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# CALCULUS II MATH 166

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This 49 page Class Notes was uploaded by Ms. Helen Sipes on Sunday September 27, 2015. The Class Notes belongs to MATH 166 at Iowa State University taught by Staff in Fall. Since its upload, it has received 22 views. For similar materials see /class/214503/math-166-iowa-state-university in Mathematics (M) at Iowa State University.

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Arc Length Here let C be a curve in the Xy plane de ned parametrically by xftygtaStSb The functions ft and gt are assumed to be continuous and to have continuous rst order derivatives f 39t and g39t on the closed interval a b Moreover we suppose that for ng gt 0 for every real number t in a b Now let Pat0 ltt1 ltt2 ltt3 ltlttn 13 be a partition of the closed interval a b of the t aXis For each integer i 123 n the length of the closed straight line segment joining the successive points tk1 gt11 and ftl gtl of the curve C is given by f a f In 2 g0 gtH 2 y x171 ft171 Lyzrl gt171 For each integeri 123 n there is by the Mean Value Theorem a real numberI in the closed I such that ft flt171 I txri and also by the Mean Value Theorem there is a real number ti in the closed interval twtl such that gt gt171 gle I t171 Thus for each integer i 123 n the length of the closed straight line segment joining the succesive points f t11 gt171 and f t1 gt of the curve C is given by interval rm V lb l rnJ The sum of the lengths of all of these line segments is given by the sum WW g39a gt12 a gt Even though the points Z and t in this sum for each given integer i may not be the same as the norm of the partition P approaches 0 this sum still approaches the integral 1 W g39r2dt We take the length L of C to be given by Lva WEVWW Consider the arc length function given by am lvvlkvlw where t is in the interval a b For each real number t in the closed interval a b 0t is the length of the portion of C from the point fa gato the point ft gt The derivative rate of change of this arc length function 0t is given by dmvv lk l If the parameter t here is time and if the equations Xfty gta St Sb are parametric equation for the motion of a particle along the curve C then the derivative oquott is the speed of this particle at time t Exercises 64 19 Consider here the cycloid C de ned parametrically by x t sinty l cost0 S t S 2H Find the length of C dt 1 costand dt Slnt g2 1 cost2 Sint2 1 2cost cos2 t 511120 2 2 cost 2 1 cost sin2 Lose 2 2 2 E Q 4sin2 611 611 2 21 a Find the length of the curve C de ned by yfxlu3 1dufor1SxS2 x3 1 dx I 1gjz 1x3 1x3 2 Zquot xszdx 2 2 zixsz 5 1 2 2 2g252 1g4212 1 11 Find the length of the curve C de ned by 4 y 1 2for 3SyS 2 E 2y 4y3 43 1y3y43 y 16 4 2 dx 1 6 1 6 61y 16y 2 y 2 in 1iy iy 6 my 16 2 Z 72 1dy 3 dy 1 81 1 17 77 8 16 18 14418729 8 144 14439 19 The region R is bounded by the graphs ofy X3 Xl and y X The graph ofy X3 Xl yx2 4x3 This is a parabola opening upward The verteX low point is where 1 0 X g 2x 4 0forX2 dx lt0forXlt2and gt0forXgt2 dX dX 2 d 2gt0fora11X 2 y decreases as X increases for X S 2 and y increases as X increases for X Z 2 The graph is concave upward The graph of y X is a straight line Intersection ofy X3 Xl and y X x2 4x3x x2 5x30 SinB 167 A 4 yX3X1 2 1 AreaofRJ5ff3 x x 3Xx ldx T 55 IE x25x 3lx 2 5 lx3 xZ 3x 2 3 2 35 E5 E3 2 2 5 2 f M 13 13 5 I 152 25JEEZ3810E195E I 5225E JEZ38 10E19 5E xE 2 5 2 5462 5 25 2amp2 2s E2 B 4 l 4 HgE 3 53352E35E2 12575E151313E 32088JE4011E SQBI 53352 E35 E2 ET 12575E1513 13E 320 88JE40 11JE 15E35 E322JE 3 2 2 AreaofR 3EE22E Exercises 61 23 x8y y2 y8 y0f0ry0y8 X8yy2 4 R 164 AreaofR J8y y2cb 8 1 4y27y3 3 0 256 E 3 3 ixv Return now to the above power series 1 Zan39W XOY 0 