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# Calculus I MATH 1210

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39 METEO 3110 Introduction to Atmospheric Sciences Prof Zhaoxia Pu Department of Meteorology University of Utah August 20 2007 METEOaMn Pvuilnauxia Pu 39 Introduction What is atmospheric sciences The science devoted to the description and understanding the phenomena in the atmospheres ofthe earth and the other planets Two major disciplines of atmospheric Sciences 39 Climatology Longterm statistical properties ofthe atmosphere that onstitute climate 2 39 Meteorology Atmospheric phenomena and their timedependent behavior METEO 31m Pvuilnauxia Pu Introduction cont Three main subidisciplines of Meteorology I Physical I Synoptic I ic Main subidisciplines of Climatology I Physical I Applied Applications of Atmospheric Sciences Fo ecast Assessment Beneficial modification Provision METanim Pvuilnauxia Pu 39 I Importance of our Atmosphere I We are always affected by the atmosphere I Many natural disasters are linked With the atmosphere I There is concern that our climate is changing I METEO 31m Pvuilnauxia Pu Genuine scie 1 ents inp ns 1545 raaars usea to observe precipitation 1560 rst weather satellite 15505 Doppler raaars 15505 other sophisticatea instruin ents Aristotle arouna 34 50 hook on natural philosophy callea Met wora meteor r errea to any particle ef nce With weather instruin ents 3 haronieter huin a 1520 fronts and air masses 1540s aaily upper air ineasurein 15505 co r la i uters fo calcu to History of Meteorology eorologlca from sky MEYEOJMUWlehaoxaVu sun P Operational For Inn il 51nmmemoul MEYEocmu miznsaasm Progress in Weather Forecast NCE ecast Skill cm and 72 lawn a must g sun ma um an AM Major Developments in Afmaspher ic Sciences Big Developments 10772003 Pixwress in vezilhri hicrush Atidniiii A lls hole Rucogllllitln ii Emcnllmwc minim New siibl iulils alums it cllclnislry p cllmme tlyllmnlcs Marzowu rraiznsaasm How acid rain affec fs s ronewor k MEYEOJMD rraiznsaasm 39 Objectives of This course I This course provides an introduction to the eld of meteorology for both meteorology majors and other scientists and engineers I This course is the rst ofa series of theoretical and practical courses that you will take to qualify yourself as a meteorologist METEO 311m Pierzhauha Pu Course Content 1 Introduction hgih cumpusmun aha gas eeheehnaimhs Dismbutiuri utiempeiaimewwa aha vaEipitatmn ZAlmnspheric Thermnuynzmiizs as iaw ydiustatic equation The 15 iaw enhehheayhamies aha adiabatic lapse rate Waieivapewei adiabatic laps rate s1aiie s1abihty Second iaw enheiheimeayhamie The 3 The h h 3Almnspheric Aernsnl and Clnud Micrnphysiczl Pmeesses Almusp ehe aemseis aha iuud hueieaiieh S m Ciuudt esampciuudmuiphuiugy Thuhaeismnhs humcanE amp eyeiehes a Radiative Transrer Eiectiumagnetic spectrum amp biackbudy Yadiati n Absuiptivityemissivetyampsuiaiabsui huh ihhaiea YadiatNEtYanSiEY amp seanehhg utsuniight Eheigy haiahee h the uppei aimespheie Stsurtace Etfects emaee gases aeiusuis Smiuuds Em 2712ng haiahee 5Almnspheric Dynamiis Easicfuices The Equatiun utmutiun The ihehhai Wiha The thEYmudynamic 2712ng equation The cuntmuity Equatiun METEO 311DFTuiZhauxia Pu e METEO 1010 MATH 1210 and 1220 PHYSC 2210 Prerequ Required Textbook Wallace J M and P V Hobbs Atmospheric Science An Introductory Survey Second edition Academic Press 2006 Recommended References Holton J Introduction to Dynamic Meteorology 4th Edition Academic Press 2004 SalbyML Fundamentals oIAtmospheric PhysicsAcademic Press 1996 ogers R R and M K Yau A Short Course in Cloud Physics 3rd Edition ButterworthHeinemann 1989 Petty G W A First Course in Atmospheric RadiationSundog Publishing 2004 METEO 311m Pierzhauha Pu 39 1 Grading policy Grades will be based upon your performance on the homework exercises exams and the class participa ion The wei hted contribution of each ofthese items to your nal grade is given below 35 Homework 15 Exam 1Atmospheric Thermodynamics 15 Exam 2 Cloud Microphysics 15 Exam 3 Radiative Transfer 15 Exam 4 Atmospheric Dynamics 5 Class participation attendance class discussionand inclass problem solving Final grades are based on the following scale gt90 guarantees an A or A guarantees a c c or c guarantees a D D or D lt60 results in an E METEO 311DFTuiZhauxia Pu 39 I Homework policy Homework may be assigned aller each lecture but will be collected in every Vou are expected to work independenty to solve the problems though discussions among classmates are allowed Pagansm