lim f(x) does not exist, such that reducing a common factor between p(x) and q(x) results in the agreeable function k(x), then

Transcription

1 AP Clculus AB/BC Formul nd Concept Chet Sheet Limit of Continuous Function If f(x) is continuous function for ll rel numers, then lim f(x) = f(c) Limits of Rtionl Functions A. If f(x) is rtionl function given y f(x) = p(x),such tht p(x) nd q(x) hve no common fctors, nd c is rel q(x) numer such tht q(c) = 0, then I. lim f(x) does not exist II. lim f(x) = ± x = c is verticl symptote B. If f(x) is rtionl function given y f(x) = p(x), such tht reducing common fctor etween p(x) nd q(x) results q(x) in the greele function k(x), then lim f(x) = lim p(x) q(x) = lim k(x) = k(c) Hole t the point (c, k(c)) Limits of Function s x Approches Infinity If f(x) is rtionl function given y (x) = p(x), such tht p(x) nd q(x) re oth polynomil functions, then q(x) A. If the degree of p(x) > q(x), lim x f(x) = B. If the degree of p(x) < q(x), lim x f(x) = 0 y = 0 is horizontl symptote C. If the degree of p(x) = q(x), lim x f(x) = c, where c is the rtio of the leding coefficients. y = c is horizontl symptote Specil Trig Limits sin x A. lim x 0 x = 1 B. lim x x 0 sin x 1 cos x = 1 C. lim = 0 x 0 x L Hospitl s Rule If results lim f(x) or lim f(x) results in n indeterminte form ( 0, x 0 f(x) = p(x), then q(x),, 0, 00, 1, 0 ), nd lim f(x) = lim = lim q(x) p(x) p (x) q (x) nd p(x) lim f(x) = lim = lim x x q(x) x p (x) q (x)

2 The Definition of Continuity A function f(x) is continuous t c if I. lim f(x) exists II. f(c) exists III. lim f(x) = f(c) Types of Discontinuities Removle Discontinuities (Holes) I. lim f(x) = L (the limit exists) II. f(c) is undefined Non-Removle Discontinuities (Jumps nd Asymptotes) A. Jumps lim f(x) = DNE ecuse lim f(x) lim f(x) + B. Asymptotes (Infinite Discontinuities) lim f(x) = ±

3 Intermedite Vlue Theorem If f is continuous function on the closed intervl [, ] nd k is ny numer etween f() nd f(), then there exists t lest one vlue of c on [, ] such tht f(c) = k. In other words, on continuous function, if f()< f(), ny y vlue greter thn f() nd less thn f() is gurnteed to exists on the function f. Averge Rte of Chnge The verge rte of chnge, m, of function f on the intervl [, ] is given y the slope of the secnt line. m = f() f() Definition of the Derivtive The derivtive of the function f, or instntneous rte of chnge, is given y converting the slope of the secnt line to the slope of the tngent line y mking the chnge is x, Δx or h, pproch zero. f f(x+h) f(x) (x) = lim h 0 h Alternte Definition f f(x) f(c) (c) = lim x c

4 Differentiility nd Continuity Properties A. If f(x) is differentile t x = c, then f(x) is continuous t x = c. B. If f(x) is not continuous t x = c, then f(x) is not differentile t x = c. C. The grph of f is continuous, ut not differentile t x = c if: I. The grph hs cusp or shrp point t x = c II. The grph hs verticl tngent line t x = c III. The grph hs n endpoint t x = c Bsic Derivtive Rules Given c is constnt, Derivtives of Trig Functions Derivtives of Inverse Trig Functions

