calc bc cheat sheet, Cheat Sheet of Calculus

Download calc bc cheat sheet and more Cheat Sheet Calculus in PDF only on Docsity! AP CALCULUS AB and BC Final Notes Trigonometric Formulas 1. 1cossin 22 =+ θθ 2. θθ 22 sectan1 =+ 3. θθ 22 csccot1 =+ 4. θθ sin)sin( −=− 5. θθ cos)cos( =− 6. θθ tan)tan( −=− 7. ABBABA cossincossin)sin( +=+ 8. ABBABA cossincossin)sin( −=− 9. BABABA sinsincoscos)cos( −=+ 10. BABABA sinsincoscos)cos( +=− 11. θθθ cossin22sin = 12. 13. θθ θθ cot 1 cos sintan == 14. θθ θθ tan 1 sin coscot == 15. θ θ cos 1sec = 16. θ θ sin 1csc = 17. ( )2 1cos 1 cos 2 2 θ θ= + 18. ( )2 1sin 1 cos 2 2 θ θ= − Differentiation Formulas 1. 1)( −= nn nxx dx d 2. fggffg dx d ′+′=)( Product rule 3. 2)( g gffg g f dx d ′−′ = Quotient rule 4. )())(())(( xgxgfxgf dx d ′′= Chain rule 5. xx dx d cos)(sin = 6. xx dx d sin)(cos −= 7. xx dx d 2sec)(tan = 8. xx dx d 2csc)(cot −= 9. xxx dx d tansec)(sec = 10. xxx dx d cotcsc)(csc −= 11. xx ee dx d =)( 12. aaa dx d xx ln)( = 13. x x dx d 1)(ln = 14. 21 1)sin( x xArc dx d − = 15. 21 1)tan( x xArc dx d + = 16. 1|| 1)sec( 2 − = xx xArc dx d 17. [ ] 0d c dx = 18. ( ) ( )'d cf x cf x dx =   2 2 2 2cos 2 cos sin 2cos 1 1 2sinθ θ θ θ θ= − = − = − Integration Formulas 1. Caxdxa +=∫ 2. ∫ −≠++= + 1 , 1 1 nC n xdxx n n 3. ∫ += Cxdxx ln 1 4. ∫ += Cedxe xx 5. ∫ += Ca adxa x x ln 6. ∫ +−= Cxxxdxx ln ln 7. ∫ +−= Cxdxx cos sin 8. ∫ += Cxdxx sin cos 9. ∫ +−+= CxCxdxx coslnor secln tan 10. ∫ += Cxdxx sinln cot 11. ∫ ++= Cxxdxx tansecln sec 12. csc x dx ln csc x cot x= − + +∫ C 13. Cxxdx +=∫ tan sec2 14. ∫ += Cxdxxx sec tansec 15. ∫ +−= Cxdxx cot csc2 16. ∫ +−= Cxdxxx csc cotcsc 17. ∫ +−= Cxxdxx tan tan 2 18. ∫ +    = + C a xArc axa dx tan122 19. ∫ +    = − C a xArc xa dx sin 22 20. ∫ +=+= − C x aArc a C a x Arc aaxx dx cos1sec1 22 14. Increasing and Decreasing: Let f be differentiable for bxa << and continuous for a bxa ≤≤ , 1. If 0)( >′ xf for every x in ( )ba, , then f is increasing on [ ]ba, . 2. If 0)( <′ xf for every x in ( )ba, , then f is decreasing on [ ]ba, . 15. Concavity: Suppose that )(xf ′′ exists on the interval ( )ba, 1. If 0)( >′′ xf in ( )ba, , then f is concave upward in ( )ba, . 2. If 0)( <′′ xf in ( )ba, , then f is concave downward in ( )ba, . To locate the points of inflection of )(xfy = , find the points where 0)( =′′ xf or where )(xf ′′ fails to exist. These are the only candidates where )(xf may have a point of inflection. Then test these points to make sure that 0)( <′′ xf on one side and 0)( >′′ xf on the other. 16a. If a function is differentiable at point ax = , it is continuous at that point. The converse is false, in other words, continuity does not imply differentiability. 16b. Local Linearity and Linear Approximations The linear approximation to )(xf near 0xx = is given by ))(()( 000 xxxfxfy −′+= for x sufficiently close to 0x . In other words, find the equation of the tangent line at ( )( )0 0,x f x and use that equation to approximate the value at the value you need an estimate for. 17. ***Dominance and Comparison of Rates of Change (BC topic only) Logarithm functions grow slower than any power function ( )nx . Among power functions, those with higher powers grow faster than those with lower powers. All power functions grow slower than any exponential function ( )xa , a 1> . Among exponential functions, those with larger bases grow faster than those with smaller bases. We say, that as x →∞ : 1. ( )xf grows faster than ( )g x if ( )( )x x lim g x f →∞ = ∞ or if ( ) ( )x g x lim 0 x→∞ = f . If ( )xf grows faster than ( )g x as x →∞ , then ( )g x grows slower than ( )xf as x →∞ . 