Suppose now that one of the following two cases occurs i The power series 1 converges for all real numbers X In this case let R 00 ii There is a positive real number R such that the power series 1 converges for each real number X satisfying lXX0lltR and diverges for all real numbers X satisfying lXX0lgtR Let I be the open interval consisting of all real numbers X such that lXX0lltR Let fX be the function with domain I such that for each real number X in Ithe corresponding functional value fX is the sum of the power series 1 It can be shown that for each real number X in the interval I the derivative fX eXists the power series 2 Zna x x n1 converges and fX is the sum of 2 Note that fXo a1 Also for each real number X in I the integral ftdt eXists the power series no 1 1 3 7 a x x m 0 nlt ogt converges and ftdt is the sum of 3 Then for each real number X in the interval I the second derivative fquotX eXists the power series 4 2M 1anx 960 H n2 converges and fquotX is the sum of 4 Note that f xo21az Continuing for each real number X in the interval I the third derivative fquotX eXists the power series 5 Znn1n2anxxo quot 3 n3 converges and fquotX is the sum of 5 Note that fquotx0 302010a3 We may continue in this manner to show that fX has derivatives of all orders throughout the interval I and that fquotx0 noan for n 1 2 3 Thus the coefficients an in the power series 1 are given by the formula n 6 anf739x for 110 123 n The power series f M x n 7 Z 0 x 3 n0 n is called the Taylor series for fX about X X0 EXample Find the Taylor series for fX cosX about X 0 f0l 3610 1 fX sinX 3 f0 0 3 a1 0 fquotX cosX 3 fquot0 l 3 a2 fquotX sinX 3 fquot0 0 3 a3 0 fquotX cosX 3 fquot013 a4 etc Thus this Taylor series is l i 2ix ix 2 4 639 iHy 1 ox 10 2 We now discuss the convergence of this Taylor series ForXiO 1 11 I x21 l lt2ltJ1gtl z 1 Hy 1 ml 2jl2j2 2139 gt 0 as j gt 00 Thus this Taylor series converges for all real numbers X We do not know that this Taylor series actually converges to cosX EXample We now consider the Taylor series about X 0 for arctanX Note rst that the geometric series 2 x 0 converges to l l l x l x for all real numbers X satisfying lltXltl Then the in nite series no no Z x2 1 Zenx Fe 0 converges to l l x2 for all real numbers X satisfying lltXltl We have that farctanx dx 1 x2 for all real numbers X satisfying lltXltl If X is a real number such that lltXlt 1 then x 1 w x imdt l Ltzfdt 1 2 no 1 I t21 g0 2l L0 1 no 1o ox 12x 2j1 Thus if X is a real number such that lltXlt 1 then 2j1 1 t 7 t 0 l 10 0x21 arc anX arc an 2j1 Hence the Taylor series for arctanX about X 0 is Z l o 1 0x21 x lx3 1165 lx7 10 2 1 3 5 7 EXample NeXt we consider the Taylor series about X 0 for fX 1 90 l x Note rst that f0 1 Then fX l1 x 2110 i 2 239 l 3 s fquot o l x A x 2 2 lt gt 103 3 fquot 7 X 22 l 3 5 7 fquotX 7 i 7 ol x A ltgt MUM 2 fquotX 3395 20202 etc In general 1352n 1 n n f 1 202022 41 90 13052n l W0 1quot 2222 Thus the Taylor series is 103052n l n 2222 n 1 1 quotZak 10203n x 103052n lxn 11n o 246 2n 1 It can be shown that this Taylor series does 1n fact converge to f x for lltXltl H x Thus the in nite series w l352n lx2 2462n n converges to for lltXltl Note that x2 1 l x iarcsin c 2 for l lt X lt 1 Thus the Taylor series for arcsinX about X0 is Z l352n l xm quot12462n2n1 l 3 13 5 l35 xix 716 x 23 245 2467 xleixsi 7 6 40 112 Example By proceeding as above in our discussion of the Taylor series for cosX about 7 X0 we can show that the Taylor series for sinX about X0 is x ix3ix5 3 539 1 1 21 El 2j1 We now use this information to nd a few terms in the Taylor series for tanX about X0 We proceed by using long division 7x x 7x 3 6 72 2 ti 15 315 3 s i 7 15 15 1x7 315 Thus the rst few terms of the Taylor series are x 1x3 3 5 l 7 3 1 5 3 1 5 Assignment due