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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

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