DIFFEREN AND INTEGRAL CALCULUS
DIFFEREN AND INTEGRAL CALCULUS M 408C
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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 24 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
UT Learning Center 5 mvwa Succead Excel M4080 Final Exam Review Day 4 Exponential Growth and Decay kt 1 The equation MO meg describes the radioactive decay of a substance Where mf is the amount remaining from an initial mass mu a er t yea1s ofdecay Ifthe halflife the amount of time required for half of the quantity to decay of a substance is 200 yea1s and We begin With 100 gmms ofthe substance When will the mass be reduced to 10 grams 2 How much money is gained from investing 1000 for 3 years at 6 continuously compounding interest How long does it take for any investment to double at 6 continuously compounding interest Area Under the Curve 3 Find the area bounded by y x andy x2 4x 4 Find the volume of the solid created by revolution of functions about an indicated axis The Disc Method 4 fx x2 6x 1S x S 2 About the xaXis The Washer Method 5 fx x25 gx3 x Abouty 2 x7 6 fx 3 gxVx l2 Abouty axis 7 Let Q be the area bound between the functions fx e y 4 x 0 A Write an integral which expresses the volume of the solid formed by rotating the area Q about the XaXis B Write an integral which expresses the volume of the solid formed by rotating the area Q about the yaXis What technique is necessary to evaluate such an integral M4080 HYPERBOLIC FUNCTIONS INTEGRATION BY PARTS AND TRIGONOMETRIC INTEGRALS November 25 2008 Let s brie y review the de nitions and properties of hyperbolic functions i I 7 m I m sinh s1nhz 1 coshm i tanhm m 2 2 cosh z 1 1 cosh z csch 7 sech 7 coth i m s1nh z m cosh z m s1nh x d d tanh z sechzm d h h 7 d cos x s1n z m d 7sinh z cosh z 7 1 dm 1 d d 2 csch z icsch z coth z sech z isech z tanh z coth z icsch z Integration by parts is one of the most useful techniques of integration we will learn It is a good technique to try when u substitution does not work The formula is simple where u m v 9m so du f zdm and do g mdz Learning what to set 741 equal to takes some experience but generally one wants to set u to be a function that is easily differentiated and do to a function that is easily integrated Thus it helps to think of setting u do instead of u 1 When there are bounds on the integral make sure to carry over those bounds b Ab m g ltzgtdz fzgzl 7 bye Hm 1 Finally we can use trigonometric identities to solve some rather thorny integrals oftrigonometric functions These integrals typically involve powers and products of trigonometric functions for which u substitution is not helpful The following formulas are generally the most helpful sin2zcos2z1 sian 17cos2m coszm 1cos2m tanzmsec2z71 tanzdmlnlsecml0 secmdzlnlsecztanml0 1 2 M4080 REVIEW AND FUNCTIONS September 27 2008 4 118 A taxi company charges two dollars for the rst mile or part of a mile and 20 cents for each succeeding tenth of a mile or part Express the cost C in dollars of a ride as a function of the distance x traveled in miles for 0 lt z lt 2 and sketch the graph of this function Solution 2 0 lt x g 1 Cmi2210m711 mlt2 Simplifying we get 2 0 lt x g 1 C 2m1zlt2 The graph of this function is a piecewise continuous function that is constant at 2 for z between 0 and 1 and is a line with slope 2 for z between 1 and 2 connecting 11 with 2 4 5 1210 Recent studies indicate that the average surface temperature of the earth has been rising steadily Some scientists have modeled the temperature by the linear function T 002t850 where T is C and t represents years since 1900 ie 84 C a What do the slope and T intercept represent b Use the equation to predict the average global surface temperature in 2100 Solution a The slope represents the amount temperature increases as time increases ie for every 1 year the temperature goes up by 002 C The T intercept represents the temperature in Celcius in 1900 b T 002 200 850 4 850 1250 C 6 1214 Jason leaves Detroit at 200pm and drives at a constant speed west along l 