2. ( )xf and ( )g x grow at the same rate as x →∞ if ( )( )x x lim L 0 g x→∞ = ≠ f (L is finite and nonzero). For example, 1. xe grows faster than 3x as x →∞ since x 3x lim x→∞ = ∞ e 2. 4x grows faster than ln x as x →∞ since 4 x xlim ln x→∞ = ∞ 3. 2x 2x+ grows at the same rate as 2x as x →∞ since 2 2x x 2xlim 1 x→∞ + = To find some of these limits as x →∞ , you may use the graphing calculator. Make sure that an appropriate viewing window is used. 18. ***L’Hôpital’s Rule (BC topic, but useful for AB) If ( )lim ( )x a f x g x→ is of the form ∞ ∞or 0 0 , and if ( )lim ( )x a f x g x→ ′ ′ exists, then ( ) ( )lim lim ( ) ( )x a x a f x f x g x g x→ → ′ = ′ . 19. Inverse function 1. If gf and are two functions such that xxgf =))(( for every x in the domain of g and xxfg =))(( for every x in the domain of f , then f and g are inverse functions of each other. 2. A function f has an inverse if and only if no horizontal line intersects its graph more than once. 3. If f is strictly either increasing or decreasing in an interval, then f has an inverse. 4. If f is differentiable at every point on an interval I , and 0)( ≠′ xf on I , then )(1 xfg −= is differentiable at every point of the interior of the interval )(If and if the point ( ),a b is on ( )f x , then the point ( ),b a is on )(1 xfg −= ; furthermore ( ) ( ) 1' ' g b f a = . 20. Properties of xey = 1. The exponential function xey = is the inverse function of xy ln= . 2. The domain is the set of all real numbers, ∞<<∞− x . 3. The range is the set of all positive numbers, 0>y . 4. xexe dx d =)( and ( )( ) ( ) ( )'f x f xd e f x edx = 5. 2121 xxexexe +=⋅ 6. xey = is continuous, increasing, and concave up for all x . 7. lim x xe →∞ = +∞ and lim 0 x xe →−∞ = . 8. xxe =ln , for xxex => )ln( ;0 for all x . 21. Properties of xy ln= 1. The domain of xy ln= is the set of all positive numbers, 0>x . 2. The range of xy ln= is the set of all real numbers, ∞<<∞− y . 3. xy ln= is continuous and increasing everywhere on its domain. 4. ( ) baab lnlnln += . 5. ba b a lnlnln −=      . 6. arra lnln = . 7. xy ln= 10 if 0 <<< x . 8. −∞= +→ +∞= +∞→ x x x x ln 0 lim and lnlim . 9. lnlog lna xx a = 10. ( )( ) ( )( ) ' ln f xd f x dx f x = and ( )( ) 1lnd x dx x = 22. Trapezoidal Rule If a function f is continuous on the closed interval [ ]ba, where [ ]ba, has been equally partitioned into n subintervals [ ] [ ] [ ]nxnxxxxx ,1... ,2,1 ,1,0 − , each length n ab − , then [ ])()1(2...)2(2)1(2)0(2 )( nxfnxfxfxfxfn abb a dxxf +−++++ − ≈∫ , which is equivalent to ( )1 2 Leftsum Rightsum+ 23a. Definition of Definite Integral as the Limit of a Sum Suppose that a function )(xf is continuous on the closed interval [ ]ba, . Divide the interval into n equal subintervals, of length n abx −=∆ . Choose one number in each subinterval, in other words, 1x in the first, 2x in the second, …, kx in the thk ,…, and nx in the thn . Then ( ) ( ) 1 lim ( ) ( ) bn kn k a f x x f x dx F b F a →∞ = ∆ = = −∑ ∫ . 23b. Properties of the Definite Integral Let )(xf and )(xg be continuous on [ ]ba, . i). ∫∫ =⋅ b a dxxfc b a dxxfc )( )( for any constant c . ii). 