Section 1.1. Calculus: Areas And Tangents. Difference Equations to Differential Equations

Transcription

1 Differece Equatios to Differetial Equatios Sectio. Calculus: Areas Ad Tagets The study of calculus begis with questios about chage. What happes to the velocity of a swigig pedulum as its positio chages? What happes to the positio of a plaet as time chages? What happes to a populatio of owls as its rate of reproductio chages? Mathematically, oe is iterested i learig to what extet chages i oe quatity affect the value of aother related quatity. Through the study of the way i which quatities chage we are able to uderstad more deeply the relatioships betwee the quatities themselves. For example, chagig the agle of elevatio of a projectile affects the distace it will travel; by cosiderig the effect of a chage i agle o distace, we are able to determie, for example, the agle which will maximize the distace. Related to questios of chage are problems of approximatio. If we desire to approximate a quatity which caot be computed directly (for example, the area of some plaar regio), we may develop a techique for approximatig its value. The accuracy of our techique will deped o how may computatios we are willig to make; calculus may the be used to aswer questios about the relatioship betwee the accuracy of the approximatio ad the umber of calculatios used. If we double the umber of computatios, how much do we gai i accuracy? As we icrease the umber of computatios, do the approximatios approach some limitig value? Ad if so, ca we use our approximatig method to arrive at a exact aswer? Note that oce agai we are askig questios about the effects of chage. Two fudametal cocepts for studyig chage are sequeces ad limits of sequeces. For our purposes, a sequece is othig more tha a list of umbers. For example,, 2, 4, 8,... might represet the begiig of a sequece, where the ellipsis idicates that the list is to cotiue o idefiitely i some patter. For example, the 5th term i this sequece might be 6 = 2 4, the 8th term 28 = 2 7, ad, i geeral, the th term 2 where =, 2, 3,.... Notice that the sequece is completely specified oly whe we have give the geeral form of a term i the sequece. Also ote that this list of umbers is Copyright c by Da Sloughter 2000

2 2 Calculus: Areas Ad Tagets Sectio. approachig 0, which we would call the limit of the sequece. I the ext sectio of this chapter we will cosider i some detail the basic questio of determiig the limit of a sequece. The followig two examples cosider these ideas i the cotext of the two fudametal problems of calculus. The first of these is to determie the area of a regio i the plae; the other is to fid the lie taget to a curve at a give poit o the curve. As the course progresses, we will fid that geeral methods for solvig these two problems are at the heart of the techiques used i calculus. Moreover, we will see that these two problems are, surprisigly, closely related, with the area problem actually beig the iverse of the taget problem. This itimate coectio was oe of the great discoveries of Isaac Newto ( ) ad Gottfried Leibiz (646-76), although aticipated by Newto s teacher Isaac Barrow ( ). Example Suppose we wish to fid the area iside a circle of radius oe cetered at the origi. Of course, we have all leared that the aswer is π. But why? Ideed, what does it mea to fid the area of a disk? Area is best defied for polygos, regios i the plae with lie segmets for sides. Oe ca start by defiig the area of a square to be oe uit. The area of ay other polygoal figure is the determied by how may squares may be fit ito it, with suitable cuttig as ecessary. For example, it is see that the area of a rectagle with base of legth b ad height a should be ab. Sice a parallelogram with base of legth b ad height a may be cut ad pasted oto a rectagle of legth b ad height a (see Problem ), it follows that the area of such a parallelogram is also ab. As a triagle with height a ad a base of legth b is oe-half of a parallelogram of height a ad base legth b (see Problem 2), it easily follows that the area of such a triagle is 2ab. The area of ay other polygo ca be calculated, at least i theory, by decomposig it ito a suitable umber of triagles. However, a circle does ot have straight sides ad so may ot be hadled so easily. Hece we resort to approximatios. (0, ) (-, 0) (, 0) (0, -) Figure.. A regular octago iscribed i a uit circle