Wnotbe toeated Please hand in your homework at beginning of the class on the due day Late homework will not be acceptedresult a 39039 grade 39 m eu on as excellent completion Vou are encouraged to correct errors in your homework ilthe homework was rated below to 90As a result you are able to make up loryour lower homework grade to a rate up to 90 Homework achievement will be contributed to your nal grade 35 in total METEO 31m PruchanxiaPu 39 100 Instructor Dr Zhaoxia Pu Office 813 WBB Office hours MWF 12401330 It s important to see me during the office hours I may not be available for you other than the office hours MEranMnmmzhanwaPu 39 1 Components of the earth system Atmosphere Oceans Cryosphere Terrestrial biosphere Crust and mantle Atmosphere is one ofthe components of earth system METEO 31m PruchanxiaPu 39 I Thinness of the atmosphere Atmosphere is a thin envelo e gases and tiny particles that surround Earth 99 of atmosphere39s mass is con ned to a layer ofthickness 1 ofthe earth39s diameter osphere around Earth is thus thin like the peel of an apple atmosphere is essential for life contains oxygen and car dioxide for life sustaining processes supplies water and shields life from harm Jl ultraviolet radlatlon from sun METEOSMDFthhanxiaPu 39 1 Composition of the atmosphere noble gases wltum mm mm mm m1 1mm cwmwn tm mt m Cantu mm mm llntttm Attml nalcmlttl um um tum w rm 5 m M c M0 t2 gamma u t r mu ltwrnnrmulzz Helttnlllc Mn mman h m m twruwmilla l 2 v Hymyrrmlhnr a Jllpl39Mx39 m METED 31m Pthhanxia w Evolution of Atmosphere Early Primeval Phase I Earth39s birth was about 46 billion years ago I Lava ashes gases from volcanoes quotoutgassingquot form Earth39s primeval atmosphere hugging planet due to gravitational eld of Earth I Atmosphere consisted of mostly COZ carbon dioxide NZ nitrogen and H20 water vapor METED 31m Pvnchanxta w 39 Early Phase I Surface temperature might have been as high as 85 to 110 C compared to 15 C today I Planet cooled water vapor condensed to form clouds and rain hence oceans I A lot of CO2 in atmosphere dissolved in rainwater I Life formed about 2 billion years ago and photoswthesis produced oxygen 02 I Ozone 03 shield formed METED 31m Pthhanxia w 39 Evolution of Atmosphere Modern Phase I Main atmospheric components are NZ 7808 by volume and 02 2095 in layer below 80 km other constituents are water vapor trace gases and aerosols water vapor concentration is highly variable ranging from 0 to 4 trace amounts of C02 03 and other gases aerosols in atmosphere are tiny liquid and solidparticles from forest res wind erosion of soil salt from ocean dust spray volcanic emission and meteonc METED 31m Pvnchanxta w 39 7 Recent changes in gas concentrations Greenhouse gases and global warming Manna Lm dots and South Pole line C02 1958 2002 3quot mm Aim ml um mttm lllll 1 an H H 12 Yllll rim mu ll il lll lllill nr l as 0 Hi n 71 m an 0 Carbon dioxide in units of part per million METEO BllO PlehanXla Pu 39 I Zonalmean mixing ratio of water vapor contoured and density of water vapor or absolute humidity shaded as functions of latitude and pressure W l u39 Vapcgt g kgquot PRESSURE mb w l v39 l gto7oreo sc to woo c VO 2v so am 50 so 70 3 LATITUDE The shaded levels correspond to 20 40 and 60 ofthe maximum value METEO BllO PlehanXla Pu Water vapor imagery METEO BllO PlehanXla Pu 39 Zonalmean mixing ratio and density of ozone wazssam mm n r m mum mm mm 41 None wnluuxcdl md may at mom shaded nlvnlli dm lllll ich l l i liullintilmlenlliulc MllyKMumoMamtdlmmllw um Inland Mmilmi ol m suiimpm 11mm mm mm in mm Imh mi Wild In m 40 lllnl we of ill mniiilllml lill Ozone lters out incoming solar radiation in the ultraviolet part of he spectrum METEO BllO PlehanXla Pu GETS6133 4444 ENquuu4444m onmuo wwommqommuo nwvwomaucmowoucmc DosonUmts GEN234ZDD5 durkgruyforlt100undgt500 Du METEO EMEI Prufzhauxwa Pu AerosolIndex a Aeroso ndex 2005 5 Goddard Space 25 10 3 Aeroso ndex FthL Center METEO 31m PerZhaEMaPu 39 Tofnl fracfionnl cloud cover annual averaged from 1983 1990 compiled using dam from fhe Infernnfional Safellife Cloud Climafology Projecf ISCCP ISCCP Tolal Cloud Amuum 19831990 Percent METEO 31m Prufzhauxwa Pu INTRODUCTION TO POLYNOMIAL CALCULUS 1 Straight Lines Given two distii39ict points in the plane there is exactly oi ie straight line that coi itains tl iei n both This is one of the important prii icipls of Plane Geoi39netry If the plane is equipped with a Cartesian coordinate system it should be possible to write down an equation for such a line in terms of