Monday March 27 Exercises 105 2 4 8 12 18 20 22 26 Exercises 106 1 6 8 11 20 Exercises 62 l R is the region bounded by the graphs ofy X2l y 0 X 0 X 2 Let S be the solid of revolution generated by revolving R about the XaXis Find the volume of S y yxzl or xy l R 2 Volume 0fSH39JXZ lzabcllx4 2x2 ldx 2 H lx5 2x3x 5 3 0 H E2 5 3 H968030 15 2061 1 15 39 We now get this volume by the shell method discussed in Section 63 1 5 Volume ofS 21 14y2aly2HJl y2 1y l7ly 2Hy2 0 2Hf2y yF7Iy 2H2H 52y y 1FFjy 2H2HfZy y l y ldy 2H2Hy2 y 1 y 1 2H2H 25 32 8 1 00 2H2H 25 E 1 5 3 2m 375 192 80 15 206H 15 39 3 Let R be the region bounded by the graphs ofy 4 7 X2 y 0 X 0 X 2 a Let S be the solid of revolution generated by revolving R about the XaXis Find the volume of S Disk method Section 62 Volume ofS HJ 4 x2 de HJl6 8x2 x4ix 2 H l6x x3 lx5 3 5 0 11 32 3 3 5 15 Shell method Section 63 Volume ofS 2HL4y 4 ydy 4 2110 4 y4 y44 y1y 4 3 1 2HL m yy44 4 w4 y 2H m w 4 m w 2H 332 8 5 3 2561 1 15 15 39 b Let S be the solid of revolution generated by revolving R about the yaXis Find the volume of S Disk method Section 62 Volume ofS I LL4 Wydy H39l4ydy 2 4 H4yL 2 0 81 1 Shell method Section 63 Volume of S 2H J2 x4 xzyix 0 2Hj024x x3zx 2 2H2xZ lx4 4 0 2H4 8H 7 Let R be the region bounded by the graphs of y 1 0x 2x 4y 0 x Let S be the solid of revolution generated by revolving R about the XaXis Find the volume of S l 4 X Diskmethod 2 4 VolumeofSHJ4l dxHJ lizdxH l 2 x 2x x 2 l l H 4 2 4 Shellmethod VolumeofS2HJ ylt4 2dy2HJ yi 2iy 1 1 2HyZE2H121 2ydy Z 2H 2Hy y2 E2nl r 4 116D i2nii 8 16 4 D IA 9 Let R be the region bounded by the graphs ofy 9 x2 and y 0 between x 2 and x 3 Let S be the solid revolution generated by revolving R about the XaXis Find the volume of S y y 9 x2 y 9 x2 xV9y2 x 9 y J R 2 I 3 V X Diskmethod Volume ofS HJ x9 x2 zdx Hf29 x21x n 9x Jll n lt279gt wsn 436 2 100H 73 Shell method Volume ofS 2Hszy 9 y2 2gtIy2HL3 y9 y2 9 y2 dy 2HJfy 9 y2 2yiy21lJj y2 9 y2 dy J 3 2Hl9y2y2 4Hl9y2 3 0 3 J 2H l85 l270D4H l0l8 3 3 3 3 3 3 10011 T R X2 J Revolve R about the yaXis 2 4 4 2 4 y Volume 11L 25 ab 4110 ydy 4117 32H 0 yZ b2 1 2 x 17 7 a2 2 y 1 13 2 2 b upper half of x2 a Revolve R about the XaXis x2 y2 y2 x2 2 x2 771371 7 b21 7 a2 b2 b2 a2 y k a2 2 3 B Volume Illa2 I x7dx HI2 16 11 0 a 3a in Hb2 a laj ala iHalal 3 3 3 yz4X 168 4 R X 2y0 Revolve R about the XaXis 2 Volume HJ64x x de HJ64x x2dx 16 n 2362i 11 512 E 12 0 3 3 n 21 Revolve R about the yaXis WNWIlz lz wEma Hly2 H2 E k 3 23 length2 4 x2 The cross section at X is a square with length ofa side equal to 2 x4 x2 The area of the cross section at X is Altxgt2m744x2 The volume of the solid is fZAxdxf244 x2lx lm3sz 88 K 33 y 4 8 3 y 354 R X4 x y a Revolve R about the line X4 2 Volume Hf4 y q39y H J168y yay 8 H l6y gy y 2H 128g32 128 210241 1 5 7 o 5 7 35 b Revolve R about the line y8 Volume H Jsz 8 16 jzjdx H 0464 82 16x 8de IT 16 x3 6 4 11 163x 1x4 5 4 0 W 8 y Volume2H08 y4y SJq39y 2Hj32 4y 8y yq39y 8 221 1 322y2 g y y 5 g 0 2H256 l28 32 256 2Hl28 96 2H224 Assignment to work on Exercises 62 Problem l4 16 18 22 25 32 34 Math 166N Jan292009 56 MOMENTS AND CENTER OF MASS 1 Momenti Momentlevel armxmass iiMoment of a system 71 i1 iii Center of mass is the balance point in other words suppose i is the center of mass then n m a 32gt i1 therefore n 2 mm 7 M i1 m 7 a 7 n In face the concept of the center of mass is that of an average of the masses factored by their distances from a reference pointwhy Eamplel particles of mass m1 27712 5 and 7713 7 are located at 1 1 2 72 and 3 3 where is the center of mass 2 Continuous Mass Distribution along a line Consider now a straight segment of thin wire of varying density 6m then the balance point i M 7 m6mdm m 6zdz Example 2A straight wire 7 units long has density 6z mil2 