5 Derivtives of Exponentil nd Logrithmic Functions Explicit nd Implicit Differentition A. Explicit Functions: Function y is written only in terms of the vrile x (y = f(x)). Apply derivtives rules normlly. B. Implicit Differentition: An expression representing the grph of curve in terms of oth vriles x nd y. I. Differentite oth sides of the eqution with respect to x. (terms with x differentite normlly, terms with y re multiplied y dy per the chin rule) dx II. Group ll terms with dy on one side of the eqution nd ll other terms on dx the other side of the eqution. III. Fctor dy dy nd express in terms of x nd y. dx dx Tngent Lines nd Norml Lines A. The eqution of the tngent line t point (, f()): y f() = f ()(x ) B. The eqution of the norml line t point (, f()): y f() = 1 (x ) f () Men Vlue Theorem for Derivtives If the function f is continuous on the close intervl [, ] nd differentile on the open intervl (, ), then there exists t lest one numer c etween nd such tht f (c) = f() f() The slope of the tngent line is equl to the slope of the secnt line.

6 Rolle s Theorem (Specil Cse of Men Vlue Theorem) If the function f is continuous on the close intervl [, ] nd differentile on the open intervl (, ), nd f() = f(), then there exists t lest one numer c etween nd such tht f (c) = f() f() = 0 Prticle Motion A velocity function is found y tking the derivtive of position. An ccelertion function is found y tking the derivtive of velocity function. x(t) Position x (t) = v(t) Velocity * v(t) = speed x (t) = v (t) = (t) Acclertion Rules: A. If velocity is positive, the prticle is moving right or up. If velocity is negtive, the prticle is moving left or down. B. If velocity nd ccelertion hve the sme sign, the prticle speed is incresing. If velocity nd ccelertion hve opposite signs, speed is decresing. C. If velocity is zero nd the sign of velocity chnges, the prticle chnges direction. Relted Rtes A. Identify the known vriles, including their rtes of chnge nd the rte of chnge tht is to e found. Construct n eqution relting the quntities whose rtes of chnge re known nd the rte of chnge to e found. B. Implicitly differentite oth sides of the eqution with respect to time. (Rememer: DO NOT sustitute the vlue of vrile tht chnges throughout the sitution efore you differentite. If the vlue is constnt, you cn sustitute it into the eqution to simplify the derivtive clcultion). C. Sustitute the known rtes of chnge nd the known vlues of the vriles into the eqution. Then solve for the required rte of chnge. *Keep in mind, the vriles present cn e relted in different wys which often involves the use of similr geometric shpes, Pythgoren Theorem, etc.

7 Extrem of Function A. Asolute Extrem: An solute mximum is the highest y vlue of function on given intervl or cross the entire domin. An solute minimum is the lowest y vlue of function on given intervl or cross the entire domin. B. Reltive Extrem I. Reltive Mximum: The y-vlue of function where the grph of the function chnges from incresing to decresing. Another wy to define reltive mximum is the y-vlue where derivtive of function chnges from positive to negtive. II. Reltive Minimum: The y-vlue of function where the grph of the function chnges from decresing to incresing. Another wy to define reltive mximum is the y-vlue where derivtive of function chnges from negtive to positive. Criticl Vlue When f(c) is defined, if f (c) = 0 or f is undefined t x = c, the vlues of the x coordinte t those points re clled criticl vlues. *If f(x) hs reltive extrem t x = c, then c is criticl vlue of f. Extreme Vlue Theorem If the function f continuous on the closed intervl [, ], then the solute extrem of the function f on the closed intervl will occur t the endpoints or criticl vlues of f. *After identifying criticl vlues, crete tle with endpoints nd criticl vlues. Clculte the y vlue t ech of these x vlues to identify the extrem.

