Math Bootcamp 2012 Calculus Refresher

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1 Mth Bootcmp 0 Clculus Refresher Exponents For ny rel number x, the powers of x re : x 0 =, x = x, x = x x, etc. Powers re lso clled exponents. Remrk: 0 0 is indeterminte. Frctionl exponents re lso clled roots. Squre root, x / = x; cube root, x /3 = 3 x. Generlly, x m/n = n x m. Negtive exponent is defined s x r = /x r. Properties of exponents:. x x b = x +b ;. x x b = x b ; 3. (x ) b = x b. Functions We sy y is function of x, written s y = f(x). This mens, for every vlue of x, there is one nd only one vlue of y. XY -plne consists of two perpendiculr xes, horizontl X-xis nd verticl Y -xis. The intersect is the origin (0, 0). The plne is prtitioned into four disjoint regions clled qudrnts. Every point P in the plne cn be represented by the ordered pir (x, y). The first vlue is the x-coordinte, nd indictes its horizontl position reltive to the origin. The second vlue is the y-coordinte, nd indictes the verticl position reltive to the origin. For given function y = f(x), the set of ll ordered pirs (x, y) tht stisfy this eqution is clled the grph of the function. Exercise: Drw the grph of x = 4, y =, y = x + 3, y = x, y = x /, y = /x, y = x, y = x.

2 The liner function y = x + b forms the grph of line in the XY -plne, where b is the Y -intercept, nd is the slope of the line. The slope is the rtio of the height chnge δy to the length chnge δx, or = δy = y y δx x x. Reltion between grphs:. The grph of f(x) is flipping the grph of f(x) long the x-xis;. The grph of f( x) is flipping the grph of f(x) long the y-xis; 3. For c > 0, the grph of f(x) + c is the grph of f(x) shifted up for c units; 4. For c > 0, the grph of f(x + c) is the grph of f(x) shifted to the left for c units; 5. For c > 0, the grph of f(x c) is the grph of f(x) shifted to the right for c units; 6. For c > 0, the grph of cf(x) nd the grph of f(x) hve the sme shpe nd re only different in the y-scle. 7. For c > 0, the grph of f(cx) nd the grph of f(x) hve the sme shpe nd re only different in the x-scle. The qudrtic function y = x + bx + c (stndrd form), 0 forms the grph of prbol. It cn be convex (holds wter), such s y = x ; or concve (spills wter), such s y = x. By complete of squres, we hve y = x + bx + c = [x + b x + ( b ) ] + (c b 4 ) = (x + b ) + (c b 4 ). From the bove vertex form, we see the xis of symmetry is x = b ;. If > 0, it s convex; if < 0, it s concve;

3 3. The discriminnt is defined to be = b 4c. If < 0, then the polynomil x +bx+c hs no rel roots; if = 0, then the polynomil hs one rel root; if > 0, then the polynomil hs two rel roots x, = b ± b 4c. It is useful tht x + x = b/ nd x x = c/. The function y = /x forms the grph of hyperbol, which hs two symmetric brnches. Exercise: Consider generl power functions y = x p with different constnt p. The function y = x is not power function. Its grph is n exponentil growth curve, which doubles in height with ech unit increse in x. The grph of y = x is n exponentil decy curve. The most useful exponentil function is y = exp(x) = e x, where the bse e = is the Euler s constnt. 3 Injection, surjection nd bijection Domin of function is the set of input or rgument vlues for which the function is defined. Codomin of function is the set into which ll of the output of the function is constrined to fll. Let f be function whose domin is set X. It is n injection, then for ll nd b in X, if f() = f(b), then = b. Let f be function with domin X nd codomin Y. It is n surjection, then for every y in the codomin Y there is t lest one x in the domin X such tht f(x) = y. If f is both n injection nd surjection, then it is bijection, or one-to-one correspondence. If there is one-to-one correspondence between x nd y, x = f (y) is the inverse function of y = f(x). x f y f x nd y f x f y. 3

