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Question | Answer | |||||||
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Parallel lines are at least | two lines that are equally | separated. | ||||||

A diagonal that cuts through 2 parallel lines is called an | Intersecting transversal. | |||||||

When a line cuts through 2 parallel lines the 2 bunches of angles are | Equal. | |||||||

When a line cuts through 2 parallel lines there are only 2 different | angles. One is acute while the other is | obtuse. | ||||||

An alternate angle is in a | ‘z’ shape where the inside angles are equal. An alternate angle can also be a ‘z’ shape | but backwards. | ||||||

Supplementary angles are a | ‘c’ or ‘u’ shape where the inside angles add up to | 180 degrees. | ||||||

Supplementary loosely translated means; | Add up to 180. | |||||||

Corresponding angles are an | ‘f’ shape where the 1st and 3rd inside angle are | Equal and possibly probably obtuse. | ||||||

Diagonally opposite angles in a rhombus are | Equal. | |||||||

It 6ts take 11d to collect n how long would it take for 12ts to do so? In fractions 12ts ÷ 6ts= | 2/1. The inverse of 2/1 is 1/2. 11d x 1/2= | 5.5d. | ||||||

To solve simultaneous Linear and Quadratic equations; first eliminate either | 'x' or 'y' by | substitution. Then if it's not already turn the equation to a | Quadratic equation using | 'rearranging formulae'. The Quadratic equation can be solved through Quadratic | factorisation or | completing the square. Then use the discovered | value(s) to find the value(s) of the remaining | term(s). |

In Cubic equations apart from manipulation it is possible to use an | approximation method of | trial and improvement, which can be refined to different degrees of | accuracy. The object of trial and improvement is to systematically find a value of | 'x' which makes the expression as close as desirable to | 0. This value is called the root. | |||

To construct a perpendicular bisector move the compass to | the end of the line, set it over [] way along the line | half way-- and draw an | arc the size of at least about half a | circle, then repeat from | the other side. Afterwards draw a line through both of the [] points of the arcs | intersection--. Things that may be labelled are the two intersecting points of the arcs and the right angle(s) formed. | ||

To bisect an angle: place the compass on the | vertex and draw an | arc that crosses | both sides of the angle. Draw an arc from each | crossing points; the second set of arcs should cross-over between | the 2 sides of the angle. If you label that crossing over point between the 2 sides of the angle: 'C', then a straight line is drawn through | the vertex and 'C'. Including 'C', the 4 thing that may be labelled are | all the crossing points and the vertex. | crossing points and the corner of the angle. |

To construct a triangle with sides of 6cm, 5cm and 4cm: first draw the side of length | 6cm (AC). If AB is the side of length 5cm: after that move the compass to A and with the compass opened to a radius of | 5cm, draw an arc [] the line | above--, if CB is the side of length 4cm: after that move the compass to C and with the compass opened to a radius of | 4cm, construct an [] above the line | arc--. Join each end of the line to the point where the arcs | cross. | ||

To construct a triangle with angles; c°, d° and a side of 6cm: draw | the 6cm line. You may label the | ends of the line. From each | end, measure either one of the | angles and mark it with a | dot. With lines connect | the lines to the dots. The lines should keep going until you have drawn a | triangle. In a test, unless asked to do otherwise, if you have any, leave the | construction lines. |

To construct a triangle with an sides of a cm (the longest side), b cm and an angle of c°: draw | the longest side (a cm). From the | end of the line, draw an | angle of | c°. From the edge of the line at an angle of c° draw a | line of length | b cm. To finish it join up all the | sides. In test you generally should leave the [] lines | construction--. |

To construct an angle of 30°: draw | a straight line, set the compass width to a length that is more than half way across the | line, draw an arc from under the line to above the line from | the end of the line, from the crossing point draw another | arc that crosses the previous arc. Then from the new crossing point created, draw another | arc that crosses the | previous arc. The line that joins with the starting line to make the 30° angle goes from the side first used on the starting line to the last created | crossing point. | |

Every member has 'an equal chance of being selected' in a type of sampling known as | random sampling. The sample could be chosen by giving every member of the population a | number and using random | number tables, or on a calculator; the random | number function. To ensure a sample is random as well as accurate, the sampling should be | repeated a few | times and then the results | averaged. | |

The amplitude of a trigonometric function is the maximum displacement on the graph of that | function. In the case of sin and cos functions, this value is the leading | coefficient of the function. If y = A sin x, then the amplitude is | 'A'. The amplitude would be infinitely large in the 4 cases | of tan, cot, sec, and csc, regardless of the value of | A unless the sought domain is | limited, where 'A' would determine the | maximum height of these | functions. (ie. Y=sine 2x.) |

The formula for experimental probability is | occurrences/total tries. | |||||||

The first non-negative point that the graph of sine passes is | 0°, it then goes up to | 1 by (in degrees) | 90° or in radians, every | pi/2 radians. | ||||

The graph of Cosine starts at a y value of | 1, meaning the first non-negative point on the x-axis it crosses is | 0°. It then goes down to | -1 until (in degrees) | 180° or (in radians) | π radians, then heads up again to 1 by the same amount of both | radians and degrees. This is continually | repeated. | |

The graph of sine is similar to the plot of | cosine. The length that they are apart is (in degrees) | 90° or (in radians) | π/2 radians. | |||||

On a graph the lines for y=tan x looks quite like the y=x.. | ^3. Instead of being together, unlike the lines for y=sine x and the lines for y=cos x, the LINES for y=tan x are | seperated. The centre points, where the lines cross the | x axis repeat every | 180°, with the first non-negative number crossed being | 0. The value of tan x is infinite starting at | 90° and after or before that every | 180°.At π/2 radians(90°)(and -π/2, 3π/2, etc) the function is officially undefined, because it could be positive Infinity or negative Infinity. The lines at which tan is 'infinite' occur in the gaps between the | curves. |

At points where the value of tan (x) is infinite, the function is officially | undefined, because it could be either | positive infinity or negative infinity. | ||||||

Pi radians converted to degrees are | 180. | |||||||

With Sine, Cosine and Tangent graphs, on the x axis, there are both (or just one of) | radians and degrees, while on the y axis there are | numbers. On cosine graphs and sine graphs, these numbers on the y-axis go from | 1 to -1, but on tan graph, the numbers are | infinite. | ||||

On inverse graphs of trigonometric functions, the labels on the axes | swap. The graph for inverse sine is similar to the plot for the | tangent function, but the middle section is | longer. The plot for inverse sine goes upwards from | left to right and is similar to the plot for the | inverse cos function but | backwards. The plot for inverse tan is like the plot for tan but with the lines made | backwards and turned (in degrees) | 90 degrees. |

The cos function can be written as | cos(x), while the sine function can be written as | sine(x) and the tan function can be written as | tan(x). To write the inverse functions for these using indices write them like that but to 'the power of' | -1 or in a fraction one of (in alphabetical order) | 1/cos(x), 1/sine(x) and 1/tan(x). | |||

N represents cos, tan or sine and a is a value. To solve the equation a* n(x)=b for c°>=x>=d°, draw the lines of both | y=a*n(x) and y=b. The answer(s) is/are the values of x where the lines | intersect between | c° and d°. | |||||

