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Question | Answer | |||||||
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When plotting absolute value graphs: you may start by finding the | corner, where f(x) = | 0. | ||||||

A differential equation is any equation with one or more differential | terms(i.e dy/dx+3y); if we differentiate y n times, we have the nth differential, notated | (d^n)y/dx^n. | ||||||

If a curve between the values of x=a and x=b is rotated about the x-axis [#A9#], the volume of the solid formed may be found with the formula: V = | V=(∫^b)a (a ss)πy²dx, but if the curve is rotated around the y-axis: V= | (∫^b)a (a ss)πx²dx; with finding volume when the shape (between the x values of a and b) is rotated about the y-axis : the limits to the integral should be, for each of x=a and x=b, the value of | y. If not finding a definite integral, use the formulae, but without their | limits. Note: after finding the value of the area or definite integral, it may be appropriate to add 'cubic units'. | ||||

∫x^(n) dx = | x^(n+1) / n+1 + c; ∫f'(x)/f(x) dx = | ln|f(x)| + c. | ||||||

sine(A+B) = | sineAcosB + cosAsineB, sine(A-B) = | sineAcosB - cosAsineB. If B=A, sine(A+B) = sine(2A) = [] = [] | sineAcosA+cosAsineA = 2sineAcosA. | |||||

2sineAsineB = | cos(A-B)-cos(A+B); 2cosAcosB = | cos(A-B) + cos(A+B); 2sineAcosB = | sine(A+B) + sine(A-B); 2cosAsineB = | sine(A+B) - sine(A-B). | ||||

If an integral has no limits it is called an | indefinite integral, and to the integral of the function f(x), which is written | F(x), add an | arbitrary constant, "usually written | C." | ||||

[Using capital letters to represent integrals] ∫f(x) dx = | F(x)+C; ∫cf(x) dx = | cF(x)+C; ∫ f(x)+g(x) dx = | F(x) + G(x) + C. | |||||

With the function y = f(x): the set of values of x that can be put into the function to calculate a value of y is the | domain. Note: a value of that the graph tends to but doesn't reach is not in the domain. | |||||||

The set of values of y that a function can take is its | range. | |||||||

If y = sine x: dy/dx = | cos x, if y = cos x: dy/dx = | -sine(x). When differentiating trigonometric functions have the x-values in | radians. {If y = sec x: dy/dx = | sec x tan x, if y = cosec x: dy/dx = | -cosec x cot x, if y = cot x: dy/dx = | |||

If y = tan x: dy/dx = | sec²x. With questions such as: differentiate cos³x with respect to x": may be best to apply | the chain rule. | ||||||

Verify that n is a solution to the differential equation: dy=f'(x); to solve: put y= | n, then find | dy/dx, into the left side of the equation substitute | n, which is a solution if both sides are | equal. | ||||

"When given a differential equation, you will often be asked to "solve" the differential equation or find the "general solution". This basically means find an expression which does not contain any | derivatives. To do this you will need to | integrate." | ||||||

1 + cot²x = | cosec²x, tan²x + 1 = | sec²x, tanx = | sinex/cosx. | |||||

We say that a function is one-to-one if, for every point y in the range of the function, there is only one value of [] | x such that y = | f(x). Is f(x) = x² one-to-one? | No, there are two values of x such that f(x) = 4 (namely –2 and 2). On a graph, a function is one to one if any horizontal line cuts the graph | once. | ||||

Functions can be graphed. If a function has no breaks: it's | continous. | |||||||

A function is even if f(x) = | f(-x) . The graph of such a function will be symmetrical in the | y-axis. Even functions which are polynomials have degrees that are | even (e.g y=x^2). An odd function f(x) is such that -f(x) = | f(-x). The graph of the function will have rotational symmetry about the | origin of order | 2, and with be symmetrical about the line | y=x. A function is neither even nor odd if it has no s[] | symmetry. |

The graph of y=|x-1| is the graph y=|x| shifted one place to | the right. | |||||||

To solve |f(x)| = g(x), solve both []=[] and []=[] | f(x)=g(x) and -f(x) = g(x). Unfortunately, as f(x) and -f(x) (from reflecting the negative parts of f(x) in the x-axis) may have intersected with g(x) more if they had gone below the x-axis, you may get some wrong solutions which need to be eliminated. | |||||||