3 Sectio. Calculus: Areas Ad Tagets 3 (0, ) 2 π 8 (-, 0) (, 0) (0, -) Figure..2 Decompositio of a regular octago ito eight isosceles triagles Let P be a regular -sided polygo iscribed i the uit circle cetered at the origi ad let A be the area of P. For example, Figure.. shows P 8 iscribed i the uit circle. We may decompose P ito cogruet isosceles triagles by drawig lie segmets from the ceter of the circle to the vertices of the polygo, as show i Figure..2 for P 8. For each of these triagles, the agle with vertex at the ceter of the circle has measure 360 degrees, or 2π radias, where π represets the ratio of the circumferece of a circle to its diameter. Hece, sice the equal sides of each of the triagles are of legth oe, each triagle has a height of ( π h = cos ) ad a base of legth ( π b = 2 si ) (see Problem 3). Thus the area of a sigle triagle is give by where we have used the fact that ( π ) ( π 2 b h = cos si = ) ( 2π ) 2 si, si(2α) = 2 si(α) cos(α) for ay agle α. Multiplyig by, we see that the area of P is A = 2 si ( 2π We ow have a sequece of umbers, A, A 2, A 3,..., each umber i the sequece beig a approximatio to the area of the circle. Moreover, although ot etirely obvious, each term i the sequece is a better approximatio tha its predecessor sice the ).

4 4 Calculus: Areas Ad Tagets Sectio. correspodig regular polygo more closely approximates the circle. For example, to five decimal places we have A 3 =.29904, ad A 4 = , A 5 = , A 6 = , A 7 = , A 8 = , A 9 = , A 0 = , A = , A 2 = Cotiuig i this maer, we fid A 20 = , A 50 = , ad A 00 = As we would expect, the sequece is icreasig ad appears to be approachig π. Ideed, if we take a polygo with 644 sides, we have A 644 = 3.459, which is π to five decimal places. Alteratively, istead of defiig π to be the ratio of the circumferece of a circle to its diameter, we could defie it to be the area of a circle of radius oe. That is, we could defie π to be the limitig value of the sequece A. Symbolically, we express this by writig π = lim A. I that case, let B be the area of a circle of radius r ad let B be the area of a regular -sided polygo Q iscribed i the circle. If we decompose Q ito isosceles triagles i the same maer as P above, the each triagle i this decompositio is similar to ay oe of the triagles i the decompositio of P. Sice the ratios of the legths of correspodig sides of similar triagles must all be the same, the sides of a triagle i the decompositio of Q must be r times the legth of the correspodig sides of ay triagle i the decompositio of P. Hece each of the triagles i the decompositio of Q must have a base of legth rb ad a height of rh, where h is the height ad b is the legth of the base of oe of the isosceles triagles i the decompositio of P. Thus the area of oe of the triagles i the decompositio of Q ito isosceles triagles will be from which it follows that 2 (rb )(rh ) = 2 r2 b h, B = 2 r2 b h = r 2 ( 2 b h ) = r 2 A. Sice r is a fixed costat, we would the expect that, i the limit as the umber of sides grows toward ifiity, B = lim B = lim r2 A = r 2 lim A = πr 2.

5 Sectio. Calculus: Areas Ad Tagets Figure..3 Parabola y = x 2 with taget lie (blue) ad a secat lie (red) Hece we arrive at the famous formula for the area of a circle of radius r, i which the costat π has bee defied to be the area of a circle of radius oe. Example I this example we wish to fid the lie taget to the curve y = x 2, a parabola, at the poit (, ). This problem may ot at first seem as useful as that of fidig the area of a plaar regio, but we shall fid that the ideas behid the solutio have may applicatios, ad are, ultimately, importat i the solutio of the area problem as well. First there is the questio of exactly what is a taget lie. At the preset it will be sufficiet to leave the otio at a ituitive level: a taget lie is a lie which just touches a give curve at a poit, givig a close approximatio betwee curve ad lie. I Chapter 3, we will see that a lie l is taget to a curve C at a poit P o C if l passes through P ad, i a sese that we will make precise at that time, gives a better approximatio to C for poits close to P tha ay other lie. Now let C be the curve with equatio y = x 2, let P = (, ), ad let l be the lie taget to C at P. Sice l passes through P, i order to fid the equatio of l we eed oly fid its slope m. Ufortuately, to fid m i the stadard way we eed to kow two poits o l, ad we kow oly oe, amely P. Hece we will agai have to resort to approximatios. For example, the lie through the poits (, ) ad (2, 4) is ot l (it is a secat lie, rather tha a taget lie), but sice it itersects C at P ad at aother poit which is close to P, its slope should approximate m (see Figure..3). Namely, we have m 4 2 = 3. Sice ( 3 2, 9 4 ) is o C ad is closer to P tha (2, 4), a better approximatio is give by the slope of the lie passig through (, ) ad ( 3 2, 9 4 ), that is, m = = 5 2.