at point x units from one end Find the distance from this end to the center of mass 3 Mass Distributions in the Plane 1Consider 71 point masses of sizes m1 m2 mn situated at points m1y1z2y2 zn y in the coordinate plane then total moments M1 and My with respect to the x axis and y axis respectively are given by 7L 7L My E 90mlth E yimi 11 i1 1 The coordinates of center of mass are 71 71 7 My 7 211 mimi Mm 7 211 yimi m 7 7 7 7 W7 m i1 ml m 2177 7y Example 3 Find the center of mass and the moments of the following system of particles with respect to the coordinate axis 17 21262472736731228 2 Consider the homogeneous lamina bounded by z 017 z by x and y gm7 assume that f lt 9 and the density 6 of the lamina is constant7 then the moments with respect to the coordinate axes and the center of mass centroid are 17 17 My 6 we 7 gltzgtgtdz7 Mm g we 7 ltgltzgtgt2idz i more egltzgtgtdz g wow 7 ltgltzgtgt2idz m ffltflta gt79ltxgtgtdx7 m ffxgmdx Example4 Find the centroid of region bounded by 1y27z 2 2 y 2x272 7z3 Example 5 pp3147 2224 Here let fX be a realvalued function with continuous derivatives of all orders on an open interval I of the XaXis Let X0 be a real number in I The Taylor series for fX about X0 is the power series no fk x 1 Z x xok k0 M For each X in I and each nonnegative integer n let SnX be the corresponding nth partial sum of 1 given by n fk x 2 SnxZxxok H k Then for each X in I and each nonnegative integer n let RnX be the error 3 Rnxfx Snx in using the value of the n3911 partial sum SnX to approximate the actual functional value fX There is a result known as Tavlor s Formula with the 39 A which says that if X is real number different from X0 in the interval I and if n is a nonnegative integer then there is a real number c in the interval I between X0 and X such that fn1c quot1 4 RAM THMU x0 EXample Let fXcosX for every real number X We have shown that the Taylor series for cosX about X0 is Find a bound on the absolute value of the error in using 1 2 1 4 1 5 l 25 45 65 to appr0Ximate cos5 Note that this partial sum is actually k 05k k Thus there is a real number c between 0 and 5 such that the error is 8 Rn558 Hence since if 8 S l for all real numbers x the absolute value of this error does not exceed 58 000000097 Wehavethat 1 2 1 4 1 6 155 165 a5 877582465 Note The calculator gives cos5 877582562 Example Here we consider the exponential function fx eX We begin by de ning the number e by the limit 1 h l elllltllzj h The following table of values for I suggests the value of e h h 11 h 1 2 10 259374246 102 270481383 103 271692393 104 271814593 105 271826824 106 271828047 107 271828169 108 271828182 109 271828183 The number e appears in situations where the amount of a certain quantity changes instantaneously at a rate proportional to the amount present An example of such a problem is the continuous compounding of interest Suppose you deposit P dollars in a bank which pays an annual interest rate of r percent compounded m times per year We assume that the m compounding periods are essentially of equal length The amount At in dollars you have in the bank after t years is given by the formula 2 AtPl m We are interested in what happens to At as m 9 00 For simplicity of notation let 3 h7 Then the formula 2 becomes 4 At P 1 if Asm ooh ooandso At gt Pe 01 The formula 5 210 gt Pe 01quot is said to provide the amount in dollars you have in the bank after t years under continuous compounding of interest We now discuss some important properties of the exponential function fX ex First we note that fx x fx1fx2 for all real numbers X1 and X2 Then we note that fr x fool for all real numbers r and X Since e gt 1 it follows that lim f x 00 Also lim f x 0 so the XaXis is a horizontal asymptote for the graph of f We now illustrate the