96 He passes Ann Arbor 40 mi from Detroit at 250pm a Express the distance traveled in terms of the time elapsed b Draw the graph of the equation in part a c What is the slope of this line What does it represent Solution a One way to express the distance Jason travels in terms of the time elapsed is to multiply his speed by the time that has elapsed We determine his speed by noting that he traveled 40 mi in 50 minutes thus his speed is 4050 45 miles per minute Thus dt 45 t where t is the time elapsed and d is the distance traveled b The graph is a line with slope 45 and d intercept 0 c The slope is 45 miles per minute which means that for every minute Jason drives he travels 08 of a mile 7 1322 Graph y itan m 7 g by starting with a graph of one of the standard functions given in Section 12 and applying the appropriate transformations not by plotting points Solution Take the graph oftanm translate it it to the left by 7r4 to get the graph of tanz7 and then scale it by 14 to get the nal graph 8 1354 A spherical balloon is being in ated and the radius of the balloon is increasing at a rate of 2 cms a Express the radius r of the balloon as a function of the time t in seconds b If V is the volume of the balloon as a function of the radius nd V o r and interpret it Solution a rt 2t where t is the time that has elapsed in seconds and r is the radius in centimeters One way to see this is to do unit analysis we look at the units of the numbers we have M4080 QUIZ 13 REVIEW November 18 2008 72 The Natural Logarithmic Function 0 lnz f1mdt for z gt 0 0 Basic properties and formulas given on cover sheet 0 limmace lnz oo limma0 lnz 7oo o Logarithmic differentiation useful for differentiating products quotients and powers of functions 1 Take ln of both sides of y fz and use laws of logs to simplify 2 lmplicitly differentiate with respect to z 3 Solve equation for y 73 The Natural Exponential Function 0 De ned as inverse of ln 51 y ltgt lny z 0 Most properties and formulas given on cover sheet 0 limmH00 em oo limmaioo em 0 74 General Logarithmic and Exponential Functions For 3 gt 0 3 7g 1 de ne 5w ewln 7 Laws basically the same as natural exponential think 3 e 7 If B gt 1 3m has graph like em 7 1f 0 lt B lt 1 3m has graph like e m o For B gt O B 73 1 de ne log as inverse of 3m log z y ltgt y z 7 Change of base formula log z 1175 76 lnverse Trigonometric Functions 0 arcsin sin 1 is de ned as inverse of sin on restricted domain sin z yltgtsiny z for y 6 Egg 7 Domain of sin 1 is 711 range is 7 7 Values of sin 1 determined by using unit circle and values of sin 7 Derivative and integral formulas given on cover sheet 0 arctan tan 1 is de ned as inverse of tan on restricted domain tan 1 z y ltgttany z for y E 7 7 Domain of tan 1 is 700 00 range is 7 7 Values of tan 1 determined by using unit circle and values of tan 7 Derivative and integral formulas given on cover sheet UT Learning Center 5 mvwa succeed Excel M4080 Final Exam Review Day 3 1 Determine Which integration technique would be used to solve the following integmtion problems Do not Work the problem completely dx A Ix274x13 2 B I2xjildx c Ilenxdx csc2 x cot3 x dx D J39 E I2 cos3xdx x2 F I G j d X H I 2x2 79x 1 J x72 dx J j dx K J39 L Ifxcos xdx 1 ex 2dx x x274x7 M J 0ch lc2 72x 3 N Ixze dx Integration by Substitution 2s 2 d 13 6 5s2 S 3 j fo1dc 4 abc x I P 1112098 QSJQAIII 03 14111213 zx ZDJ ZocII xlgum magma eSJQAIII D x 19 x 914ugs j Z x14ugs x xi I I7 91113 QSJQAUI suoyoun 3119IIIOII03LIL asmAuI GILL gtHH Qx1oo xosou1 39936030 Qxumxoesu1 xpxoesJ Qxuysu1xpxlooj Qxoesu1 Qx soou1 x pxuelj Qxoso xpx100xosoj Qxpxumxoesj Qx100 xpxzosoJ Qxu21xpxzoesj Qxsoo xpxust Qxuysxpxsooj suopoun 311311101103H L JO uopm aml x Zoso 36103 x Zoes xu21 xumxoes xoes xuys x soc x100xoso xoso xsooxuys suonoun summouo ul JO seszAueq 5 Jcscz xdx I cot3x 6 d m Integration by Parts Reminder Iu dv uv Ivdu 8 I lnxdx Z 4 9 Ixcosxdx 0 10 Ia cos3xdx 1119862de Integration Using Partial Fractions J x2 4x7 x 39 x1x2 2x3 2x2 9x I x 23 13
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