the 1 and y coordinates In this section we shall show how to do this The notion of slope will be very useful in this task The slope of a line is defined to be the change in the y coordii iate the rise divided by the change in the 1 coordii39iate the run we move from one point on the line to a secoi39id point on the line see Figure 11 That is if 131111 and 132112 are two points on the line then the slope of the line is the 1 1111 Ill 1 m wl39iere m 12 11 62 61 This number does not depend on which two points on the line are cl39iosen In fact if two other points 63 13 and 64 14 are chosen then it follows from the similar triangles in Figure 12 that 12 11 7 14 13 32 31 34 33 and so the points 513 13 and 514 14 give the same value for the slope m do the points 11y1 and 5621112 The slope measures wl39ietl39ier a line rises or falls we i39nove to the right and how steeply it does so Positive slope means the line to the right while negative slope means it fall to the right Slope zero means the line is horizontal since it m 39g y coordii39iate does not change at all from point to point on the line A vert line has no slope some would say it has ii39ifii39iite slope This is becaus the 1 coordii ia of points on such a line are all the same and so the d1 1o1 ni1 1ator is zero in the equatioi i defining slope gt3 Y p F g C C 2 Example 1 Find the slope of the line which contains the points 1 2 and 3 5 7 Solution Here the rise is 5 2 3 and the run is 3 1 2 and so the slope is 32 The pointslope form of the equation of a line Using slope we can easily write down an equation for any line that is not vertical Let in be the slope of the line and let 69 yo be some fixed point on the line If 51 y is a variable point on the line then 1 1o 55 550 m and so 1 10 misc cm 1 This is called the pointslope form of the equatioi i of a straight line It gives the equatioi39i of the line in teri39ns of the slope in of the line and a point 179 yo on the line Example 2 Find the equatioi39i of the line which passes through the point 1 and has slope 2 Solution Here the slope m is 2 and the point 510 yo is 1 and so the equatioi39i of the line is y 32zr 1 or y2zr1 Example 3 Find the equatioi39i of the line which coi itains the points 1 2 and 0 5 Solution The slope of this line is 5 2 0 13 m and the line contains the point 0 5 so the equatioi39i of the line is y 53zr 0 or y3zr5 The slopeintercept form of the equation of a line If a line crosses the y axis at the point 0 1 then the nui nlmr b is called the y ii itercept of the line If the line has slope m then the poii it slope form of its equatioi39i using the point 0 5 is y b mm 0 or y ma I apt form of the equatioi39i of the line Note that every line except a ept form for its equatioi39i quot is the pointinter vertical line does cros the y axis and thus has a slope inte Example 4 What is the equatioi39i of the line with slope 5 and y ii itercept 4 2 Solution 1 5z 4 The general equation of a line Every equatioi39i of the form ActrlByl C0 is the equatioi39i of a line and coi39iversely every line has an equatioi i that can be put in this form A non vertical line has an equatioi39i of the form 3 mar I which can be written my 1 1 0 wl39iich is of the form Act By C with A m B 1 and C I On the other hand a Vertical line has the form 1 k wl39iere k is a constant This can be rewritten 1 k 0 which is of the form Act By C 0 with A 1 B 0 and C k Example 5 Find the slope and y intercept for the line which is described by the equation 21 31 6 0 Solution If we solve for y in this equatiol39i we get the equatior39i 2 2 r 7 y 3 The equatior39i of the line is now in slope ir itercept form and we can just read off the slope and the y ir itercept 2 Parallel and perpendicular lines If the two lines in Figure 13 are parallel then the two triangles are similar from which it follows that the two lines have the same slope On the other hand if the two lines have the same slope tl39ier i the two triangl in Figure tl39iree must b 39nilar sir ice they contain right angles with proportional ac ent This implies that the two lines are parallel since they make the same angle with a horizontal line Thus two lines are parallel if and only if they have the same slope What about perper idicular lines Tl39iere is an old carpenter s trick for checking that an angle is a right angle One measures off three units along one leg and makes a mark and tl39ier i measures four units along the other leg and makes another mark The angle is then a right angle if and only if the stance tween the two marks is