limit eh l 6 11301 h 1 h h e l h l 1718281828 1 1051709181 01 1005016708 001 1000500167 0001 100005 00001 1000005 So now let X be some given real number and let h be a nonzero real number Then fxh fxe h e h x e e e eh l h h gte lash gt0 Hence for each real number X the derivative fX exists and is given by the formula 7 7 e e Then fX 6 has derivatives of all orders on the entire real XaXis and for each real number X we have that MK e for n 123 In particular W0 1 forn 123 Thus the Taylor series for fX ex about X 0 is lxix2 ix3 ix4 2 3 4 We previously used the ratio test to show that this power series converges for all real numbers X We now discuss why this power series actually converges to ex for all real numbers X We treat the case where X is a positive real number The case where X is a negative real number can be treated similarly For each positive integer n let sltxgt1iixh k1 k and let Rn x fx Snx For each positive integer n there is a real number c depending on n between 0 and X such that R xfn1cnxn1 n1 ecquot n1 nl x Thus n1 R s K x quotxi e n1 Because the power series converges gt0asn gtoo n 1 gggSxfx so this power series actually converges to ex Thus EXample We return now to consideration of the function 1 7y x l x 2 forlltXltlWeshowedbeforethat f m135 nlt2n1gt1x for each positive integer n Then fmm135n2n 1 2 for each positive integer n The Taylor series for fX about X 0 is no n 2 f 016 n n1 1 1Zl352n 1xn quot1 2462n llx x2 ix3 x4 2 8 16 128 We can use the ratio test to show that this power series does actually converge for l lt X lt lNow if X is real number different from 0 and satisfying 1 lt X lt l and if n is a positive integer let Snx 1Z 1 3 5quot392k1xk k1 24 62k and then let Rn x fx Sn 6 For eXample l 3 5 35 S 5 1i 5 7 5 27 5 37 5 4 4 2816128 1399902344 which is an apprOXimation to l l f 5 7 7 x1 5 J3 Now again let X be a nonzero real number satisfying 1 lt X lt l and let n be a positive integer There is a real number cn depending on n between 0 and X such that n1 Ru 0 f c x1 n 1 Note that 135 2nl 1CK 2 f c gt 1 3 5gf lt1 cW lt1 cgt Consider rst the case where 0ltXlt1 Then 0 ltcn ltX 1 c gt 1 x Hence 1 Thus n1 06 1 1352n1 x 1cn 2462n2 1 0 1 1352n1 x quot1 xl x 2462n2 39 lt 1 x Suppose we have i lt 1 1 x x lt 1 x 2x lt 1 1 x lt 7 2 In this case the in nite series w 1352j1 x 1 11 2462j2 l x converges by the ratio test and hence 13 5 I i gt 0 as gt oo 2462j2 l x Therefore Rnx gt0asn gtoo Hence Snx gtfxasn gtoo It is not clear how to handle the case where X 5 However we discuss a bound on the error in using Sn5 to approximate f5 To begin note that Also note that Thus For example Note that 5lt84 1 5gt1 84 1 1 lt 1 5 1 84 i1 5 0ltRn5lt25M 2462n2 13579111315171921 246810121416182022 25168188095 420470239 0 lt 11105 lt 25 1 3 2 5 3 35 4 slo5155 5 E5 75 128 2635 7231 5 7429 57 6435 58 256 1024 2048 32768 10939559 415701 510 589824 2359296 1414055485 Observe that 1 1 f5 m so that f52 2 Also observe that sm52 1999552915 Consider next the case where 1 lt X lt 0 Then X lt cn lt 0 Now lcn gt1 0 1 lt1 lcn Again n1 Rnxlcn1352n1 x I 2462n2 l cn Hence Rnwk1352n1 x mi 2462n2 Thus Rnx gt0asn gtoo Quadratic Functions and Parabolas Here let a b c be given real numbers such that a 7i 0 Let f be the function with domain the set of all real numbers and with the property that fx ax2 bx c for every real number X We call f a quadratic function The graph of f is an example of the type of curve called a parabola We now discuss some basic properties of f and its graph First note that fxcvc2 bxca x2 xjc a 2 b W W a x 7x 7 c a 7 a 2a 2a upon completing the square in X Thus b 2 4610 b 2 x a x 7 7 2a 4a It then follows that the graph of f is symmetric about the vertical straight line X 2b If a gt 0 then a f 2b lt fX for every real number X