When working with continuous data, in the first column, the intervals are arranged in order of | size from | smallest to biggest. The other 3 columns are | 'f' 'x' and 'fx'. 'F' stands for | frequency, 'x' represents | the midpoints of the interval and 'fx' can be found by multiply | the frequency and the mid-point. The '∑' of 'f' and 'fx' can be found on the row at the | bottom. | |

To draw a cumulative frequency graph from a table of data, on the x-axis: plot the | top value of each | class interval and on the y-axis: plot the | cumulative frequency. Afterwards plot points on the graph appropriate to the | data on the | table. Join the points with a smooth | curve. The line starts at the | lowest value. | |

To prove the sum of angles in triangle 'abc'(labelled left to right): draw a straight line(OP) through | 'b'. Angle 'OBA' is equal to angle | 'a' because of | alternate angles and angle 'OBC' is equal to angle | 'c' because of | alternate angle. Hence angles: 'a' + 'b' + 'c'= (with 3 letter:) | OPB + ABC + CBP which equals (in degrees) | 180°. Angle 'ABC' may be shortened to angle | 'b'. |

The sine and cosine rules can be used for triangles that do not contain a | right angle. If the length of 1 side and the sizes of 2 angles are known, then to find the length of the missing side(s) or the value of the angle(s): use the | sine rule. To find a side when 2 sides and the angle between them is known: use the | cosine rule(a²=b²+b²-(2bc x cos A). To find an angle when you know the lengths of the 3 sides: use the | cosine rule in the form cos A= | (b²+c²-a²)/2bc or cos B= | (c²+a²-b²)/2ca or cos C= | (a²+b²-c²)/2ab. If the lengths of 2 sides and the size of the angle opposite one of these 2 sides are known, then to find the value of the missing angle(s) or the length of the missing side: use the | sine rule. |

To find a side in a triangle when you are given 2 sides and the angle between them, then use the | cosine rule with the formula: a²= | b²+c²-(2bc x cos A) or (b²= | c²+a²-(2ca x cos B) or c²= | a²+b²-(2ab x cos C). | ||||

The hypothetical trapezium has top side: a, buttom side: b and length between: h. To prove the area of the triangle: draw an adjacent and connected | trapezium (of the same | size, with the same | labeling) but with the difference that it is inverted. The new shape is now a | parallelogram(#1). The formula for the area of a parallelogram is height x base length. 'H' replaces height and base length is replaced by | a+b thus the new formula is | hab, but since that's for two trapeziums, the formula for one is hab | /2 or ab | /2 x h. |

Proof that the angle subtended by an arc at the centre of a circle is twice the angle subtended at any point on the circumference. Construct the situation: mark two points that are below the [] and that are on the [] | centre-- circumference, then mark another point that is above the [] and that is on the [] | centre--circumference-- (for the angle 'b'). Join each point below the centre to both the | point at the circumference and the centre (making angle 'a'). [#A25]. Add a radius from the centre to | b, making two triangles that are isoceles due to the | radii. Label: label the equal angles in the triangle to the left 'w' and the angle at the origin 'x', then label the equal angles in the triangle to the right 'y' and the angle at the origin 'z' [#A26]. Using the angles in their triangles: x=[] and z=[] | x=180°-2w and z=180°-2y. Using those values of x and z, and the sum of angles around a point: 360° = a+x+z =[...], which simplies to 0 =[...] | 360°=a+(180°-2w)+(180°-2y), which simplifies to 0 = a-2w-2y, therefore (with brackets) a = [], thus a = []b | a = 2(w+y), thus a = 2b. Q.E.D. |

A chord forms two | segments. The two types of segments are known as the | major and minor segments with the major segment being the | bigger one. | |||||

Proof that angles subtended by the same arc, and are at the circumference are equal. Construct the situation: mark the circle's [] and draw a c[] | centre and draw a chord. Draw two triangles in the same segment that share the [] as one of their sides | chord--. In each triangle: label the angle opposite the | chord. Label the angle at the left: a and label the other angle: b [#A27]. Using the points where the chord touches the circumference as vertices: draw a | triangle with the other vertex at | the centre of the circle. Label the angle at the centre - c [#A28#]. c = []a = []b | c = 2a = 2b (based on the circle theorem that states: | the angle subtended by an arc at the centre of a circle is twice the angle subtended at any point on the circumference). c = 2a = 2b, therefore 2a=2b, thus | a = b. Q.E.D |

To proof the sum of opposite angles of a cyclic quadrilateral(in relation to the circle) is 180°. In the circle: mark the | centre. Create a cyclic quadrilateral, with the vertices at the | circumference. Label the two | opposite angles (a and b). To each of the other vertices: draw a | radius. As an angle subtended at the circumference by an arc is half that | subtended at the centre: the angles at the origin are | 2a and 2b [#A51#]. 2a + 2b = | 360° (which is the sum of angles around a point). Therefore a + b = | 180°. Q.E.D. |

[#A24]We want to prove the alternate angle theorem that the angle between a tangent and a chord at the point of contact is equal to the angle in the alternate segment. Construct an alternate segment situation in a | semicircle. Remember to add a tangent to the diameter enclosing the semicircle. Label as shown in [#A24]. When tangents meet with radii: an angle is formed of | 90°. Thus BCA + z = | 90°. As it is subtended at the circumference by the diameter, angle ABC = | 90°. ABC + BCA + y = | 180° (due to the sum of interior angles in a triangle). Simplifying the equation gives BCA + y = | 90°. We have established two addition equations that result in 90°: | BCA + y = 90° and BCA + z = 90°. You can state that in order for them to be true,: [] = [] | z = y. Q.E.D. (That can be shown through more rearrangement.) |

If we have 2 curves: o and p, we can put (the equation of) o=(the equation of) | p. Rearrange it into the form of | ax²+bx+c=0 (or higher polynomial if such is the case), then | factorise. Seperate the 2 pairs of | brackets and for each of the seperate equations(which equal 0), find the value of | x, then input each of the values into any of the | original equations (any as they'll have the same result since o= | p), to get the value of | y. |

To solve the a quadratic inequality, you can sketch its | graph. You can replace (>, =>, < or =<)0 with | =y. Then change the other side of the equation through | factorisation. If it factorises to (x+n)(x-o), then y=0 at both x= | -n and o. As the equation was in the form: y=ax²+bx+c, it is possible to find the y- | intercept. It is also possible to find the vertex of a quadratic function, which is found where x = | -b/2a. Then knowing the roots, the y intercept and the vertex: sketch the | graph, then note where the graph is (>, =>, < or =<) | 0 using inequalities. |

When trying to solve graphical inequalities graphically and there isn't (>, ≥, < or ≤)0, you can make it like that through | rearrangement. | |||||||

When solving a quadratic inequality graphically, with an inequality that includes -x²: the graph is like the graph of an equation that includes x², but | inverse (upside down). | |||||||

If the equation of a graph can, through factorisation, turn to (x+n)(-x+o), then (-x+o) can be changed to (o | -x). The roots of the graph (with equation: (x+n)(o-x) are | -n and o. | ||||||

To solve a simultaneous equation: 1)rearrange the equations into the form(where 'a', 'b' and'c' are (possibly negative) numbers): | ax+by=c, . 2) Match up the | coefficients of either the | x's or y's, by multiplying 1 or more of the | equations by a suitable | number. 3) Find the difference between both equations and divide it by | the difference between the | non-matching cooefficient to find the value of one of the variables, which can be used to find the value of the other | variable. |