live This works because 32 42 52 and an angle in a triangle is a right angle if and only if ya a2 b2 2 wl39iere a and b are the lengths of the two agacent sides of the angle and c is the length of the opposit side the Fytl iagorear i Tl39ieorer n Let 1 how we can apply this to tell when two lii ies are perpei idicular We may well assur39 that r39ieitl39ier line is horizontal since it is easy to see wh 1 a lin is perpendicular to a horizontal line it must be Vertical Then the first line inte i ctr axis in a point 511 0 and the second line i1 lt its the ctr axis in a point 512 0 Let suppose the two rms are not parallel and so they int t each other in a point 510 go see Figure 14 Then they intersect at a right angle if and only if a2 2 2 wl39iere a is the distance from 30 go to 511 0 I is the distance from 30 go to 512 0 and C is the distance from 31 0 to 512 0 This leads to the equation 61 602 115cv2 602 15 62 612 Wl liCl l sii39nplilies to 235 215 26051 23305132 2z1z2 On dividing by 2 and rearranging teri39ns this becomes 15 5 0551 0552 1552 or after factoring 2 10 3 0 Clalll39o 51 or 10 7 l o 171 60 62 go which that 1 7722 7 7721 Wl iere m1 is the slope of the first line and 7712 is the slope of the second line Thus two non horizontal lines are perpendicular if and only if the slope of the second is the negative reciprocal of the slope of the rst Example 6 Find the equation of the line which is parallel to the line with equation 2 31 5 and pa through the point 1 2 Solution We solve for y in the equation of the first line in order to put its equation in the form 2 5 r 7 7 y 3 3 This shows that the line has slope The line which has the sai39ne slope and passes through the point 1 2 has the equation 2 2 8 y 2 gz 1 or y gargar or 2ar3y8 Example 7 Find the equation of the line which is perpmidieular to the line 21 3 3 and meets it at the point 1 1 Solution If we solve the first equation for y it becomes 1 3 g y 2 2 Wl liCl l tells us that its slope is A line perpmidieular to this line will tlmrefore have slope 2 Since the line we seek must pass through the point 1 1 its equation is y 12z 1 or y2z 1 2 Slope of 3 Curve We know about the slope of a straight line It is the change in the y coordinate divided by the change in the f coordii39iate rise divided by run we i39nove from a given point on the line to any other point on the line The law of similar triangles that this ratio is independent of the two points on the line that are cl39iosei39i What about the slope of a curve that is not a straight line Does this make sei39ise If so how do we calculate it Let s look at an example say the curve y The graph of this curve Figure 21 makes it clear that if its slope i39nakes sense it cannot be a fixed nui39nber To the left of 1 0 the curve slopes downward ative slope while to the right of 1 0 the curve slopes upward positive slope and it rises i39nore reply the furtl39ier to the right we go Thus if it i39nakes sei39ise at all the slope must depend on wl39iere we are on the curve In fact the slope of the curve 3 62 does make 1se at eacl391 point 51 g on the curve but it cl39iai39iges the point cl39iai39iges This s sugg by the fact that if a microscopically ce of the curve is i39nagi39iilied ei iough to be visible then it looks like a straight line illustrated by the graphs in Figure 21 w quot 39 39 39 39 3 gtrs In other words the at near 1 1 the i39nore the curve looks like a straight line The slope of this straight line should be what we mean by the slope of the curve at the point 1 1 How can we calculate this slope We do the same thing we would do if the curve were a straight line We calculate the cl39iai39ige in 1 divided by the change in 1 we i39nove from one point on the curve to al39iotl39ier However now we cl39ioose special points We let the first one be 1 1 itself and we cl39ioose the second one to be near 1 1 The nearer to 1 1 we make it the more the curve betweei39i tl39iese two points looks like a straight line and the closer our ratio will be to the slope of this li Let s try this for soi ne cl39ioices of points on the curve 3 512 see Figure 22 In each case we will be calculatii39ig the rise divided by the run betweei39i a first point 1 1 and soi ne nearby secoi39id point on the curve In other words we will be calculatii39ig the slope of the line joining tl39iese two points With secoi id point 3 9 we get A 110 9 1 8 il a 7 7 4 s opr 3 1 2 With secoi id point 2 we get 4 1 3 3 7 7 s op 2 1 1 5 With secoi id point 15 225 we get sloe 22 L 1 1a With second point 11 121 we get 