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typed by Sean Bird, Covenant Christian High School                updated August 15, 2009                                 ...
BC TOPICS and important TRIG identities and values              l’Hôpital’s Rule                           Slope of a Para...
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Stuff You Must Know Cold for the AP Calculus BC Exam!

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Stuff You Must Know Cold for the AP Calculus BC Exam!

  1. 1. typed by Sean Bird, Covenant Christian High School updated August 15, 2009 AP CALCULUS Stuff you MUST know Cold * means topic only on BC Curve sketching and analysis Differentiation Rules Approx. Methods for Integration y = f(x) must be continuous at each: Chain Rule Trapezoidal Rule dy d du dy dy du bcritical point: = 0 or undefined dx and look out for endpoints dx [ f (u )] = f (u ) dx OR dx = du dx ∫a f ( x)dx = 1 b− a 2 n [ f ( x0 ) + 2 f ( x1 ) + ...local minimum: + 2 f ( xn −1 ) + f ( xn )] dy goes (–,0,+) or (–,und,+) or d 2 y >0 Product Rule Simpson’s Rule dx dx 2 blocal maximum: d dx (uv) = du dx v+u dv dx OR u v + uv ∫a f ( x)dx = 2 1 dy goes (+,0,–) or (+,und,–) or d y <0 3 ∆x[ f ( x0 ) + 4 f ( x1 ) + 2 f ( x2 ) + ... dx dx 2 2 f ( xn −2 ) + 4 f ( xn−1 ) + f ( xn )]point of inflection: concavity changes Quotient Rule d 2 y goes from (+,0,–), (–,0,+), d u du dx v −u dv dx u v − uv  = dx  v  v2 OR v2 dx 2 (+,und,–), or (–,und,+) Theorem of the Mean Value i.e. AVERAGE VALUE If the function f(x) is continuous on [a, b] Basic Derivatives “PLUS A CONSTANT” and the first derivative exists on the d n interval (a, b), then there exists a number dx ( ) x = nx n−1 The Fundamental Theorem of x = c on (a, b) such that Calculus b d ( sin x ) = cos x f (c ) = ∫ a f ( x)dx dx b (b − a) d ( cos x ) = − sin x ∫a f ( x)dx = F (b) − F (a) This value f(c) is the “average value” of dx where F ( x) = f ( x) the function on the interval [a, b]. d ( tan x ) = sec2 x dx Corollary to FTC Solids of Revolution and friends d ( cot x ) = − csc2 x Disk Method dx d b( x) x =b V =π∫ 2 [ R ( x)] dx ∫a ( x ) d f (t )dt = dx ( sec x ) = sec x tan x x=a dx Washer Method d f (b( x))b ( x) − f (a ( x))a ( x) dx ( csc x ) = − csc x cot x V =π∫ b a ([ R( x)] − [r ( x)] ) dx 2 2 Intermediate Value Theorem General volume equation (not rotated) d 1 du ( ln u ) = If the function f(x) is continuous on [a, b], b dx u dx and y is a number between f(a) and f(b), V = ∫ Area ( x) dx a d u du then there exists at least one number x= c dx ( ) e = eu dx in the open interval (a, b) such that *Arc Length L = ∫ 1 + [ f ( x)] dx b 2 a where u is a function of x, f (c ) = y . b =∫ 2 2 and a is a constant. a [ x (t )] + [ y (t )] dt More Derivatives Distance, Velocity, and Acceleration d  −1 u  1 du Mean Value Theorem velocity = d (position) sin = dx   a  2 a −u 2 dx dt If the function f(x) is continuous on [a, b], d (velocity) d −1 acceleration = AND the first derivative exists on the dx ( cos −1 x = ) interval (a, b), then there is at least one dt 1 − x2 dx dy number x = c in (a, b) such that *velocity vector = , d  −1 u  a du dt dt tan = ⋅ f (b) − f (a ) dx  a  a 2 + u 2 dx  f (c) = . speed = v = ( x )2 + ( y ) 2 * b−a d −1 dx ( cot −1 