A step graph is a special type of | line graph that is made up of | lines in several horizontal (either named) | intervals or ‘steps’. The end of the previous horizontal step lies on the same vertical line for the start of the next | step. In a step graph: each horizontal step has a opened and closed | circle. A point is excluded from that step if there is an | from that step when there is a | closed circle. | |

It is a horizontal asymptote when: as x goes to infinity (or to -infinity) then the curve approaches some fixed | 'constant' value | "b" on the axis of | y. It is a Vertical asymptote when: as x approaches some constant value "c" on the axis of | x(from the left or right) then the curve goes towards either | infinity (or -infinity). It is an Oblique asymptote when: as x goes to infinity (or to -infinity) then the curve goes towards a line defined by | y=ax+c (note: a is not | zero as the line would be | horizontal). |

To find the n-point moving average find the average of the first | n numbers. Repeat through all the data values, until you have included the | last data value, each time move along one | data value (number k amount of time: first data value included is data value | k. You may then make a list of the discovered | averages. The averages may then be plotted onto a | graph and roughly through the points as a sort of line of best fit line, you may draw a straight | line. | |

The trapezium rule is a numerical method for estimating | integrals. It is most useful when, to an integral, there is no | analytical answer and all that is needed ((not)) is a | number. It works by approximating the area under the curve by a series of | trapezia, then adding the evaluated | areas. The formula for the trapezium rule is (#a6#)where y0 (0 SS) = | f(x0) (0 SS) and y1 (1 SS) = f(x1) (1 SS)etc, and the step size or the increment by which the values of x increase is | h. For ease of calculation it may be a good idea to have the values of y | tabulated. |

In general, for the integral of concave functions, the trapezium method gives an | underestimate, while it gives an overestimate for | concave functions. | ||||||

Plotting lines from exponential equations: "The most convenient form in which to analyse the relationship between two variables x and y is the | linear form, and we must see if, given a suspected relationship between two variables, we can construct a linear graph from the suspected relationship." Eg the relationship between mass(m) and time(t) is m=ab^t. To make the graph linear, of both sides, | take the log. to get log m=t*log b + log a With the data previously given, if not, it may be best to present it in a form that's | tabular. As the log of both sides have been takes, have the table | recalculated. Using the data in the table, the value of log a and lob b can be deduced, giving the line's | equation. If log a = p: a = | 10^p, if log b = q: b = | 10^q. As a = 10^p and b = 10^q log m = t*log b + log a: m = | 10^(t*10^q + 10^p). |

Proof of the sine rule: the sine rule states that for a triangle (labelled like #a7#): | sine A/a = sine B/b = sine C/c. "To prove it, start by drawing a | perpendicular line from a | vertex to the opposite side, say from B to b. Label this perpendicular "x". Then x is equal to both | c sine A and a sine C. Thus c sine A = a sine C and sine A/a = | sine C/c. Draw another line from a vertex to the | opposite side: y, thus y is equal to both | c sine B and b Sine c. Thus sine B/b = sine C/c, and as sine A/a = sine B/b: | sine A/a = sine B/b = sine C/c. #A8#. |

Example: The first three terms of a geometric sequence are k+1, 3k-2 and 4k+4. Find k, the first term and the common ratio. The common ration is the value when the terms are | divided by the terms before. [First terms: k+1, 3k-2, 4k+4] Hence the common ratio may be deduced with the equation: | 3k-2/k+1 = 4k/3k-2. K may be deduced through | rearrangement. The value k = 0 or k = 4; If k=4: the first three terms are 5, 10, 20. The first term is 5 and the common ratio is 2; if k=0: the first three terms are 1,-2, 4. The first term is 1 and the common ratio is | -2. Example: The 1st, 2nd and 4th terms of a geometric sequence are the 1st, 2nd and 3rd terms of an arithmetic sequence. Find the common ratio of the geometric sequence. The 1st, 2nd and 4th terms of a geometric sequence may be written | a, ar, ar^3. The difference of the sequence is equal to both | ar-a and ar^3-ar, thus d=ar-a=ar^3-ar. The value of r may be deduced from ar-a=ar^3-ar through rearrangement; r=-1±√5/2, but if r<0, then the signs would | alternate and it wouldn't be | arithmetic; so r(the common ratio)=1+√5/2. |

Typically we have to solve equations of the form: e^4x-7e^x+10=0. If we substitute e^2x = | y; we obtain the normal quadratic equation: y²-7y+10, which factorises to | (y-2)(y-5), so y=2,5. If y=2: e^2x=2; 2x= | ln(2); x=ln(2)/2. If y=5: e^2x=5; x= | ln(5)/2. | ||||

Example: 4^6x - 5*4^3x -4 = 0, have y= | 4^3x. If y=4^3x in 4^6x - 5*4^3x -4: to find y solve the equation: | y²-5y-14; y=7,-2. If 4^3x=7: 3x= | log4(ss)7; x= | (log4(ss)7)/3; if 4^3x=-2: x= | (log4(ss)-2)/3. As there is no value of | log4(ss)-2: the solution is x= | (log4(ss)7)/3. | |

The formula for the difference quotient for a function f(x) at a point p=(x,y) is | f'(x)= (lim Δx→ 0) f(x+Δx) - f(x)/Δx [= dy/dx = df/dx = d(f(x)/dx]. Find the difference quotient for the function f(x)=x²+2: f'(x)= | (lim Δx→0) (x+Δx)²+2-x-2/Δx= | (lim Δx→0) 2xΔx/Δx = 2x. | |||||

The sum of the first n terms of a geometric progression is: | a(1 - r^n )/ 1 – r. This can be proved as follows: Sn(ss)= [the first 3 and the last value] | a+ar+ar²+...+ar^(n-1). Sn(ss) = a+ar+ar²+...+ar^(n-1) | ||||||

The sum of the first n terms of a geometric progression is: | a(1 - r^n )/ 1 – r. If r is between | 1 and -1, thus |r| < | 1, then we may sum an infinite number of terms and obtain a proper answer. Since in the expression for Sn(ss), for |r| < 1: r^∞ = | 0. If r to the power of infinity is 0, and Sn(ss) = a(1 - r^n )/1–r, then S∞(ss) = | a/1-r. | |||

The first term in a geometric sequence is 4 and the 4th term is 0.0625, find the least value of n, such that the difference between Sn(ss) and S∞(ss) is less than 10^-6. 0.0625 = 1/16 = 4r^ | 3; r = | ∞(ss) - Sn(ss)|. |S∞(ss) - Sn(ss)| = |a | /(1-r) - a(1-r^n)/(1-r)| = (through simplification) |ar^n/1-r|. As "r" = 1/4, "a" = 4 and |ar^n/1-r| < 10^-6: through substitution |ar^n/1-r| may be changed to 4(1/4)^n/1 - 1/4 = 4(1/4)^n / (3/4). Rearranging gives (1/4)^n < | 3/16 * 10^-6. You may rearrange for n by taking | the logs to get | n > log(3/16 * 10^-6) / log(1/4) > 11.17, but as n is an | integer, n = | 12. |

