text:
User:Adamsaleel

Infinite product
$$\frac{A^{24}}{2^{\frac{2}{3}}\pi^2{e^2}}=\left(\frac{2}{1}\right)^{\frac{1}{2}} \left(\frac{2^2}{1\cdot3}\right)^{\frac{2}{2}}\left(\frac{2^3\cdot4}{1\cdot3^3}\right)^{\frac{3}{4}}\cdots$$

Euler constant
$$\gamma=\lim_{x\to\infty}\frac{xe^{\frac{\alpha\gamma}{x}}-e^{\frac{\beta\gamma}{x}}\Gamma\left(\frac{1}{x}\right)}{\alpha-\beta+1}$$

Special condition

$$\gamma=\lim_{x\to\infty}\frac{xe^{\frac{1}{x}}-e^{-\frac{1}{x}}\Gamma\left(\frac{1}{x}\right)}{3}$$

pi
\

$$\sqrt{\frac{\pi}{2}}=\lim_{n \to \infty}\left[(8n-1)(8n-3)\right]^{\frac{1}{4}}\prod_{k=1}^{2n-1} \left(\frac{1}{2k}\right) ^{(-1)^{k-1}}$$

$$\sqrt{\frac{\pi}{2}}=\lim_{n \to \infty}\frac{\left[(8n+1)(8n+3)\right]^{\frac{1}{4}}}{4n+1}\prod_{k=1}^{2n} \left(\frac{1}{2k}\right) ^{(-1)^{k-1}}$$

Others
$$\lim_{n \to \infty}{\sum_{k=2^n}^{2^{n+r}-2^{r-1}}\left(\frac{1}{k}\right)-\frac{1}{2^{n+r+1}-1}}=r\ln2$$

$$\lim_{n \to \infty}\sum_{k=3^n}^{3^{n+1}-2}{\frac{1}{k}}=\ln3$$

$$\lim_{n \to \infty}\sum_{k=(2i+1)^n}^{(2i+1)^{n+1}-(i+1)}{\frac{1}{k}}=\ln(2i+1)$$

$$\frac{\pi^2}{6}=\sum_{n=1}^{\infty}\frac{1}{\sum_{i=1}^{2n-1}(-1)^{i-1}\left(2n+1-i\right)^2i^3}$$

$$=\sum_{n=1}^{\infty}\frac{1}{\sum_{i=1}^{2n-1}(-1)^{i-1}\left(2n+1-i\right)i^2}$$

$$\lim_{n \to \infty}\left[2+\frac{2}{3}+\frac{2}{5}+\cdots+\frac{2}{2^n-1}+\ln\left(\frac{1}{2^{n+1}-1}\right)-\frac{1}{2^{n+1}-1}\right]=\gamma$$

Special case aln3
$$\lim_{n \to \infty}\sum_{k=3^n}^{3^{n+1}-2}{\frac{1}{k}}=1\ln3$$

$$\lim_{n \to \infty}\sum_{k=3^n}^{3^{n+2}-[2+3]}{\frac{1}{k}}=2\ln3$$

$$\lim_{n \to \infty}\sum_{k=3^n}^{3^{n+3}-[2+3+3^2]}{\frac{1}{k}}=3\ln3$$

$$\lim_{n \to \infty}\sum_{k=3^n}^{3^{n+4}-[2+3+3^2+3^3]}{\frac{1}{k}}=4\ln3$$

$$\lim_{n \to \infty}\sum_{k=3^n}^{3^{n+a}-\left(2+\sum_{i=1}^{a-1}3^i\right)}{\frac{1}{k}}=a\ln3$$

General case aln(2i+1)
$$\lim_{n \to \infty}\sum_{k=(2i+1)^n}^{(2i+1)^{n+a}-\left(i+1+i\sum_{j=1}^{a-1}(2i+1)^j\right)}{\frac{1}{k}}=a\ln(2i+1)$$

General case ln(2i)
$$\sum_{k=(2i)^n}^{(2i)^{n+1}-(i+1)}\frac{1}{k}=\ln(2i)$$

$$\sum_{k=(2)^n}^{(2)^{n+a}-(1+2^{a-1})}\frac{1}{k}=a\ln(2)$$

$$\sum_{k=(4)^n}^{(4)^{n+a}-(1+2^{2a-1})}\frac{1}{k}=a\ln(4)$$

Continued fractions
$$\frac{e+1}{e-1}-{\frac{2}{e-1}}\left(\sum_{j=1}^{2n-1}(-1)^{j+1}\sqrt[2n]{e^j}\right)=\frac{1}{4n+\frac{1}{12n+\frac{1}{20n+\frac{1}{28n+\cdots}}}}$$

Let n = 1, yields

$$\frac{e+1}{e-1}-{\frac{2\sqrt{e}}{e-1}}=\frac{1}{4+\frac{1}{12+\frac{1}{20+\frac{1}{28+\cdots}}}}$$

Radical 2 odd
$$\frac{a-\sqrt[2k-1]{a}}{\sqrt[2k-1]{a^2}-1}=\sqrt[2k-1]a+\sqrt[2k-1]{a^3}+\sqrt[2k-1]{a^5}+\cdots+\sqrt[2k-1]{a^{2k-3}}$$

$$\frac{a\sqrt[2k-1]{a}-\sqrt[2k-1]{a^2}}{\sqrt[2k-1]{a^2}-1}=\sqrt[2k-1]{a^2}+\sqrt[2k-1]{a^4}+\sqrt[2k-1]{a^6}+\cdots+\sqrt[2k-1]{a^{2k-2}}$$

$$\frac{a+(a-1)\sqrt[2k-1]{a}-\sqrt[2k-1]{a^2}}{\sqrt[2k-1]{a^2}-1}=\sum_{j=1}^{2k-2}{\sqrt[2k-1]{a^j}}$$

$$\frac{a-(a+1)\sqrt[2k-1]{a}+\sqrt[2k-1]{a^2}}{\sqrt[2k-1]{a^2}-1}=\sum_{j=1}^{2k-2}(-1)^{j+1}{\sqrt[2k-1]{a^j}}$$

Radical 3 even
$$\frac{a-\sqrt[2n]{a^2}}{\sqrt[2n]{a^2}-1}=\sqrt[2n]{a^2}+\sqrt[2n]{a^4}+\sqrt[2n]{a^6}+\cdots+\sqrt[2n]{a^{2n-2}}$$

$$\frac{(a-1)\sqrt[2n]{a}}{\sqrt[2n]{a^2}-1}=\sqrt[2n]{a}+\sqrt[2n]{a^3}+\sqrt[2n]{a^5}+\cdots+\sqrt[2n]{a^{2n-1}}$$

Radical 1 (mixed)
$$\frac{1}{2}{\times}\frac{\sqrt[5]2+1}{\sqrt[5]2-1}-\frac{5}{2}=\sqrt[5]4+\left(\sqrt[5]{\frac{1}{5}}+\sqrt[5]{\frac{4}{5}}\right)^{\frac{1}{2}}{\times\sqrt[5]{125}}$$

$$\frac{1}{2}{\times}\frac{\sqrt[3]2+1}{\sqrt[3]2-1}-\frac{1}{2}=2\sqrt[3]2+\left(4\sqrt[3]{\frac{2}{3}}+5\sqrt[3]{\frac{1}{3}}\right)^{\frac{1}{8}}{\times\sqrt[3]{9}}$$

$$\frac{1}{2}{\times}\frac{\sqrt[3]2+1}{\sqrt[3]2-1}-\frac{1}{2}=\left(\sqrt[3]{\frac{1}{9}}+\sqrt[3]{\frac{4}{9}}+\sqrt[3]{\frac{16}{9}}\right)^{\frac{3}{2}}$$

$$X=\frac{1}{4}\left(1-\frac{\sqrt[3]2+1}{\sqrt[3]2-1}\right)+\frac{2}{\sqrt[3]2}=-\frac{16813}{500000}$$; Rational number!!!

$$Y=\left(\frac{9}{2\sqrt[3]4}-\frac{9}{4}\right)^{\frac{1}{3}}=\frac{41813}{500000}$$; Rational number!!!

$$X+Y=\frac{1}{2}$$

$$\left(1-\frac{\sqrt[3]2-1}{\sqrt[3]2+1}\right)^3=\frac{8}{3}\left(\sqrt[3]2-1\right)$$

$$\frac{a-1}{2}\left(\frac{\sqrt[3]a+1}{\sqrt[3]a-1}-1\right)=2\sqrt[3]a+\left[\left(a^2-7a+1\right)+\left(6a-3\right)\sqrt[3]a+\left(6-3a\right)\sqrt[3]{a^2}\right]^{\frac{1}{3}}$$

$$\frac{a-1}{2}\left(\frac{\sqrt[3]a+1}{\sqrt[3]a-1}-1\right)=2+2\sqrt[3]{a^2}-\left[\left(a^2-7a+1\right)+\left(6a-3\right)\sqrt[3]a+\left(6-3a\right)\sqrt[3]{a^2}\right]^{\frac{1}{3}}$$

$$\frac{a+1}{2}\left(1-\frac{\sqrt[3]a-1}{\sqrt[3]a+1}\right)=\left[\left(a^2-7a+1\right)+\left(6a-3\right)\sqrt[3]a+\left(6-3a\right)\sqrt[3]{a^2}\right]^{\frac{1}{3}}$$

$$\frac{a-1}{4}\left(\frac{\sqrt[3]a+1}{\sqrt[3]a-1}-1\right)-\frac{a+1}{4}\left(1-\frac{\sqrt[3]a-1}{\sqrt[3]a+1}\right)=\sqrt[3]a$$

$$\frac{a-1}{4}\left(\frac{\sqrt[3]a+1}{\sqrt[3]a-1}-1\right)+\frac{a+1}{4}\left(1-\frac{\sqrt[3]a-1}{\sqrt[3]a+1}\right)=1+\sqrt[3]{a^2}$$

Ramanujan
General of ramanujan's nested radical

Stage:1

$$\frac{\sqrt[n]k+1}{\sqrt[n]k-1}=\frac{2}{k-1}\left(\frac{k+1}{2}+\sqrt[n]{k}+\sqrt[n]{k^2}+\sqrt[n]{k^3}+\cdots+\sqrt[n]{k^{n-1}}\right)$$

$$\frac{\sqrt[2n]k-1}{\sqrt[2n]k+1}=\frac{2}{k-1}\left(\frac{k+1}{2}-\sqrt[2n]{k}+\sqrt[2n]{k^2}-\sqrt[2n]{k^3}+\cdots+\sqrt[2n]{k^{2n-1}}\right)$$

$$\frac{\sqrt[2n-1]k-1}{\sqrt[2n-1]k+1}=\frac{2}{k+1}\left(\frac{k-1}{2}+\sqrt[2n-1]{k}-\sqrt[2n-1]{k^2}+\sqrt[2n-1]{k^3}+\cdots+\sqrt[2n-1]{k^{2n-2}}\right)$$