8 Incresing nd Decresing Functions For differentile function f A. If f (x) > 0 in (, ), then f is incresing on (, ) Tngent line hs positive slope B. If f (x) < 0 in (, ), then f is decresing on (, ) Tngent line hs negtive slope C. If f (x) = 0 in (, ), then f is constnt on (, ) Tngent line hs zero slope (horizontl) First Derivtive Test After clculting ny discontinuities of function f nd clculting the criticl vlues of function f, crete sign chrt for f, reflecting the domin, discontinuities, nd criticl vlues of function f. A. If f (x) chnges sign from negtive to positive t x = c, then f(c) is reltive minimum of f. B. If f (x) chnges sign from positive to negtive t x = c, then f(c) is reltive mximum of f. *If there is no sign chnge of f (x), there exists shelf point Concvity For differentile function f(x), A. If f (x) > 0, the grph of f(x) is concve up This mens f (x) is incresing B. If f (x) < 0, the grph of f(x) is concve down This mens f (x) is decresing Second Derivtive Test For function f(x) tht is continuous t x = c A. If f (c) = 0 nd f (c) > 0, then f(c) is reltive minimum. B. If f (c) = 0 nd f (c) < 0, then f(c) is reltive mximum. * If f (c) = 0 nd f (c) = 0, you must use the first derivtive test to determine extrem

9 Point of Inflection Let f e functions whose second derivtive exists on ny intervl. If f is continuous t x = c, f (c) = 0 or f (c) is undefined, nd f (x) chnges sign t x = c, then the point (c, f(c)) is point of inflection. Optimiztion Finding the lrgest or smllest vlue of function suject to some kind of constrints. A. Define the primry eqution for the quntity to e mximized or minimized. Define fesile domin for the vriles present in the eqution. B. If necessry, define secondry eqution tht reltes the vriles present in the primry eqution. Solve this eqution for one of the vriles nd sustitute into the primry eqution. C. Once the primry eqution is represented in single vrile, tke the derivtive of the primry eqution. D. Find the criticl vlues using the derivtive clculted. E. The optiml solution will more thn likely e found t criticl vlue from D. Keep in mind, if the criticl vlues do not represent minimum or mximum, the optiml solution my e found t n endpoint of the fesile domin. Derivtive of n Inverse If f nd its inverse g re differentile, nd the point (c, f(c)) exists on the function f mening the point (f(c), c) exists on the function g then d dx [g(x)] = 1 f (f 1 (x)) = 1 f (f(c)) BC Only: Derivtives of Prmetric Functions If f nd g re continuous functions of t on n intervl, then the equtions x = f(t) nd y = g(t) re clled prmetric equtions, providing the position in the coordinte plne, nd t is clled the prmeter. A. The slope of the curve t the point (x, y) is dy dx = dy/dt, provided dx/dt 0 dx/dt B. The second derivtive t the point (x, y) is d 2 y dx 2 = d dt (dy dx ) dx dt

10 Antiderivtives If F (x) = f(x) for ll x, F(x) is n ntiderivtive of f. f(x) = F(x) + C * The ntiderivtive is lso clled the Indefinite Integrl Bsic Integrtion Rules Let k e constnt. Definite Integrls (The Fundmentl Theorem of Clculus) A definite integrl is n integrl with upper nd lower limits, nd, respectively, tht define specific intervl on the grph. A definite integrl is used to find the re ounded y the curve nd n xis on the specified intervl (, ). If F(x) is the ntiderivtive of continuous function f(x), the evlution of the definite integrl to clculte the re on the specified intervl (, ) is the First Fundmentl Theorem of Clculus: f(x)dx = F() F() Integrtion Rules for Definite Integrls *This mens tht c is vlue of x, lying etween nd

