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by: Reyes Glover


Marketplace > University of Texas at Austin > Mathematics (M) > M 408C > DIFFEREN AND INTEGRAL CALCULUS
Reyes Glover
GPA 3.67


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This 14 page Class Notes was uploaded by Reyes Glover on Sunday September 6, 2015. The Class Notes belongs to M 408C at University of Texas at Austin taught by Staff in Fall. Since its upload, it has received 10 views. For similar materials see /class/181489/m-408c-university-of-texas-at-austin in Mathematics (M) at University of Texas at Austin.

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Date Created: 09/06/15
M4080 THE EXPONENTIAL FUNCTION November 13 2008 We de ne the natural exponential function as the inverse of the natural logarithm function We can do this since In x is an increasing function so it is oneto one and thus has an inverse The two functions undo each other ie The domain of the exp is the range of ln ie foo 00 The range of exp is the domain of ln ie 0 00 Since lnem zlne z and exp is the inverse of ln we have that e1 expm This is the notation we will typically use for expm The cancellation laws then become eh1m z for z gt 0 and lnem z for all m The limiting behavior of e1 is well understood We also have the laws of exponents which follow from the laws of logarithms em emey em y 3 and emT 5 Finally differentiating and integrating the exponential is straightfor ward For the general exponential function we build on top of our previous work We will discuss the general logarithm function on Tuesday So if a gt 0 and r is any number then atT eh aT Th1 Thus we can de ne a1 em a This is the exponential function with base a The laws of exponents hold for a1 as well awry away am y g a y am and abm ambm We also have the derivative and integral formulas 5 M4080 INVERSE FUNCTIONS November 6 2008 We say a function f is onetoone if it never takes on the same value twice In other words if 1 7 m2 then fz1 7 zz for a one to one function 1 One more way of saying this is that if fa fb then a b for a oneto one function f Graphically we can determine if a function is oneto one by using the horizontal line test If f is a one to one function we can properly think of its inverse f l Note that f 1 is not the reciprocal of f le We will often write x for an element of the domain of f l even though z is usually used for the domain of f The critical thing about a function and its inverse is the following fact To determine the inverse of a oneto one function 1 write y Then solve this equation for z in terms of y if possible Finally interchange z and y to get y as a function of m This function is the inverse of f y f 1z Geometrically the graph of f 1 is the re ection of the graph of 1 across the line y m lnverse functions are well behaved if f is oneto one and continuous then its inverse f 1 is also continuous Moreover if f is a one to one function with inverse f l g with f ga 7 0 then the inverse is differentiable at a and M4080 RELATED RATES LINEAR APPROXIMATION amp DIFFERENTIALS September 307 2008 4 3817 A man starts walking north at 4 fts from a point P Five minutes later7 a woman starts walking south at 5 fts from a point 500 ft due east of P At what rate are the people moving apart 15 minutes after the woman starts walking Solution Let x mty yt denote the distance travelled by the man and woman7 re spectively Thus dzdt 4 fts and dydt 5 fts Let 2 2t denote the distance between them7 which is the hypotenuse of the right triangle with height z y and base 500 ft Thus 22 m y2 5002 Differentiating this with respect to the independent variable t via im plicit dil llerentiation7 we get 22 in 2m y lt gt We want dzdt since that is the rate at which the man and woman are walking apart We know dzdt and lgalt7 so we need to determine m y and 2 Now7 15 minutes after the woman starts walking7 the man has been walk ing for 20 min So z 4 20mTm 60mgm 4800 ft After 15 minutes7 the woman has walked y 5 20mlm 60 m5 4500 ft Plugging this in for z m y2 50027 we get 2 x86740000 in ft after 15 minutes Solving for dzdt we get dz z y dz dy 4800 4500 837 i 77 45 899t dt 2 dt dt x86740000 x8674 f S 5 3928 Use linear approximation to estimate V998 Solution Set fz so that we want to a linear approximation to f at a 100 We quickly compute f z 7 so Lz 7 m fam 7 a 7 m n 710010 1 Plugging in m 998 99 we get x99 z L99810 275 10 7 999 6 3936 One side of a right triangle is known to be 20 cm long and the opposite angle is measured as 30 7 with a possible error of i1quot Use differentials to estimate the error in computing the length of the hypotenuse What is the percent error Solution Let 9 be the angle in question7 and let 340 be the length of the