Sum

$$\frac{\sqrt[n]k+1}{\sqrt[n]k-1}=\frac{2}{k-1}\left(\frac{k+1}{2}+\sum_{j=1}^{n-1}\sqrt[n]{k^j}\right)$$

$$\frac{\sqrt[2n]k-1}{\sqrt[2n]k+1}=\frac{2}{k-1}\left(\frac{k+1}{2}+\sum_{j=1}^{2n-1}(-1)^{2j-1}\sqrt[2n]{k^j}\right)$$

k = 5

Yields ramanujan's equation

$$\frac{\sqrt[4]5+1}{\sqrt[4]5-1}=\frac{1}{2}\left(3+\sqrt[4]{5}+\sqrt[4]{5^2}+\sqrt[4]{5^3}\right)$$

Fibonacci
$$a^2+b^2=c^2$$

$$a^4+b^4=2\left(c^2+ab\right)^2-\left(a+b\right)^4$$

$$a^4+b^4=\left(a+b\right)^4-2ab\left(2c^2+3ab\right)$$

$$\left(a+b\right)^2=ab+\frac{a^4+b^4+\left(a+b\right)^4}{a^2+b^2+\left(a+b\right)^2}$$

$$2\left(a+b\right)^2=ab+\frac{\left(a+b\right)^4-\left(a^4+b^4\right)}{\left(a+b\right)^2-\left(a^2+b^2\right)}$$

$$G_n=G_{n-1}+G_{n-2}$$

G1 = i and G2 = j

i & j ɛ {1,2,3,4,...}

$$G_1+G_2+G_3+\cdots+G_n=G_nG_{n+1}+G_1(G_1-G_2)$$

Formulae

$$\frac{F_n^4+F_{n+1}^4+F_{n+2}^4}{2}=\left(F_n^2+F_{n+1}F_{n+2}\right)^2$$

$$\sum_{i=1}^{k}F_i^4+\sum_{i=1}^{k}F_{i+1}^4+\sum_{i=1}^{k}F_{i+2}^4=2\left(F_{i-1}F_{i}+\sum_{i=1}^{k}F_iF_{i+1}\right)^2$$

Assessement for years 1 - 2
Q1.

Add

4 1 8       9 2 5 4         3 6 0 4   + 2 3 9      + 6 4 3 4       + 1 2 0 3   ---      -       -

Q2.

Subtract

4 1 8       9 9 5 4         3 0 4 2   - 1 1 8      - 6 4 3 4       - 1 0 2 3   ---      -       -

Q3.

Mulitply

2 × 4 =

5 × 6 =

2 × 4 × 1 =

1 × 1 =

3 × 0 =

Q4.

Division

6 ÷ 2 =

10 ÷ 5 =

8 ÷ 8 =

Q5.

Put the numbers in order of smallest size first.

a. 6, 1, 0, 3

b. 18, 10, 15, 11

Q6.

Shade three portions of the shape

a. _______ | | | | |  | | | | |  |_|_|_|_| Shade 8 portions of the shape

b. _______________ |_|_|_|_|_|_|_|_|  |_|_|_|_|_|_|_|_|

Assessement for years 7 - 8
Q1. a. Find the sum of -10 and 20

b. Find the product of zero and six

c. Find the difference of -2 and 3

Q2.

Work out the following a. - 2 + 8 =       b. - 5 + 5 =        c. - 5 + 0 = d. - 2 - 8 =       e. - 1 + 2 - 1 + 3 - 3 = f. - 2 × 8 =       g. - 2 × (-4) =     h. 12 ÷ (-3) =

Q3.

Work out the following

a. $$\frac{1}{4}+\frac{2}{5}$$

b. $$\frac{3}{4}-\frac{2}{5}$$

c. $$\frac{2}{3}\times{\frac{1}{3}}$$

d. $$\frac{1}{4}\div{\frac{2}{5}}$$

Q4.

Write the following

a. Four square numbers greater than 1

b. Three cube numbers less than 100

c. Five prime numbers between 4 and 20

Q5.

Work out the following

a. 23

b. 50

c. $$\sqrt{9}$$

d. 2-3

e. $$4^{\frac{1}{2}}$$

f. $$\sqrt[3]{8}$$

g. $$8^{\frac{2}{3}}$$

Q6.

Find the answer

a. 15 - 9 ÷ 3 =

b. 2 ÷ 2 - 2 =

c. (16-6) ÷ 2 + 5

Q7.

Work out the following

a. 0.4 × 0.6

b. 12 ÷ 2.4 =

c. 12.0054 + 0.00542

Q8.

£20 is divided in the ratios 2 : 3 : 5

a. Work out the amount of the smallest ratio

b. Find how much more is the largest amount from the smallest amount

Q9.

a. What is 20% of £600.00?

Q10.

Draw the following triangles and label with its appropriate symbols

a. An isoscle triangle

b. An equlateral triangle

c. A scalene triangle

d. A right angle triangle

Q11. .    | .  8m |. 10m |    .     |_______.        6m

a. Find the area of the triangle

b. Find the perimter of the triangle

Q12.

Simplify the following

a. 5y + 5x - y

b. $$2x\times4y$$

Q13.

Expand and simplify

a. $$2(3x+3)+4$$

b. $$4(4y+1)+2(2y+3)$$

Q14.

Solve the for x

a. x - 8 = 1

b. 2x + 4 = 10

c. x2 = 16

Q15.

a. True or false 3 > 4

b. True or false 3 ≤ 3

Q16. e   / 60 ________ /_________  →               /  f    / ____/______________ →    /

a. Find angles e and f

Q17.

y = 2x - 1

a. Fill in the table

b. Plot the graph y ↑ |    8-|       |     6-|       |     4-|       |     2-|    ___|__________________________________→       |   1   2   3   4   5   6   7      x

Q18.

Shoe sizes 5, 2, 2, 1, 3, 4

a. Find the mean size

b. Find the median size

c. Find the modal size

d. Find the range of size

Q19.

John tosses a coin

a. Calculate the probability that it lnads on a head first

Q20.

The probability Henry comes to school late on monday morning is 65%

a. Find the probability that Henry comes to school on monday early.

Q21.

John did a survey, he wants to find out how many people like the following colours

Here is a table and graph showing John's survey

People ↑ 11-|        |      10-|                             |--|          |                             |  |       8-|                             |  |         |                             |  |       6-|        |--|                 |  |         |        |  |                 |  |       4-|  |--|  |  |                 |  |         |  |  |  |  |                 |  |       2-|  |  |  |  |                 |  |    _____|__|__|__|__|_________________|__|______________  →         |  Red   Blue  Green  Purple  Yellow     Colours

a. Fill in the table and the graph.

b. How many more people like yellow colours than red colours

c. How many people are there in John's survey

Assessement for years 9 - 10
Q1.

Work out

a. 31.2 × 0.12

Q2. Computer _______________     |               |       _________      |               |      |         |      |               |      |  Offer  | |              |      | 20% Off | |_______________|     |_________|         __________         \_- - - _/

Price £2400.00

a. Work out the sale price

Q3.

a. Change 0.12 into fraction with the lowest term

Q4.

Evaluate

a. 90        b. 9-1        c. 92       d. $$\sqrt9$$

e. $$\sqrt[3]8$$       f. 271/3

Q5.

Work out

a. $$1\frac{4}{5}-1\frac{2}{5}$$

b. $$\frac{1}{5}+\frac{2}{3}$$

c. $$1\frac{1}{2}\times{\frac{2}{3}}$$

Q6.

Work out

a. $$\left(3\times4-4\div2\right)^2+1$$

Q7.

Simplify

a. 6x + y -5x + 3y

b. 6x2 - 5 + 3x2 + x + 2

Q8.

Expand and simplify

a. $$2(5x-1)+3(x+1)$$

b. $$(2x+3)^2$$

Q9.

Factorise the following

a. x - xy

b. 2x2 - 4x

Q10.

Solve for x

a. 5x - 2 = 8

b. -3x - 5 = -5x + 5

c. $$\frac{x}{3}-1 = 2$$

Q11.

List all the possible values of x

a. - 1 < x < 3

Q12.

Solve the following inequalities

a. 3y - 5 > 4

Q13.

Solve the simultaneous equation

x - y = 3 2x + y = 12

Q14.

The formula c2 = a2 + b2

Given a = 3m and b = 4m

a. Find the value of c

Q15.

Rearrange the formula y = mx + c to make x the subject of the formula

Q16.

Triangle ABC is an isoscle B. .  .             .       .           .           .         .               . ___ X        .___________________.__|____ A                    C Angle A =  30°

a. Find angles X and B

Q17. 6m ________                 |        |              |        | 4m |       |          10m |        |_____ |             |              |              |              |______________|

8m

a. Calculate the the area of the L shape

b. Calculate the perimeter of the L shape

Q18.

a. Name the follwing shapes.

a                                             _____________ .-.                     .                 |             |       .     |     .                    .|.                |             |      .      |h. . | .            a |             | .      |       .                .  |h. |            |    .________|________.              .___|___.             |_____________|             b                           b                        b

--                           --

b. Write down the formulae for each shape to work out the area.

a                                             _____________ .-.                     .                 |             |       .     |     .                    .|.                |             |      .      |h. . | .            a |             | .      |       .                .  |h. |            |    .________|________.              .___|___.             |_____________|             b                           b                        b

Area =                        Area =                   Area =

Q19.

A circle has a radius of 4mm

a. Calculate the area of the circle

b. Calculate the circumference of the circle

Q20.

Here is a list of the size of a quantity

10, 6, 8, 4, 2

a. Calculate the mean size

b. Calculate the median size

c. Calculate the range of the size

Q21.

a. Calculate the mean weight

b. Find the modal weight

Q22.

A bag contain 5 green, 4 blue and 2 black marbles

a. Calculate the probability that a black marble is chosen at random.

b. Calculate the probability that a green and a blue marbles chosen at random

Assessement for years 11
Q1.

Fig.1 below is a right angle triangle

a. Find lenght c. Round the answer to 3 d.p. |. 5m |. c     |. |___.      3m Fig.1

b. Work out the area of this triangle

Q2.