121 1 21 cl 17721 0 11 1 1 With second point 101 10201 we get 10201 1 0201 slope 7 201 101 1 01 A graphic descriptiol39i of what is l39iappei39iii ig l39iere is sl39iowi39i in Fig 2 One can now guess that if we continue doing this for SOCOHd points that are closer and closer to 1 1 th l we are calculating will simply get closer and closer to the number 2 This would mean that the slope of the curve 1 12 at the point 1 1 is 2 Let us now take a perfectly gei ieral secoi39id point one that is obtained by Cl lal lgil lg the 1 coordii iate of the first point by an ai39nount 11 so that the second point becoi39nes 1 1391 1 I12 The si39naller 1391 is ch the closer th cond point is to 1 1 The slope we get for the line joining 1 1 to tl nd point is then slope 2 11 1 11 1 i 211h 391 11 Now if 1391 is cl39iosei39i very small then this number will be Very close to 2 In other words I1 approaclms 0 our slope approaclms 2 This should COI IViI lC us that the slope of the curve 1 12 at the point 1 1 is 2 The fact that 2 11 approaclms 2 I1 appoaclms 0 is COIHIHOI lly written lim2 11 2 h gt Here lim2h is sl39iortl39iai id for the limit of 21 1391 approaches 0 and it simply means the rulinlmr that 2 11 approaclms I1 approac as 0 The method used above works for otl39ier curves well Example 1 Find the slope of the curve 3 12 31 at the point wl39iere 1 2 Solution For the function f 1 12 31 we have f2 2 f2H12H12 3121 11 2hh2 Thus the slope of the curve when 1 2 is 2 f2 139 I 2 lim 7 I lim1 11 1 I1 39 h gt slope lim h gt h gt 1 Returning to the curve 3 512 we would like a fori nula for the slope of this curve at any point of the curve not just at the point 1 1 We use the same tecl39inique Given a point CIT 512 on the curve we move to a nearby secoi39id point 1 h 1 102 obtained by changing the 1 coordinate of the first point by an amount h Then we calculate the slope of the line joining these two points It is 1 h 2 512 2611 h2 slope 21 h h h As h approaches zero this slope approaches the number 251 In other words lim2z h 21 h gt We conclude that the slope of the curve 3 172 at the point 6562 is 251 The pr gling discussioi39i leads to the following delii39iitioi39i of the slope of a curve which is given the graph of a function f De nition A The graph of a function f is said to have slope m at a point CIT fzr on the graph provided in I in lii n m h gt h That is provided the xpression approaches m as h approaches 0 In this case we call the number m the derivative off at 1 and denote it by f zr Thus the derivative of a function f is another function f of the same variable 51 Its value at 1 is the slope of the graph of f at the point CIT f 37 In other words it is the instantaneous rate of cl39iai39ige of y with respect to 1 we pass through the point CIT fzr while moving along the graph of f For now we will take the statei39nent approaclms m h approaclms 0 intuitively understood but later in the course we will not 1 to make this idea i39nore precise We will do this when we study limits As we shall see in the next section when f is a polyi ioi nial it is quite easy to see what happens to the expressioi i h approaclms 0 and so to study derivatives of polyi39ioi nials we do not need a sopl iisticated study of limits The discussion pr c ling the above delii iitioi i shows that the derivative of 512 is 251 We give two other examples of the calculation of a derivative Example 2 Find the derivative of a constant fui39iction fz 2 Solution I f zr lii39n lii n 391 I lii39n 0 0 h gt h h gt 39 h gt Thus the derivative of a constant function is 0 This just re ects the fact that the graph of a constant function is a horizontal line and thus has slope 0 Example 3 Find the derivative of the fui39ictioi39i f 1 31 2 Solution y y 2 2 f zr lim a h g lim 53 h 33 lim 3 3 h gt h h gt h h gt Thus the derivative of the fui39iction zz 31 2 is the coi39istant 3 This is not surprising since the graph of the fui39iction f 1 31 2 is a straight line with slope 3 3 Derivative of a Polynomial In the last section we delii ied the derivative f 1 of a function f 1 to be the slope of the curve 3 f 1 at the point 51 fz This in turn is defined to be the number that W approaches h approaclms 0 In other words f l h it I 7 39H fWm h the expression We will now use this definition to calculate the derivative of any polyi ioi nial We begin by calculatii39ig the derivative of the i39noi39ioi39nial 51 for each value of n For n 0 1 23 this was done in the previous section and its problem set However we will repeat tl39iese calculatioi39is l39iere in