hypotenuse Thus sin0 so y0 20 sin0 and set y f0 wher f0 20 sin 0 We are told the error is i1quot7 so 10 i1 iamp radians Now7 to determine the error in computing the length7 we need to compute dy f0dt92 7 7 72 L L080 dy 7 20 s1n0 cos0 i180 7 igltsin02 Evaluating this at 9 30 39Ir67 we get x3 7ri 23 d i 2 ii zi121 y 9122 9 7139 cm The percent error is given by recall y 20 sin 0 lg L ggn39 2 1 i ii7 4003i3 9 W 40 1807r M4080 PARTIAL FRACTIONS December 4 2008 We will integrate rational functions by writing them as a sum of sim ler fractions we know how to integrate The rst step is given some rational function x 62 perform long division so that we can write Sm 57 where Sm is a polynomial and degR lt degQ lf degP lt degQ no long division is needed If is a rational function we can integrate we are done If it is not we factor Qm as far as possible This will give us a factorization of Q into a product linear factors those of the form ax b and irreducible quadratic factors those of the form azz bx c The next step is to express as a sum of partial fractions Doing so requires an analysis of the different types of factors found in the factorization of Q Case 1 Qm is a product of distinct linear factors Case 2 Qm is a product of linear factors some of which are repeated Case 3 Qm contains distinct irreducible quadratic factors Case 4 Qm contains repeated irreducible quadratic factors Once we have rewritten RzQm we then solve for the unknowns A B etc by getting a common denominator of the partial fraction sum and comparing the numerator of this fraction to One can use comparison of coefficients or plug in special values of z to determine the unknowns Once have determined the unknowns we have our partial fraction expansion and we can integrate our original fraction term by term 1 2 2 8420 Evaluate I dem Solution Notice the degree of the numerator 2 is less than the degree of the denominator 37 so apply partial fraction expansion 275z16 7 A B C Ax722Bz722m1C2z1 2m1m722 7 2m1 72 722 2m1m722 We thus conclude that x2 7 5x 16 Az 7 22 Bz 7 2x235 1 C2z 1 Plugging in z 2 tells us 10 507 so C 2 Plugging for C7 our equation becomes 275m16Az274m4B2m273z724m2 Comparing coe icients we nd x2 1A2B x7574A73B4 1164A72B2 The 2 equation tells us A 17 2B7 which can be plugged into the z equation to conclude B 71 so A 3 Thus 3 71 2 3 d 3 Eirll2gg1l71nlgg72l2 77lnl2z7171l7lnlz72l7 LL 2 0 2 m 7 2 2 Solution The degree of numerator is smaller than the degree of the denominator7 so we apply partial fraction expansion x272z71 A B CmDAm71m21Bm21CmDx71239 71221z711712 21 712z21 We thus conclude 3 8428 Evaluate I f dz m2 7 2m 71 Az 71m2 1 Bx2 1 Oz Dx 712 Plugging in z 17 we get 72 2B7 so B 71 Plugging in for B7 our equation becomes 272z71Az37m2x717m271CzDz272m1 Expanding all this out7 we can compare coefficients to conclude zSOAC 3521771471720 z72AC72D 1717A71D The 3 equation tells us C 7A and the constant equation tells us D A Plugging these into the 2 equation and solVing for A7 we nd A 17 so C 71 and D 1 Thus 1 71 7x1 du 71 I d d d l 717 77 d x71m712 mz21 n m u2 z21m 1 m 1 1 du 1 71 77 7d 7d 1 71 77 7 t 1 n m H5x71 m21mm21m n m H5x71 2uan 1 m lnlz71l 17lnlz21ltan 1z0 M4080 AREAS AND VOLUME November 4 2008 The area between the curves y fz and y gz and between z a and z b is An analogous formula holds for the area between the curves y c and y d and z y and z Qty For a solid S lying between z a and z b with cross sectional area function Am A continuous7 the volume of S is Again7 an analogous formula holds for the volume of a solid S lying between y c and y d with cross sectional area function Ay A continuous In this case7 3 6150 Find the number a such that the line x a bisects the area under the curve y 1x2 for 1 g x g 4 Also7 nd the number I such that the line y b bisects the same area Solution We want our line x a to split our areas That is7 we want an a so that flu dx f dx Taking the integrals7 we nd 17 i i 7 Solving for at7 we nd that E ie a E For the second part7 we want to split the areas using a horizontal line y b A horizontal line will give us a wedge above the line and a region below that includes a rectangle sitting below the line y 116 The area of the lower region is 3 1716 flbm 7 1 lg The rst term comes from the rectangle of length 3 between 1 and 4 and height 116 The second terms comes from determining the area between the curve x 1y and x 1 The upper region has area given by fbl 7 1 lg We set these two equal to each other and integrate to get 3 1 b 1 1 3 1 1 7 2 7 27l 7 212712727 4271725 b 16ly2 ylle W ybgt16 416 1 We simplify this to get 2574l 0 To solve for b7 we set 02 b and consider 202 740 0 We solve for c using the quadratic formula to get 0 1 i Now 13 lt 1 since I lt 1 Thus 0 llt 17 