Work out the following

a. $$2\frac{2}{5}+\frac{2}{3}$$

b. $$2\frac{2}{5}\times{1\frac{2}{3}}$$

Q3.

Expand and simplify

a. $$\left(x-2\right)^2-x-2$$

b. $$2\left(2x-1\right)-2x\left(x-1\right)$$

Q4.

Factorise the following

a. $$x^2-2x+1$$

b. $$x^2-1$$

c. $$x\left(x-2\right)+2\left(x-2\right)$$

Q5.

Find the HCF and LCM of 24, 32 and 42

Q6.

Add

a. $$\frac{2}{x+1}+\frac{3}{x-1}$$

Solve for x

b. $$\frac{2}{x+1}+\frac{3}{x-1}=2$$

Q7.

£34.00 travel fair has increased by 23% from last year. Work out the price before the increase.

Q8.

The sequence 3, 7, 11, 15, ...

a. Find the nth of the sequence

b. Work out the 25th of the sequence

c. Show that 50 is not a term in the sequence

Q9.

Diagrams below are mathematically similiar .       |.        | .                 .        27m |. |.        |   .             y |. |   .              |  .        |_____.             |___.          15cm               5m

a. Find the lenght of y

Q10.

Three children of David, Bob, Steve and John shared a sum in the ratios 2 : 5 : 8

Steve received £450 from the sum.

a. Calculate the sum of the sum money shared by the children.

b. How much more did John get more than Bob?

c. Calculate the amount of John in percentage.

Q11.

Work out and write the answer in standard form

a. $$\left(12.4\times10^4\right)+\left(15\times10^3\right)$$

Q12.

a. Change 360km/h into m/s

b. 4m2 into mm2

Q13.

Solve the following simultaneous equations 3x - 2y = 5 2x + 4y = 14

Q14.

Rearrange the formula to make angle B the subject.

a. $$A=\frac{1}{2}acsinB$$

b. Evaluate the angle B, given A = 10, a = 5 and c = 8

Q15.

Solve the equtions to fin the value for x

a. 2x - 5 = 15

b. -3x - 4 = -6x + 2

c. -3(x - 1) + 2(2x + 3) = 10

d. $$\frac{2}{3}+\frac{2x + 1}{6}=7$$

Q16.

Below is a straight line graph

Y     ↑ |  12-|            .      |              10-|         . B(x2,y2) |           8-|      .            |    6-|   .  A(x1,y1) |   4-|      |    2-|         |    0-|---1---2---3---4---5---6---7---8---→ X      |

a. Write down the co-ordinates of points A and B

b. Find the equation of this straight graph

Q17.

Simplify the following indices

a. $$3^{-5}\times3^{12}$$

b. $$3^{2}\div3^{-2}$$

c. $$(2^{-2}x^{-3}y^2)^{-2}$$

d. $$\frac{x^4\times{x^{-2}}}{x^{-4}\times{x^2}}$$

Q18.

Evaluate the following

a. 40

b. 4-1

c. 41/2

d. 43/2

e. 82/3

f. $$\left(4^{\frac{3}{2}}\right)^2$$

Q19.

Solve the following inequalities

a. 2x - 3 ≥ 5

b. 5x + 3 > 7x + 5

Q20.

List all the possible values of x in this inequality

a. -2 < x ≤ 1

Q21.

Expand and simplify the following

a. $$\sqrt{3}\left(\sqrt{3}-1\right)+2\left(\sqrt{3}+1\right)$$

b. $$\left(\sqrt{2}+1\right)\left(2\sqrt{2}-1\right)$$

Q22.

Use the quadratic formula to solve the quadratic equation, round the answers to 2 s.f.

Quadratic formula

$$x=\frac{-b\pm{\sqrt{b^2-4ac}}}{2a}$$

a. 2x2 - 4x + 1 = 0

Q23.

Below is a right angle triangle

A                  .| . |               .  |            c. | b                 .    | .    |            .______|           B   a    C

Given angle B = 30°, lenght b = 10m.

a. Find lenght a, give the answer to 2 s.f

Q24.

Steve invested £2000.00 in a saving account at a compound interest of 1.5% per year.

a. Work out the interest he earns after 10 years period.

Q25.

The rectangle below has an area of 35m2

(2x-1)m _________    |         |     |         |     |         | (2x+1)m |        |     |_________|

a. Show that 4x2 - 36 = 0

b. Work out the lenght and height of the rectangle

Q26.

In a bag there are 6 colur sweets. Three are reds and the rest are blues

a. Calculate the probability that Steve eats both red sweets.

Q27.

Below is table shows record time in seconds for an event

a. Find the mean time for the event

Q28.

Below is a triangular shape

.|                   . | 4m . |               .| .___|              . |    /             .  |  /  10m .___|/            3m

a. Calculate the surface area

b. Calculate the volume

Q29.

Estimate

a. $$\frac{5.9\times4.01}{11.98}$$

Q30.

Use a calculator, work out and write down all the digits on your calculator

a. $$\frac{\sqrt3-\sqrt2}{\sqrt5}$$

Q31.

y = 2x - 3 and y1 = 7 - 3x1

a. Fill in the table and plot the eqautions on the graph below

↑   y | |  10-|      |    8-|      |    6-|      |    4-|      |    2-|    ____|_____________________________→ -1   |   1   2   3   4   5   6    x   -2 | |  -4 |      |   -6 |      |

title: User:Adamsaleel


text:
User:SamHB/MVCalc2

This is the second lecture in multivariable calculus.

=Submanifolds=

In this lecture we will discuss lower-dimensional regions embedded in a higher-dimensional space, such as lines within 2- or 3-dimensional space, or surfaces within 3-dimensional space. The proper name for the "spaces" we have been discussing is manifold, which means "a set that can have coordinate systems".

The points in an embedded submanifold are described by sets of numbers, commonly called "parameters", though we will just call them coordinates. The assignment of numbers to the points is commonly called "parameterization", though they are really just coordinates. For example, a surface in 3-dimensional space might be described by two parameters or coordinates $$u\,$$ and $$v\,$$, so that the surface is described by functions:
 * $$x(u, v) \ \ \ y(u, v)\ \ \ z(u, v)\,$$

Similarly, a particle's path through space might be "parameterized by time", so that its position at any instant is given by 3 functions:
 * $$x(t) \ \ \ y(t)\ \ \ z(t)\,$$

We will treat the subject of describing submanifolds simply by choosing a coordinate system for the larger space, and then removing some coordinates from the system by setting them to constants. Once we have developed the formulas for working with submanifolds, we will be able to dispense with the extra coordinates.

For example, a circle of radius $$R\,$$, in a 2-dimensional plane, can be described by choosing polar coordinates $$(r, \theta)\,$$ on the plane, centered at the circle's center. Set the $$r\,$$ coordinate to a constant: $$r = R\,$$. It drops out of the coordinate system, leaving $$\theta\,$$ as the single coordinate.

The surface of a sphere, within 3-dimensional space, can be described by choosing spherical coordinates $$(r, \theta, \phi)\,$$, and setting $$r = R\,$$. The coordinates of the surface are $$(\theta, \phi)\,$$. An ellipse can be defined by choosing confocal coordinates and setting $$\rho\,$$ to some constant.

Out With the Jacobian, In With the Metric
There is some unfinished business from lecture 1 that needs to be attended to. We defined the Jacobian matrix in terms of the transformation between two coordinate systems, and we came close to considering the Jacobian between Cartesian coordinates and some coordinate system $$(u, v, w)\,$$ as "the Jacobian of the system $$(u, v, w)\,$$". We stopped short of doing that because it isn't true. The usual Jacobian $$[J\langle xyz/r\theta\phi\rangle]\,$$ for spherical coordinates is:
 * $$J_1\langle xyz/r\theta\phi\rangle = \begin{bmatrix} \sin\theta \cos\phi & r \cos\theta \cos\phi & - r \sin\theta \sin\phi \\ \sin\theta \sin\phi & r \cos\theta \sin\phi & r \sin\theta \cos\phi \\ \cos\theta & - r \sin\theta & 0 \end{bmatrix}$$

But a different Cartesian system, created, for example, by permuting $$x\,$$, $$y\,$$, and $$z\,$$, would have:
 * $$x = r \cos\theta\,$$
 * $$y = r \sin\theta \cos\phi\,$$
 * $$z = r \sin\theta \sin\phi\,$$

with a Jacobian matrix of:
 * $$J_2\langle xyz/r\theta\phi\rangle = \begin{bmatrix} \cos\theta & - r \sin\theta & 0 \\ \sin\theta \cos\phi & r \cos\theta \cos\phi & - r \sin\theta \sin\phi \\ \sin\theta \sin\phi & r \cos\theta \sin\phi & r \sin\theta \cos\phi \end{bmatrix}$$

There is no "Jacobian matrix of spherical coordinates $$(r, \theta, \phi)\,$$". But, fortunately for integration, there is a "Jacobian determinant of $$(r, \theta, \phi)\,$$". The determinants $$\mathbf{J_1}\,$$ and $$\mathbf{J_2}\,$$ are both $$r^2 \sin\theta\,$$.

This explains why we were able to pretend that a given coordinate system had a "Jacobian" when we were developing the integration formulas—we only used the determinant. If a coordinate system $$(u, v, w)\,$$ has two Jacobian matrices $$[J_1]\,$$ and $$[J_2]\,$$ for its transformations to two Cartesian coordinate systems, we have
 * $$[J_1] = [C][J_2]\,$$

where $$[C]\,$$ is the orthogonal matrix connecting the two Cartesian systems.

Now the determinant of an orthogonal matrix is $$\pm 1\,$$.
 * Why?   $$\ \ \ \ C^t = C^{-1}\,$$, so $$det(C) = det(C^t) = det(C^{-1}) = \frac{1}{det(C)}\,$$

Since we are being careful to maintain orientation, the determinant is always $$+1\,$$.
 * Some textbooks don't bother with that, and have to take the absolute value of the Jacobian determinant in integration formulas.

This means that $$det(J_1) = det(J_2)\,$$, so the integrations formulas work.

But taking the determinant of $$\mathbf{J}\,$$ throws away too much information. Is there some manipulation that we can perform that preserves more information but is still impervious to the choice of the "reference" Cartesian system?

This is sometimes called the "metric tensor" (for reasons that we definitely won't go into), or just the "metric".