0 )( =∫ a a dxxf iii). b a f (x) dx f (x) dx a b = −∫ ∫ iv). ∫∫∫ += b c dxxf c a dxxf b a dxxf )( )( )( , where f is continuous on an interval containing the numbers cba and , , . v). If )(xf is an odd function, then 0 )( = − ∫ a a dxxf vi). If )(xf is an even function, then ∫∫ = − a dxxf a a dxxf 0 )(2 )( vii). If 0)( ≥xf on [ ]ba, , then 0 )( ≥∫ b a dxxf viii). If )()( xfxg ≥ on [ ]ba, , then ∫∫ ≥ b a dxxf b a dxxg )( )( • Isosceles Right Triangle cross-sections (leg in the xy plane): ( )21 2 b a V top function bottom function dx= −∫ • Semi-circular cross-sections: ( )2 8 b a V top function bottom function dxπ= −∫ • Rectangular cross-sections (height function or value must be given or articulated somehow – notice no “square” on the {top – bottom} part): ( ) ( / ) b a V top function bottom function height function value dx= −∫  • Circular cross-sections with the diameter in the xy plane: ( )2 4 b a V top function bottom function dxπ= −∫ • Square cross-sections with the diagonal in the xy plane: ( )21 2 b a V top function bottom function dx= −∫ 2. For cross sections of area )(yA , taken perpendicular to the y-axis, volume = ∫ b a dyyA )( . 30c. ***Shell Method (used if function is in terms of x and revolving around a vertical line) where a x b≤ ≤ : 2 ( ) ( ) ( ) if a.r. is y-axis ( 0) ( ) ( . .) if a.r. is to the left of the region ( ) ( . . ) if a.r. is to the right of the region ( ) ( ) if only revolving with one f b a V r x h x dx r x x x r x x a r r x a r x h x f x π= = = = − = − = ∫ ( ) unction ( ) if revolving the region between two functionsh x top bottom= − 31. Solving Differential Equations: Graphically and Numerically Slope Fields At every point ( )x, y a differential equation of the form ( )dy x, y dx = f gives the slope of the member of the family of solutions that contains that point. A slope field is a graphical representation of this family of curves. At each point in the plane, a short segment is drawn whose slope is equal to the value of the derivative at that point. These segments are tangent to the solution’s graph at the point. The slope field allows you to sketch the graph of the solution curve even though you do not have its equation. This is done by starting at any point (usually the point given by the initial condition), and moving from one point to the next in the direction indicated by the segments of the slope field. Some calculators have built in operations for drawing slope fields; for calculators without this feature there are programs available for drawing them. ***Euler’s Method (BC topic) Euler’s Method is a way of approximating points on the solution of a differential equation ( )dy x, y dx = f . The calculation uses the tangent line approximation to move from one point to the next. That is, starting with the given point ( )1 1x , y – the initial condition, the point ( )( )1 1 1 1x x, y ' x , y x+ ∆ + ∆f approximates a nearby point on the solution graph. This aproximation may then be used as the starting point to calculate a third point and so on. The accuracy of the method decreases with large values of x∆ . The error increases as each successive point is used to find the next. ( ), :x y given :dy given dx :x given∆ dyy x dx ∆ = ∆ ( ),x x y y+ ∆ + ∆ Start again 32. ***Logistics (BC topic) 1. Rate is jointly proportional to its size and the difference between a fixed positive number (L) and its size. dy yky 1 dt L  = −    OR ( )dy ky M y dt = − which yields kt Ly 1 Ce− = + through separation of variables 2. t lim y L →∞ = ; L = carrying capacity (Maximum); horizontal asymptote 3. y-coordinate of inflection point is L 2 , i.e. when it is growing the fastest (or max rate). 32(a). ***Decomposition: Steps: 1. Use Long Division first if the degree of the Numerator is equal or more than the Denominator to get ( ) ( ) ( ) ( ) ( ) N x r x dx q x dx dx D x D x = +∫ ∫ ∫ 2. For the second integral, factor ( )D x completely into Linear factors to get ( ) ( ) ... #1 #2 r x A B D x linearfactor linearfactor = + + 3. Multiply both sides by ( )D x to eliminate the fractions 4. Choose your x-values wisely so that you can easily solve for A, B, C, etc 5. Rewrite your integral that has been decomposed and integrate everything. 