4 If we drw the grph of function nd its inverse function together, they re going to be symmetric bout the digonl line y = x. The composition of two functions is defined to be f g(x) = f(g(x)). Thus we hve f f (x) = f f(x) = x. The inverse function of y = b x is the logrithm function y = log b (x). For exmple, we hve 0 = 00, 9 / = 3, 99 0 =, 5 = 0.; log 0 (00) =, log 9 3 = /, log 99 () = 0, log 5 0. =. The choice of bse b = 0 is clled the common logrithm log 0 (x); the bse b = e is the nturl logrithm log e (x), nd is usully denoted by ln(x). Then we hve e ln(x) = x nd ln(e x ) = x. Bsic properties of logs. log b (xy) = log b x + log b y;. log b (x/y) = log b x log b y; 3. log b (x n ) = n log b x; 4. log b x = log x/ log b. As result of 4, we hve log b = / log b. 4 Limits nd derivtives We hve seen tht s x becomes lrger, the vlue of /x become smller. It cn not rech 0, but we cn mke /x s close to 0 s possible by mking x lrge enough. This is limit sttement: lim x = 0. x The (ɛ, δ)-definition of the limit of function is s follows: Let f be function defined on n open intervl contining c nd let L be rel number. Then lim f(x) = L x c 4

5 mens for ech rel ɛ > 0, there exists rel δ > 0 such tht for ll x with 0 < x c < δ, we hve f(x) L < ɛ. Symboliclly, ɛ > 0 δ > 0 : x(0 < x c < δ f(x) L < ɛ). The first order derivtive of function y = f(x) is dy dx = f δy (x) = lim δ 0 δx = lim f(x + δ) f(x). δ 0 δ The process of clculte the derivtive is clled differentition. We cll dy nd dx differentils. Geometriclly, f (x 0 ) is the instnt rte of chnge t point (x 0, f(x 0 )), or the slope of the tngent line to the curve y = f(x) t point (x 0, f(x 0 )). For exmple, y = x, then dy dx = lim (x + δ) x δ 0 δ Generlly, we hve the power rule: = lim δ 0 (x + δ) = x. dx p dx = pxp. We hve dex = dx ex, or the derivtive of e x is itself. The generl exponentil rule is: db x dx = bx ln(b). We hve d[ln(x)] dx =. The generl logrithm rule is: x d[log b (x)] dx = x ln(b) If y = f(x) = c is constnt for ny x, then f (x) = 0 for ll x. Not every function hs derivtive everywhere! Properties of derivtives:. For ny constnt c nd differentible function f(x), [cf(x)] = cf (x); For ny two differentible functions f(x) nd g(x). Sum nd difference rule, [f(x) ± g(x)] = f (x) ± g (x); 5

6 3. Product rule, [f(x)g(x) ] = f (x)g(x) + f(x)g (x); [ ] 4. Quotient rule, f(x) g(x) = f (x)g(x) f(x)g (x) for g(x) 0; [g(x)] 5. Chin rule, {f[g(x)]} = f [g(x)]g (x); 6. If f(x) is the inverse function of g(x), then g (x) = f (g(x)). Problems from the pre-exm. Find the derivtives of the following functions: ) f(x) = (6x 3 x)(0 0x). b) f(x) = 3x+9 x. c) f(x) = x.. Determine whether the following sttements re right or wrong: cos(θ) ) lim θ 0 θ θ = 0 nd lim =. sin(θ) θ 0 b) lim θ 0 cos(π/θ) does not exist. c) lim x 0 x = nd lim x 0 x does not exist. 3. Find eqution of tngent line to f(x) = 4x 8 x t x = Determine f (0) for f(x) = x. Exmples from clss Find y for ) y = x x ; b) x 3 y 5 + 3x = 8y 3 +. Plese lso red more.pdf, pges 7 to. The dditionl mteril is due to Prof. Ismor Fischer from University of Wisconsin Mdison (http : //pges.stt.wisc.edu/ if ischer/). 6