The set of all values that a function may take is called either the | range or codomain of the function. Note: a value of y that the graph tends to but doesn't reach is not in the range. | |||||||

A function of x is any formula with x as the only | variable, so that a value may be chosen for x , and for this value of x, | the value of the function may be calculated. A function is usually given in the form of an | equation. | |||||

If you want to compose functions f(x) and g(x): as they are both functions of x: you may notate it as f[]=f[]=f[] or g[]=g[]=g[] | f(g(x))=f[g(x)]=fg(x) or g(f(x))=g[f(x)]=gf(x) (note f[g(x)] may have a different value to g[f(x)]). With the function f(g(x)) (for example), apply first | g, then f (i.e apply the function f to the function g). With the function g[f(x)], the pronounciation is [] of [] of [] | g of f of x. | |||||

If one graphs y=f^-1(x) and y=f(x), there is a line of symmetry with equation y= | x. The graph y=f^-1(x) can be obtained by reflecting f(x) in [...] | reflecting y=f(x) in the line y=x; note: the as[]s are also reflected. | asymptotes-- | |||||

If one is to estimate the value of an item to three decimal places, calculate values to | four decimal places; The final answer is quoted to | three decimal places. If the estimated value is e and the actual value is a, the percentage error may be calculated with the equation: percentage error = | |e-a|/a * 100%. | |||||

{Exponential decay is a real phenomenon with many practical applications – radioactive decay, Newton's Law of Cooling, measuring and controlling the thickness of metals – since the intensity of radiation exponentially | decays with penetration into the metal.. Radioactivity is used in smoke alarms to – too low a level of radiation detected will | set off the smoke alarm. Because the alarm must have a certain lifespan, the activity must be calculated at this point and the detector calibrated accordingly. The general equation for exponential decay is | N = [A+]N0e^(kt), A is | the background temperature or background level of radioactivity for example; N0 is | the initial excess temperature or level of radioactivity for example; k is | the decay/growth constant. | ||

Integration by parts is used to integrate complicated | products. | |||||||

{Integrating f(x) between x=a and x=b may also be written, where F is the integral of f, as | F(b)-F(a). | |||||||

With odd functions: the definite integral ∫^a(-a) f(x) dx= | 0. | |||||||

Even function*odd function= | odd function; even function*even function= | even function; odd function*odd function= | even function. A function is neither odd nor even if it has no | symmetry. | ||||

R{Odd or even: y=x: | odd; y=x^2: | even; y=cos x: | even; y = x^3: | odd; y= sine x: | odd; y=|x|: | even. | ||

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, use | the 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 | the radius (the equation is) | dv/dt = dV/dr*dr/dt, where 'V' represents [], 'r' - [] and 't' - []. | volume--radius--time--. Suppose then that the radius is increasing at 1 cm per second, so dr/dt = | 1. If asked to find how fast the volume is increasing when the radius is exactly n cm: | input n into dV/dt. | |

Two functions can be combined to make a more complicated function through multiplication. Then they can be differentiated using the | product rule: (where y=f*g, and f and g are functions of x) | dy/dx = f*dg/dx + g*df/dx = (f·g)' = f'·g + f·g'. R{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 = (e·f·g)' = (with apostrophes) | e'·f·g + e·f'·g + e·f·g'. | ||||

The chain rule: if y=f(u), where u is a function of x - dy/dx= | dy/du * du/dx. Make sure that you final answer is in terms of | x. Example differentiate cos(x²+x): u = [], y=[] | x²+x--cos(u)--, dy/du = | -sine(u), du/dx = | 2x+1. The answer is thus | -(2x+1)sineu, but as the final answer should be in terms of | x, substitute for u: | 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." | |||

{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 cos²yi+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. |

{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 cylinders with radius | y and thickness | Δx, and thus volume: ΔV = | πy²Δx; 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. |

Simpson's rule allows approximate calculations of [] integrals | definite integrals that might otherwise not be easily | calculable. The approximation is given by | I = (∫^b)a(a ss)f(x) dx ≃ h/3((y0 + yn + 4(y1+y3+y5+...) + 2(y2+y4+y6+...))(values after y subscripted), where h is the | width of each strip, and the last ordinate (the value of a coordinate on the vertical axis) is | yn (n ss), where n must be | even. (It works by considering the top of curves as a series of quadratic curves). | ||

