Mathematics calculus cheat sheet, Cheat Sheet of Mathematics for Computing

Download Mathematics calculus cheat sheet and more Cheat Sheet Mathematics for Computing in PDF only on Docsity! Calculus Cheat Sheet Limits Definitions Precise Definition : We say lim f(x)=L if xa for every ¢ > 0 there is a 5 > 0 such that whenever 0 <|x-a|<6 then |f(x)-L]<e. “Working” Definition : We say lim f(x)=L xa ifwe canmake f(x) as close to L as we want by taking x sufficiently close to a (on either side of a) without letting x= a. Right hand limit : lim f(x)=L. This has xa the same definition as the limit except it tequires x >a. Left hand limit: lim f(x)=L. This has the xa same definition as the limit except it requires x<a. Limit at Infinity : We say lim f(x) =L ifwe x90 canmake f(x) as close to L as we want by taking x large enough and positive. There is a similar definition for lim f(x)=L Xo 0 except we require x large and negative. Infinite Limit : We say lim f (x)= if we xa canmake f(x) arbitrarily large (and positive) by taking x sufficiently close to a (on either side of a) without letting x= a. There is a similar definition for lim f(x) =—0 xa except we make f(.x) arbitrarily large and negative. Relationship between the limit and one-sided limits lim f(x)=L > lim f(x)= lim f(x)=L xa lim f(x)= lim f(x)=L > lim f(x)=L lim f (x)# lim f(x) > lim f(x) Does Not Exist Properties Assume lim f(x) and lim g(x) both exist and c is any number then, xa 1. lim[ef (x) ]=clim f(x) xa 2. lim[ f(x)+ g(x) ]=lim f(x)+lim g(x) xa xa 3. lim[ f(x) g(x) = lim f(x) limg(x) 6. xa xa xa xa 5. lim f (x) xa provided lim g(x) #0 im f(x) = im lim g(x) lim f(x) ]' =[lim f(x) xa tim| AC) ]= efi (9) Basic Limit Evaluations at + © Note : sgn(a)=1 if a>0 and sgn(a)=-1 if a<0. lim e* =0 Xo 1. lme’=0 & x0 2. limIn(x)=0 & x0 lim In(x)=- x0* 3. if r> Othen im = 0 x0 x! 4. If r>0 and x’is real for negative x then lim J. 0 x= x Visit http://tutorial.math lamar.edu for a complete set of Calculus notes. 5. neven: lim x"=% x bo 6. nodd: limx"=0 & lim x"=-0 x0 xo 7. neven: lim ax"+---+bx+c=sgn(a)o x bo 8. nodd: lim ax"+---+bx+c=sgn(a)o x90 9. nodd: lim ax"+---+ex+d=-sgn(a)o xo © 2005 Paul Dawkins Calculus Cheat Sheet Evaluation Techniques Continuous Functions L’Hospital’s Rule If f (x)is continuous at athenlim f (x)= f(a) If lim f(x) = 2 or lim f(x) +? the wa g(x) 0. g(x) to Continuous Functions and Composition _ f(x), f(x). f (x) is continuous at b and lim g(x)=5 then hm g(x) ~ im g'(x) ais anumber, or —%9 xa gla gl 5 5 Polynomials at Infinity lm f(g(x))=f{limg(x))= f(b 3 . tim F( (*)) # (im ( ) f() p(x) and q(x) are polynomials. To compute Factor and Cancel (x) _ x? +4x-12_.. (x-2)(x+6) lim 2” factor largest power of x in q(x) out lim ——— = lim ———>—_— x20 g(x) x92 x 2x 92 x(x-2) of both p(x) and g(x) then compute limit. _ xt+6_ 8 =lim =—=4 eX 2 _ 3x7 - . Rationalize Numerator/Denominator im 3p de = im im 32M = jm dV Bt hm ag) Ser Sade Piecewise Function . =lim 9-x =lim zl lim g(x) where ao) {" : mo (x2 81)(34Vx) 79 (x+9)(3+Vx) : Use Bees Compute two one sided limits, ew! . lim g(x)= lim x°+5=9 (18)(6) 108 im ‘km Combine Rational Expressions a g()= wo 13x57 ki “( 1 *) __1{ x-(x+h) One sided limits are different so lim g(x) im— -—|=lim—| —~—— x2 moh\x+h x) toh x(x+h) doesn’t exist. If the two one sided limits had 4 1 been equal then lim g(x) would have existed . -h . mil tg) ect x: andhad the same value. Some Continuous Functions Partial list of continuous functions and the values of x for which they are continuous. 