7 5 Applictions 5. Newton-Rphson method Suppose we wnt to solve for the roots of eqution f(x) = 0.. We strt with initil guess x 0.. Clculte x i+ = x i f(x i) f (x i, for i = 0,,. ) 3. Stop the itertion until x i+ x i < ɛ, where ɛ is prespecified smll positive number, sy 0 4. Otherwise go bck to Step. Exercise: Solve for x: f(x) = x 3 x + 35x 0 = 0. Hint: Check tht f() = 6 nd f(3) = 3. Use initil vlue x 0 =. Note: Solve for x /3 = 0 using Newton-Rphson. We will get x i+ = x i, nd the lgorithm will not converge. This mens Newton-Rphson my not lwys work! 5. Locl extrem If function f(x) hs either locl mximum or locl minimum t some vlue x 0, nd f(x) is differentible t x 0, then the tngent line must be horizontl, or f (x 0 ) = 0. We cll x 0 locl extrem if it is either locl mximum or locl minimum. On the other hnd, not ll solutions of f (x) = 0 is locl extrem. Properties:. If f(x) hs locl extrem t x = x 0 nd f(x) is differentible t x 0, then f (x 0 ) = 0.. If f (x) > 0 for every x on some intervl I, then f(x) is incresing on the intervl. 3. If f (x) < 0 for every x on some intervl I, then f(x) is decresing on the intervl. 4. If f (x) = 0 for every x on some intervl I, then f(x) is constnt on the intervl. 5. If f (x) > 0 for ll x on some intervl I, then f(x) is concve up on I. 7

8 6. If f (x) < 0 for ll x on some intervl I, then f(x) is concve down on I. 7. If f (x 0 ) = 0, f (x 0 ) < 0 nd f (x) is continuous in region round x = x 0, then x = x 0 is locl mximum. 8. If f (x 0 ) = 0, f (x 0 ) > 0 nd f (x) is continuous in region round x = x 0, then x = x 0 is locl minimum. 9. If f (x 0 ) = 0, f (x 0 ) = 0 nd f (x) is continuous in region round x = x 0, then x = x 0 cn be locl mximum, locl minimum, or neither. Exercise: Study f(x) = x 3 x + 35x 0 = 0. Clculte the first nd second order derivtive t x = 5, 6, 7, 8, 9. Describe wht you find. 5.3 Tylor series Derivtive of the derivtive is clled the second-order derivtive f () (x) = f (x) = [f (x)] = d ( ) dy. dx dx Similrly, we cn define the n-th order derivtive nd denote it by f (n). The Tylor series of function f(x), which is infinitely differentible in neighborhood of, is power series: f(x) f() + f ()(x ) + f ()! Some importnt series: e x = log( + x) = n=0 (x ) + f (3) () (x ) 3 + 3! x n n! = + x + x! + x3 3! + n+ xn ( ) n for < x n= x = x n for x < n=0 8

9 5.4 Some exmples Exmple : Air is being pumped into sphericl blloon t rte of 5 cm 3 /min. Determine the rte t which the rdius of the blloon is incresing when the dimeter of the blloon is 0 cm. Solution: Both the volume of the blloon nd the rdius of the blloon will vry with time nd so re functions of time. Denote them s V (t) nd r(t) respectively. Then we hve V (t) = 4 3 πr3 (t). Tke derivtives with respect to t on both sides nd we get V (t) = 4 3 π3r (t)r (t). Plug in V (t) = 5 nd r(t) = 0/ = 0, we get r (t) = V (t) 4π r (t) = 5 4π 0 = 80π cm/min. Exmple : A 5 foot ldder is resting ginst the wll. The bottom is initilly 0 feet wy from the wll nd is being pushed towrds the wll t rte of /4 ft/sec. How fst is the top of the ldder moving up the wll seconds fter we strt pushing? Solution: Define the distnce of the bottom of the ldder from the wll to be x(t) nd the distnce of the top of the ldder from the floor to be y(t). Becuse x(t) decreses s t increses, we hve x (t) = /4. We lso hve x (t) + y (t) = 5. Tke derivtives with respect to t on both sides nd we get x(t)x (t) + y(t)y (t) = 0, nd it follows tht y (t) = x(t)x (t)/y(t). We hve t t =, x(t) = 0 /4 = 7, nd y(t) = 5 x (t) = 5 7 = 76. Together we hve y (t) = x(t)x (t)/y(t) = 7 ( /4)/ 76 = 0.39ft/sec. 9