To solve cos2x = sine (x), use one of the quadratic trigonometric formulae(cos2x=1-2sine²x) and substitute to turn the equation into sine (x)= | 1-2sine²x, rearrange to get 0= | 2sine²x+sine(x)-1, which factorises to | (2sine(x)-1)(sine(x)+1). The value of x is either | sine^-1(0.5) or sine^-1(-1). If an equation may not be factorised, it may be solved using the | quadratic formula. Example: solve -.3cos2x=2sine(x); by using one of the formulae (cos2x=1-2sine²x) there may be the obtaining of the equation: 0= | 2sine²x+2sinex-1; this may be solved by using the quadratic formula, one way through substitution of sinex with a variable. If p = sine x: 0 = | 2p²-2p-1. | |

Solve sine2x=cosx. The equation sine2x=cosx leads to 0= | 2sinxcosx-cosx, which factorises to give 0 = | cosx(2sinex-1). There may be some trigonometric equations that are not actually quadratic, but still require manipulation: 3cosx=7sinx, to solve divide both sides by | 7sinex to get 3 | /7 = tanx: x= | tan^-1(3/7). | |||

PFT(1): Let p(x) be a polynomial of degree | n; (x-a) is a factor of p(x) if and only if (x-a) divides | p(x) or is, of p(x), a | factor. Let p(x) be a | polynomial and let a be a | number. | |||

If (x-a) divides | p(x), then the remainder on division of p(x) by | (x-a) is | 0 and there is q(x) which is a | polynomial such that p(x) = | (x-a)q(x), so that p(x)= | 0 and (x-a), of p(x), is a | root. Now assume that x=a is a root of p(x), so that p(x) = | 0. |

Perform on p(x) long | division by | (x-a) to obtain quotient | q(x) and remainder | r(x) to have p(x)= | (x-a)q(x)+r(x). The degree of r(x) is less than the degree of | (x-a)[?] so r(x) is a | constant. Write r(x)= | c. |

Substitute x=a into | p(x) = (x-a)q(x)+r(x) to get p(a) = | 0 (the value of c is thus | 0). PRT: f(x) = (x-a)·q(x) + r(x), but as r(x) is a constant substitute it for | c, so f(x)= | (x-a)q(x)+r(x); with x being equal c: f(c)= | r. | ||

The cosine rule states that a²=b²+c²-2bccos(A), to prove this draw (in#a7#) a line from | B to b. Split b into half, to the left: label | x, and to the right: label | x. Pythagoras' theorem gives x²+y²= | c², thus y²=c²-x², a²= | (b-x)² + y²: y²=a²-(b-x)²=c²-x²: a²-(b-x)²=c²-x²: c²-x² = a²-b²-x²+2bx, hence a² = | b²+c²-2bx, as x = | c cos A: a² = | b²+c²-2bccosA. |

A quantity is said to grow or decay exponentially if the quantity at the start of each time period is multiplied, to obtain the quantity at the end of the period or the start of the next period, by a c... | constant factor.If the constant factor is less than 1, then the quantity is exponentially | decaying. If the quantity is greater than 1, the quantity is exponentially | growing. | |||||

We can estimate an integral using the [] or [] rule | least in each interval (the first three being | [n,n+1]{n+1,n+2][n+2,n+3]) and add up the areas of the | rectangles with these | heights. If the value of the sum is equal to o and i is the actual value of the integral: n | <= i. A similar method may be used when finding overestimates, but in the intervals find the maximum value of | y and adding the areas, with these heights, of the | triangles. If the value of the sum is p: the the value of i may be written with inequalities as | p>=i=>o. |

It is possible to define the trigonometric ratios for angles of any size using | coordinates. Draw a | circle with radius | 1 unit. The point P with coordinates ( | x,y) moves round the circle's | circumference. OP(with O as the origin) makes an angle θ with the positive | x-axis. The angles increases as P rotates anti-clockwise. For any angle θ the sine, cosine and tangent are given by the coordinates of | P. Consider a triangle is formed, with a point at P, O, and the x-axis; as implied, θ by O and the hyp' is 1; the opp' is y and adj' is x; using how trig' functions are defined in a right-angle triangle: the value of sineθ, cosθ and tanθ respectively are | (sineθ=)y/1=y, (cosθ=)x/1=x, (tanθ=)y/x. |

Sine, tangent and cosine are positive between (in degrees) | 0 and 90; between 90 and 180 degrees the trigonometric function that's positive is | sine; between 180 and 270 degrees, the trigonometric function that's positive is | tangent; between 270 and 0 degrees, the trigonometric function that's positive is | cosine. | ||||

Partial fractions: a question may seek for the the expression of fractions such as 1/x²-2x-8 in the form | A/Cx+F + B/Ex+F; as x²+2x-8 factorises in (x+4)(x-2): 1/x²+2x-8 = 1/(x+4)(x-2) = A/ | x+4 + B/x-2; to get 1=A(x+0)+B(x+p), multiply through by | (x+4)(x-2) to get 1=A(x-2)+B(x+4) after cancellation. Eliminate A by having x equal to | 2 to get b= | 1/6, then eliminate B by having x equal to | -4 to get A= | -0.5. Thus 1/x²+2x-8 = | -1/2(x+4) + 1/6(x-2). |

If we have a denominator which includes an quadratic term which does not factorise, include, in the answer a term | Dx+E/Ax²+Bx+C. In general we have one term for every term in the denominator which does not factorise, and for that term, the highest exponent of x in the numerator is less than the highest exponent of x in the denominator by | 1. When trying to find the value of variables, variables of which the value is known may be replaced by their | value. | |||||

Two functions can be combined to make a more complicated function through multiplication. Then they can be differentiated using the | product rule: | dh/dx = f*dg/dx + g*df/dx = (f· | g)' = | f'·g + f·g', if a function h is comprised of the multiplication of the functions | f and g. The product rule can be used repeatedly with any number of | products. If a function h consists of three simpler functions and multiplied together, then dh/dx = (with apostrophes) | (e·f·g)' = (with apostrophes) | e'·f·g + e·f'·g + e·f·g'. |

The chain rule: d(u(v))/dx = | du/dv * dv/dx. Example differentiate cos(x²+x): u = | cosv, v= | x²+x, du/dv = | -sinev, dv/dx = | 2x+1. The answer is thus | -(2x+1)sinev, but as the final answer should be in terms of | x, substitute for v: | x²+x. |

Linear combinations of trigonometric formulae are very important: in fact any continuous function can be expressed as a sum of [] and [] terms under certain conditions: | sine and cosine. Any function of the form Acos(x)± Bsine(x) can be expressed in the form, either, | Csine(x±a) or Ccos(x±a). C can be found in any case by the equation: C= | √(A²+B²), but for a there are 4 possibilities. If we are expressing Acos(x)±Bsine(x) in the form C(sin±a) then a = | tan^-1(A/B), if Acos(x)±Bsine(x) is expressed in the form Ccos(x±a) then a = | tan^-1(B/A). Where Acosx - Bsinex = Csine(x-a): a = | -tan^-1(A/B), where Acosx - Bsinex = Ccos(x+a): a = | -tan^-1(B/A). | |

"Exponential growth means growth without | limit." The rate of growth of a quantity is directly proportional to the | quantity and this leads to equations of the form | Q = Ae^ct, where A and c are | constants, Q represents the | quantity and t represents the | time. If we know the exponential function we can find the quantity present at any time by substituting, into the expression for the quantity, the value of | t. | |