The metric matrix, like the Jacobian, is referenced to two coordinate systems: $$[g\langle xyz/uvw\rangle]\,$$, though we will soon see that we don't need to name the Cartesian system.

Like the Jacobian, its determinant is indicated by a boldface lowercase $$\mathbf{g}\,$$ or $$\mathbf{g}\langle xyz/uvw\rangle\,$$.

Theorem: The metric matrix between a given coordinate system and some "reference" Cartesian system is independent of that Cartesian system.

Proof: The Jacobians relating the given "test" system to the two "reference" Cartesian systems are $$[J_1]\,$$ and $$[J_2]\,$$. We showed above (in a "digression") that $$[J_1] = [C][J_2]\,$$ where $$[C]\,$$ is an orthogonal matrix.

So
 * $$[g_1] = [J_1]^t [J_1] = ([C][J_2])^t ([C][J_2]) = [J_2]^t [C]^t [C] [J_2] = [J_2]^t [C]^{-1} [C] [J_2] = [J_2]^t [J_2] = [g_2]\,$$

This means that we really are justified in calling some matrix $$[g]\,$$ the "metric of the coordinate system", as soon as we can calculate it relative to any Cartesian coordinate system. And we no longer need to state what the "reference" system is. $$[g\langle uvw\rangle]\,$$ means the same as $$[g\langle xyz/uvw\rangle]\,$$.

Theorem: $$[g]\,$$ is a symmetric matrix.

Proof:
 * $$[g]^t = ([J]^t [J])^t = [J]^t {[J]^t}^t = [J]^t [J] = [g]\,$$

Theorem: $$\mathbf{J} = \sqrt{\mathbf{g}}\,$$

Proof: $$\det[g] = \det([J]^t [J]) = \det[J]^t \det[J] = (\det[J])^2\,$$

The square root of the metric determinant is much more commonly used than the determinant itself.

We can now reformulate the integration formula from the previous lecture:

Or, letting the $$q_i\,$$ be some Cartesian system, we get the formula for integrating in arbitrary coordinates relative to the metric:

And similarly in other dimensions.

Here are some metrics:

2-dimensional polar coordinates:
 * $$[g] = \begin{bmatrix} 1 & 0 \\[2.5ex] 0 & r^2 \end{bmatrix}\ \ \ \ \mathbf{g} = r^2\ \ \ \ \sqrt{\mathbf{g}} = r\,$$

3-dimensional spherical coordinates:
 * $$[g] = \begin{bmatrix} 1 & 0 & 0 \\[2.5ex] 0 & r^2 & 0 \\[2.5ex] 0 & 0 & r^2 \sin^2\theta \end{bmatrix}\ \ \ \ \mathbf{g} = r^4 \sin^2\theta\ \ \ \ \sqrt{\mathbf{g}} = r^2 \sin\theta\,$$

Toroidal coordinates:
 * $$[g] = \begin{bmatrix} 1 & 0 & 0 \\[2.5ex] 0 & (S + r \cos\phi)^2 & 0 \\[2.5ex] 0 & 0 & r^2 \end{bmatrix}\ \ \ \ \mathbf{g} = r^2 (S + r \cos\phi)^2\ \ \ \ \sqrt{\mathbf{g}} = r (S + r \cos\phi)\,$$

Confocal coordinates:
 * $$[g] = \begin{bmatrix} A^2 (\sinh^2\rho + \sin^2\theta) & 0 \\[2.5ex] 0 & A^2 (\sinh^2\rho + \sin^2\theta) \end{bmatrix}\ \ \ \ \mathbf{g} = A^4 (\sinh^2\rho + \sin^2\theta)^2\ \ \ \ \sqrt{\mathbf{g}} = A^2 (\sinh^2\rho + \sin^2\theta)\,$$

Submanifolds
Now that the preliminaries are out of the way, we can get to the subject of this lecture.

A submanifold is the result of setting some coordinates, in some coordinate system of the larger manifold, equal to some constants. The remaining coordinates are the coordinates of the submanifold.

A surface embedded in 3-dimensional space is the result of holding one coordinate constant. If we hold one coordinate constant in 3-dimensional Cartesian coordinates, we get the plane determined by the other 2 coordinates. But we can get more interesting surfaces by using more interesting coordinate systems. And we can get curves in 2- or 3-dimensional spaces by holding constant all cordinates except one. Here are some examples:

 * If the original coordinate system was spherical coordinates, the result of holding $$r = R\,$$ is the surface of a sphere of radius $$R\,$$.

 * A toroidal surface can be the result of holding $$r = R\,$$ in toroidal coordinates $$(r, \theta, \phi)\,$$. Its coordinates are $$\theta\,$$ and $$\phi\,$$.

 * A circle in the plane can be the result of holding $$r = R\,$$ in polar coordinates.

 * A helix in 3 dimensions can be made by introducing "helical coordinates" $$(r, \theta, s)\,$$:
 * $$x = r \cos\theta\,$$
 * $$y = r \sin\theta\,$$
 * $$z = P \theta + s\,$$
 * One then sets $$r = R\,$$ and $$s = 0\,$$. The result is a 1-dimensional manifold, that is, a curve, with coordinate $$\theta\,$$.  $$P\,$$ is the "pitch" of the helix, per radian.

 * Since we will need these soon, here are the Jacobian and metric for helical coordinates:
 * $$[J\langle xyz/r\theta s\rangle] = \begin{bmatrix} \cos\theta & - r \sin\theta & 0 \\[2.5ex] \sin\theta & r \cos\theta & 0 \\[2.5ex] 0 & P & 1 \end{bmatrix}$$

 * $$[g\langle r\theta s\rangle] = \begin{bmatrix} 1 & 0 & 0 \\[2.5ex] 0 & r^2 + P^2 & P \\[2.5ex] 0 & P & 1 \end{bmatrix}\ \ \ \ \mathbf{g} = r^2\ \ \ \ \sqrt{\mathbf{g}} = r\,$$

Does This Really Give Us Freedom to Choose the Coordinates on the Submanifold?
Yes, as long as we have freedom to choose any coordinates on the main manifold. And we do have that freedom—that's why we did not restrict ourselves to orthogonal coordinate systems.

Here is an example. Suppose we are interested in a surface which is a hemisphere of unit radius, centered at the origin, and above the $$x-y\,$$ plane. But instead of the obvious $$(\theta, \phi)\,$$ coordinates, we want to express a point's position in terms of its $$x\,$$ and $$y\,$$ coordinates, that is, the coordinates of the point on the $$x-y\,$$ plane directly below it. We create an artificial coordinate system $$(u, v, w)\,$$:
 * $$x = u\qquad\qquad\qquad\ \ \ \ u = x\,$$
 * $$y = v\qquad\qquad\qquad\ \ \ \ v = y\,$$
 * $$z = \sqrt{w^2 - u^2 - v^2}\ \ \ \ w = \sqrt{x^2 + y^2 + z^2}\,$$

This system is obviously nonstandard. We calculate:
 * $$[J\langle xyz/uvw\rangle] = \begin{bmatrix} 1 & 0 & 0 \\[2.5ex] 0 & 1 & 0 \\[2.5ex] \displaystyle\frac{- u}{\sqrt{w^2 - u^2 - v^2}} & \displaystyle\frac{- v}{\sqrt{w^2 - u^2 - v^2}} & \displaystyle\frac{w}{\sqrt{w^2 - u^2 - v^2}} \end{bmatrix}$$

 * $$[g\langle uvw\rangle] = \begin{bmatrix} \displaystyle\frac{w^2 - v^2}{w^2 - u^2 - v^2} & \displaystyle\frac{uv}{w^2 - u^2 - v^2} & \displaystyle\frac{- uw}{w^2 - u^2 - v^2} \\[2.5ex] \displaystyle\frac{uv}{w^2 - u^2 - v^2} & \displaystyle\frac{w^2 - u^2}{w^2 - u^2 - v^2} & \displaystyle\frac{- vw}{w^2 - u^2 - v^2} \\[2.5ex] \displaystyle\frac{- uw}{w^2 - u^2 - v^2} & \displaystyle\frac{- vw}{w^2 - u^2 - v^2} & \displaystyle\frac{w^2}{w^2 - u^2 - v^2} \end{bmatrix}$$

 * $$\mathbf{g} = \displaystyle\frac{w^2}{w^2 - u^2 - v^2}\ \ \ \ \sqrt{\mathbf{g}} = \displaystyle\frac{w}{\sqrt{w^2 - u^2 - v^2}}\,$$

This is a very peculiar coordinate system. But it might be just what we need if we had to find the electric charge on a hemispherical object in which the charge density was given as a function of $$x\,$$ and $$y\,$$.

Now we set $$w = 1\,$$. We have:
 * $$x = u\,$$
 * $$y = v\,$$
 * $$z = \sqrt{1 - u^2 - v^2}\,$$

We get a 2-dimensional manifold, with a coordinates $$(u, v)\,$$. For all points of the manifold, we have $$\sqrt{x^2 + y^2 + z^2} = 1\,$$, that is, the points lie on the hemisphere.

Now we create the 2-dimensional metric for the hemisphere. To do this, we simply remove the rows and columns of the metric that referred to the coordinates that were removed from the submanifold. We get this metric:
 * $$[g\langle uv\rangle] = \begin{bmatrix} \displaystyle\frac{1 - v^2}{1 - u^2 - v^2} & \displaystyle\frac{uv}{1 - u^2 - v^2} & \\[2.5ex] \displaystyle\frac{uv}{1 - u^2 - v^2} & \displaystyle\frac{1 - u^2}{1 - u^2 - v^2} \end{bmatrix}\ \ \ \ \mathbf{g} = \displaystyle\frac{1}{1 - u^2 - v^2}\ \ \ \ \sqrt{\mathbf{g}} = \displaystyle\frac{1}{\sqrt{1 - u^2 - v^2}}\,$$

This is the (2-dimensional) metric on the hemisphere, effectively in $$(x, y)\,$$ coordinates. We might say that the hemisphere has been parameterized by $$x\,$$ and $$y\,$$.

We might want to express these in the more familiar Cartesian Coordinates. In fact, the only reason we used completely different names $$(u, v, w)\,$$ for the new coordinate system, even though the two coordinates that we were actually interested in are $$x\,$$ and $$y\,$$, is that one must be extremely careful in calculating partial derivatives to obtain the Jacobian.