order to dei39noi39istrate the Iattern that gai39nergc If at at 1 then 1 1 I 7 39 H 7 39 H f CL h 0 If f 1 511 1 then 7 CITlIt lei 7 fWwgwiwbl If f 1 512 then 1 102 512 I 7 f f f 51 lilting h 111523 11 21 If f 1 513 then 3 3 p p f z lim h a 312 3611 119 3512 h gt I lim h gt The pattern here suggests that for any natural number n the derivative of 51 should be mun 1 This is in fact true The proof is not dif cult From the de nition of derivative we have that I n n 1 39I 1 1 hm h gt h If we expand a h using the binomial theorem we obtain it n 1 it h 1 turn 1h 7 2 a 2li2 TLCITI39In 1 h It is not ii39nportai39it l39iere to know ex tly each term in this expansion It is importai it to know the first two terms and the fac that every term except the first two has a factor of h raised to a power of at least two When we subtact at from this expai isioi i and divide by h we obtain a h at I con 2h hath 2 fin 1 39I 1 mf rl nn2 Here every term except the first one has a factor of h Thus when we take the limit h appoaclms 0 all terms except the first one will vanish We conclude that 1 n n a h a my 39I a lim h gt This proves the following tl39ieorei39n Theorem A If n is any nonnegative integer then a mun 1 We now know the derivatives of a large nui nber of fui39ictions For example 64 433 5110 101 3 5195 955694 Next we would like to be able to find the derivative of a polyi39iomial like 513 4512 a 5 which is a linear coi nbii iatioi39i of moi iomials We ned the following tl39ieorei39n Theorem B If f and g are functions and C is a constant then 1 Cf Cf 2 fg f g That is the derivative of a constant times afvnetion is that eonstant times the derivative of the function and the derivative of the sum of two functions is the sum of their derivatives We won t prove this theorem now Its proof is in the textbook and will be done later in the course Using this theorem and the preceding one we can now find the derivative of any polyi ioi nial Example 1 Find the derivative of 3512 21 5 Solution 332 2x 5 3x2 my 5 Th 132 3g2 2x 5 Th 131 32czr210 ThA 61 2 Example 2 Find the derivative of 515 11333 91 2 Solution 515 11a3 91 2 515 11z3 9zr 2 534 113zzr2l91l0 5x4 3332 9 Example 3 Find the slope of the curve 3 174 3512 2 at the point 1 0 Solution We first find the derivative of 514 35132 2 514 3512 2 4513 3 2 0 433 61 We then evaluate the derivative at 1 1 to get the slope of the curve at the point 1 0 Thus the answer is 4 6 2 Example 4 Find the rate of cl iai ige of the function f 1 173 41 with respect to 1 Wl39ien 1 2 Solution We first find the derivative of f 51 f zr 513 450 3512 4 We then evaluate the derivative at 1 2 to find the rate of cl iai ige of fz with respect to 1 at 1 2 Thus the answer is 322 48 Velocity and Acceleration Much of Physics and Ei39igiimering is coi iceri ied with the i39natl39imnatics of moving bodies Suppose a body moves along a straight line If we x an origin for this line and units for measuring distal lce on the line then the position of the body at any time t is described by its coordinate on the line often denoted by 813 The velocity vt of the body is then de ned to be the rate of change of 813 with respect to t that is the derivative 3 03 of 813 Similarly the 39 leration 113 of the body is de ned to be the rate of change of vt with respect to t that is the derivative v t of vt In summary Example 5 If a ball if dropped off the top of a 64 foot high building then its height 803 above the ground 13 seconds later is described by the formula set 16t2 64 What is its velocity when it hits the ground What is its acceleration at any time Solution The ball hits the groui id when 16t2 64 0 This happei39is Wl iei i t2 4 that is wl39iei39i t 2 Thus we need to know the velocity of the ball when t 2 But vt 8 03 162t0 32t At time t 2 this is 64 Thus the ball hits the ground With velocity 64 ft sec The acceleration at any time t is 113 vim 32t 32 Thus the acceleration is a constant 32 ftsecsec 4 Antiderivatives of Polynomials We have ii itroduced the operatioi39i of dil arentiation finding the derivative of a fui39ictioi39i and shown how to dil arentiate any polynomial In this section we will study the reverse operation De nition A If f at is a function then an antiderivative for f is a function having f at as its derivative es dil ar ei itiatioi39i If f is the derivative of g tl39iei39i g is an ai itiderivative of f that we said an ai itiderivative not the ai itiderivative Thi 39s l ause a fui39ictioi39i wh has an ai itiderivative always has ii39ilii iitely many In fact if g is an ai