Exponential functions: "In the long term of course, nothing ever grows without | limit, ... and exponential growth functions can only apply over certain | ranges. If a question ever asks, why is this wrong, and you have arrived at an exponential function, the answer is probably because in the long term exponential growth functions are | impossible." Exponential decay are | posibble. "The best example is probably the exponential decay | curve. The quantity of a radioactive material decays smoothly to zero, and zero is a very plausible quantity to have." | |||

In calculus, the chain rule is a formula for computing the derivative of the composition of two or | more | functions. That is, if f is a function and g is a function, then the chain rule expresses the derivative of the composite function | f ∘ g in terms of the derivatives of | f and g. For example, the chain rule for (f ∘ g)(x) is df/dx = | df/dg x dg/dx. d(f(g(x)/dx = (with apostrophes) | f'(g(x)g'x), possibly think of them as layer: differentiate the outer layer first, then the inner layer: to differentiate (3x+1)²: first differentiate the outer layer to get | 2(3x+1)^(2-1), then differentiate the inner layer to get | 3, multiplying both terms together gives 6(3x+1). |

To find dy/dx when x is a fuction of y or there are several occurrence of both x and y, in these cases differentiate | implicitly. | |||||||

Example: differentiate x = sine(y): differentiate both sides with respect to | x obtaining 1 on the left hand side, but when differentiating the right hand side we must remember that you are differentiating a function of | y with respect to | x and so use the | chain rule, to get in this case(with working out) d(sine(y))/dx = | d(sine(y))/dy * dy/dx = | cosy(dy/dx). As the left hand side if the equation is equal to one: 1 = cosy(dy/dx): dy/dx= | 1/cosy. If you are to express y' (dy/dx) in terms of x, in this case it is possible: by using cos²t+sine²y=1, you may get the equation: cosy = √(1-sine²y): dy/dx = 1/√(1-sine²y): dy/dx = 1/ | √(1-x²). |

Find y'(dy/dx) for x² + xy + y² = 1, to both sides differentiate to get (with apostrophes) | 2x + xy' + y + 2yy' = 0, put to the right side the terms without | y' to get | xy'+2yy' = -2x-y, then | factorise to recieve | y'(x+2y) = -2x -y, to find the value of y': to both sides | divide by | x+2y to get | y' = -2x-y/x+2y. |

When wanting to find the tangent or normal to a curve with an equation where y is not explicitly a function of y: in these cases typically: differentiate | implicitly and find dy/dx as a function of | x and y, and then substitute a point | (x,y) into the equation to find, at that point, the | gradient. You may find the equation of the LINE with the equation: | y-y1 = m(x-x1) (1 ss). | |||

Solids or volumes of revolution: We start with a graph y=f(x). If the graph is rotated about the x-axis it traces out a surfaces as shown. (#A9#) Between the surface and the x–axis we may form a | solid. We show here how to find the volume of this solid. We may picture the solid as being made up of slices of | y and thickness | Δx, and has volume: ΔV = | πy²Δx; y summing these obtain ΣΔV = | Σπy²Δx, we get an approximate value for the volume. The value becomes exact as Δx tends towards | 0, making the summation an | integral. Hence, if a curve between the values of x=a and x=b is rotated about the x-axis, the volume of the solid formed is | V=(∫^b)a (a ss)πy²dx. |

Integration by substitution: suppose we are integrating a difficult integral which is with respect to | x; you may be able give f(for example) a value incorporating | x, to make the integral | easier. As long as we change "dx" to dx/df df, where dx/df is | f differentiated. By integrating with respect to | t(integrate like usual) "we will get the same answer as if we had done the original integral", afterwards substitute f for its | value. The substitution rule may be defined, where the substitution x = f has been made: ∫u(x) dx = | ∫u(f)dx/df df, but if x = f(theta): dx = | df/dθ dθ. |

Simpson's rule allows approximate calculations of d... | definite integrals that might otherwise not be easily | calculable. The approximation is given by | I = (∫^b)a(a ss)f(x) dx ≃ (b-a)/n((y0 + 4y1 + 2y2 + 4y3 +...+ 4y(n−3) + 2y(n−2) + 4y(n−1)+y(n))(values after y subscripted), where n, the number of | strips is | even and y = | f(x). | ||

cos(90-x) = | sine x; cos(90+x) = | -sine x; sine(90-x) = | cos x; sine(90+x) = | cos x; tan(90-x) = | cot x; tan(90+x) = | -cot x; cot(90°-x) = | tan x; cot(90°+x) = | -tan x. |

sine(90°-x) = | cos x; cot(90°-x) = | tan x; sec(90° -x) = | csc x; csc(90° - x) = | sec x. | ||||

Interchanging the limits of an integral, changes the | sign: (a ss) (∫^b)af(x) dx = | -(∫^b)af(x) dx; (∫^b)acf(x) dx = | c(∫^b)af(x) dx: an integral can be factorised with a | constant. If the limits of an integral are the same: the integral is | 0; (∫^b)af(x)+g(x) dx = | (∫^b)af(x) dx + (∫^b)ag(x) dx: the integral of the sum of two functions is the sum of | ||

(∫^b)aK dx = | K(b-a); if the function is odd: (∫^b)af(x) dx = | -(∫^b)af(-x) dx, if the function is even: (∫^b)af(x) dx = | (∫^b)af(-x) dx; (∫^b)af(x) dx + (∫^c)bf(x) dx = | (∫^c)af(x) dx, the limits follow in a natural way; if (∫^b)af(x) dx = 0 for all a and b, then for all x: f(x) = | 0. | |||

If f(x) = ln x, domain and range (respectively): | x>0, real numbers; sine x or cos x, domain and range (respectively): | ℝ, -1<y<1; tan x, domain and range (respectively): | ℝ - {nπ,n∈ℕ), ℝ; e^x, domain and range (respectively): | ℝ, y>0; | |x|, domain and range (respectively): | ℝ, y=>0; 1/x, domain and range (respectively): | x≠0, y≠0. Where x is multiplied by something or added to something for example: ln(3x-1) rearrange: | 3x-1>0. |

Iterative formulae. The equation x²+2=e^x has a solution somewhere between 1 and 2. If this solution is called a, then a satisfies | a²+2=e^a. "We can rearrange this equation in various ways." Two such are a = √ | (e^a -2) and a = ln | (a²+2)."We can use these rearrangements as iteration formulae to attempt to find a to 2 decimal places say." The iteration formulae are (n and n+1 ss) f(xn) = x | n+1 = √(e^(xn) - 2 and g(xn) = xn+1 = ln(xn²+2), 1.5 = | x0(ss). The progress of the iterates may be shown in g...[] or t...[] form or on a | graph or tabular form. In the graph xn+1 = √(e^(xn) - 2, the gradient of the graph is greater than 1 and the graph crosses the line y=x from below. The iterates | diverge. In the graph below right the gradient of the graph is less than 1 and the graph crosses the line from above. The iterates | converge to a. |