different from 2b the value of fX decreases as X increases a a b b b for X S 2 and the value of fX1ncreases as X 1ncreases for X Z 2 If a lt 0 then f 2 gt fX a a a b for every real number X dlfferent from 2 the value of fX1ncreases as X 1ncreases a b and the value of fX decreases as X 1ncreases for X Z 2 for X S 2b The point of the parabola a a a where X 2b is called the verteX If a gt 0 then the parabola opens upward and the verteX is the a low point of the parabola if a lt 0 then the parabola opens downward and the verteX is the high point of the parabola The X intercepts of the parabola are given by b 2 4ac b2 a x7 70 2a 4a so these X intercepts are given by the formula 1 X bib2 4ac 2a Formula l for the X intercepts of the parabola is known as the quadratic formula for the solutions of the quadratic equation ax2 bx c 0 It should be pointed out that the parabola may have 2 X intercepts just 1 X intercept or no X intercepts Often it is easier to find the X intercepts by factoring rather than using the quadratic formula We now present some eXamples of sketching graphs of quadratic functions and applications of quadratic functions EXample 1 Sketch each of the following parabolas showing the verteX of the parabola in each case ay fX X2 2X8 Here a lb 2c 8The X coordinate of the verteXis given by X i 2 2a 2 1 and the y coordinate of the verteX is fl l2 2 l 8 9 The X intercepts are given by X 4X 2 0 X 4 X 2 The y intercept is given by f0 8 We now determine a few more points on the parabola to aid in the sketching x 54 32 1012 3 fX705898507 Since a lt 0 the parabola opens downward and the verteX is the high point Figure lis a sketch of this parabola b y fX X2 6x 16 Here a l b 6 c l6 The X coordinate of the verteXis given by X 261 3 and the y coordinate of the verteX is f3 32 6 3 16 25 The X intercepts are given by X8X2 0 X 8 X 2 The y intercept is given by f0 16 Since a gt 0 the parabola opens upward and the verteX is the low point We now determine a few more points on this parabola x 3 2 1 0 1 2 3 4 5 6 7 8 9 fX 11 0 9 16 2124 25 24 2116 9 0 11 n y Vertex 19 yfXX2 2X8 Figure 1 Example Here let F be the parabola which is the graph of the equation y axz where a 7i 0 This is a special case of the quadratic function y axz bx c where b0 and c0 In the calculus textbooks this parabola has the equation x2 4py Thus 1 7 a 4p or i p 4a I The vertex V ofF is the origin 00 The focus F of F is the point 02 or 0p The a directrix L of F is the horizontal straight line de ned by the equation y 4a or y P Recall now that F is the set of all points P x y such that the distance between P and the focus F is equal to the distance between P and the directrix L We now check this using the equation x2 4 py for F the focus F given by the point 0p and the directrix L the horizontal straight line defined by the equation y P m let P xy be a point of F The distance between P and the focus F is given by x 0V y p2 y2 2py P2 y p2 b ly pl This is the distance between P and the horizontal straight line L de ned by the equation y P NeXt let P Xy be a point such that the distance between P and the point F0p is equal to the distance between P and the horizontal straight line L de ned by the equation y p Then xOY yP2 lypl x2 y p2 yp2 x2 y2 2pyp2 y2 2pyp2 x2 4 py Hence P is on the parabola F We now consider the optical property of a parabola This property is used for example in designing searchlights with light source at the focus To illustrate this property we use the parabola F de ned by the equation y2 4px with pgt0 This is a horizontal parabola opening to the right with vertex at the origin and with focus the point F p0 Exercises 65 3 A force of 06 newton is required to keep a spring of natural length 008 meter compressed to a length of 007 meter Find the work