itiderivative of f then g c is also an anti rivative of f for any constant 2 This follows from the fact that the derivative of a constant is zero Given a function f does f have an ai itiderivative This is a dif cult question in gei ieral and will be studied in soi39ne atail later in the course However for polynomials it is quite easy to see that the answer is yes We begin by looking at the moi iomial so where n is a i ioi i i igatiV ii39iteger Is this the derivative of soi ne otl39ier polyi39ioi39nial If we dil arentiate 56 we get a 1 n Dar If we divide by n 1 this leads to the equation C3H4 7 71 n1 339 Thus we have found an ai itiderivative for so i39iamely This is not the only ai itideriv ative for so since we can add any constant to a fui39ictioi39i without chai iging its derivative Thus any fui39ictioi39i of the form 4 man n1 n 1 where 2 is a coi istai it will have derivative 51 2 De nition B Iffzr is a function the set of all antiderivatives for at is denoted f so do and is called the inde nite integral offzr We have shown that every fui39ictioi39i of the form c is an ai itiderivative for so Convex e very anti ivative for so is of this form as long we oi39isider the domain for a to be the wl39iole real line or at least an interval on the real line However to prove this requi 39 of the big Cl l 0l39 is of calculus the Mean Value Tl l 0l39 1 Ii wl39iich will not be discussed until later in the course For now we will simply assume that this is true Then we can state the following Cl l 0l39 1 112 Theorem A If n is a nonnegative integer then an i j a a 71 1 Rei nei nlmr this Cl ieorei n siinply that the set of antiderivatives for so is the set of functions of the form 2 If we apply this for n 0 1 2 we get ldlf 510 dd a C 2 arriara1riar 6 3 a2riar C If h is an ai39itiderivative for f and a is a constant tl iei i where C ranges over all constants ah ah af and so ah is an antiderivative for a f Similarly if h is an antiderivative for f and k is an antiderivative for g tl iei i hlk h k fg and so h k is an antideyivative for f g Thus we have proved the following Cl ieorei n Theorem B If f and g are functions and a is a constant then 1 fa a da afzrdar 2 am wows fmdw amaze In other words the integral of a constant times afunction is the constant times the integral of the function and the integral of the sum of two functions is the sum of the integrals of the two functions Tl ieorei ns A and B combined allow us to ii39itegrate find the indefinite ii39itegral of any polyi ioi nial Example 1 Ii itegrate 133 31 6 Solution 333 31 6 d3 513 d3 d3 6dzr Th 132 argdzzr3zzrdzzr 61dzz Th 131 514 3512 Note that we can always check our answer to an intgration problem by dil arentiating if we get back the original fui39iction For example let s dil arentiate the answer in the preceding example 514 3512 I I 433 3251 3 2 Ii ideed we do get back our original fui39iction and this verifies that our answer was correct Example 2 Find the antiderivative of 513 3512 that has value 0 when 1 1 Solution All antiderivatives of 513 3512 are given by 4 73 3512 d3 513 C VVl iei i 1 1 this becormas 14 10 34C Thus if we want the antiderivative that has value 0 when 1 1 we should cl39ioose 2 34 The answer is then 514 x3 3 4 I 4 Example 3 The acclration experiei iced by an object due to the force of gravity is 32ftsec2 A proj t39l is red straight up with an initial velocity of 128 ft after which the only forc ng on it is gravity What is its velocity 13 seconds later VVl lOll does it reach its maximum height Solution Acceleratioi39i is the rate of change of velocity with respect to time that is it is the derivative of velocity a function of time We know the a leration is 32ftsec2 Thus the velocity vt is an antiderivative of the constant function 32 So m vt 32 32t C VVl iei i t 0 the velocity is the initial velocity 128 ft sec Thus the coi39istant 2 must be 128 Tlmrefore the velocity at time t is vt 32t 128 The object reaches its i39naxii39nui39n height wl39iei39i its velocity drops to zero it is not going up ai39iyi39nore This l39iappei39is wl39iei39i 32t 128 0 that is wl39iei39i t 4 seconds Example 4 For the pre ading example assui39ne that the object was fired from an initial height of 10 feet off tl39 ground and tl n find the height above the ground of the object at any time 13 Find the i39naxii39nal height ach39 ved by the obj Solution Velocity is the rate of change of distance in this 1 height with respect to time That is it is the derivative of height with respect to time Thus the height st we are seeking