Given the sequence that is defined by the iterative formula: (2x+5)^(1/3) with x0(ss) "= 2" converges to a. 1: find a correct to 4 decimal places, 2: find an equation that has a as a root, 3: does the equation have any more roots? 1: by plugging in | x0 first to get | x1, eventually there is a repetition if the answers are put to | four decimal places, the answer to question 1 is the | repeated value. 2: rearrange | x = (2x+5)^(1/3) to get | x^2-2x-5 = 0. 3: to determine if the function has any roots just graph of both | the highest power of x and the rest (in this case: y=x^3 and y = 2x+5), the number of roots is equal to | the number of times the graphs cross(in this case once). |

The chain rule (function of | a function) is very important in differential calculus and states that: dy/dx = | dy/dx x dt/dx. Example: If y = (1 + x²)³ , find dy/dx, let t= | 1+x², therefore y= | t^3; dy/dt= | 3x² and dt/dx = | 2x, dy/dx = 3x² x 2x: turn t into | 1-x² to get dy/dx = | 6x(1+x²)². |

If y^3 = x, how would you differentiate this with respect to x? There are three ways: rewrite it as y= | x^(1/3) and differentiate as normal "(in harder cases, this is not possible!)"; find d | x/dy: dx/dy = | 3y² and use the fact: dy/dx = dx/dy^ | -1 to get dy/dx= | 1/3y²; | |||

Example: Differentiate a^x with respect to x. The value of y would be xa^(x-1) if we were differentiating with respect to | a not x. Put a^x = | y, then of both sides take the | log, to get | log y = log (a^x): log y = | x log(a), differentiating implicitly gives log(a) = | (dy/dx)/y: dy/dx = | y log a = | a^x log a. |

Forming simple differential equations: (possibly almost) "the velocity of a body is proportional to its distance from O. The body starts at 1. If x is the distance from O, then the velocity is the rate of change of distance = | dx/dt, hence dx/dt= | kx, where k is the | constant of proportionality. Now, integrating gives us: | ln x = kt + c, and we know that x(0) = | 1, hence c= | 0, so ln x= | kt: x = | e^kt." |

The quotient rule: if a function h consists of two simpler functions f and g with h = f/g, then dh/dx = | (g*df/dx - f*dg/dx)/g². Proof: dh/dx = (lim h → 0) h( | x+h)-h(x)/h = (using h=f/g) (lim h→0) | ((f(x+h)/g(x+h) - f(x)/g(x)) / h = | (lim h→0) f(x+h)g(x)-f(x)g(x+h) / g(x+h)g(x)h = (by, to the numerator, adding and subtracting | f(x)g(x)) f(x+h)g(x) -f(x)g(x)+f(x)g(x) - f(x)g(x+h) / g(x+h)g(x)h, which factorises to | (lim h→0) g(x)(f(x+h)-f(x)) - f(x)(g(x+h)-g(x)), (lim h→0) f(x+h)f(x)/h and g(x+h)g(x) respectively equal | df/dx and dg/dx, and then if we let h=0 in the denominator g(x)g(x+h) = | g²(x), we the obtain g*f'(x) - f*g'(x)/g², the quotient rule. |

For y=f(x), show that y has a zero a between x=o and x=p: show that between x=o and p, y has a change of | sign(one value bigger than 0, one value smaller), thus somewhere in between x=o and p, there is a value of x,a for which 0= | y(a); to show that a possible solution is given by a=s: in f(x), change x for | a and have the equation equating to | 0, then | rearrange for a; if s, for example is equal to √(1/3-a), the iterative formula is (n+1,n ss) | xn+1=√(1/3-xn); starting from the initial value x0=0.6, find x1, x2 and x3 to four decimal places, then solve the equation and give the solution to three decimal places: to solve the equation continue from x3 until two succesive iterations | agree to | three decimal places, then give the answer to three decimal places. |

Using the multiple angle formulae to find values of trigonometric functions: using #A11# and the multiple angle formulae: cos(A±B)= | cosAcosB±sineAsineB and sine(A±B)= | sineAcosB±cosAsineB; we may find cos 75, sine 75 and tan 75 for example, if we chose the values of A and B properly. To find cos 75, we can choose, say, A=45 and B=30; cos(45+30)= | cos45cos30-sine45sine30: cos 75 = | 1/√2*√3/2 - 1/√2*1/2 = √3/2√2 - 1/2√2 = (√3)-1/2√2 = | √6-√2 / 4; sine(45+30)=sine45cos30 + cos45sine30: sine75 = 1/ | √2 * √3/2 + 1/√2*1/2 = √3/2√2 + 1/2√2 = | √6+√2 / 4; tan 75 = sine 75/cos 75 = (√6+√2 / 4) / (√6+√2 / 4) = √6+√2 / √6-√2, multiply the numerator and denominator by the | conjugate root (√6+√2). |

If you draw the graphs of y=f(x) and y=f^-1(x) on the same axis: y=x is a | line of symmetry: to obtain y=f^-1(x), reflect y=f(x) in | the x-axis(this may be as y and x interchanged). A problem may arise if you have a function f(x) which, for more than one value of x, | gives the same value of y. When you try to invert the function and find y=f^-1(x), a value of x may return either no value or more than one value of y. "It is necessary in a case like this to restrict the inverse function's | domain to eliminate those “impossible” x's and “duplicate” y's. For example, if y=f(x)=x^2 and f^-1(x)= √x, to have only one value of y for each value of x, only take square roots that are | positive. If f(x)=sine xm then f^-1(x) = | sine^1(x), so there is only one value of y for each value of x: take the range to be | [pi,-pi]. | |

The mid ordinate rule: also called the | midpoint rule, it is another method that is for numerically | estimating integrals. It states: if an area of integration is divided into | n strips, the area between | xi and x(i+1)(i,i+1 ss) is given by | (x(i+1)-xi)f((xi + xi+1)/2), so that the width of the strip is multiplied by the y–value at | the midpoint. We do this for all n strips obtaining I≃ | ∑(i=0)^n(x(i+1)-xi)f((xi + xi+1)/2); if all the strips are of the same width: h=b-a / n, where a and b are the limit of integration, then I≃ | ∑(i=0)^n(x(i+1)-xi)f((xi + xi+1)/2). |

The mid ordinate rule states: if an area of integration is divided into | n strips, the area between | xi and x(i+1)(i,i+1 ss) is given by | (x(i+1)-xi)f((xi + xi+1)/2), so that the width of the strip is multiplied by the y–value at | the midpoint. We do this for all n strips obtaining I≃ | ∑(i=0)^n(x(i+1)-xi)f((xi + xi+1)/2); if all the strips are of the same width: h= | b-a / n, where a and b are the | limit of integration, then I≃ | ∑(i=0)^n(x(i+1)-xi)f((xi + xi+1)/2). |

If N0(ss) is the initial population and the growth/decay rate is k: the population N at time t is N = | [A+]N0(ss)e^kt, where k is less than or greater than | 0, sometimes the equation may have a background level which is | A. Example: The temperature of a cup of coffee is give at any time t minutes by the equation N=20+80e^0.1t: a)Write down the room temperature, b)Find the time when the coffee has cooled halfway; a: The room temperature is | 20 and when the temperature when the coffee has cooled completely ie t= | "infinity"....; b: when the coffee has cooled halfway, the excess temperature(possibly the temperature added to the room temperature ) has fallen by half; find it and add it to | 20 and put it in the equation as equal to | N,. then solve for | t. |