done in compressing the spring from its natural length to a length of 006 meter Let X denote the change in meters of the length of the spring from its natural length and let FX denote the corresponding force in newtons required to produce this change Using Hoose s Law we have that there is a constant k such that FX k X We are given that 06 k 01 so that k 60 newtons meters The work done is given by L 2Fxdx 0 60xdx 30x2 2 012 5 Consider a spring obeying Hooke s Law Again let X denote the change of the length of the spring from its natural length and let FX denote the corresponding force required to produce this change There is a constant k such that FX k X Find the work done in stretching the spring a distance d Work done J FXdx d 0 Idkxdx 1ka lde o 2 2 7 A spring obeying Hooke s Law is such that the force required to keep it stretched s feet is given by 9s pounds How much work is done in stretching it 2 feet Work done L29sds 2 2s2 18 2 0 9 y y10 05 5 2 ix53x2 i y 2 5y i X 20 20 The above region is the vertical cross section of a tank with the lengths in feet The tank is 10 feet long The tank is full of water Find the work done in emptying the tank by pumping the water to a height 5 feet above the top of the tank 5 2 W0rkJ06241022gy10 yq32 1248 520 6 2 2 y y 2 5 1248 20y 3y27y3 15 0 1248100 75 j 21248E 3 52000 11 A y lt y 7 9 34 34 39 LengthX y y 8 2 x 3 t X 3 0 lt0 2 2 The above region is the vertical cross section of a tank with the lengths in feet The tank is 10 feet long Find the work done in emptying the tank by pumping the water to a height 5 feet above the top of the tank Consider a typical horizontal slice of the water in the tank to be lifted with bottom at height y feet above the base of the tank and with thickness Ay feet The bottom width ofthis slice is 2 3y and the top width ofthis slice is 2 Ay Thus the volume ofthis slice is not less than 8 3 3 102 7 7 A lsy 2 y and is not more than 102 yAyAy Hence there is a real number y between y and y Ay such that the volume of this slice is 3 3 102 7 397 A 8 y 2i y The work done in lifting the water in this slice is not less than the work done in lifting this water a distance 9 y Ay feet and is not more than the work done in lifting this water a distance 9y feet Thus there is a real number y between y and y Ay such that the work done in lifting the water in this slice is equal to the work done in lifting this water a distance 9y feet Hence the work done in lifting the water in this slice is equal to 3 3 624102 7 397 A 9 quot 8y 2 y y 2 V height density volume Using partitions of the interval 04 on the yaXis approximating sums as indicated above and the limit as the norms of the partitions go to 0 we have that the work done in emptying the tank is given by the integral 4 3 3 L624102 y59 ycbx Thus this work is 12485415 8 1248 61 76128 Exercises 65 y The above region R is a vertical cross section of a cylindrical tank with the radius of the base equal to 4 feet and with the height equal to 10 feet The tank is full of oil with density 50 pounds per cubic foot We want to nd the work done in pumping the oil over the top of the tank Work done 05016 o10 yq39y 800 o10y y2 400007 125664ft lb 10 0 27 Consider a real number y between 0 and 4 When the depth of the water still left in the bottom tank is y the volume of the water that has been pumped is 47y 10 440 47y Let d denote the depth of the water now in the top tank The volume of the water now in the top tank is n 62 d Then 36nd 4004 7 y so that 10 d 70 4 9 y Thus the work done is 462410410B4yydy 0 97139 4 40 10 2496o 107 7 d L 9 g y yy 40 5 1 4 2496o 10 7 7 2 7 2 y g y g y 2y0 2496o40 8 97139 97139 80 2496 o 32 7 97139 86934 ft 1b PX0 y B G K a Q0 Rpm NV The goal here is to show that the angles or and in the figure are equal We