{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 - {(2n+1)π/2,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: in ln(3x-1) for example, for the domain, 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^3-2x-5 = 0. 3: to determine if the function has any roots just graph of both | the highest power of x and the rest put to the other side (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, so the answer is no). |

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. |

Using the multiple angle formulae to find values of trigonometric functions: using A11 and the multiple angle formulae: cos(A±B)= | cosAcosB±sineAsineB [meaning the operation used on one side is used on the other] and sine(A±B)= | sineAcosB∓cosAsineB [if '+' is used on one side, '-' is used on the other, and vice-versa]; 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). |

Mid-ordinate rule: if an area of integration is divided into n strips, the area of the strip between xi and xi+1 (i,i+1 ss) = | x(i+1)-xi)f((xi + xi+1)/2), so that the [] of the strip is multiplied by the y–value at the [] | width--midpoint--. We do this for all n strips obtaining I≃∑(i=0)^n... | (xi+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≃ | h∑(i=0)^n 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= |

The graph of y = ln x is a reflection of y=e^x in the line | y=x, and has vertical asymptote | x=0. | ||||||

The equation of the vertical asymptote of the graph of ln(x) is x= | 0, as at that point ln(x) would be | undefined. | ||||||

To find the inverse of the an exponential function y=f(x): find [] in terms of [] | x in terms of y, then have the terms of x and y | swapped (e.g for y=e^x - the inverse is y = ln x [it prior gave x = ln y]). The inverse of function f(x) is notated | f^-1(x). | |||||

If the domain of f(x) is d and its range is r, then the domain an range respecitvely of f^-1(x) is | r and d. | |||||||

The general formula for growth is | y=A*e^kt, where A and k are | constants and t is | time (e is Euler's number). (In real life, growth of bacteria and other natural phenomena such as population growth can be modelled around this exponential formula). Note: you generally set the first point you have as t = | 0, and all times before can have 't' as negative (e.g a murder victim was found at 4 a.m. with a temperature of 27 degrees: if that's the earliest point given, you can put that as where t = | 0 (but you can also put it elsewhere if appropriate). | |||

The general formula for decay is | y=A*e^-kt. | |||||||

f(x) = g(x) can be written as f: | x → g(x). | |||||||

If a function is such that there is at least one value of f(x) generated by more than one value of x: the function is a []-[]-[] function | many-to-one. Note: one-to-many mappings are not | functions. | ||||||

If x is a member of the set of all real numbers: one may note it (using set notation) as | x ∈ ℝ. | |||||||

The inverse of a many-to-one function f(x) is a | one-to-many mapping. But, as one-to-many mappings are not functions, f has no | inverse function. However, f can have a valid inverse function if one [] its [] | restricts its domain (e.g put x ∈ ℝ,x<=n), to make it no longer a | many-to-one mapping (thus making it's inverse no longer a one-to-many mapping). 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 x then f^-1(x) = | sine^-1(x); so that sine^-1(x) is not a one-to-many mapping: limit the domain of f(x) to | [π/2,-π/2] (i.e -π/2<=x<=π/2) [#A34#] - those limits are chosen as they span the full range of [], without [] any | values--repeating. |

If f(x)=x+2 and g(x)=x^2, gf(x)= | (x+2)^2 and ff(x)= | x+4. Note: ff(x) = f^2(x). | ||||||

With even functions: ∫^a(-a) f(x) dx = 2∫^a([]) f(x) dx = 2∫^0([]) f(x) dx. | 2∫^a(0) f(x) dx = 2∫^0(-a) f(x) dx. | |||||||

A function which keeps repeating it after a set interval is called a | periodic function; the set interval is called the | period. A function g(x) is periodic with period p if, for all values of x: g(x) = | g(x+p) (this means that the function repeats itself every interval p). | |||||

If asked to solve |f(x)|=c and the value which solves f(x)=c has been found: solve c= | -f(x). When finding the intersections of |f(x)| and g(x), and one has found the solutions to f(x)=g(x) and -f(x)=g(x), and more solutions than possible are given: [] the 'solutions' with [] values of [] | exclude the 'solutions' with negative values of y. (Those graphs of f(x) and -f(x) would have intersected with g(x) at the pseudo-solutions if carried on.) | ||||||