1. Polynomials for all x. 7. cos(x) and sin(x) for all x. 2. Rational function, except for x’s that give . division by zero. 8. tan(x) and sec(x) provided 3. ¥en odd) for all x. xe 2,2 7 3a 4. x (n even) for all x>0. 20222 . 5. @ forall x. 9. cot(x) and cse(x) provided 6. Inx for x>0. x#---,-2,-1,0,7,27,- Intermediate Value Theorem Suppose that f(x) is continuous on [a, b] and let M be any number between f(a) and f(b). Then there exists a number c such that a<c<b and f(c)=M. Visit http://tutorial.math lamar.edu for a complete set of Calculus notes. © 2005 Paul Dawkins Calculus Cheat Sheet Extrema Absolute Extrema 1. x=cis an absolute maximum of f(x) iff (c) = f(x) for all x in the domain. 2. x=e is an absolute minimum of f (x) if f (c) < f(x) for all x in the domain. Fermat’s Theorem If f(x) has a relative (or local) extrema at x=c, then x=c isa critical point of f(x). Extreme Value Theorem If f(x) is continuous on the closed interval [a,5] then there exist numbers c and d so that, 1. a<c,d<b,2. f(c) is the abs. max. in [a,b], 3. f(d) is the abs. min. in [a,b]. Finding Absolute Extrema To find the absolute extrema of the continuous function f(x) on the interval [a,b] use the following process. 1. Findall critical points of f(x) in [a,b]. . Evaluate f(x) at all points found in Step 1. 2 3. Evaluate f(a) and f(b). 4 . Identify the abs. max. (largest function value) and the abs. min.(smallest function value) from the evaluations in Steps 2 & 3. Relative (local) Extrema 1. x=c isarelative (or local) maximum of f (x) if f (c) = f(x) for all x near ec. 2. x=c isarelative (or local) minimum of F(x) if f(c) < f(x) for all x near c. 1* Derivative Test If x=c isa critical point of f(x) then x=c is 1. a rel. max. of f(x) if f'(x) >0 to the left of x=c and f'(x) <0 to the right of x=c. 2. arel. min. of f(x) if f'(x)<0 to the left of x= cand f'(x) > Oto the right of x=c. 3. nota relative extrema of f(x) if f’(x) is the same sign on both sides of x=c. 2"4 Derivative Test If x=c isa critical point of f(x) such that f'(c)=0 then x=c 1. isarelative maximum of f (x) if f"(c)<0. 2. isarelative minimum of f(x) if f"(c)>0. 3. may be a relative maximum, relative minimum, or neither if f"(c)=0. Finding Relative Extrema and/or Classify Critical Points 1. Find all critical points of f(x). 2. Use the 1* derivative test or the 2" derivative test on each critical point. Mean Value Theorem If f(x) is continuous on the closed interval [a,b] and differentiable on the open interval (a,b) b)- then there is a number a<c<b such that f'(c)= fe) fa) . —a Newton’s Method If x, is the n™ guess for the root/solution of f (x)= 0 then (n+1)* guess is x,,, =x, - a xy provided f'(x,) exists. Visit http://tutorial.math lamar.edu for a complete set of Calculus notes. © 2005 Paul Dawkins Calculus Cheat Sheet Related Rates Sketch picture and identify known/unknown quantities. Write down equation relating quantities and differentiate with respect to ¢ using implicit differentiation (i.e. add on a derivative every time you differentiate a function of #). Plug in known quantities and solve for the unknown quantity. Ex. A 15 foot ladder is resting against a wall. The bottom is initially 10 ft away and is being pushed towards the wall at + ft/sec. How fast is the top moving after 12 sec? + 15 =e eH x’ is negative because x is decreasing. Using Pythagorean Theorem and differentiating, rty=lS > Ixx'4+2yy'=0 After 12 sec we have x =10-12(4)=7and so y= V15°-7? = V176 . Plugin and solve for y’. 