x = 1 + x2 ) displacement = ∫t tf v dt o d  −1 u  a du Rolle’s Theorem final time dx   sec = a  u u 2 − a 2 dx  ⋅ distance = ∫initial time v dt If the function f(x) is continuous on [a, b], tf d ( csc −1 x = −1 ) AND the first derivative exists on the ∫t o ( x ) 2 + ( y ) 2 dt * dx x x2 − 1 interval (a, b), AND f(a) = f(b), then there average velocity = is at least one number x = c in (a, b) such final position − initial position d u( x) du dx ( ) a = a ( ) ln a ⋅ u x dx that f (c) = 0 . = total time ∆x d 1 = ( log a x ) = ∆t dx x ln a
  2. 2. BC TOPICS and important TRIG identities and values l’Hôpital’s Rule Slope of a Parametric equation Values of Trigonometric f (a) 0 ∞ Given a x(t) and a y(t) the slope is Functions for Common Angles If = or = , g (b) 0 ∞ dy dy = dt θ sin θ cos θ tan θ f ( x) f ( x) dx dxthen lim = lim dt x →a g ( x) x → a g ( x ) 0° 0 1 0 Euler’s Method Polar Curve π 1 3 3If given that dy = f ( x, y ) and that For a polar curve r(θ), the 6 2 2 3 dx AREA inside a “leaf” isthe solution passes through (xo, yo), θ2 2 π 2 2 y ( xo ) = yo ∫θ1 1 2  r (θ )  dθ   4 2 2 1 where θ1 and θ2 are the “first” two times that r = π 3 1 0. 3 y ( xn ) = y ( xn−1 ) + f ( xn−1 , yn−1 ) ⋅ ∆x 3 2 2 The SLOPE of r(θ) at a given θ isIn other words: dy dy / dθ π = 1 0 “∞ ” xnew = xold + ∆x dx dx / dθ 2 dy π 0 −1 0 dθ  r (θ ) sin θ  d = d   ynew = yold + ⋅ ∆x Know both the inverse trig and the trig dx ( xold , yold ) dθ  r (θ ) cos θ    values. E.g. tan(π/4)=1 & tan-1(1)= π/4 Integration by Parts Ratio Test Trig Identities ∞ Double Argument ∫ udv = uv − ∫ vdu The series ∑ ak converges if k =0 sin 2 x = 2sin x cos xIntegral of Log cos 2 x = cos 2 x − sin 2 x = 1 − 2sin 2 xUse IBP and let u = ln x (Recall ak +1 lim <1 1u=LIPET) k →∞ ak cos 2 x = (1 + cos 2 x ) 2 ∫ ln x dx = x ln x − x + C If the limit equal 1, you know nothing. Taylor Series 1If the function f is “smooth” at x = Lagrange Error Bound sin 2 x = (1 − cos 2 x ) 2a, then it can be approximated by If Pn ( x) is the nth degree Taylor polynomial Pythagoreanthe nth degree polynomial of f(x) about c and f ( n+1) (t ) ≤ M for all t sin 2 x + cos 2 x = 1 f ( x) ≈ f (a ) + f (a )( x − a) (others are easily derivable by between x and c, then + f (a) ( x − a)2 + … dividing by sin2x or cos2x) M n+1 2! f ( x) − Pn ( x) ≤ x−c 1 + tan 2 x = sec 2 x f ( n ) (a) ( n + 1)! + ( x − a)n . cot 2 x + 1 = csc 2 x n! Reciprocal Maclaurin Series 1A Taylor Series about x = 0 is Alternating Series Error Bound sec x = or cos x sec x = 1 cos xcalled Maclaurin. 1 x 2 x3 N csc x = or sin x csc x = 1 If S N = ∑ ( −1) an is the Nth partial sum of a n ex = 1 + x + + +… sin x 2! 3! k =1 Odd-Even 2 x x4 convergent alternating series, then sin(–x) = – sin x (odd) cos x = 1 − + − … 2! 4! S∞ − S N ≤ aN +1 cos(–x) = cos x (even) x3 x5 Some more handy INTEGRALS: sin x = x − + − … 1 3! 5! Geometric Series ∞ ∫ tan x dx = ln sec x + C = 1 + x + x2 + x3 + … a + ar + ar 2 + ar 3 + … + ar n −1 + … = ∑ ar n −1 = − ln cos x + C 1− x n =1 x 2 x3 x 4 a ln( x + 1) = x − + − + … 2 3 4 diverges if |r|≥1; converges to 1− r if |r|<1 ∫ sec x dx = ln sec x + tan x + CThis is available at http://covenantchristian.org/bird/Smart/Calc1/StuffMUSTknowColdNew.htm

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