This document contains some important questions on vector integration like greens theorems,gauss divergence theorem, Stokes theorems and volume.

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A New way, The Area of Trapezium by Piyush Goel
Lot of mathematicians have proved Pythagoras theorem in their own ways. If you google it you will indeed found hundred of ways.
Meanwhile I was also sure that maybe one day I could find something new out of this incredible Pythagoras theorem and Recently I got something which I would like to share with you.
To Prove: Deriving the equation of area of trapezium using Arcs
Proof: There is a triangle ABC with sides a b and c as shown in the figure.
Now,
Area of ∆ BCEG = Area of ∆ BDC +Area of ⌂ DCEF + Area of ∆ EFG
c^2=ac/2+ Area of ⌂ DCEF + (c-b) c/2
(2c^2– ac –c^2+ bc )/2=Area of ⌂ DCEF
(c^2– ac+ bc )/2=Area of ⌂ DCEF
c(c– a+ b)/2=Area of ⌂ DCEF
Area of ⌂ DCEF=BC(DE+CF)/2
Copyrighted©PiyushGoel

A note on the Factorial Function
Write down 0,1,2,3,4,5 and put parallel square of each number like 0,1,4,9,16,25, then start to subtract the bigger one to the lower one (1–0),(4–1),(9–4),(16–9) and (25–16) to get 1,3,5,7,9 and again subtract the bigger one to the lower one (3–1),(5–3),(7–5) and (9–7) to get (2,2,2,2).
Again we squared each number, at the same time we cubed each number (0,1,8,27,64,125,216) and the same procedure follows, subtract the bigger one to the lower one (1–0),(8–1),(27–8),(64–27) ,(125–64) and(216–125) to get (1,7,19,37,61,91) and again (7–1),(19–7),(37–19),(61–37) and (91–61), to get (6,12,18,24,30) same again(12–6),(18–12),24–18) and (30–24) till the result come out here we get (6,6,6,6).
At the same time if we do it again for 4 and 5 (power).When we get 2 for 2 ,6 for 3 ,24 for 4 and 120 for 5. The result is the factorial function.
https://piyushtheorem.wordpress.com/2017/02/08/a-note-on-the-factorial-function/

Graph is a mathematical representation of a network and it describes the relationship between lines and points. A graph consists of some points and lines between them. The length of the lines and position of the points do not matter. Each object in a graph is called a node. A graph ‘G’ is a set of vertex, called nodes ‘v’ which are connected by edges, called links ‘e'. Thus G= (v , e).
Vertex (Node): A node v is an intersection point of a graph. It denotes a location such as a city, a road intersection, or a transport terminal (stations, harbours, and airports).
Edge (Link): An edge e is a link between two nodes. A link denotes movements between nodes. It has a direction that is generally represented as an arrow. If an arrow is not used, it means the link is bi-directional.Transport geography can be defined by a graph. Most networks, namely road, transit, and rail networks, are defined more by their links than by nodes. But it is not true for all transportation networks. For instance, air networks are defined more by their nodes than by their links since links are mostly not clearly defined. A telecommunication system can also be represented as a network. Mobile telephone networks or the internet is the considered the most complex graph. However, cell phones and antennas can be represented as nodes whereas links could be individual phone calls. The core of the internet or servers can also be represented as nodes while the physical infrastructure between them, like fiber optic cables, can act as links. This suggests that all transport networks can be represented by graph theory in some way.

Number theory is a branch of pure mathematics devoted to the study of the natural numbers and the integers. It is the study of the set of positive whole numbers which are usually called the set of natural numbers. As it holds the foundational place in the discipline, Number theory is also called "The Queen of Mathematics".The number theory helps discover interesting relationships between different sorts of numbers and to prove that these are true . Number Theory is partly experimental and partly theoretical. Experimental part leads to questions and suggests ways to answer them. The theoretical part tries to devise an argument which gives a conclusive answer to the questions.
Here are the steps to follow:
1. Accumulate numerical data
2. Examine the data and find the patterns and relationships.
3. Formulate conjectures that explain the patterns and relationships.
4. Test the conjectures by collecting additional data and check whether the new information fits or not
5. Devise an argument that conjectures are correct. All the steps are important in number theory and in mathematics. A scientific theory is an ability to predict the outcome of experiments. In mathematics one requires the step of a proof, that is, a logical sequence of assertions, starting from known facts and ending at the desired statement.