11 Riemnn Sum (Approximtions) A Riemnn Sum is the use of geometric shpes (rectngles nd trpezoids) to pproximte the re under curve, therefore pproximting the vlue of definite integrl. If the intervl [, ] is prtitioned into n suintervls, then ech suintervl, Δx, hs width: x = n. Therefore, you find the sum of the geometric shpes, which pproximtes the re y the following formuls: A. Right Riemnn Sum Are x [f(x 0 ) + f(x 1 ) + f(x 2 ) + + f(x n 1 )] B. Left Riemnn Sum Are x [f(x 1 ) + f(x 2 ) + f(x 3 ) + + f(x n )] C. Midpoint Riemnn Sum Are x [f(x 1/2 ) + f(x 3/2 ) + f(x 5/2 ) + + f(x (2n 1)/2 )] D. Trpezoidl Sum Are 1 2 x [f(x 0) + 2 f(x 1 ) + 2 f(x 2 ) + + 2f(x n 1 ) + f(x n )] Properties of Riemnn Sums A. The re under the curve is under pproximted when I.A Left Riemnn sum is used on n incresing function. II. A Right Riemnn sum is used on decresing function. III. A Trpezoidl sum is used on concve down function. B. The re under the curve is over pproximted when I.A Left Riemnn sum is used on decresing function. II. A Right Riemnn sum is used on n incresing function. III. A Trpezoidl sum is used on concve up function.

12 Riemnn Sum (Limit Definition of Are) Let f e continuous function on the intervl [, ]. The re of the region ounded y the grph of the function f nd the x xis (i.e. the vlue of the definite integrl) cn e found using f(x)dx n = lim f(c i ) x n Where c i is either the left endpoint (c i = + (i 1) x) or right endpoint (c i = + i x) nd x = ( )/n. i=1 Averge Vlue of Function If function f is continuous on the intervl [, ], the verge vlue of tht function f is given y 1 f(x)dx Second Fundmentl Theorem of Clculus If function f is continuous on the intervl [, ], let u represent function of x, then A. B. C. d dx d dx x [ f(t)dt] = f(x) [ f(t)dt] = f(x) x u(x) d dx [ f(t)dt ] = f(u(x)) u (x) Integrtion of Exponentil nd Logrithmic Formuls

13 Integrtion of Trig nd Inverse Trig BC Only: Integrtion y Prts If u nd v re differentile functions of x, then u dv = uv v du Tips: For your choice of the function u, mke the selection following: A. LIPET: Logrithmic, Inverse Trig, Polynomil, Exponentil, Trig B. LIATE: Logrithmic, Inverse Trig, Algeric, Trig, Exponentil Comes from Integrtion y Prts. MEMORIZE ln x dx = x ln x x + C

14 BC Only: Prtil Frctions Let R(x) represent rtionl function of the form R(x) = N(x). If D(x) is fctorle polynomil, Prtil Frctions cn D(x) e used to rewrite R(x) s the sum or difference of simpler rtionl functions. Then, integrtion using nturl log. A. Constnt Numertor B. Polynomil Numertor

15 BC Only: Improper Integrls An improper integrl is chrcterized y hving limits of integrtion tht is infinite or the function f hving n infinite discontinuity (symptote) on the intervl [, ]. A. Infinite Upper Limit (continuous function) B. Infinite Lower Limit (continuous function) C. Both Infinite Limits (continuous function) f(x)dx = lim f(x)dx f(x)dx = lim f(x)dx c f(x)dx = lim f(x)dx + lim f(x)dx, where c is n x vlue nywhere on f. D. Infinite Discontinuity (Let x = k represent n infinite discontinuity on [, ]) f(x)dx c k = lim f(x)dx + lim f(x)dx x k x k + k BC Only: Arc Length (Length of Curve) A. If the function y = f(x)is differentile function, then the length of the rc on [, ] is 1 + [f (x)] 2 dx B. If the function x = f(y)is differentile function, then the length of the rc on [, ] is 1 + [f (y)] 2 dy C. Prmetric Arc Length: If smooth curve is given y x(t) nd y(t), then the rc length over the intervl t is ( dx 2 dt ) + ( dy 2 dt ) dt