If you want to find the equation of an ellipse, you may find it by transforming the equation of a | circle (e.g you could find the equation of a circle within it and transform the circle). | |||||||

The equation of an oval with width 2w and length 2l is | x²/w² + y²/l² = 1. | |||||||

If f(x)=sine x then f^-1(x) = | sine^-1(x); so that sine^-1(x) is not a one-to-many mapping: limit the domain of f(x) to | [π/2,-π/2] (i.e -π/2<=x<=π/2) [#A34#] - those limits are chosen as they span the full range of [], without [] any | values--repeating. With f(x) = cos x, to make f^-1(x) valid: limit the domain to | 0<=x<=π [#A35#]. With f(x) = tan x, to make f^-1(x) valid: limit the domain to | -π/2<=x<=π/2 [#A36#]. | |||

The principal value of y=sin^-1(x) is defined as that value lying in the range | -π/2<=y<π/2. | |||||||

The principal value of y=cos^-1(x) is defined as that value lying in the range | 0<=y<=π. | |||||||

The principal value of y=tan^-1(x) is defined as that value lying in the range | -π/2<y<π/2. | |||||||

1/sin x = | cosec x; 1/cos x = | sec x; 1/tan x = | cot x. As tan x = sin x/cos x: cot x = | cos x/sin x. | ||||

Taking t(x) to be a representation of sin(x), cos(x) and tan(x): the points at which the graph of t(x) equal 0 are the [] of the graph of 1/t(x) | asymptotes--. Where t(x)=1: 1/t(x) = | 1, and where t(x)=-1: 1/t(x)= | -1. If t(x) represented cos(x) and sin(x) only: the range of t(x) is | 1/t(x)>=1,1/t(x)<=-1, the maximum points on t(x) are the [...] on 1/t(x) | minimum points--, and the minimum points on t(x) are the [...] on 1/t(x) | maximum points--. [#A36#]. Where tan x has asymptotes, cot x = | 0. | |

By rearranging sin²x + cos²x = 1, one can get tan²x + 1 = | sec²x, 1 + cot²x = | cosec²x. | ||||||

The graph of |f(x)| is as the graph of f(x), except the part(s) of the graph that have a negative y-value (i.e are under the x-axis) are | reflected in the x-axis. | |||||||

If y=f(x): x= | f^-1(y). | |||||||

cos(A+B) = | cosAcosB - sineAsineB, cos(A-B) = | cosAcosB + sineAsineB (latter found by swapping B for -B in the former). If B=A, cos2A = [...] = [...] | cosAcosA-sinAsinA = cos²A-sin²A, as in sin²A+cos²A=1 - we have cos²A=1-sin²A, thus additionally cos2A = [...] = [...] | 1-sin²A-sin²A = 1-2sin²A; also sin²A = 1-cos²A, thus also cos2A = [...] = [...] | cos²A-1+cos²A = 2cos²A - 1. In conclusion cos2A = [] = [] = [] | cos²A-sin²A = 1-2sin²A = 2cos²A - 1. | ||

tan(A+B) = | tanA+tanB/1-tanAtanB, (through replacement of B with -B:) tan(A-B) = | tanA-tanB/1+tanAtanB. If B=A: tan(A+B)= tan(2A) = [] = [] | tanA+tanA/1-tanAtanA = 2tanA/1-tan²A. | |||||

When trying to find intersections in graphs (esp trigonometric), sometimes (e.g. in 3sinxcosx = sinx), instead of cancelling out (in this case sinx), make the equation so that you can | factorise (3sinxcosx-sinx = 0 = sinx(3cosx-1), thus to find the full range of intersections solve both sinx = 0 and cosx = 1/3). | |||||||

The R-alpha method involves writing expressions of the form ncos(x) ± msin(x) as [] or [] | Rcos(x±a) or Rsin(x±a). | |||||||