do this by showing that the triangle FQP in the figure is isoceles Thus we want to show that the distance between the focus F and the point P xo ya on the parabola is equal to the distance between the focus F and the point Q 70 on the x axis We begin by finding the x coordinate of the point Q Note that Q is on the tangent line to the parabola at the point P xo ya We now find the slope of this tangent line Observe that 2yo 4p so that dy 2p E 7 I39 Thus this tangent line has the equation yyo Zipkn ya We then set y 0 in this equation and solve for x 2 yo ipxixa ya We have that yoz 41096 Then 2p 2X0 x xt7 Thus finally we get X X0 The distance between the focus F p0 and the point P X0 yt7 is given by X0 pf 0 02 xon 2ng p2 y2 Xopl Xop Note that X0 pis indeed the distance between the focus F and the point Q Suppose now that a light source is placed at the focus F Consider then a light ray emanating from this source and hitting the parabola at the point P Then this ray us re ected outward parallel to the aXis of the parabola Polar Coordinates Let P Xy be a point of the plane different from the origin Let r 6 be an ordered pair of real numbers such that x rcos9 y rsin We say that r 6 is a set ofpolar coordinates for P Example Consider the curve de ned in polar coordinates by ed r lecos6 60 where e and d are positive real numbers and where 60 is a real number Note that cos6 6390 cos6 cos6 0 sin6sin60 H In the spec1al case where 60 3 we have that cos6 sin6 We have that r ed 1 ecos6 cos6 0 sin6 sin6 0 r 1 6 cos6 cos6 0 sin6 sin6 0 ed r e r cos639cos6390 r sin6 sin6390 ed r e cos6 0 xsin6 0 y ed r ed e cos6 0 x sin6 0 r e d cos6 0 x sin6 0 y Next we get that d cos6 0 x sin6 0 y2 d cos6 0 x sin60yd cos6 0 x sin6 0 d2 acos6 0x asin6 0y alcos6390xcos26390x2 cos6 0 sin6 0 xy d sin60 ysin60cos60 xysin2 60 y2 d2 2acos6 0x 2alsin6390ycos26390x2 2cos60sin60xysin260y2 Thus r2 e2 cos2 60x2 2cos6 0sin6390xysin26390y2 2dcos60x 2dsin60yd2 Hence x2 y2 e2 cos26390x2 2cos6 0sin6 0xysin26390y2 2dcos60x 2dsin60yd2 Therefore 62 cos26390 lx2 282 cos6390 sin 0 xy e2 sin26390 ly2 282 clcos6390 x 282 clsin6390 yezal2 0 Now let A e2 cos260 LB 282 cos60sin60 C 62 sin2 60 1 Note that AC 84 cos260sin260 62 cos2 60 e2 sin260l 64 cos2 6390 sin26390 e2 l Hence 32 4AC 4e2 1 We now consider the possible cases for e M e1 Then Bl 4AC 0 Now the curve is a parabola or a limiting form w eltl Then B2 4AC lt0 Now the curve is an ellipse or a limiting form egtl Then Bl 4AC gt0 Now the curve is a hyperbola or a limiting form Next we discuss using rotation of axes to remove the product term Recall that the angle of rotation to be used is an angle 6 such that A C cot26 We have here that A C 62 cos2 6390 l e2 sin26390 l e2 cos26390 sin2 60 e2 cos260 and B 282 cos60 sin60 e2 sin2z90 Thus e cfs26 cot260 B e s1n260 Hence we may take 60 as our angle of rotation The equations of rotation are then x u cos6 0 vsin60 y u sin6 0 vcos6 0 Then we get 62 cos2 6390 l u cos6 0 vsin6 0 2 262 cos6390 sin6390 u cos6 0 vsin6 0 u sin6 0 vcos6 0 62 sin2 6 l u sin6 0 vcos6 0 2 282d cos6390 u cos6 0 vsin6 0 Zezalsin6390 u sin6 0 vcos6 0 62612 Thus 62 cos2 6 l u2 cos2 6390 2uvcos6390 sin6 0 v2 sin2 60 262 cos6390 sin6390 u2 cos6390 sin60 uvcos2 6390 sin2 60 v2 sin6 0 cos60 62 sin2 6 l u2 sin2 60 2uvsinlt 90 cos60 v2 cos2 6390 282 alucos2 60 sin2 60 Zezdv cos6390 sin6 0 sin6 0 cos60 2612 0 Collecting coef cients we get 22 cos4 60 005260 222 cos2 60 sin2 60 22 sin4 60 sin2 60 u 2 222 cos3 60 sin60 200560 sin60 222 00560 sin60 005260 sin 260 222 sin3 60 cos60 25in60 cos60 uv 22 cos2 60 sin2 60 sin2 60 222 cos2 60 sin2 60 22 sin2 60 cos2 60 cos2 60 v2 262du 62512 0 s0 e2 cosz60 sin2602 lu2 0uv l v2 Zezdu 2ch2 0 Thus nally we get le2u2 v2 Zezdu ezdz

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