16 Exponentil Growth nd Decy When the rte of chnge of vrile y is directly proportionl to the vlue of y, the function y = f(x) is sid to grow/decy exponentilly. A. Differentil Eqution for rte of chnge: dy dt = ky B. Generl Solution: y = Ce kt I. If k > 0, then exponentil growth occurs. II. If k < 0, then exponentil decy occurs. BC Only: Logistic Growth A popultion, P, tht experiences limit fctor in the growth of the popultion sed upon the ville resources to support the popultion is sid to experience logistic growth. A. Differentil Eqution: dp dt = kp (1 P L ) B. Generl Solution: P(t) = L 1+e kt P = popultion k = constnt growth fctor L = crrying cpcity t = time, = constnt (found with intitl condition) Grph Chrcteristics of Logistics I. The popultion is growing the fstest where P = L 2 II. The point where P = L represents point of inflection 2 III. lim t P(t) = L

17 Are Between Two Curves A. Let y = f(x) nd y = g(x)represent two functions such tht f(x) g(x)(mening the function f is lwys ove the function g on the grph) for every x on the intervl [, ]. Are Between Curves = [f(x) g(x)] dx B. Let x = f(y) nd x = g(y)represent two functions such tht f(y) g(y)(mening the function f is lwys to the right of the function g on the grph) for every y on the intervl [, ]. Are Between Curves = [f(y) g(y)] dy Volumes of Solid of Revolution: Disk Method If defined region, ounded y differentile function f, on grph is rotted out line, the resulting solid is clled solid of revolution nd the line is clled the xis of revolution. The disk method is used when the defined region orders the xis of revolution over the entire intervl [, ] A. Revolving round the x xis Volume = π (f(x)) 2 dx B. Revolving round the y xis Volume = π (f(y)) 2 dy C. Revolving round horizontl line y = k Volume = π (f(x) k) 2 dx D. Revolving round verticl line x = m Volume = π (f(y) m) 2 dy

18 Volumes of Solid of Revolution: Wsher Method If defined region, ounded y differentile function f, on grph is rotted out line, the resulting solid is clled solid of revolution nd the line is clled the xis of revolution. The wsher method is used when the defined region hs spce etween the xis of revolution on the intervl [, ] A. Revolving round the x xis, where f(x) g(x)(mening the function f is lwys ove the function g on the grph) for every x on the intervl [, ]. Volume = π ([f(x)] 2 [g(x)] 2 )dx B. Revolving round the y xis, where f(y) g(y)(mening the function f is lwys to the right of the function g on the grph) Volume = π ([f(y)] 2 [g(y)] 2 )dy C. Revolving round horizontl line y = k, where f(x) g(x)(mening the function f is lwys ove the function g on the grph) for every x on the intervl [, ]. Volume = π ([f(x) k] 2 [g(x) k] 2 )dx D. Revolving round verticl line x = m, where f(y) g(y)(mening the function f is lwys to the right of the function g on the grph) Volume = π ([f(y) m] 2 [g(y) m] 2 )dy

19 Volumes of Known Cross Sections If defined region, ounded y differentile function f, is used t the se of solid, then the volume of the solid cn e found y integrted using known re formuls. For the cross sections perpendiculr to the x xis nd region ounded y function f, on the intervl [, ], nd the xis. I. Cross sections re squres II. Cross sections re equilterl tringles Volume = [f(x)] 2 dx Volume = 3 4 [f(x)]2 dx III. Cross sections re isosceles right tringles with leg in the se Volume = 1 2 [f(x)]2 dx IV. Cross sections re isosceles right tringles with the hypotenuse in the se Volume = 1 4 [f(x)]2 dx V. Cross sections re semicircles (with dimeter in se) Volume = π 8 [f(x)]2 dx VI. Cross sections re semicircles (with rdius in se) Volume = π 2 [f(x)]2 dx