10 Exmple 3 : A spot light is on the ground 0 ft wy from wll nd 6 ft tll person is wlking towrds the wll t rte of.5 ft/sec. How fst is the height of the shdow chnging when the person is 8 feet from the wll? Is the shdow incresing or decresing in height t this time? Solution: Define the distnce of the person to the spot light to be x(t) nd the height of the shdow to be y(t). Then we hve x(t) 0 = 6, or y(t) = 0/x(t). y(t) Tke derivtives with respect to t on both sides nd we get y (t) = 0x (t)/x (t). Becuse x(t) increses s t increses, we hve x (t) =.5. x(t) = 0 8 =. Together we hve We lso hve y (t) = 0.5/ =.083ft/sec. The negtive sigh mens tht y(t) decreses s t increses, or the shdow becomes shorter s the person wlks closer to the wll. Exmple 4 : An prtment complex hs 50 prtments to rent. If they rent x prtments then their monthly profit, in dollrs, is given by, P (x) = 8x + 300x How mny prtments should they rent in order to mximize their profit? Solution: P (x) = 6x = 6(x 00). Thus if x < 00, P (x) > 0 nd P (x) increses. If x > 00, P (x) < 0 nd P (x) decreses. This mens for 0 x 50, P (x) becomes the lrgest when x = 00. Thus 00 prtments should be rented to mximize the profit. Exmple 5 : Determine ll the intervls where the following function is incresing or decresing f(x) = x x x Solution: Tke first order derivtive nd we get f (x) = 5x 4 + 0x x = 5x (x 4)(x + ). 0

11 Then it is esy to see tht f (x) > 0, thus f(x) increses if < x < 0 or 0 < x < 4. f (x) < 0, thus f(x) decreses if x < or x > 4. Exmple 6 : A light is on the top of ft tll pole nd 5ft 6in tll person is wlking wy from the pole t rte of ft/sec. ) At wht rte is the tip of the shdow moving wy from the pole when the person is 5 ft from the pole? b) At wht rte is the tip of the shdow moving wy from the person when the person is 5 ft from the pole? Solution:. Define the distnce of the person from the pole to be x p, the length of the shdow s x s. Then we hve x s x s + x p = 5.5 nd x s = x p /3. Define the distnce between the tip of the shdow nd the pole s x. Then x = x p + x s = x p + x p /3 = 4x p /3. It follows tht b. Becuse x = x p + x s, we hve x = 4 3 x p = 4 3 = 48 3 ft/sec. x = x p + x s nd x s = x x p = 48/3 = /3. 6 Integrls Suppose we hve generl function y = f(x). For simplicity, let f(x) > 0 nd f(x) continuous. Denote F (x) = re under the grph of f in the intervl [,x]. Then we hve, for some vlue z in the intervl [x, x + δ] F (x + δ) F (x) = f(z)δ, or F (x + δ) F (x) δ = f(z).

12 As δ goes to 0, z goes to x, nd we hve F (x) = lim δ 0 F (x + δ) F (x) δ So F is ntiderivtive of f, nd we denote F (x) = x = lim z x f(z) = f(x). f(t)dt. This is definite integrl of the function f from to x. f is clled the integrnd. We lso hve indefinite integrl. Tht is, for n rbitrry constnt C, Properties of integrls: f(x) = F (x) + C.. For ny constnt c nd ny integrble function f(x), cf(x)dx = c f(x)dx.. For ny two integrble functions f(x) nd g(x), [f(x) ± g(x)]dx = f(x)dx ± g(x)dx. 3. Integrtion by prts. Given the existence of ll integrtions, we hve b Exmple : 3y dy 5y +4 Solution: 3y dy = 3 5y +4 0 Exmple : 3 dy 5y +4 Sol: 3 dy = 3/4 dy = 3 5y +4 5y /4+ 4 c f(x)g (x)dx = f(x)g(x) b 0y dy = 3 5y Exmple 3 : 0 e x dx, 0 xe x dx, 0 x e x dx b f (x)g(x)dx. 5y +4 d(5y +4) = 3 0 ln(5y +4)+c. ( d( 5/y) = 3 5y/) + rc tn( 5y/)+ 5