 * $$[g\langle xy\rangle] = \begin{bmatrix} \displaystyle\frac{1 - y^2}{1 - x^2 - y^2} & \displaystyle\frac{xy}{1 - x^2 - y^2} & \\[2.5ex] \displaystyle\frac{xy}{1 - x^2 - y^2} & \displaystyle\frac{1 - x^2}{1 - x^2 - y^2} \end{bmatrix}\ \ \ \ \mathbf{g} = \displaystyle\frac{1}{1 - x^2 - y^2}\ \ \ \ \sqrt{\mathbf{g}} = \displaystyle\frac{1}{\sqrt{1 - x^2 - y^2}}\,$$

We normally think of $$x\,$$ and $$y\,$$ as Cartesian coordinates, but, in this case, on the surface of a sphere, the coordinates are most definitely not Cartesian.

Of course, we can see the 3-dimensional space, including the $$z\,$$ coordinate, and we know that
 * $$\sqrt{\mathbf{g}} = \displaystyle\frac{1}{z}\,$$

but tiny creatures living on the hemispherical surface don't know that. They only see the $$x\,$$ and $$y\,$$ coordinates.

What is accomplished by getting the metric on the 2-dimensional manifold? We can integrate! The square root of the metric determinant is the correction factor that we have to put into an integral to integrate over the hemisphere. The integration is in terms of $$u\,$$ and $$v\,$$, or $$x\,$$ and $$y\,$$ if we clearly understand that these coordinates are not the actual Cartesian ones.

The area of a tiny patch of the hemisphere in these coordinates is
 * $$\sqrt{\mathbf{g}} \Delta u\ \Delta v = \displaystyle\frac{1}{\sqrt{1 - u^2 - v^2}} \Delta u\ \Delta v\,$$

So we can use this to obtain the surface area of the hemisphere by integration. Use the usual rule—plug in the square root of the metric matrix:
 * $$A = \int_{-1}^1 \int_{-\sqrt{1 - u^2}}^{\sqrt{1 - u^2}} \displaystyle\frac{1}{\sqrt{1 - u^2 - v^2}}\ dv\ du\,$$

but
 * $$\int \displaystyle\frac{1}{\sqrt{1 - u^2 - v^2}}\ dv = \tan^{-1}\frac{v}{\sqrt{1 - u^2 - v^2}}\,$$

so
 * $$\int_{-\sqrt{1 - u^2}}^{\sqrt{1 - u^2}} \displaystyle\frac{1}{\sqrt{1 - u^2 - v^2}}\ dv = \tan^{-1}(\infty) - \tan^{-1}(-\infty) = \pi\,$$

so
 * $$A = \int_{-1}^1 \pi\ du = 2 \pi\,$$

Before proving this, we will look at a few more examples.

For the torus, set $$r = R\,$$ in toroidal coordinates, so the 2-dimensional patch has coordinates $$\theta\,$$ and $$\phi\,$$. After removing $$r\,$$, and setting it to the constant $$R\,$$, we have:
 * $$[g] = \begin{bmatrix} (S + R \cos\phi)^2 & 0 \\[2.5ex] 0 & R^2 \end{bmatrix}\ \ \ \ \mathbf{g} = R^2 (S + R \cos\phi)^2\ \ \ \ \sqrt{\mathbf{g}} = R (S + R \cos\phi)\,$$

The reader should convince himself/herself that, for a small "rectangular" patch on the surface of a torus, the "height" around the cross-sectional circle is $$R \Delta \phi\,$$, the "length" around the main circle is $$(S + R \cos\phi) \Delta \theta\,$$, and the area is $$R (S + R \cos\phi) \Delta\theta \Delta\phi\,$$.

For the helix, set $$r = R\,$$ and $$s = 0\,$$ in helical coordinates. It can be seen that the horizontal distance traversed for a small change in $$\theta\,$$ is $$R \Delta\theta\,$$, and the vertical distance is $$P \Delta\theta\,$$. (Remember that $$P\,$$ is the pitch, or vertical rise, per radian of $$\theta\,$$.) The actual distance along the helix itself is $$\sqrt{R^2 + P^2} \Delta\theta\,$$.

Here is the metric, once we have removed the 1st and 3rd rows and columns. The metric matrix is a 1x1 matrix:
 * $$[g] = \begin{bmatrix} R^2 + P^2 \end{bmatrix}\ \ \ \ \mathbf{g} = R^2 + P^2\ \ \ \ \sqrt{\mathbf{g}} = \sqrt{R^2 + P^2}\,$$

Geometrical Meaning of the Metric
To prove what we have been saying about the metric (and to prepare for the next lecture), we need to work out the geometric significance of the metric. It should come as no surprise that something with the wonderful properties we saw in the previous sections should have a simple and elegant geometrical interpretation.

Vector Components Defined Relative to a Coordinate System
Consider the 3 infinitesimal vectors, starting out tangent to the coordinate lines, going from the point $$(u, v, w)\,$$ to $$(u+\Delta u, v, w)\,$$, $$(u, v+\Delta v, w)\,$$, and $$(u, v, w+\Delta w)\,$$.

Well, they can't go precisely to those points, because the coordinate lines are curved, and vectors have to be straight. But the notion of such a vector works in the infinitesimal limit. We will define "natural components" for these vectors, relative to our coordinate system, that will make them, in effect, "natural basis vectors".

We will say that:
 * The vector $$[\Delta u, 0, 0]\,$$ goes from $$(u, v, w)\,$$ to $$(u+\Delta u, v, w)\,$$

 * The vector $$[0, \Delta v, 0]\,$$ goes from $$(u, v, w)\,$$ to $$(u, v+\Delta v, w)\,$$

 * The vector $$[0, 0, \Delta w]\,$$ goes from $$(u, v, w)\,$$ to $$(u, v, w+\Delta w)\,$$

That is, the 3 components of tiny vectors tell how much the 3 coordinates change when we move from a given point to another point. In the language of vector algebra, we have chosen a new basis for infinitesimal vectors, that is, a new description of vectors in terms of sets of 3 numbers. We have shown just the basis vectors, but of course a vector can point anywhere, and have any or all of its components nonzero.

Now these are actual geometric vectors. As such, they have actual $$x\,$$, $$y\,$$, and $$z\,$$ Cartesian components. The "natural curvilinear vector components" are different from the Cartesian components.

 * We need to emphasize this. Vector lengths, angles, dot products, and cross products are physical geometric concepts that are independent of the coordinate system.  But vectors have different components when measured according to different coordinate systems.  We will explain how, given the coordinate system, these geometric concepts may be calculated from the vector components, such that one always gets the right answer; in fact, this is the main result of this course.  One thing that we can say right now is that you must not assume that the length of a vector is the square root of the sum of the squares of its components, unless the coordinate system is Cartesian.  The length (and everything else!) comes from the knowledge of the coordinate system, which can be found in the metric.

Given a vector's natural curvilinear components, we can get its Cartesian components by using the Jacobian. When we moved from $$(u, v, w)\,$$ to $$(u+\Delta u, v, w)\,$$, the $$x\,$$, $$y\,$$, and $$z\,$$ coordinates changed thusly:
 * $$\Delta x = \displaystyle\frac{\partial x}{\partial u} \Delta u = J_{11} \Delta u\,$$

 * $$\Delta y = \displaystyle\frac{\partial y}{\partial u} \Delta u = J_{21} \Delta u\,$$

 * $$\Delta z = \displaystyle\frac{\partial z}{\partial u} \Delta u = J_{31} \Delta u\,$$

So the Cartesian components of the vector with natural curvilinear components $$[\Delta u, 0, 0]\,$$ are $$[J_{11}\Delta u, J_{21}\Delta u, J_{31}\Delta u]\,$$.

Similarly, $$[0, \Delta v, 0]\,$$ has Cartesian components $$[J_{12}\Delta v, J_{22}\Delta v, J_{32}\Delta v]\,$$, and $$[0, 0, \Delta w]\,$$ has Cartesian components $$[J_{13}\Delta w, J_{23}\Delta w, J_{33}\Delta w]\,$$.

Now let's calculate the dot product of the natural vectors $$[\Delta u, 0, 0]\,$$ and $$[0, \Delta v, 0]\,$$.

 * The reader should look at the dot product page if necessary. That particular properties we need are linearity, and the "sum of pairwise products" rule—in Cartesian coordinates:
 * $$\vec{P} \cdot \vec{Q} = P_x Q_x + P_y Q_y + P_z Q_z\,$$.

Once the components of a vector are expressed in Cartesian terms, the answer is obtained by the sum of pairwise products rule:
 * $$[\Delta u, 0, 0] \cdot[0, \Delta v, 0] = J_{11}\Delta u\ J_{12}\Delta v\ +\ J_{21}\Delta u\ J_{22}\Delta v\ +\ J_{31}\Delta u\ J_{32}\Delta v\,$$
 * $$= \left(J_{11} J_{12} + J_{21} J_{22} + J_{31} J_{32}\right)\Delta u\ \Delta v = [J^t J]_{12}\ \Delta u\ \Delta v = g_{12}\ \Delta u\ \Delta v\,$$

Similarly:
 * $$[\Delta u, 0, 0] \cdot[0, 0, \Delta w] = g_{13}\Delta u \Delta w\,$$
 * $$[\Delta u, 0, 0] \cdot[\Delta u, 0, 0] = g_{11}(\Delta u)^2\,$$

And so on. We can get the dot products of any pairs of natural coordinate basis vectors directly from the metric!

We can only do this for vectors that are of infinitesimal length, because, as we have seen, the vectors only follow the coordinate lines for infinitesimal distances. The vectors start out tangent to the coordinate lines, but the coordinate lines curve while the vectors stay straight. The rule here is: when vectors are considered to "point to" a point on the manifold, they must be infinitesimal. This doesn't mean that we can never deal with large vectors—when doing physical problems involving electric and magnetic fields, for example, the vectors can be any size we like. We just can't interpret such vectors as "pointing to" some location on the manifold.

A few things that we noticed before now become obvious.
 * The symmetry of the metric matrix follows from the symmetry of the dot product.
 * If the coordinate system is orthogonal, so that the coordinate lines are orthogonal at any point, then the tiny basis vectors are orthogonal, so their dot products are zero. This means that the off-diagonal entries in the metric matrix are zero in that case.
 * The length of a vector is the square root of its dot product with itself, so the length of the vector $$[\Delta u, 0, 0]\,$$ is $$\sqrt{g_{11}}\Delta u\,$$. If the coordinate system is orthogonal, so that we have Lamé coefficients, this is just $$h_1\Delta u\,$$.  The Lamé coefficients are the scale factors for the lengths of infinitesimal basis vectors, if the coordinate system is orthogonal.