7(-1)+V176 y'=0 => y- ft/sec 4V176 Ex. Two people are 50 ft apart when one starts walking north. The angle @ changes at 0.01 rad/min. At what rate is the distance between them changing when @ = 0.5 rad? t Moving Person x Stationary Person We have 6’ = 0.01 rad/min. and want to find x’. We can use various trig fens but easiest is, x ' secO=—— => sectand 6’ =— 50 50 We know 6 = 0.5 so plug in 6’ and solve. seo((0.5) tan(0.5)(0.01) = = x = 0.3112 ft/min Remember to have calculator in radians! Optimization Sketch picture ifneeded, write down equation to be optimized and constraint. Solve constraint for one of the two variables and plug into first equation. Find critical points of equation in range of variables and verify that they are min/max as needed. Ex. We’re enclosing a rectangular field with 500 ft of fence material and one side of the field is a building. Determine dimensions that will maximize the enclosed area. Building ¥ x Maximize A= xy subject to constraint of x+2y=500. Solve constraint for x and plug into area. A=y ( 500-2 y) =500y-2y" Differentiate and find critical point(s). A =500-4y => y=125 By 2™ deriv. test this is a rel. max. and so is the answer we’re after. Finally, find x. x= 500—2(125)=250 The dimensions are then 250 x 125. x=500-2y => Ex. Determine point(s) on y= x? +1 that are closest to (0,2). 3 (0,2) (ony “ty) Sa EEL n Minimize f = d? =(x-0)' +(y-2)° and the constraint is y =x’ +1. Solve constraint for x and plug into the function. x=y-l=> f=x'+(y-2) = y-1+(y-2) =y'-3y4+3 Differentiate and find critical point(s). f=2y-3 > y=3 By the 2™ derivative test this is a rel. min. and so all we need to do is find x value(s). x=i-l=t > The 2 points are then Visit http://tutorial.math lamar.edu for a complete set of Calculus notes. © 2005 Paul Dawkins Calculus Cheat Sheet Integrals Definitions Definite Integral: Suppose f(x) is continuous Anti-Derivative : An anti-derivative of f(x) on [a,b]. Divide [a,b] into n subintervals of is a function, F(x), such that F'(x)= f(x). width Ax and choose x; from each interval. Indefinite Integral : Jf (x)ax =F(x)+e Then fi fls) dx= lim S F(x “Ax where F (x) is an anti-derivative of f(x). noo Fundamental Theorem of Calculus Part I: If f(x) is continuous on [a,b] then Variants of Part I: = E70 dt is also continuous on [a,b] ay “F(0) dt=u'( (x) F[u( x ) and e(x)=2f. far f(x). ite -v (x) fLv(*)] Part II: f(x)is continuous on[a, ‘ F(x) is Sp par (x) f[ua]-v (x) foo] an anti-derivative of f(x) (ie. F(. x)= [f(x x)dx) then f° f(x) de =F (b)- F(a). Properties Jf (xt a(x Jdx= [f(x )dvx [g(x)dv ols )dv= ef f(x) x)dx , cis a constant pre (x)tg (x)de= f(x (x)drt J? o(x)de p of (x x)dv= ef” f (x)dx, cis a constant eal x)dx=0 fiedv=e(b-a) [i f(x)ar= -f f (x)dx If. £0) a <f'|#(x)|ax [fle If f(x Iff(x)20 onasx<b then f° f (x)dx=0 )dx= fr x)dv+ f° f (x)dx for any value of c. )2e(x x) ona<x<bthen f° f(x (x)av2 f" e(x)de If ms f(x)<M onas<x<b then m(b-a)<f" f(x)de<M (b-a) Common Integrals [kava keve Joose du =sinu+e ftanu du = Insecu|+e fx’ dx = att x" +en#-1 fsinu du =-cosu+c Jsccu du =In|secw + tanu|+e fx‘ de= [tde=In|s|+e Jseo? udu = tanu+e Jeu du= ttan'(4)+e Jz dx = Flnjax+5|+¢ fsccutanu du = secu +e \yeqree si’ (4)+e finu du=uln(u)-ut+e Joscu cotudu =—escu+e fe" du=e'+c Jose? udu =—cotu+e Visit http://tutorial.math lamar.edu for a complete set of Calculus notes. © 2005 Paul Dawkins Calculus Cheat Sheet Applications of Integrals Net Area: fF) dx represents the net area between f(x) and the 4 a x-axis with area above x-axis positive and area below x-axis negative. = Area Between Curves : The general formulas for the two main cases