To write ncos(x) + msin(x) as Rcos(x-a), expand Rcos(x-a) to get [] = [] | R(cosxcosa+sinxsina) = Rcosxcosa + Rsinxsina (=RHS). Rewrite the RHS, so that it is in the same [] as the LHS | order--. (E.g for 3sinx+4cosx, write Rcosxcosa+Rsinxsina as | Rsinasinx + Rcosacosx). Compare the parts to get ncos(x) + msin(x) = Rcosacosx + Rsinasinx, in which n=[] and m=[] | [(1)]n=Rcosa and [(2)]m=Rsina; write both in a simultaneous equation. Divide equation [] by equation [] | 2--1(the aim is to get to tan)--, to get tan(a) = m/n, whece a can be deduced. We can find R by doing []²+[]² | (1)²+(2)², which gives R²cos²A + R²sin²A = m² + n²; factorise to get R²(cos²A+sin²A) = m²+n², meaning R= | √(m²+n²). As you now have the value of 'R' and the value of 'a', write down ncos(x) + msin(x) = | Rcos(x-a). NB: this method can be also applied to cases with Rcos(x+a), Rsin(x-a) and Rsin(x+a), though they all expand differently. |

With the R-alpha method, if the arithmetic operation used in the starting equation is '+', use R[] or R[] | Rcos(x-a) or Rsin(x+a), and if the arithmetic operation is '-': use R[] or R[] | Rcos(x+a) or Rsin(x-a). | ||||||

{dy/dx = 1/ | dx/dy. If y=ln x: x=e[] | ^y: dx/dy = [] = [] | e^y = x: dy/dx = | 1/x. | ||||

If y=ln x: dy/dx = | 1/x. | |||||||

The derivative of ln c, where c is a constant is | 0. | |||||||

Shortcuts - if y=(ax+b)^n: dy/dx= | na(ax+b)^n-1, if y=(ax^p + b)^n = u^n: dy/dx= | n*u'*u^(n-1). If y=e^f(x): dy/dx = | f'(x)*e^f(x). If y=ln(f(x)), dy/dx= | f'(x)/f(x). | ||||

If y=u/v, where u and v are functions of x, using the quotient rule: y' = | vu'-uv'/v². {Proof: y=u/v, can be rewritten as | u=yv; you can now use the [] rule | product rule to give u'=yv'+vy', thus dy/dx=y'= | u'-yv'/v, afterwards substitute y for | u/v to give y'=vu'-uv'/v². | |||

Rates of change. If for every unit that r changes, s changes by a: a = | ds/dr. The amount that t changes for every unit that s changes equals | dt/ds. The amount that t changes for every unit that r changes equals | dt/dr. dt/dr = | dt/ds*ds/dr (the chain rule). If asked to find how fast t is changing, when r=6: | input r=6 into dt/dr. | |||

The radius of a balloon is increasing by 4cm per second. Q - How fast is the volume increasing when the radius is 2cm? V - volume, r - radius, t - time. What we are looking for is | dV/dt; dV/dt = | dV/dr * dr/dt. From the primary information: dr/dt = | 4. dV/dr is the derivative of [], differentiated with respect to [] | V(=4/3*πr^3), differentiated with respect to r, thus 4πr². Hence dV/dt = 4 * 4πr² = 16πr². When the balloon is 2cm in radius: the volume is increasing at a rate of | 64π cm^3/sec. | |||

Integration by inspection involves deducting and integrating an expresion whose integral is near the answer you seek. Generally, if f(x) = au^n, where u is a function of x, one would differentiate | u^n+1, which would give b* | u^n, meaning bu^n integrates to | u^n+1, hence ∫au^n = | a * u^n+1/b + c. In the case of ∫x(x²+2)^3 (for example), it's wise to deduct and integrate | (x²+2)^4, | |||

∫e^x dx = | e^x + c; ∫e^ax dx = | e^ax/a + c. | ||||||

∫1/x dx = | ln|x| + c. ln |x| is used as ln of a negative number is | undefined. ∫a/x dx = | aln|x| + c. | |||||

∫1/ax+b dx = | ln|ax+b|/a + c. | |||||||

Integration by substitution. We want to find ∫y dx, where y equals the function f with input u. Put y= | f(u). Find []/dx | du/dx (=n), then rearrange for | dx = du/n. Thus ∫y dx = | ∫f(u) du/n, which can be simplified to | 1/n*∫f(u) du. If, after implementing 'n', you have a term with a variable that isn't 'u': | write the term using 'u' (e.g u=x-1, thus ∫2xu² du = ∫2(u+1)u² = 2∫(u+1)u² = 2∫u^3 + u². | ||