But we can do even better. The metric matrix gives us directly (albeit laboriously) the dot products of any vectors, in terms of their natural components. This is because the dot product is linear in each argument:
 * $$\vec{P} \cdot (\lambda \vec{Q}) = \lambda (\vec{P} \cdot \vec{Q})\,$$
 * $$\vec{P} \cdot (\vec{Q} + \vec{R}) = \vec{P} \cdot \vec{Q} + \vec{P} \cdot \vec{R}\,$$

By breaking a vector into the sum of the basis vectors times its various components, and using linearity, we have:
 * $$\vec{P} \cdot \vec{Q} = \sum_{i, j = 1}^3 g_{ij} P_i Q_j\,$$

where $$P_i\,$$ and $$Q_j\,$$ are the natural components of the two vectors.

This is more easily calculated using matrix manipulations. Express one vector as a "row matrix", followed by the metric matrix, followed by the other vector as a "column matrix". When the three matrices are multiplied, the result is a 1x1 matrix. The value in that matrix is the dot product.
 * $$\vec{P} \cdot \vec{Q} = \begin{bmatrix} P & \text{as} & \text{row} \end{bmatrix} \begin{bmatrix} & & \\ & g & \\ & & \end{bmatrix} \begin{bmatrix} Q \\ \text{as} \\ \text{column} \end{bmatrix}\,$$.

The Metric on a Submanifold
It is now easy to see why we can use the method of striking rows and columns out of the metric matrix to get the metric for a submanifold. The elements of the metric matrix are just the dot products of basis vectors. A basis vector in a submanifold is also a basis vector in the larger manifold, with all of the dropped components equal to zero. The dot product of two vectors is the same whether we use the large metric matrix or the small one, because the extra matrix elements are multiplied by zero.

In fact, any vector in the submanifold is a vector in the larger manifold, with the extra components in the larger manifold equal to zero. So the smaller metric matrix gives the dot products of vectors in the submanifold, as required.

Finally, let's give a direct proof that the square root of the metric determinant gives the correct "area/length/volume element correction factor" in integrals.

We have tiny basis vectors
 * $$\vec{U} = [\Delta u, 0, 0]\,$$
 * $$\vec{V} = [0, \Delta v, 0]\,$$
 * $$\vec{W} = [0, 0, \Delta w]\,$$

We need to find the actual geometrical volume of the parallelepiped that they span. If they were orthogonal (which they would be if the coordinate system were orthogonal), they would span a rectangular region, and the volume would be simply the product of their geometrical lengths. But in the general case, we need to use the "parallelogram trick" to make them all perpendicular to each other. This is done by adding appropriate multiples of one vector to another. This does not change the volume, but the correct choice of multiplier can make the vectors orthogonal.

First, we subtract a multiple of $$\vec{U}\,$$ from $$\vec{V}\,$$, obtaining $$\vec{P}\,$$. Let that multiple be:
 * $$A \frac{\Delta v}{\Delta u}\,$$

We have:
 * $$\vec{P} = \vec{V} - A \frac{\Delta v}{\Delta u} \vec{U} = [0, \Delta v, 0] - A [\Delta v, 0, 0] = [-A, 1, 0] \Delta v\,$$

Then
 * $$\vec{P} \cdot \vec{U} = \begin{bmatrix} -A, 1, 0 \end{bmatrix} \begin{bmatrix} & & \\ & g & \\ & & \end{bmatrix} \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} \Delta u \Delta v = \begin{bmatrix} -A, 1, 0 \end{bmatrix} \begin{bmatrix} g_{11} \\ g_{12} \\ g_{13} \end{bmatrix} \Delta u \Delta v = (g_{12} - A g_{11}) \Delta u \Delta v\,$$.

So, to get $$\vec{P} \perp \vec{U}\,$$, we need this equal to zero, so:
 * $$A = \displaystyle\frac{g_{12}}{g_{11}}\,$$

While we are "straightening" the vectors by subtracting multiples of one vector from another without changing the volume that they span, we perform a similar operation on the metric matrix, subtracting multiples of one row from another. A theorem of determinants tells us that this does not change the value of the determinant. We subtract $$A\,$$ times the first row from the second row.

 * $$\mathbf{g} = \begin{vmatrix} g_{11} & g_{12} & g_{13} \\ g_{12} & g_{22} & g_{23} \\ g_{13} & g_{23} & g_{33} \end{vmatrix} = \begin{vmatrix} g_{11} & g_{12} & g_{13} \\ g_{12} - A g_{11} & g_{22} - A g_{12} & g_{23} - A g_{13} \\ g_{13} & g_{23} & g_{33} \end{vmatrix} = \begin{vmatrix} g_{11} & g_{12} & g_{13} \\ 0 & g_{22} - A g_{12} & g_{23} - A g_{13} \\ g_{13} & g_{23} & g_{33} \end{vmatrix}\,$$

A very important cancelation occurs.

Similarly, we create a vector $$\vec{Q}\,$$ by subtracting a multiple of $$\vec{U}\,$$ from $$\vec{W}\,$$.
 * $$\vec{Q} = \vec{W} - B \frac{\Delta w}{\Delta u} \vec{U} = [-B, 0, 1] \Delta w\,$$

The value to make $$\vec{Q}\,$$ perpendicular to $$\vec{U}\,$$ is:
 * $$B = \displaystyle\frac{g_{13}}{g_{11}}\,$$

Now $$\vec{U}\,$$, $$\vec{P}\,$$, and $$\vec{Q}\,$$ span the necessary volume, and $$\vec{P}\,$$ and $$\vec{Q}\,$$ are both perpendicular to $$\vec{U}\,$$, but not to each other.

We also subtract $$B\,$$ times the first row of the matrix that we are working on from the third row. As before, the determinant is still $$\mathbf{g}\,$$:
 * $$\mathbf{g} = \begin{vmatrix} g_{11} & g_{12} & g_{13} \\ 0 & g_{22} - A g_{12} & g_{23} - A g_{13} \\ g_{13} - B g_{11} & g_{23} - B g_{12} & g_{33} - B g_{13} \end{vmatrix} = \begin{vmatrix} g_{11} & g_{12} & g_{13} \\ 0 & g_{22} - A g_{12} & g_{23} - A g_{13} \\ 0 & g_{23} - B g_{12} & g_{33} - B g_{13} \end{vmatrix}\,$$

But the rules for calculating determinants say that we can now remove the first column:
 * $$\mathbf{g} = (g_{11})\ \begin{vmatrix} g_{22} - A g_{12} & g_{23} - A g_{13} \\ g_{23} - B g_{12} & g_{33} - B g_{13} \end{vmatrix}\,$$

Finally, we "straighten" $$\vec{Q}\,$$, to make it perpendicular to $$\vec{P}\,$$. We define:
 * $$\vec{R} = \vec{Q} - C \frac{\Delta w}{\Delta v} \vec{P} = [AC-B, -C, 1] \Delta w\,$$

Now:
 * $$\vec{R} \cdot \vec{P} = \begin{bmatrix} AC-B, -C, 1 \end{bmatrix} \begin{bmatrix} & & \\ & g & \\ & & \end{bmatrix} \begin{bmatrix} -A \\ 1 \\ 0 \end{bmatrix} \Delta v \Delta w\,$$

 * $$= \begin{bmatrix} AC-B,1 -C, 1 \end{bmatrix} \begin{bmatrix} g_{12} - A g_{11} \\ g_{22} - A g_{12} \\ g_{23} - A g_{13} \end{bmatrix} \Delta v \Delta w = \begin{bmatrix} AC-B, -C, 1 \end{bmatrix} \begin{bmatrix} 0 \\ g_{22} - A g_{12} \\ g_{23} - A g_{13} \end{bmatrix} \Delta v \Delta w\,$$

 * $$= (-C (g_{22} - A g{12}) + (g_{23} - A g{13}) ) \Delta v \Delta w\,$$

To get $$\vec{R} \perp \vec{P}\,$$, this must be zero, so:
 * $$C = \displaystyle\frac{g_{23} - A g_{13}}{g_{22} - A g_{12}}\,$$

Also, $$\vec{U} \cdot \vec{R} = \vec{U} \cdot (\vec{Q} - C \frac{\Delta w}{\Delta v} \vec{P}) = \vec{U} \cdot \vec{Q} - C \frac{\Delta w}{\Delta v} \vec{U} \cdot \vec{P} = 0\,$$

So $$\vec{U}\,$$, $$\vec{P}\,$$, and $$\vec{R}\,$$ are all orthogonal, so the volume is the product of their lengths.

We will need these facts presently:
 * $$A g_{13} = B g_{12}\,$$
 * $$C(g_{22} - A g_{12}) = g_{23} - A g_{13}\,$$

We subtract $$C\,$$ times the first row of the (now 2x2) matrix that we are working on from the second row.
 * $$\mathbf{g} = (g_{11})\ \begin{vmatrix} g_{22} - A g_{12} & g_{23} - A g_{13} \\ g_{23} - B g_{12} - C(g_{22} - A g_{12}) & g_{33} - B g_{13} - C(g_{23} - A g_{13}) \end{vmatrix}\,$$

 * $$= (g_{11})\ \begin{vmatrix} g_{22} - A g_{12} & g_{23} - A g_{13} \\ 0 & g_{33} - B g_{13} - C(g_{23} - A g_{13}) \end{vmatrix} = (g_{11})\ (g_{22} - A g_{12})\ (g_{33} - B g_{13} - C(g_{23} - A g_{13}))\,$$

 * $$= (g_{11})\ (g_{22} - A g_{12})\ ((AC-B) g_{13} - C g_{23} + g_{33})\,$$

Now calculate the lengths of $$\vec{U}\,$$, $$\vec{P}\,$$, and $$\vec{R}\,$$.