33. ***Definition of Arc Length If the function given by )(xfy = represents a smooth curve on the interval [ ]ba, , then the arc length of f between a and b is given by [ ] dx b a xfs 2)(1∫ ′+= . 34. ***Improper Integral ∫ b a dxxf )( is an improper integral if 1. f becomes infinite at one or more points of the interval of integration, or 2. one or both of the limits of integration is infinite, or 3. both (1) and (2) hold. 35. ***Parametric Form of the Derivative If a smooth curve C is given by the parametric equations )( and )( tgyxfx == , then the slope of the curve C at 0 , is ),( ≠÷= dt dx dt dx dt dy dx dyyx . Note: The second derivative, 2 2 d y d dy d dy dx dx dx dx dt dx dt    = = ÷       . 36. ***Arc Length in Parametric Form If a smooth curve C is given by )( and )( tgytfx == and these functions have continuous first derivatives with respect to t for bta ≤≤ , and if the point ),( yxP traces the curve exactly once as t moves from btat == to , then the length of the curve is given by ( )( ) ( )( ) 2 2 2 2 ' ' b b a a dx dys dt f t g t dt dt dt    = + = +       ∫ ∫ . ( )( ) ( )( )2 2speed f ' t g ' t= + 37. ***Vectors Velocity, speed, acceleration, and direction of motion in Vector form • position vector is ( ) ( ) ( ),r t x t y t= • velocity vector is ( ) ,dx dyv t dt dt = • speed is the magnitude of velocity because ( ) 2 2dx dyspeed v t dx dt    = = +        • acceleration vector is ( ) 2 2 2 2, d x d ya t dt dt = • the direction of motion is based on the velocity vector and the signs on its components Displacement and distance travelled in vector form • Displacement in vector form ( ) ( )1 2, b b a a v t dt v t dt∫ ∫ • Final position in vector form ( ) ( )1 1 2 2, b b a a x v t dt x v t dt   + +    ∫ ∫ • Distance travelled from ( ) ( )( ) ( )( )2 21 2 to is given by b b a a t a t b v t dt v t v t dt= = = +∫ ∫ 10. Ratio Test: Let ∑ na be a series with nonzero terms. i) If 11lim <+∞→ na na n , then the series converges absolutely. ii) If 11lim >+∞→ na na n , then the series is divergent. iii) If 11lim =+∞→ na na n , then the test is inconclusive (and another test must be used). 11. Power Series: A power series is a series of the form ......22100 +++++= ∞ = ∑ nxncxcxccnx n nc or ...)(...2)(2)(100 )( +−++−+−+= ∞ = −∑ naxncaxcaxcc n naxnc in which the center a and the coefficients ,...,...,2,1,0 ncccc are constants. The set of all numbers x for which the power series converges is called the interval of convergence. 12. Taylor Series: Let f be a function with derivatives of all orders throughout some intervale containing a as an interior point. Then the Taylor series generated by f at a is ∑ ∞ = +−++− ′′ +−′+=− 0 ...)( ! )()(...2)( !2 )())(()()( ! )()( k nax n anfaxafaxafafkax k akf The remaining terms after the term containing the thn derivative can be expressed as a remainder to Taylor’s Theorem: ∫∑ +−=+−+= x a dttnfntx n xnR n xnR naxanfafxf )()1()( ! 1)( where 1 )())(()()()( Lagrange’s form of the remainder: ( ) ( ) ( 1) 1( ) | ( ) || | | | ( 1)!n n nf c x af x P x R xn n + +− − = = + , where xca << . The series will converge for all values of x for which the remainder approaches zero as x →∞ . 13. Frequently Used Series and their Interval of Convergence ∑ ∞ = =+++++= − 0 ......21 1 1 n nxnxxx x , 1<x ∑ ∞ = =+++++= 0 ! ... ! ... !2 2 1 n n nx n nxxxxe , ∞<x ∑ ∞ = + +− =+ + + −+−+−= 0 )!12( 12)1(... )!12( 12 )1(... !5 5 !3 3 sin n n nxn n nxnxxxx , ∞<x ∑ ∞ = − =+−+−+−= 0 )!2( 2)1(... )!2( 2 )1(... !4 4 !2 2 1cos n n nx n nxnxxx , ∞<x