaration is also known as fundamental theorem of calculus.

We identify `b` the upper limit of `x` and a the lower limit.

If in place of `F(x)` we take` F(x) +c` as the value of the integral, we have

`int_a^b f(x)dx = [F(x) + C ]^b_a`

`= [F (b) + c] - [F (a) + c]`

`= F (b) + c - F (a) - c`

`= F (b) - F (a)`

Therefore, the value of a definite integral is unique. It does not depend on the constant c and hence in the evaluation of a definite integral the constant of integration does not play any role.

Let `int f(x) dx = F(x) +C`

Then `int_a^b f(x)dx = F(b)-F(a)`

Note down:-

From the above two theorem, we infer the following

`intf(x)dx` =(Anti derivative of the function `f(x) ` at `b` ) - (Anti derivative of the function `f(x) ` at `a` )

The fundamental theorem of integral calculus gives you an idea about a close relationship between differentiation and integration. These theorems give an exchange method evaluating definite integral, without calculating the limit of a sum.


Example on Fundamental Theorem of Integral Calculus:-

Evaluate the definite integral of the following

`int^(pi/4)_0(3sec^2 x + x^2 + 3)`

Solution:

`int_0^(pi/4) (3sec^2 x + x^2 + 3)`

`= 3 int_0^(pi/4)sec^2 xdx + int_0^(pi/4)x^2dx + int_0^(pi/4)3dx`

`= 3[tanx] ^ (pi/4) _0 + [(x^3)/ (3)] ^ (pi/4) _0 + [3x] ^ (pi/4) _0`

`= 3 (tan (pi/4) - tan0) + 1/3 (pi/4) ^3 - 0+ [(pi/2) - 0]`

`= 3 + (pi^3)/192+ (pi)/ (2).`

Cumulative Probability

In probability theory the cumulative distribution function (CDF) or just distribution function, completely describes the probability distribution of a real valued random variable X. Cumulative distribution functions are used to specify the distribution of multivariate random variables.

For every real number x, the CDF of real valued random variable is X its given by
                              f(x)=  P[X <= x]

where the function right hand side is represents the probability of  the random variable X takes on the value less than or equal to x. The probability of  X is lies in the interval (a, b) is therefore F_X (b) − F_X (a).

Cumulative Probability Example:

Consider a coin flip experiment. If we flip a coin two times, we might ask that what is the probability that the coin flips would result in one or fewer heads? The answer would be a cumulative probability. It would be the probability when that the coin is flip results in zero heads plus the probability that the coin flip results in one head. Thus, the cumulative probability would equal:

P(X < 1) = P(X = 0) + P(X = 1) = 0.25 + 0.50 = 0.75

The table below shows that the both of the probabilities and the cumulative probabilities associated with this experiment.

Number of heads              Probability       Cumulative Probability

0                                            0.25                                0.25

1                                             0.50                                0.75

2                                             0.25                                1.00

 Assume we have a random variable X. Cumulative probabilities that are provide for each value x, the probability of a result less than or equal to X, P[X <= x].

Example:

Here's the probability distribution and for a discrete random variable X
X                  f(x)
1                   0.1
2                   0.2
3                    0.4
4                    0.3
The cumulative distribution function tables and for each value x = 1, 2, 3, 4, the probability of a result less than or equal

For example:
* P[ X <= 1 ] = 0.1

* P[ X <= 2 ] = 0.1 + 0.2 = 0.3

* P[ X <= 3 ] = 0.1 + 0.2 + 0.4 = 0.7

* P[ X <= 4 ] = 0.1 + 0.2 + 0.4 + 0.3 = 1

 These probabilities can be tabled

 X                    P[X <= x]
 1                        0.1
 2                        0.3
 3                        0.7
 4                        1.0

Decrease in Percentages

To understand this concept, let us start with an agenda or a plan to go further which will make us stay organized all through the explanation. The explanation will begin with the introduction to the concept followed by the introduction to the concept of percent-ages, which will be followed by the change in the same and then the polarity or the direction of the change will be seen, then what do we with just finding out the solution the next step will obviously be the interpretation part which requires the complete understanding of the concept.

The concept which will be dealt is the same given in the header. The change can always be an increase or a Percentage Decrease; it can be either way round. So it is necessary for us to learn both the method of calculations, actually both require the same formula which is the Percent Decrease Formula, the only difference will be the final answer, that too in the interpretation part, since obviously without understanding the number one can never say if it is an increase or a decrease in the change occurred. Before starting with, one has to be clear with the concepts of pct to solve the questions easily without much strain.

Let us now try to answer the question of how to Calculate Percentage Decrease. Answering this question is never a great task, it is all about the numbers we get for solving, and one can easily solve such type of questions, if one understands the problem and interprets the Percent Decrease. Let us consider an example to get a better understand Calculating Percent Decrease of the numbers whose initial value that is the original value is 50 and the final value is 25; the answer for this question is 50 – 25 / 50 * 100 = 50%, the answer shows us that there is an 50 % decrease in the final value from the original value and that shows the decrease.

The fact is that anyone can solve such problems of increase or decrease, but the interpretation part can only be done by the person who completely understood the concept. The final interpretation shows us the direction of change that is it shows whether the change is positive or negative and increase or decrease in the number in terms of percentages which makes the concept complete with the final interpretation.