Proof of Simpson's rule(1): Let P be a partition of | [a,b] into n | subintervals of equal | width, P: a=x | 0<x1<...<xn = | b, where xi - x(i-1) = | (b-a)/n for i = | 1,2,...,n. Here we require that n be | even. |

Proof of Simpson's rule(2): Over each interval [x( | i-2),xi], for i = | 2,4,...n, we approximate | f(x) with a | quadratic curve that interpolates the points | (x(i-2),f(x(i-2)), (x(i-1),f(x(i-1)), (xi,f(xi))(#A12#). Since only one quadratic function can interpolate any three (non-colinear) | points, we see that the approximating function must be unique for each | interval[xi-2,xi]. The following quadratic function interpolates the three points: | (x(i-2),f(x(i-2)), (x(i-1),f(x(i-1)), (xi,f(xi)): y= (#A13#). |

Proof of Simpson's rule(3): since this function is unique, this must be the quadratic function with which we approximate f(x) on | [x(i-2),xi]. Also, if the three interpolating points all lie on the same line, then this function reduces to function that's | linear. Therefore xi - x(i-1) = | Δx for each | i, integrating f(x) between x(i-2) and xi gives | (#A14#), by evaluation the integral on the right, we obtain: | (#A15#), summing the definite integrals over each | integral | [x(i-2),xi], for i= |

[3]2,4,...,n, provides the approximation | (#A16#),(#A17#). The approximation scheme may be achieved by having the sum | simplified. | ||||||

When quantities depend on other quantities that are changing, for example the volume of a sphere depends on the radius which is increasing at 1 cm per second, we have to be very methodical in our | approach if we want to find the rate of change of the sphere's volume. We use | chain rule, which in this case can be made to relate the rate of change of volume to the rate of change of [] with [] | volume with radius and rate | of change of radius: (the equation is) | dv/dt = dV/dr*dr/dt. Suppose then that the radius is increasing at 1 cm per second, so dr/dt = | 1 and suppose at some instant the radius is 5. As V=4/3πr^3: dV/dr= | 4πr^2 so at the instant when r = 5, dV/dr = 4πr^2 = π, substitute these values into | dV/dt to get dV/dt = 100π *dr/dt = 100π. |

The cartesian form of a plane is | ax+by+cz=d, where a, b and c are | constants. The amount of points used to find the equation of a plane is | 3. Each point determines an equation in | a,b,c,d. Solve the | simultaneous equations to find [] in terms of [] | the constants a,b,c, in terms of d and write down the plane's | equation, finally have the constant d, which appears throughout | cancelled. |

To find the equation of a plane we need three points. Each point determines an equation in (1,2,3),(0,3,1) and (-1,-2,1). Find the equation of the plane. The three simultaneous equations are | a+2b+2c=d(1), 3b+c=d(2), -a-2b+c=d(3). (1)+(3) get c=d/2, substitute c=d/2 into (2) to get b=d/6, substitute b=d/6 and c=d/2 into (1)to get a=-5d/6. The equation of the plane is then | (-5d/6)x + (d/6)y + (d/2)z = d, which simplifies to | (-5/6)x + (1/6)y + (1/2)z = 1: -5x + 6y + 3z = 6 | |||||

There is an alternative form for the equation of a plane, to terms of vectors: f( | t,s) = A+ut+vs, where s and t are | parameters and A is a | point on the plane. For the plane given above, find u and v by, having points on the plane, | subtracted from each other: (the plane passes the points (1,2,3),(0,3,1),(-1,-2,1): u and v are respectively equal to | [u=](1,2,3)-(0,3,1)=(1,-1,2); v=(1,2,3)-(-1,-2,1) = (2,4,2): r(t,s) = | (1,2,3) + t(1,-1,2) + s(2,4,2). "The vector form is not unique since any | points in the plane can be used." | |

(#A19#) If a vector n is perpendicular to a plane p, then it is perpendicular to every | vector u and | line drawn in the plane, hence u•n = | 0. We can find an expression for u by having one point in the plane | subtracted from another. This is shown on the diagram #A19# as as | (#A20#), where p(overbar) is an | arbitrary point and 1(ss) p(overbar)1 is | given, so the equation of the plane may be gotten, by finding either | (1 ss) (p(overbar) - p(overbar)1))•n = 0 or ax+by+cz=ax1 +by1 +cz1. |

Example: Find the equation of the plane perpendicular to the vector (3,2,4) passing through the point (5,2,1). (1 ss) (p(overbar) - p(overbar)1))•n = []•[] | (x-5,y-2,z-1)•(3,2,4) = s(t+u)... | 3(x-5)+ 2(y-2) +4(z-1) = 3x + 2y + 4z - 23 = 0, thus the equation in the form ax+by+cz=d is | 3x+2y+4z =23. | |||||

If two lines intersect, they are both in the same place at the same time, so to speak. We don't know what the point is, but because they both meet at the same point, we can put the equations of the lines equal to | each other. This will result in equations that are | simultaneous for the | parameters. Solve them, hen substitute back in to the lines to find the point. Example: (1,2 ss) r1(t)=(5,2,3) + t(2,1,1) and r2(s)=(2,1,1) + s(5,2,3). Find if the lines intersect and find the coordinates of the point of intersection. r1(t)= | r2(S): (5,2,3) + t(2,1,1) = (2,1,1) + s(5,2,3): put each component of r1(t) equal to the corresponding component of | r2(s). "We obtain" x:5+2t=2+5s: (1)3=5s-2t; y: 2+t=1+2s: (2)1=2s-t; z: 3+t=1+3s: (3)2=3s-t. Solving the simultaeneous equations (1),(2) and (3) gives the value of s as 1. Substitute s=1 into | (1),(2) and (3) to give t=1. If t=1, r2(s) = r1(t) = (5,2,3) + t(2,1,1) = | (5,2,3)+(2,1,1) = (7,3,4), as if s=1: (2,1,1)+s(5,2,3) = (7,3,4). Hence both equations meet at the same point: | (7,3,4). If the lines do not intersect at a point then the two points will not be the same or equivalently there will not be values of s ant t that satisfy all three equations simultaneously and perhaps contradictions when solving. |

Integrating Quotients of Algebraic Expressions: Integrating any quotient of the form Ax+B/Cx+D can be done by making the substitution u= | Cx+D. For example to find ∫2x+1/x-4:u= | x-4: x= | u+4: 2x+1= | 2u+9; dx= | du, so the integral becomes [] = ∫n+... | ∫2u+9/u du = ∫2+ 9/u du = | 2u + 9lnu + C, change the equation by | substituting u for x-4 to get 2(x-4) + 9ln(x-4) + C = 2x + 9ln(x-4) + C - 8. |

There is however an alternative method which is also useful for integrating quotients of higher order polynomials. On the quotient carry out | long division and have the result | integrated. For the example above, 2x+1/x-4 = 2x | -8+9/x-4 = | 2(x-4)+9/x-4 = | 2 + 9/x-4: the integral becomes | ∫2 + 9/x-4 = | 2x+9ln(x-4)+C, which is equal to 2x+9ln(x-4)+C-8 as | C is arbitrary. |

The quotient rule. Example: if y=x^3/x+4, find dy/dx. Let u= | x^3 and v=x+4. Using the quotient rule, dy/dx= [] = [] | (x+4)(3x^2) - x^3(1) / (x+4)^2 = | 2x^3 + 12x^2 / (x+4)^2. | |||||