is an antiderivative of of the velocity vt 32t 128 This means set 3226 128 dt 16t2 128 2 VVl iei i t 0 the height is the initial height 10 feet Thus 3 10 and set 16t2 128 10 The i39naxii39nui39n height occurs wl39iei39i t 4 by the previous example So the i39naxii39nal height acl39iieVl is 16 16 128 4 10 266 feet 5 De nite integrals Tl39iere are two i39natters which we will take intuitively and operationally acceptable for the time being Later in the course we will study these issues at greater depth 1 If two differentiable functions f and g have the same derivative in an interval I then they differ by a constant This is the same saying that if a function It has derivative equal to 0 everywl39iere in I then it is constant take h f g Now this is ii39ituitively obvious if we paraphas it saying that if a fui39ictioi39i 1391 V X cl39iai39iges ai39iywl39iere in the ii39iterval I then it is is constant However that parapl irase has probl the derivative is the instai39itaimous rate of cl39iai39ige and the assertioi39i a fui39iction wl39iich appears not to cl39iai39ige at any given instant in fact 1391 V X cl39iai39iges when i nade prec se loses its ii39ituitive clarity H 2 Area is wellde ned regions we shall consider in this course have area Again calculation of areas has been an ongoing problem since the begii39n39iing of recorded history and the Calculus provides a most el active way of doing this if we know that it makes sense As we shall see below area of a curviliimar figure is foui id by approxii39natii ig the figure by collections of r ctai39igles That the approximations lead to a good concept of area is a subtle problem which for the time being we shall igi iore Froi39n time to time in this course we will touch on the dif culties with tl iese assui nptions but for the present we shall assume them true statermants Suppose that y f 1 is a nommgatiw function on the interval 11 The area under the curve 1 fz from a to I is denoted Ab fzrdzr Theorem The area under 3 fz is equal to F F01 for any indefinite integral Fz of fz Demoi39istration For 1 any point betwmn a and b let Aw be the area under the curVe from a to 1 Let s calculate A zr For a small incrermant h the a 39 43 h up to 1 h is approximated by Amt h f at where Ir f 1 is the area of the tangle of base It and heigl it f 6 Thus 1 Amt h Azr fzrh error where error is boui ided by f 1 h f 56h if for example f is an increasing function Thus x is boui ided by f 1 h f 1 which will go to zero if f is at all a reasoi iable function we say f is continuous We COI lCllldQ AM h Amt h AM h Amt h Thus AM is an indefinite integral of f If F is any other indefinite integral 41910 2 Am Fm C wl39iere C is soi ne constant which we can easily find since AW 0 Substituting in 2 we obtain C Fa Thus the area under the curve from a to I is 405 171 Fm Examples 1 Find the area under 3 51 from 0 to 1 An indefinite integral is F 1 aw so the area is F 1 F 0 2 Find the area under the curve 39 172 2513 from 2 to 4 1 4 1 1 f 512 2513dzz 763 2764 1 1 1 1 44 72P 7 713941quot 36206 38 26 63 The delii39iite integral will be delii ied later in this course the limit of a sum follows Suppo that y f 1 is l lll l jl over the interval 11 Let us subdivide the interval by a S 1111 1C of partition points 519 a lt 511 lt 512 lt lt len1 lt a 1 Coi39isider the sum f lf1 l1 60 f l 1 lf2 CU1 f In ln gin 1 which can be writtei39i using the sui ni natioi i notation 21f339i55i Chi 1 If for example we were calculating the area under the graph of a positive fui39iction this would be the area of a polygonal figure approxii39natii39ig the curviliimar gure and would thus give an approximate value for the area If we divide the figure more lii iely using more points we would expect to get a better apr OXll39HaClOH to the area Under suitable coi39iditioi39is the coi itii iuity of the fui39iction f these sums do approacl39i a limit which is the delii39iite integral of the function f from a to I The Fundamental Theorem of the Calculus will tell us that the delii39iite integral can be coi39np11tl 173 F01 wl39iere F is any i11dlii 1it integral of f Example Suppose a particle moves in a straight horizontal line so that its velocity directed to the right at tii39ne t is vt t2 133 meters per minute How far to the left or right of the initial position of the particle will it be after 2 139ninuts Ai39iswer 2 2 1 1 8 16 4 viii 262 263 73 7 7 i D U vr 0 df 3f 41 0 3 4 0 3 the particle is 3 of a meter to the left of its initial position

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