∫1 du = | u. | |||||||

When integrating by substitution, if y=f(u) and u is a function of x: if using definite integrals and integrating with respect to 'u', input the value of each limit into | u(x) to get | the new limits. Example: with (∫^3)0 (2x+1)^5 dx, if u=2x+1: the limits change from 3 and 0 respectively to | 7 and 1, giving | (∫^6)1 u^5 du/2 (afterwards do as usual). | ||||

When one reaches cases such as ∫u+1/u du, it may be best to | split the fraction into two, to get | ∫1 + 1/u du = u + ln|u|. | ||||||

∫cos(x) dx = | sin(x) + c. | |||||||

∫sin(x) dx = | -cos(x) + c | |||||||

∫cos(ax) dx = | sin(ax)/a + c. | |||||||

∫sin(ax) dx = | -cos(ax)/a + c. | |||||||

One can find ∫ln(x) dx by putting it equal to | ∫ln(x)*1 dx and using | integration by parts, to reach ∫ln(x) dx = | xln(x) - x + c. | |||||

Often, if f(a)>0 and f(b)<0, then between the x=a and x=b, f(x)= | 0. (It is a good way to verify that there's a root between two given values). To show that f(x)=g(x) has a solution between a and b: show that h(x) does, where h(x) is 'f(x)=g(x)' rearranged with all the terms on one | side (i.e. h(x)=f(x)-g(x)=0 or h(x)=g(x)-f(x)=0). NB: to find a root through change of sign, in the relevant section, the amount of roots should number | one; additionally, in the relevant section, the function must be c[] | continuous (i.e it must not stop). | ||||

To find a point using iteration: rearrange the equation f(x)= | 0 so that on one side is the term | x and on the other side - a function of | x [i.e. g(x)], finally change x=g(x) to | xn+1 (n+1 ss) = g(xn) (n ss) - where xn+1 (n+1 ss) is the next | approximation, and the previous approximation is | xn (n ss). If the root is between a and b - x1(ss) (the first one) is often either | a or b (usually a; the start point may be given). Continue to iterate with the formula until the results converge to a value (to d decimal places [or whatever appropriate]). Note: if you have a rearrangement that diverges - try a [] [] | different rearrangement. (Cobweb diagram [the 'staircase' formed by the iteration] [#A38#]). |

In general, the Newton-Raphson formula is | xn+1 = xn - f(xn)/f'(xn) (n,n+1 ss), where the previous root approximation (i.e. the one you already have) is | xn (n ss), while the next approximation is | xn+1 (n+1 ss). It works by drawing a tangent from x=x0 (0 ss), the point that it touches the x-axis is x1 (1 ss), afterwards draw in a tangent from x=x1 (1 ss) and so on... ; this process takes us to the | root [#A39#]. If the answer (a) is to d decimal places: the error bound is | a ± 5*10^-(d+1) (e.g. if the answer was 0.2, the error bound would be 0.2±0.05); write your answer, which you can check using change of sign, with an error bound (or n s.f./ n d.p.). The Newton-Raphson method fails if you take x0 (0 ss) as a [] point | stationary-- (you can change the value of x0 [0 ss] to somewhere where there isn't a stationary point, also, it will be slow to converge if x0 (0 ss) is near a stationary point). The method may not work: it may diverge or lead to the wrong root. | ||

{When you take the inverse trig function of a positive number, the principal value is in the [] [] | first quadrant. When you take the inverse trig function of a negative number, with arcsin and arctan, the principal value is in the | forth quadrant, but with arccos, the principal value is in the | second quadrant. | |||||

The absolute value of a number is the distance that it is from | 0. The absolute value of x may be notated as | |x|; |-x|= | x. The absolute value is also the m[] | modulus. | ||||

{Integration by parts is used to integrate a | product. It is derived from the | product rule for differentiating a product: if y = uv: dy/dx=vu' + uv', subtract a term from the right hand side to give: | dy/dx - uv' = vu' and then | integrate to give ∫dy/dx dx - ∫uv' dx = | ∫vu' dx, which is usually written as | ∫uv' dx = uv - ∫vu' dx + c. |

Created by:
Toluo