 * $$\frac{\vec{U} \cdot \vec{U}}{(\Delta u)^2} = \begin{bmatrix} 1, 0, 0 \end{bmatrix} \begin{bmatrix} & & \\ & g & \\ & & \end{bmatrix} \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} = \begin{bmatrix} 1, 0, 0 \end{bmatrix} \begin{bmatrix} g_{11} \\ g_{12} \\ g_{13} \end{bmatrix} = g_{11}\,$$

 * $$\frac{\vec{P} \cdot \vec{P}}{(\Delta v)^2} = \begin{bmatrix} -A, 1, 0 \end{bmatrix} \begin{bmatrix} & & \\ & g & \\ & & \end{bmatrix} \begin{bmatrix} -A \\ 1 \\ 0 \end{bmatrix} = \begin{bmatrix} -A, 1, 0 \end{bmatrix} \begin{bmatrix} g_{12} - A g_{11} \\ g_{22} - A g_{12} \\ g_{23} - A g_{13} \end{bmatrix} = \begin{bmatrix} -A, 1, 0 \end{bmatrix} \begin{bmatrix} 0 \\ g_{22} - A g_{12} \\ g_{23} - A g_{13} \end{bmatrix} = g_{22} - A g_{12}\,$$

 * $$\frac{\vec{R} \cdot \vec{R}}{(\Delta w)^2} = \begin{bmatrix} AC - B, -C, 1 \end{bmatrix} \begin{bmatrix} & & \\ & g & \\ & & \end{bmatrix} \begin{bmatrix} AC - B \\ -C \\ 1 \end{bmatrix} = \begin{bmatrix} AC - B, -C, 1 \end{bmatrix} \begin{bmatrix} (AC - B) g_{11} - C g_{12} + g_{13} \\ (AC - B) g_{12} - C g_{22} + g_{23} \\ (AC - B) g_{13} - C g_{23} + g_{33} \end{bmatrix}\,$$

Making use of the previous facts:
 * $$\frac{\vec{R} \cdot \vec{R}}{(\Delta w)^2} = \begin{bmatrix} AC - B, -C, 1 \end{bmatrix} \begin{bmatrix} 0 \\ 0 \\ (AC - B) g_{13} - C g_{23} + g_{33} \end{bmatrix} = (AC - B) g_{13} - C g_{23} + g_{33}\,$$

But these three dot products are just the factors of $$\mathbf{g}\,$$ that we worked out previously. So
 * $$\|\vec{U}\| \|\vec{P}\| \|\vec{R}\| = \sqrt{\mathbf{g}}\ \Delta u\ \Delta v\ \Delta w\,$$

Example: What is the surface area of a torus?
 * $$\int_0^{2\pi} \int_0^{2\pi} R (S + R \cos \phi)\ d\theta\ d\phi = 2\pi\int_0^{2\pi} (RS + R^2 \cos \phi)\ d\phi = 4 \pi^2 RS$$

Exercise: Here is one in which the integrand isn't just 1. The Gaussian curvature, per unit area, of a torus is $$K = \displaystyle\frac{\cos\phi}{R (S + R \cos \phi)}\,$$. (The calculation of this is rather advanced. It involves either Riemann's tensor or the "second fundamental form".)  We have expressed this in toroidal coordinates, of course. Note that it is positive on the outer (bulging) part, and negative on the inner (saddle-like) part. What is the total surface integral of the Gaussian curvature, $$\iint_T K dA\,$$?

=Footnotes and References=

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

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User:Harro242

 * Textbook Chapter

 * Multimedia

http://screenr.com/F7c

 * E-Portfolio

Introduction

So this is my E-portfolio for Motivation and Emotion. I used this as simply a way of writing down the thoughts I had during the lectures and tuts. I thought it would be boring to simply restate what was said in the lectures and tuts so instead I expanded on thoughts about the topic that I had during them. All the lectures made me think of some very interesting things I had never thought of before, so I simply wrote them down. I hope you enjoy it.

Week 1

The intro lecture was very helpful and insightful. It explained quiet clearly the assignment items we are going to have do and in a way that seem not to freak everyone out. The historical background to the unit was also very well done and discussed so many things I had no idea about. The evolution of the topic was something I had never heard or talked about so it was a very good start to the unit.

Week 2

This week’s lecture was extremely helpful in gaining a better understanding of wiki and the things u can do with it and that’s all that can really be said about that. The tut for this week though was very fun, exciting and helpful. Starting the tut with the exercises of getting in to all those different groups according to different things has always been one of my favourite things about most of the first psychology tuts I have been in. It really breaks the ice and just gets people moving and interacting with each other and gives them a chance to find some common ground. Also breaking everyone into groups has always been a thing I like to do in tuts, its just more personal sometimes and it makes it much easier to work together and if you happen to have friends from past tuts in this tuts it allows you to continue that friendship. The discussion we had during this tut was very interesting. We discussed our group member’s interests and their ideas of the textbook chapter, which allowed me to expand on my own interests and ideas too. I really did like the activity of defining motivation and emotion; I really liked how it brought together so many units and subjects being done to come up with these very clever definitions for such hard topics. I really did enjoy this first tut and I hope the rest are exactly like this one.

Week 3

This week’s discussion about what motivation actually is was intriguing to me. What actually drives us, what makes us do the things we do? There must be a reason for our behaviour; everything has a reason in my opinion. So what I think is that the list of these reasons is endless and it would be impossible to name them all because whenever you think you have finally listed them all someone else will come along with a new one. However what drives us to these reasons, what has made us arrive at a point in time where we act or behave a certain way. It is our past and the experiences we have that make us who we are, they shape what our personality will be. For example, in very general terms, a person who has had a hard life and is angry at the world may be motivated to only look after themselves and no one else. To this person they have a very good reason to look out for only number one, however on the other side a person who has had a good and easy live may be very nice and polite to the people around them. Of course there are major exceptions to these kinds of example, sometimes it can be completely the other way round and that is why the list of reasons is endless, because everyone has a slightly different past. There are of course behaviours that are driven by physiological needs like that discussed in the lecture. Every organism has them but the thing to note is that there are different motivations behind different ways to get what you need; some people’s motivation is so strong they steal and some aren’t so strong. Again it comes to the fact the list of motivations is endless, at least in my opinion.

Week 4

Last week was about physiological needs but this week we went into psychological needs. I’d never heard the psychological need that people have broken down into those categories before. It was interesting to release during the lectures about how much of what we do is actually kind of subconscious. We have all these psychological need that on some conscious level we are probably aware of but still we act sometimes totally because of these need and we may not even know it. These needs can include feeling like we have a place in the world or we have a place within a group of friends. Of course we know we like it once we have it but the question is, subconsciously, did we know we would like it before we even had it. Are these needs built into us from birth and as we grow more of them come out and we subconsciously or consciously begin to want them and are we totally motivated by them at least in the area that they focus on. Of course we all know that having friends is a nice thing, however most people would not think of the idea of having friends as a psychological need. So although some of these needs may be conscious thoughts we still might not even know they can be classified as a psychological need.

Week 5

This week’s lecture really spoke to me. I am the type of person that only really works hard at something if I am enjoying it, like this course. So the whole part of the lecture about being motivated to expand and work on ones interests and desires made sense to me cause everyone has something they are interested in and they are sometimes drawn to it and to do something with it, be it to make another model or develop skills in a sport. However our interests do certainly change over time, we might just lose interest or over time it may develop into something else. Although not all our interest change, some people know from a very early age exactly what they want to do with there life’s and some people just never lose interest of a topic they know and love. So it is very different for everyone. The other question in the lecture that really interested me was does punishment work? If we were to look at a dog for example, it may live in a very happy home where it is loved and well taken care of, but it may do something wrong and it is punished by being put outside or smacked. Now usually in this kind of situation is treatment works, the dog slowly learns that that behaviour is unacceptable and starts not longer do it. So what about with a child, in my opinion it completely depends on their age. For example, there is probably no point trying to teach an 5 months year old baby they have done something wrong cause they probably wont understand you. But if a 5 year old does something it is very important to teach them tat it is wrong to make sure they are learning right from wrong. The method of punishment however is a big question mark, I do not believe in smacking but sending them to their room or giving them a time out are examples of punishments that may work. But also this also depends on the home and family environment they have.

Week 9

Today we started on the topic of emotions. My main thoughts were the main questions of the lecture. What are emotions, can they be defined or measured. I remember in one of the first tuts we tried to define emotion and my group came up with a fairly scientific definition. Something like emotions are chemical and physical states of the brain that cause certain bodily changes and behaviours that people interpret as emotions. They are usually brought on by some stimulus but can also sometimes come about by simply thinking certain thoughts. For example, if you were to think back on something sad that happened to you, like the loss of a family member, you will begin to feel the emotion of sadness again. The other question asked in the lecture were also very interesting, like can they be measured, how many different emotions are there and so on. In my opinion they can only really be measured by the person feeling them, only they can say how they are really feeling, I also think there is not an infinite list of emotions its just that people feel emotions differently or can handle them better then other people but they are still the same emotions.

Week 10

What good are emotion seemed to be the main topic that stuck in my head this week, what do they actually do for us? Some people who have been hurt emotionally in some way what say that all they do is make us suffer and feel nothing but pain. On the other hand they also add to the enjoyment of situations like Christmas day for example. They give us the feeling of happiness and joy. So you could say that emotions are a gift and a curse, they can make us feel fantastic and happy but they can also make us feel like we are alone and are slowly being crushed by the world. Also like almost everything it depends on our situation, if you are a generally happy person who enjoys life then you may be quite thankful for emotion, but if you’re having a bit of a hard time in life then you may say that you would give anything not to have emotions. The other thing I liked about this week was the facial feedback hypothesis, I have always been a great fan of this kind of work and have always wanted to know more about how its done. The theory that, regardless of race or origin, everyone in the world has the same facial expressions for all the different emotions is fascinating and makes sense. We are all human, and as humans our bodily functions when it comes to emotion would be the same. So if our bodies do the same thing then we probably feel the same thing and therefore it no surprise that we react in the same way.

Week 11

I really really love this week’s topic; it really rang true for me and was my favourite lecture so far. Personalities have always been one of my favourite topics, just the difference between people and also how easy it is too read someone once you really know what kind of person they are. I believe there are a finite number of personalities in the world and that can be measured and categorised and that is what was said in the lecture. There are measurable parts to personalities and once you have measured them you can start to predict how someone will act in certain situations. The idea of our personalities controlling our motivations is so true. A shy person will have a closed off personality and it may be easy to predict that in a situation where this person is faced with trying something new they may be very relucted to do so and its just the same with every other type of personality. Of course there may be exceptions but usually that just means they may have a bit more of a complicated personality, but once you know every aspect to it you can predict them just the same.

Week 12

Unconscious motivation is a important topic to talk about, it can give us further insight into ourselves and allow us to gain a little more control over our actions. Although we may not be able to stop from being motivated by an unconscious thought, we may be able to recognise the behaviour and stop ourselves. I have always liked freud’s model of the Id and ego, it makes a lot of sense that we still have those basic wants but we have learned to control them and also some are now not socially acceptable. It can maybe be said that because we have a much larger higher brain function then another animal we have the ability to see these wants but control them instead of them controlling us. But what about the people who just let them come out, why aren’t they in control. I think they are perfectly in control they just don’t care enough to keep them inside and they just think I want that so I’ll take it.