13 Sol: 0 e x dx = 0 e x d( x) = e x 0 = e 0 xe x dx = 0 xde x = xe x 0 ( 0 e x dx) = e + e = e 0 x e x dx = 0 x de x = x e x 0 ( 0 e x xdx) = e + ( e ) = 5e Exmple 4 : Determine the re of the region enclosed by y = x nd y = x. Sol:Drw the grphs of two functions in the sme figure. Then we see the re equls ( ( x x )dx = 3 x3/ ) 3 x3 0 = /3. 0 Exmple 5 : Determine the volume of the solid obtined by rotting the region bounded by y = 3 x, x = 8, nd the x-xis bout the x-xis. Sol: Drw horizontl line, which will be rotted to cylinder. We hve the volume equls πy(8 0 y3 )dy = 96π. 5 Or we cn drw verticl line, which will be rotted to circle. We hve the volume equls 8 π( 3 x) dx = 96π Prtil frctions Exmple : Determine 8 3x+ dx. 4 x x 6 Sol: Becuse x x 6 = (x 3)(x + ), let 3x + x x 6 = x 3 + b x + = (x + ) + b(x 3) (x 3)(x + ) Solve for nd b from { + b = 3 3b = nd we get = 4, b =. Thus 8 4 3x + x x 6 dx = Exmple : Determine ( 4 x 3 ) dx x + = (4ln(x 3) ln(x + )) 8 4 = 4ln5 ln0 + ln6 x x dx. 3

14 Sol: x = Solve for nd b from x + b x + nd we get = /, b = /. Thus 3 = { + b = 0 b = = (x + ) + b(x ) x x 3 ( x x dx = x + ) 3 ( x dx = x x + / x + / ) dx x + (x + ln(x ) ) ln(x + ) 3 = + ln ln4 + ln3 8 Improper integrls We hve the following properties. t f(x)dx = lim f(x)dx if the limit on the right hnd side exists nd t is finite, in which cse we sy the integrl converges. Otherwise we sy the integrl diverges.. f(x)dx = f(x)dx if the limit on the right hnd side exists t lim t nd is finite, in which cse we sy the integrl converges. Otherwise we sy the integrl diverges. 3. f(x)dx = f(x)dx+ f(x)dx if both integrls on the right hnd side converge. Exmple : For > 0, consider x p dx. Then the integrl converges if p >, nd diverges if p. Exmple : Does the following integrl diverge or converge? 0 3 x dx. 4

15 Sol: 0 0 dx = lim dx = 3 + lim 3 t. 3 x t t 3 x t Thus the integrl diverges s the limit goes to positive infinity. Exmple 3 : Does the following integrl diverge or converge? sin xdx. Sol: t sin xdx = lim sin xdx = lim ( cos t + cos ). t t Thus the integrl diverges s the limit does not exist. Comprison test: For f(x) g(x) 0. if. if f(x)dx converges, then g(x)dx converges. g(x)dx diverges, then f(x)dx diverges. Exmple 4 : Does the following integrl diverge or converge? cos x dx. x Sol: Let f(x) = /x nd g(x) = cos x/x. Then f(x) g(x) 0 nd f(x)dx = /x dx converges. It follows from the comprison test tht cos x/x dx = g(x)dx converges s well. Exmple 5 : Does the following integrl diverge or converge? 3 x + e x dx. Sol: Let g(x) = /(x + e x ) nd f(x) = /e x. Then f(x) g(x) 0 for x > 3. Consider e dx = x e x 3 = e 3 3 It follows from the comprison test tht 3 /(x + e x )dx converges. 5

16 9 Prtil derivtives Exmple: Find ll the first nd second order derivtives for f(x, y) = cos(x) x e 5y + 3y Sol: The first order prtil derivtives re f(x, y) = sin(x) xe5y x f(x, y) = y 5x e 5y + 6y The second order { prtil derivtives re f(x, y) = f(x, y)} = 4 cos(x) e 5y x x x f(x, y) = x y x y x { } f(x, y) y { = 0xe 5y f(x, y) = f(x, y)} = 0xe 5y y x { } f(x, y) = 5x e 5y + 6 y y f(x, y) = y 0 Double integrls Exmple: Find the double integrtion over A = [, 4] [, ] 6xy dxdy A Sol: If we first integrte with respect to x, then we hve ( 4 ) ( ) 6xy dx dy = 6y x 4 dy = 36y dy = y 3 = 84 x= Or we cn first integrte with respect to y nd get 4 ( ) ( 4 ) 6xy dy dx = 6x y3 3 dx = x= 4 4xdx = 7x 4 = 84 6

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