20 Differentil Equtions A differentil eqution is n eqution involving n unknown function nd one or more of its derivtives dy dx = f(x, y) Usully expressed s derivtive equl to n expression in terms of x nd/or y. To solve differentil equtions, use the technique of seprtion of vriles. Given the differentil eqution dy dx = xy (x 2 +1) Step 1: Seprte the vriles, putting ll y s on one side, with dy in the numertor, nd ll x s on the other side, with dx in the numertor. Step 2: Integrte oth sides of the eqution. Step 3: Solve the eqution for y. 1 y dy = x (x 2 + 1) dx ln y = 1 2 ln x C y = C x Given the differentil eqution dy dx = 2x2 with the initil condition y(3) = 10. A. The generl solution to differentil eqution is left with the constnt of integrtion, C, undefined. dy = 2x 2 dx dy = 2x 2 dx y = 2 3 x3 + C B. The prticulr solution uses the given initil condition to clculte the vlue of C. 10 = 2 3 (3)3 + C C = 8 y = 2 3 x3 8 BC Only: Euler s Method for Approximting the Solution of Differentil Eqution Euler s method uses liner pproximtion with increments (steps), h, for pproximting the solution to given differentil eqution, dy = F(x, y), with given initil vlue. dx Process: Initil vlue (x 0, y 0 ) x 1 = x 0 + h y 1 = y 0 + h F(x 0, y 0 ) x 2 = x 1 + h y 2 = y 1 + h F(x 1, y 1 ) x 3 = x 2 + h y 3 = y 2 + h F(x 2, y 2 ) * This process repets until the desired y vlue is given.

21 Slope Field The derivtive of function gives the vlue of the slope of the function t ech point (x, y). A slope field is grphicl representtion of ll of the possile solutions to given differentil eqution. The slope field is generted y plugging in the coordintes of every point (x, y) into the differentil eqution nd drwing smll segment of the tngent line t ech point. Given the differentil eqution dy dx = x y dy dx = 0 (0,0) 0 undefined dy dx = 0 (0,±1) dy dx (1,2) = 1 2 *These re only three exmple points. You would do this for every point in the given region of the grph. BC Only: Testing for Convergence/Divergence of Series Given the series n = Sequence of Prtil Sums The sequence of prtil sums for the series is S 1 = 1 S 2 = S 3 = S n = n If lim n S n = S, then n converges to S. If the terms of sequence do not converge to 0, then the series must diverge. Nth Term I. If lim n n 0, then n diverges. II. If lim n n = 0, then the test is inconclusive. The form of p series is 1 n p P Series I. If p > 1, then the series converges. II. If p < 1, then the series diverges.

22 Geometric Series A geometric series is ny series of the form r n n=0 I. If r < 1, then the series converges to 1 r *Series must e indexed t n = 0 II. If r > 1, then the series diverges. A telescoping series is ny series of the form n n+1 Telescoping Series *Convergence nd divergence is found using sequence of prtil sums *Prtil decomposition my e used to rek single rtionl series into the difference of two series tht form the telescoping series. Integrl If f is positive, continuous, nd decresing for x 1, then n nd f(x)dx n=1 1 either oth converge or oth diverge. A series, contining oth positive terms, negtive terms, nd n > 0, of the form ( 1) n n n=1 or ( 1) n+1 n n=1 Alternting Series The series converge if oth of the following conditions re met I. n+1 n for ll n II. lim n n = 0 When compring two series, if n n for ll n, Direct Comprison I. If n diverges, then n diverges. II. If n converges, then n converges. *The convergence or divergence of the series chosen for comprison should e known

23 Limit Comprison If n > 0 nd n > 0 nd lim n = L, where L is finite nd positive, then the series n n n nd n either oth converge or oth diverge. *The convergence or divergence of the series chosen for comprison should e known *When choosing series to compre to, disregrd ll ut the highest powers (growth fctor) in the numertor nd denomintor Given series n Root n I. If lim n < 1, then n converges. n n II. If lim n > 1, then n diverges. n n III. If lim n = 1, then the root test is inconclusive. n *This is test for solute convergence Given series n Rtio I. If lim n+1 < 1, then n converges. n n II. If lim n+1 > 1, then n diverges. n n III. If lim n+1 = 1, then the rtio test is inconclusive. n n *This is test for solute convergence BC Only: Asolute vs Conditionl Convergence For series, n, with oth positive nd negtive terms A. If n converges, then n lso converges. n is sid to e solutely convergent. n=1 n=1 n=1 B. If n diverges, ut n converges, n is sid to e conditionlly convergent. n=1 n=1 n=1 BC Only: Alternting Series Reminder Theorem Given n is convergent lternting series, the error ssocited with pproximting the sum of the series y the first n terms is less thn or equl to the first omitted term. ( 1) n+1 n = S S n = ( 1) n+1 n n=1 Error = S S n n+1