thus 0 17 Squaring this we get I 02 7 76 x 0150 4 6214 Find the volume obtained by rotating the region bounded by y x7 y 07 x 2 and x 4 about the line x 1 Solution The solid looks like the space between a funnel and a cylinder We split this solid into two different regions the lower one consists of a cylinder with a hole and the upper region is a triangle rotated around the line x 1 The lower region has cross sectional area A y 7r92 7 12 87139 and the upper region has cross sectional area AUy 7r9 7 y 7 12 Then we nd V 02ALydyA4AUydyf87rdyf7r9y712dy 2 4 1 4 56 76 Sing 7r 82y7y2dy167r7r 83471734277343 167r7r 2877 77139 0 2 3 2 3 3 5 Suppose 2 S fx 4 for all x Which of the following is false 4 5 A x dx 2 0 B The antiderivative of f is always positive C 10 g x dx 20 2 0 1 D The antiderivative of f is always negative x dx 2 2 0 Solution A is true since x 2 2 for all x C is true since 2 x 4 implies that 25 7 0 f05 x dx 45 7 0 E is true since fol x dx 2 fol 2dx 2 For B and D note that an antiderivative of f is gx x dx by the Fundamental Theorem of Calculus Since x gt 07 gx gt 0 for any x Thus B is true while D is false Hence the answer is M4080 LIMITS September 4 2008 M4080 TRIGONOMETRIC SUBSTITUTIONS December 27 2008 Last week7 we learned how to take integrals of certain trigonometric functions by exploiting some trigonometric identities Today7 we will practice converting a non trigonometric integral into a trigonometric one and then using some trigonometric identities to evaluate the integral The following table provides a solid framework to solving these types of problems Expression Substitution ldentity 3 8310 Evaluate f dt Solution Begin by setting 25 x2tan 0 so 255 4x2tan5 t9 and dt x2sec2 0010 tan5 9 sec2 5 wit 0 sec20d08 0d0tan50sec0d0 t5 I idt t2 2 a2tan20 1 x2sec0 Rewriting 8x2 and tan4 0 and then making the substitution u sec 0 so du sec t9tant9dt97 we get 82 sec20i12tan0sec0d042u2712du42u472u21du Evaluating this integral and plugging back in for u we get 5 3 4 2 8 2 I42 7 2 14 0 Tfsec50i Tfsec304 2sec00 Now7 since we set if x2tan 0 we know tan0 Thus we can form a right triangle with an angle 9 so that the side opposite to 9 has length if and the side adjacent to 9 has length x2 Then the hypotenuse of this triangle has length x t2 l 2 by the Pythagorean theorem Thus sec0 1cos0 1 t2 2 at2 22 Plugging this in7 we get 5 3 4 2 t2 2 8 2 It2 2 t2 2 Taking the powers and simplifying7 we eventually get 4 2 Imlt 3t 8t 32gta 15 4 8320 Evaluate fx dt Solution Begin by setting 25 5 sin 0 so dt 5 cos0d0 t 5 sin 9 sin 9 cos 9 I7dt 5cos0d057d05sin0d0 25 7 22 2517 sin2 0 6080 Integrating this7 we see I 75cos0 C We need to have a solution in terms of our original variable 25 though Recall that we set if 5sin 0 so we know sin0 t5 Thus we can form a right triangle with angle 9 such that the side opposite to 9 has length if and the hypotenuse has length 5 By the Pythagorean theorem7 the side adjacent to 9 must have length 25 7 2527 thus cos0 v25 7 2525 Plugging this in7 we get 257252 75 ltTgtC7V257t20 5 8322 Evaluate fol a 1dz Solution Begin by setting z tan 0 so dz sec2 0010 and adjust the bounds on the integral 1 7r4 7r4 I xm21dm xtan201sec20d0 sec30d0 0 0 0 Integrate by parts by setting u sec 0 dv secz 9 10 so du sec 0tan0 d0 1 tan 0 7T4 7r4 7r4 Isec0tan0 7 tan20sec0d0 7 se02071sec0d0 0 0 0 Expanding the integral 0n the right out7 we see 7r4 7r4 7r4 I i sec30d0 sec0d0 71 sec0d0 0 0 0 Collecting the 1 s and recalling that f sec 0d0 ln l sec 9 tan 0 we get 7r4 1 21 lnlsec0tan0l I lt lnl 1lilnl1lgt 0 Finally 1 gt 0 and ln1 07 so 1 lt lnlt 1gtgt M4080 THE GENERAL LOGARITHM AND lNVERSE TRIGONOMETRIC FUNCTIONS November 187 2008 For a gt 0 and a 7 17 am is a oneto one function We call its inverse the logarithmic function with base a So7 for example7 if a 5 then loge z lnm Recall what it means for a1 and loga z to be inverses aloga m z and logaam z One of the important formulas to know is the change of base formula The derivative of loga z is straightforward jim oga z Mia The trigonometric functions are not one to one if we consider their full domain of de nition ln stead7 we can restrict their domain7 in which case we have oneto one functions We will concentrate on the arcsin also denoted sin l and arctan also denoted tan l functions Note that we have taken care in determining the domain and range of sin 1 and tan l The derivative and integral rules for our inverse trigonometric functions are M4080 DERIVATIVES AND DIFFERENTIATION RULES September 187 2008 Recall the de nition of continuity a function f is continuous at a number a iflimmaa x fa This de nition implicitly requires three things to be true for f to be continuous at a


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