We want to find the minimum distance between a point P and a line l. If you have the point which makes this distance a minimum | labelled(A), then we "must" find the distance | AP. This distance will be a minimum when the line, to l, is at | a right-angle. This means that the dot product of AP with | the tangent vector v to | l is | 0: 0 = []•[] | v•AP(arrow above). You may use this to find [] and then [] | A and then AP. |

Suppose P= | (1,2,3) and the equation of l is | r(t) = (4,5,6) + t(0,-4,-2) = | (4,5-4t,6-2t). Take the point A as having coordinates | (4,5-4t,6-2t) so the vecor AP is []-[] | P-A = | (1-4,2-(5-4t),3-(6-2t)) = (-3,-3+4t,-3+2t). The tangent vector of the line l is | (0,4,2). The dot product of AP and the tangent vector is []•[] | (-3,-3+4t,-3+2t)•(0,-4,-2)= 18-20t = 0: t = 0.9. |

A = (4,5-4t,6-2t)=(as t = 0.9) | 4,1.4,4.2). AP = | (P,A ss) √(xP-xA)²+( yP-yA)²+IzP-zA)² = | √((1-4)² +(2-14)² +(3-4.2)²) = √10.8 = 3√(6/5) = ([]/[])√[] | (3/5)√30. | ||||

Integrals of the form ∫cos kx dx, ∫sine kx dx or ∫tan kx dx can be evaluated instantly. For example, ∫cos kx dx = | (1/k)sine kx _ C. When the integral is now multiple angle trigonometric terms multiplied together, use a | trigonometric identity to express the integral,using the identities, as | the sum of two trigonometric terms. sine nx cos mx = | (sine(n+m)x + sine(n-m)x)/2 (1), cos nx sine mx = | (sine(n+m)x - sine(n-m)x)/2 (2), cos nx cos mx = | (cos(n+m)x + cos(n-m)x)/2 (3), sine nx sine mx = | (cos(n-m)x - cos(n+m)x)/2 (4), each term of which is seperately | integrated. |

To evaluate the ∫sine 5x sine 3x dx, use the identity | sine nx sine mx = (cos(n-m)x - cos(n+m)x)/2, the integral becomes ∫[] = [] | ∫1/2 cos 2x - 1/2 cos 8x dx = 1/4 sine 2x - 1/16 sine 8x + C. To evaluate ∫cos 7x sine 3x dx use ∫cos nx sine mx = | (sine(n+m)x - sine(n-m)x)/2 to get the integral | ∫1/2 sine 10x - 1/2 sine 4x dx = -1/20 cos 2x + 1/8 cos 8x + C. To evaluate ∫cos 3x cos 5x dx use cos nx cos mx = | (cos(n+m)x + cos(n-m)x)/2. The integral becomes | ∫1/2 cos 8x + 1/2 cos(-2x) dx = 1/16 sine 8x - 1/4 sine 2x + C. To evaluate ∫sine 5x cos 3x dx use sine nx cos mx = | (sine(n+m)x + sine(n-m)x)/2. The integral becomes | ∫1/2 sine 8x + 1/2 sine 2x dx = -1/16 cos 8x - 1/4 cos 2x + C. |

We can find the vector equation of a three dimensional plane given either three points in | the plane, two vectors and | a point on the plane, a point on the plane and | two points or the equation of the plane in some | formulation. The vector equation of a plane is of the form c=[]=[] | (0 ss)r0(arrow overhead)+μv(arrow overhead) = | λw(overhead). By some means obtain both | v(arrow overhead) and w(arrow overhead). If you have three points in the plane A,B and C then you can take r0 to be any of | (arrows overhead)OA,OB or OC with (arrow overhead)v = |

B-A and (arrow overhead)W = | C-A or similar. If A=(2,1,4),B=(4,1,8) and C=(0,-2,-1) then B-A=[],C-A=[] | (2,0,4),(-2,-3,-5) so that with (arrow overhead?)r0 as (arrow overhead)OA, r(μ,λ) = | (2,1,4) + μ(2,0,4) + λ(-2,-3,-5). Another case follows since we can construct a second vector by subtracting | the two points to give a second | vector in the | plane so that we have [] and [] | two vectors and a point in | the plane. |

...All equations of a plane define a | plane, and from some formulation we must be able to find | three points in the plane, then use the method before. Suppose for example that we have the plane with cartesian equation 3x+2y+z=6. This plane contains the points A(2,0,0),B(0,3,0) and C(0,0,6), then B-A =[],C-A=[] | B-A=(-2,3,0),C-A=(-2,0,6), so that with (arrow above?)r0 as OA(arrow above): r(μ,λ) = | (2,0,0) + μ(-2,3,0) + λ(-2,0,6). | ||||

Suppose we want to express a vector v(arrow above) in terms of components perpendicular and parallel to some other vector u(arrow above). We can peg the vectors to | the origin.(#A21#) If the angle between them is θ, where θ is the direction from (ars a) u to v, where the direction is | anti-clockwise, we can write the direction vector of (ars a)u as | u/|u|. The component of (ars a) v in the direction of u is then | (1 ss)v1=v • u/|u|. To find the component of (ars a) v perpendicular to u draw in the | perpendicular to u(#A22#). The component of v(a a) in the direction perpendicular to u(a a) is then | (#A23#). If (ars a)u•v < 0 take | the negative square root and if (ars a)u•v > 0 | take the positive square root. |

To find the angle between two lines, use the same formula as the formula for | finding the angle between vectors. The vectors u and v are the | tangent vectors, which are easily identifiable since the general equation of a line is given by the equation y=mx+c: f(t)=A+tu, where a point on the line is | A(or c) and u or m is the | tangent vector. Example: Find the angle between the lines r1 = 2i+3j-k+t(4i-j-2k) and r2=i+5j-2k+t(i-2j-k). Take u=[] and v=[] | u=4i-j-2k and v=i-2j-k. cosθ = (4i-j-2k)•(i-2j-k)/| | |||

The conditional probability of event a in relationship to event b is the probability that event a occurs given that event b has already occurred. This can be written as P( | a|b) which is equal to | P(b and a)/P(b). Thus P(a|b)=P(b and a)/P(b) is a formula for | conditional probability. | |||||

Usually, with dot plots, when an event occurs corresponding to a value occurs, on the dot plot; over that value there is placed either a | dot or circle. If an event related to a value happens more than once, in terms of the previous one, the new dot or circle is placed | on top. | ||||||

If the events aren exhaustive P(c or d) = | P(c) + P(d) – P(c and d). | |||||||

To solve the Quadratic equation of a'x'^2+b'x'+c=0: first divide | all terms by 'a' then move the term | c/a to the right side of the equation which would change its | sign. Afterwards complete the square on the | left side of the equation and balance it by adding the same value(s) to the | right side of the equation. To complete the square on the left side; square | half of 'b' and add it to | both sides. → | → |

After that simplify the left side by | factorising it into; | (x+p)². Then | square root both sides so that if the equation was previously (x+p)^2=q, it then becomes; | x+p=±√q Afterwards subtract | 'p' from | both sides, leaving 2 answers in the formulae; x= | -p±√q |