PiyushTheorem
PiyushTheorem: In a Right-Angled Triangle with sides in A.P. Series, the distance between the point of intersection of median & altitude at the base is 1/10 Th the sum of other two sides. This Theorem applies in Two Conditions:
1.The Triangle must be Right-Angled.
2.Its Sides are in A.P. Series.
1.Proof with Trigonometry
Tan α =AD/DC AD= DC Tan α —————–1 Tan α = AD/DE AD= DE Tan2 α —————-2 DC Tan α = DE Tan 2 α (DE+EC) Tan α = DE Tan 2 α DE Tan α + EC Tan α = DE Tan 2 α DE Tan α + EC Tan α = 2 DE Tan α / (1- Tan2 α ) DE Tan α – DE Tan3 α + EC Tan α –EC Tan3 α = 2DE Tan α EC Tan α –EC Tan3 α – DE Tan3 α = 2DE Tan α – DE Tan α Tan α (EC – EC Tan2 α – DE T an2 α )= DE Tan α DE Tan2 α – DE = EC Tan2 α – EC -DE ( Tan2 α + 1) = -EC (1 – Tan2 α ) DE (sin2 α /cos2 α + 1) = EC (1- sin2 α /cos2 α ) DE (sin2 α + cos2 α /cos2 α ) = EC (cos2 α – sin2 α /cos2 α ) DE (sin2 α + cos2 α ) = EC(cos2 α –sin2 α ) DE (sin2 α + cos2 α ) = EC (cos2 α –sin2 α ) where (sin2 α + cos2 α =1) & (cos2 α –sin2 α = cos2 α ) DE= EC cos2 α cos α =a/a+d & sin α = (a-d)/ (a +d) cos2 α = a2/ (a +b) 2 sin2 α = (a-d) 2/ (a+ d) 2 DE= EC (cos2 α – sin2 α ) = EC (a2 / (a +b) 2 – (a-d) 2/ (a +d) 2 = EC (a2 – (a-d) 2/ (a +d) 2 = EC (a –a +d) (a+ a-d)/ (a+ d) 2 = EC (d) (2a -d)/ (a+ d) 2 = (a +d)/2(d) (2a -d)/ (a +d) 2 ————- where EC= (a +d)/2 = (d) (2a -d)/2(a +d) = (d) (8d -d)/2(4d+d) ——————where a= 4d (as per the Theorem) = 7d2 /2(5d) = 7d /10 = (3d+4d)/10= (AB+AC)/10
2.Proof with Obtuse Triangle Theorem
AC2=EC2 +AE2 +2CE.DE where EC = ( a +d) /2,AE=( a +d)/2 a2 = (a +d/2)2 + (a+ d/2)2 + 2(a +d)/2DE = (a +d/2) (a+d+2DE) = (a +d/2) (a+d+2DE) where a=4d 16d2 = (5d/2) (5d+2DE) 32d/5 = 5d + 2DE 32d/5 – 5d = 2DE 32d -25d/5 = 2DE DE =7d/10 = (3d+4d)/10 = (AB+AC)/10
3.Proof with Acute Triangle Theorem
AB2= AC2+BC2 – 2BC.DC (a-d) 2= a2 + (a+ d) 2 -2(a+ d) (DE+EC) where AB= (a-d), AC=a, BC =( a +d) & EC= (a +d)/2 (a-d) 2 – (a +d)2 = a2 -2(a +d)(DE+EC) (a- d –a-d) (a -d +a +d) = a2 -2(a+ d) (2DE+a+d)/2 2(-2d) (2a) = 2a2 -2(a +d) (2DE+a+d) -8ad – 2a2 = -2(a +d) (2DE+a+d) -2a (4d +a) = -2(a +d) (2DE+a+d) a (4d + a) = (a +d)(2DE+a+d) 4d (4 d + 4d) = (4d+d) (2DE+4d+d) 4d (8d) = (5d) (2DE+5d) 32d2/5d = (2DE+5d) 32d/5 = (2DE+5d) 32d/5 – 5d = 2DE (32d – 25d)/5 = 2 DE DE = 7d/10 = (3d+4d)/10 = (AB+AC)/10
4. Proof with Co-ordinates Geometry
Equation of BE Y – 0 =b-0/0-a(X – a) Y = -b/a(X) + b——————- (1) M1 = -b/a For perpendicular M1M2= -1 So M2=a/b Equation of AC Y – 0 = a/b(X-0) Y=a/b(X) —————— (2) Put Y value in equation (1) a/b(X) + b/a(X) =b X (a2+b2/a b) = b X = ab2/ (a2 + b2) To get Value of Y, put X value in equation (2) Y = a/b (ab2/ (a2+b2) Y = a2b/ (a2+b2) Here we got co-ordinates of Point C – ab2/ (a2 + b2), a2b/ (a2+b2) and co-ordinates of point d is (a/2, b/2) because d is midpoint. As per the “Theorem” a=z-d, b=z, c = z+ d (z +d) 2= (z-d) 2+z2 from here z=4d so a=3d and b=4d Put value of a & b ab2/ (a2 + b2), a2b/ (a2+b2) & (a/2, b/2) ab2/ (a2 + b2) = 48d/25 a2b/ (a2+b2) = 36d/25 a/ 2=3d/2 b/ 2 =4d/2 CD2= (48d/25 -3d/2)2-(36d/25-4d/2)2 = (96d-75d/50)2 + (72d-100d/50)2 = (21d/50)2 + (-28d/50)2 = (441d2/2500) + (784d2/2500) = (1225d2/2500) CD= 35d/50 = 7d/10 = 7d/10 = (3d+4d)/10 = (AB+AE)/10
https://piyushtheorem.wordpress.com/2017/02/08/a-theorem-on-right-angled-triangles/

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