6 6 Calculus: Areas Ad Tagets Sectio. More geerally, let be a positive iteger ad let m be the slope of the lie through the poits ( + (, + ) ) 2 ad P. For example, we have just see that m = 3 ad m 2 = 5 2. Now, i geeral, m = ( + ) 2 ( + ) + 2 = + 2 ( 2 = + ) 2 = 2 + for =, 2, 3,.... Hece m 3 = = 7 3, m 4 = = 9 4, m 5 = = 5, ad so o. Moreover, as icreases, decreases toward 0, ad so we would expect that as icreases, m decreases toward 2. At the same time, as icreases m more closely approximates m. Thus we should have ( m = lim m = lim 2 + ) = 2. That is, the slope of the lie taget to C at P is 2. The the taget lie l has equatio or y = 2(x ), y = 2x. Here we have used the fact that the equatio of a lie with slope m ad passig through the poit (a, b) is give by y b = m(x a). The rest of this chapter will be cocered with the study of sequeces ad their limits. The ext sectio will cosider the basic defiitios ad computatioal techiques, while

7 Sectio. Calculus: Areas Ad Tagets 7 the remaiig sectios will discuss some applicatios. We will retur to the problem of fidig taget lies i Chapter 3 ad the problem of computig areas i Chapter 4. Problems. Use Figure..4 to verify that a parallelogram with height a ad base of legth b has area ab. a b Figure..4 A parallelogram 2. Explai how ay triagle is oe-half of a parallelogram, ad use this to verify the formula for the area of a triagle. 3. Use Figure..5 to verify the formulas give for the height ad base of oe of the isosceles triagles i the decompositio of P. h π b 2 Figure..5 A isosceles triagle from the decompositio of P 4. Try the procedure of the taget example to fid the equatio of the lie taget to the followig curves at the idicated poit. (a) y = 2x 2 at (, 2) (b) y = x 2 + at (, 2) (c) y = x 3 at (, ) (d) y = x 2 at (2, 4) 5. For the area example, fid the umber of sides ecessary for the area of the iscribed polygo to approximate π to 6, 7, 8, 9, ad 0 digits after the decimal poit. 6. For the taget example, how large would have to be i order for m 2 to be less tha 0.005?

8 8 Calculus: Areas Ad Tagets Sectio. 7. For the taget example, let p be the smallest positive iteger such that m p 2 < 0.0. (a) What is p? (b) What ca you say about m 2 for values of greater tha p? 8. For each of the followig sequeces {a }, compute a 0, a 20, a 00, a 500, ad a 000. ( (a) a = si ) ( (b) a = + ) (c) a = 0, where! = ( )( 2) (2)()! 9. As we saw i the area example, there is more tha oe way to defie the umber π. For example, we ca defie it either as the area of a circle of uit radius or as the ratio of the circumferece of a circle to its diameter (of course, if the latter approach is take, oe has to show that this ratio is the same for every circle). Suppose we defie π as the area of a circle of uit radius. Cosider a circle with radius r, diameter d, circumferece C, ad area A. The we have see that A = πr 2. The followig steps show that we also have π = C d. (a) Let P be a regular -sided polygo iscribed i the circle. Let s be the legth of a side of P. By dividig P ito equal isosceles triagles as we did i the area example, argue that A rs 2. (b) Ca you see why as goes to ifiity, s approaches C? (c) Now ca you see why (d) Use the result i part (c) to show that rs A = lim 2 = rc 2? π = C d. 0. You may fid a iterestig discussio of techiques for computig areas ad volumes up to the time of Archimedes ( B.C.) i the first two chapters of The Historical Developmet of Calculus by C. H. Edwards (Spriger-Verlag New York Ic., 979). I particular, there is a discussio o pages 3-35 of Archimedes proof that the two defiitios of π metioed i the area example yield the same umber.