for each are, b d ye f(x) > A=[) {upper function] —[lower function] dx & x= f(y) => A= j. [right function ]— [left function | dy If the curves intersect then the area of each portion must be found individually. Here are some sketches of a couple possible situations and formulas for a couple of possible cases. y v= slx) y yvasls) Volumes of Revolution : The two main formulas are V = A(x) ae and V = fav) dy. Here is some general information about each method of computing and some examples. Rings Cylinders A= 7 ( (cue radius)” — (inner radu)” ) A= 2M (radius) (width / height) Limits: x/y of right/bot ring to x/y of left/top ring Limits : x/y of inner cyl. to x/y of outer cyl. Horz. Axisuse f(x), Vert. Axisuse f(y), | Horz. Axisuse f(y), | Vert. Axis use f(x), g(x),A(x) and dy. g(y),A(y) and dy. g(y),A(y) and dy. g(x),A(x) and dy. Ex. Axis: y=a>0 Ex. Axis: y=a<0 Ex. Axis: y=a>0 x outer radius :a- f(x) outer radius:|al+ g(x) tadius :a—y radius : |a|+ y inner radius : a- g(x) inner radius: |a|+ f (x) width: f(y)-g(¥) width: f(y)-g(y) These are only a few cases for horizontal axis of rotation. If axis of rotation is the x-axis use the y=a<0 case with a=0. For vertical axis of rotation (x= a>0 and x=a< 0) interchange x and y to get appropriate formulas. Visit http://tutorial.math lamar.edu for a complete set of Calculus notes. © 2005 Paul Dawkins Calculus Cheat Sheet Work: Ifa force of F ( x) moves an object Average Function Value : The average value b ina <x <b, the work done is W = [F (x)de of f(x) on as xsb is fing = ral, f(x)e Arc Length Surface Area : Note that this is often a Calc II topic. The three basic formulas are, L= g ds SA= p 2ayds (rotate about x-axis) SA= p 2axds (rotate about y-axis) a a a where ds is dependent upon the form of the function being worked with as follows. ds=fl+(2) de if y= f(x), axxsb ds=(S)+(2) at if x= f(0),y= a(t), ath ds= 14(4) ay ifx= f(y), asy<b ds= r+(£ in): d@ if r=f(6), as@<b With surface area you may have to substitute in for the x or y depending on your choice of ds to match the differential in the ds. With parametric and polar you will always need to substitute. Improper Integral An improper integral is an integral with one or more infinite limits and/or discontinuous integrands. Integral is called convergent if the limit exists and has a finite value and divergent if the limit doesn’t exist or has infinite value. This is typically a Calc II topic. Infinite Limit Jo F(x) delim f! f(x) de 2. J) f(x)ae= lim f f(x 3. c f(x)dx= f S(x)de+ f° f(x) dx provided BOTH integrals are convergent. Discontinuous Integrand 1. Discont. at af" f(x)de = lim [ f(x)de 2. Discont. at b F(x) dx= lim ff (x)dx a toa’ ot a fob va : an b ¢ b . 3. Discontinuity ata<c<b: J Sf (x)& =f Ff (x)dx+ f f (x)dx provided both are convergent. Comparison Test for Improper Integrals : If f(x) > g(x) >0on [a,«) then, 1. ref” f(x) dx conv. then” g x)dx conv. 2. ff g x)dx divg. then f° f(x)d& divg. Useful fact : If a> 0 then i ee converges if p >1 and diverges for p<1. ax Approximating Definite Integrals For given integral p f (x)d& and an (must be even for Simpson’s Rule) define Ax = -* and divide [a,b] into n subintervals [x,,x,], [,x,]..-. . [x,1.%,] with x, =a and x, =5 then, Midpoint Rule : fF (x) dew Ax| f(x; )+ f(x2)+- +f(x)], xj is midpoint [x, ,] Trapezoid Rule : [EF lsjetem SL (a) +27 (m) 42 (a) +2(x,,)+f(%,)] Simpson’s Rule : [i Ale) ate SF (a) +4f(m) +2 (2 at + DF (Hy 2) + 4F (m4) + F(x.) ] Visit http://tutorial.math lamar.edu for a complete set of Calculus notes. © 2005 Paul Dawkins