Week 13

I have personally seen the difference between people who have had nothing but positive growth from positive psychology, feedback and encouragement and negative growth resulting from negative psychology, feedback and encouragement. I have seen the damage of negative psychology and false self actualisation, its terrible and no person should have to go through it. The worst thing is when this is a result of bad parenting. Parents have a big responsibility to support their children and make sure they are happy, but when they fail to do so, it is almost criminal in my opinion and should be stopped. People who do have positive psychology and true self-actualisation on the other hand can do great things. They are usually very happy with themselves and with life and because of this can pass it on to other people and try to improve other people psychology about themselves and life. There are people who may appear to have high positive psychology and seem very happy but in truth they have problems and are choosing to deal with them later or just ignore them all together. It is often very hard to tell true positive psychology from false but it is a very important difference and one that holds so many people back from their true potential.

Week 14

I took part this course to learn how to motivate other people and I have enhanced my knowledge of how to do so. The combination of unconscious thoughts and wants, out side forces, positive and negative psychology, personality type and so much more are all things that one must take into account before even thinking about changing someone. For example maybe the way to change someone is to remove something else from their lifes and then they will change naturally. What ever they way, it will be a long and hard road because most people don’t want to change naturally, they need a push and sometimes it is made hard by the fact they don’t realise that they need to change.

Conclusion

I have very much enjoyed this course and I hope this journal has been useful to those that read it. A big thank you to James for running the subject and teaching us so much and I hope future students find it as interesting as I did.

title: User:Harro242


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 Hi , I am Percy. Nice to meet you. I love the beach, swimming, fishing, and I am currently doing some research at Wikiversity on oceanography. I am also learning to play piano. What do you like? Can I be your mascot?

Motivation and emotion
Hi. Welcome to the unit and to Wikiversity! I look forward to learning with you and I hope you find the topic and the unit rewarding. Please feel free to let me know if you have any suggestions or would like help at any point along the way. Sincerely, James Neill, -- Jtneill - Talk - c 09:07, 30 August 2010 (UTC)

-- Jtneill - Talk - c 06:03, 15 December 2010 (UTC)

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And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. --Harrypotter 05:10, 29 August 2010 (UTC)

Introduction to political science
I think you will find this course is currently inactive.Harrypotter 05:16, 29 August 2010 (UTC)

title: User talk:Mirwan24


text:
User:Soundman~enwikiversity

I'm SINOJ, a sound engineering student from India. Had worked with a Carnatic Rock Band named "Motherjane". Sorry for my poor English.

title: User:Soundman~enwikiversity


text:
Talk:How to use SPSS

PASW vs. SPSS
After SPSS Inc. sold the PASW to IBM. IBM renamed it back to "SPSS". It was only called "PASW" for about 3 months. Therefore I suggest we just stick to "SPSS" to avoid confusing everyone. Statlearn 09:26, 29 August 2010 (UTC)
 * Thanks - done. -- Jtneill - Talk - c 21:53, 29 August 2010 (UTC)

To clarify, PASW was added to the name Statistics to help differentiate between the software SPSS and the company SPSS Inc. IBM acquired SPSS Inc in 2009 and decided to drop the PASW and use SPSS as a prefix for the suite of products in the portfolio. Bradhill14 (talk) 06:13, 19 November 2012 (UTC)

title: Talk:How to use SPSS


text:
User talk:Statlearn~enwikiversity

 Hello Statlearn, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

 * Take a guided tour and learn to edit.
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And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. --Harrypotter 10:47, 29 August 2010 (UTC)

Your account will be renamed
Hello,

The developer team at Wikimedia is making some changes to how accounts work, as part of our on-going efforts to provide new and better tools for our users like cross-wiki notifications. These changes will mean you have the same account name everywhere. This will let us give you new features that will help you edit and discuss better, and allow more flexible user permissions for tools. One of the side-effects of this is that user accounts will now have to be unique across all 900 Wikimedia wikis. See the announcement for more information.

Unfortunately, your account clashes with another account also called Statlearn. To make sure that both of you can use all Wikimedia projects in future, we have reserved the name Statlearn~enwikiversity that only you will have. If you like it, you don't have to do anything. If you do not like it, you can pick out a different name.

Your account will still work as before, and you will be credited for all your edits made so far, but you will have to use the new account name when you log in.

Sorry for the inconvenience.

Yours, Keegan Peterzell Community Liaison, Wikimedia Foundation 23:44, 17 March 2015 (UTC)

Renamed
 This account has been renamed as part of single-user login finalisation. If you own this account you can |log in using your previous username and password for more information. If you do not like this account's new name, you can choose your own using this form after logging in: . -- Keegan (WMF) (talk) 06:23, 19 April 2015 (UTC)

title: User talk:Statlearn~enwikiversity


text:
User talk:Shazrules

 Hello Shazrules, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

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And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. --Harrypotter 10:48, 29 August 2010 (UTC)

title: User talk:Shazrules


text:
User talk:Soundman~enwikiversity

 Hello Soundman, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

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And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. --Harrypotter 10:48, 29 August 2010 (UTC)

Your account will be renamed
Hello,

The developer team at Wikimedia is making some changes to how accounts work, as part of our on-going efforts to provide new and better tools for our users like cross-wiki notifications. These changes will mean you have the same account name everywhere. This will let us give you new features that will help you edit and discuss better, and allow more flexible user permissions for tools. One of the side-effects of this is that user accounts will now have to be unique across all 900 Wikimedia wikis. See the announcement for more information.

Unfortunately, your account clashes with another account also called Soundman. To make sure that both of you can use all Wikimedia projects in future, we have reserved the name Soundman~enwikiversity that only you will have. If you like it, you don't have to do anything. If you do not like it, you can pick out a different name.

Your account will still work as before, and you will be credited for all your edits made so far, but you will have to use the new account name when you log in.

Sorry for the inconvenience.

Yours, Keegan Peterzell Community Liaison, Wikimedia Foundation 23:43, 17 March 2015 (UTC)

Renamed
 This account has been renamed as part of single-user login finalisation. If you own this account you can |log in using your previous username and password for more information. If you do not like this account's new name, you can choose your own using this form after logging in: . -- Keegan (WMF) (talk) 06:23, 19 April 2015 (UTC)

title: User talk:Soundman~enwikiversity


text:
User talk:Pdesai475

 Hello Pdesai475, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

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And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. --Harrypotter 10:49, 29 August 2010 (UTC)

title: User talk:Pdesai475


text:
User talk:Mfaddersb

 Hello Mfaddersb, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

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And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. --Harrypotter 10:49, 29 August 2010 (UTC)

title: User talk:Mfaddersb


text:
User talk:Hammadahmad

 Hello Hammadahmad, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

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And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. --Harrypotter 10:50, 29 August 2010 (UTC)

title: User talk:Hammadahmad


text:
User talk:Evglassett~enwikiversity

 Hello Evglassett, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

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 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. --Harrypotter 10:50, 29 August 2010 (UTC)

Your account will be renamed
Hello,

The developer team at Wikimedia is making some changes to how accounts work, as part of our on-going efforts to provide new and better tools for our users like cross-wiki notifications. These changes will mean you have the same account name everywhere. This will let us give you new features that will help you edit and discuss better, and allow more flexible user permissions for tools. One of the side-effects of this is that user accounts will now have to be unique across all 900 Wikimedia wikis. See the announcement for more information.

Unfortunately, your account clashes with another account also called Evglassett. To make sure that both of you can use all Wikimedia projects in future, we have reserved the name Evglassett~enwikiversity that only you will have. If you like it, you don't have to do anything. If you do not like it, you can pick out a different name.

Your account will still work as before, and you will be credited for all your edits made so far, but you will have to use the new account name when you log in.

Sorry for the inconvenience.

Yours, Keegan Peterzell Community Liaison, Wikimedia Foundation 23:35, 17 March 2015 (UTC)

Renamed
 This account has been renamed as part of single-user login finalisation. If you own this account you can |log in using your previous username and password for more information. If you do not like this account's new name, you can choose your own using this form after logging in: . -- Keegan (WMF) (talk) 06:08, 19 April 2015 (UTC)

title: User talk:Evglassett~enwikiversity


text:
User talk:Pravin2010

 Hello Pravin2010, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

 * Take a guided tour and learn to edit.
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 * Find out about research activities on Wikiversity.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
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 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. --Harrypotter 11:29, 29 August 2010 (UTC)

title: User talk:Pravin2010


text:
User talk:Alito~enwikiversity

 Hello Alito, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random project.
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 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. --Harrypotter 11:29, 29 August 2010 (UTC)

Your account will be renamed
Hello,

The developer team at Wikimedia is making some changes to how accounts work, as part of our on-going efforts to provide new and better tools for our users like cross-wiki notifications. These changes will mean you have the same account name everywhere. This will let us give you new features that will help you edit and discuss better, and allow more flexible user permissions for tools. One of the side-effects of this is that user accounts will now have to be unique across all 900 Wikimedia wikis. See the announcement for more information.

Unfortunately, your account clashes with another account also called Alito. To make sure that both of you can use all Wikimedia projects in future, we have reserved the name Alito~enwikiversity that only you will have. If you like it, you don't have to do anything. If you do not like it, you can pick out a different name.

Your account will still work as before, and you will be credited for all your edits made so far, but you will have to use the new account name when you log in.

Sorry for the inconvenience.

Yours, Keegan Peterzell Community Liaison, Wikimedia Foundation 23:27, 17 March 2015 (UTC)

Renamed
 This account has been renamed as part of single-user login finalisation. If you own this account you can |log in using your previous username and password for more information. If you do not like this account's new name, you can choose your own using this form after logging in: . -- Keegan (WMF) (talk) 06:02, 19 April 2015 (UTC)

title: User talk:Alito~enwikiversity


text:
User talk:RobRob

 Hello RobRob, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random project.
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 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
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And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. --Harrypotter 15:09, 29 August 2010 (UTC)

title: User talk:RobRob


text:
User talk:Narmi

 Hello Narmi, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random project.
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 * Give feedback about your initial observations
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And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. --Harrypotter 15:09, 29 August 2010 (UTC)

title: User talk:Narmi


text:
User talk:Cprimera

 Hello Cprimera, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

 * Take a guided tour and learn to edit.
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And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. --Harrypotter 15:09, 29 August 2010 (UTC)

title: User talk:Cprimera