24 BC Only: Power Series A. Power Series Structure nd Chrcteristics n x n = x + 2 x n x n + power series centered t x = 0 n=0 n (x c) n = (x c) + 2 (x c) n (x c) n + power series centered t x = c n=0 A function f cn e represented y power series, where the power series converges to the function in one of three wys: I. The power series only converges t the center x = c. II. The power series converges for ll rel vlues of x. III. The power series converges for some intervl of vlues such tht x c < R, where R is the rdius of convergence of the power series. B. Intervl of Convergence: Find this y pplying the Rtio to the given series. I. If R = 0, then the series converges only t x = c. II. If R =, then the series converges for ll rel vlues of x. III. If the Rtio Test results in n expression of the form x c < R, then the intervl of convergence is of the form c R < x < c + R. *The convergence t the endpoints of the intervl of convergence should e tested seprtely. BC Only: Tylor nd Mclurin Series (specific power series) If function of f hs derivtives of ll orders t x = c, then the series is clled Tylor Series for f centered t c. A Tylor series centered t 0 is lso known s Mclurin Series. A. Mclurin Series f(x) = f(0) + f (0)x + f (0) x2 2! + f (0) x3 3! + + f(n) (0) xn n! + = x n f(n) (0) n! B. Tylor Series f(x) = f(c) + f (c)(x c) + f (x c)2 (c) + f (x c)3 (c) + + f (n) (x c)n (c) 2! 3! n! n=0 + = f (n) (x c) n (c) n! n=0

25 BC Only: Common Series to MEMORIZE Series A. B. 1 1 x = 1 + x + x2 + + x n + = x n n=0 e x = 1 + x + x2 2! + + xn n! + = xn n! n=0 C. cos x = 1 x2 2! + x4 4! + x2n (2n)! + = ( 1)n n=0 x2n (2n)! D. sin x = x x3 3! + x5 5! + x2n+1 (2n + 1)! + = ( 1)n E. ln(1 + x) = x x2 2 + x ( 1)n x n n F. rctn x = x x3 3 + x ( 1)n x 2n+1 2n + 1 n=0 xn n = ( 1) n n=0 x 2n+1 (2n + 1)! x2n+1 n = ( 1) 2n + 1 n=0 Intervl of Convergence 1 < x < 1 < x < < x < < x < 1 < x 1 < x < BC Only: Lgrnge Reminder of Tylor Polynomil When pproximting function f(x) using n nth degree Tylor polynomil, P n (x), the ssocited error, R n (x), is ounded y (x c)n+1 R n (x) = f(x) P n (x) mx f (n+1) (z) where c z x (n + 1)! BC Only: Polr Coordintes A. The polr coordintes (r, θ)of point re relted to the rectngulr coordintes (x, y) s follows x = r cos θ y = r sin θ r 2 = x 2 + y 2 tn θ = x y B. If f is differentile function of θ (smooth curve), then the slope of the line tngent to the grph of r = f(θ) t the point (r, θ) is dy dx = dy/dθ dx/dθ = r sin θ + r cos θ r cos θ r sin θ = f (θ) sin θ + f(θ) cos θ f (θ) cos θ f(θ) sin θ C. If r = f(θ) is smooth curve on the intervl [α, β], where α nd β re rdil lines, then the re enclosed y the grph is Are = 1 β 2 r2 dθ α = 1 2 [f(θ)]2 dθ β α