Math 101 Dec 00 Solutions

Math 101 Dec 00 Solutions - Math 101 December 2000...

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Unformatted text preview: Math 101 December 2000 Solutions 1. a) Use Pythagoras to write r/Z x/2 / cos5 :cdz =/ (1 — sin2 z)2 cosz dz. 0 0 Now substitute u = sin :3; du = cosxdz. When :6 = 0, u = 0 and when a: u = 1. This gives t/Z 1 f 0065 z dz =/ (1 — u2)2(du) 0 o 1 :/ (1—2u2+u4)du o 1 = [u — 2113 +1115] 3 15 ° =[1_§+E]—[0—0+0]=fi. b) The integral represents the area under y = VS) — $5 from a: = 0 to a: = 3. But that is the region inside 1:2 + y2 = 32 with a: 2 0 and y 2 0; i.e., a quarter of a circle of radius 3. Hence /3 \/9 — 2:2 dz = $143)?) = c) Integrate by parts with u = ln(1 + 2:2), dv = dx; du = 1:32(2zdz), 12:2. Then 1 1 1 2zda: 2 _ 2 _ [o ln(1+z )dm— [(ln(1+z ))(Z)]o f0 21+32 1 2 2:: -[1n2—0]—/0 1+x2 dz: and using long division gives 1 2 _ln2-/(; ( —— 1+1?) dz 1 =ln2— [2:i:—2tan‘1 z] o =ln2— [(2~2§)—(0-0)] =1n2+§—2. . . . a: A B C d) Use partial fractions to wrlte m __ z _ 1 + (z _ U2 + x _ 2, and cross multiply to find that 2: = A(z —— 1)(z — 2) + B(:c — 2) + C(z — 1)2. Takingzz 1,1=B(—1)soB=—-1. Takingz=2,2=C(1)2soC=2. Taking a: = 0, say, 0 = A(—1)(—2) + B(—2) + C(l)2 = 2A + 2 + 2 so A = —2. Consider the indefinite integral: fde/(i—z-ri-mfnz)“ =2ln(:—2)—21n(2-— 1)+ =21n(”_2)+ 1 +0 z—l x—l The integrand does not have any singularities in the interval (3,00), so (ignoring C) we may apply the fundamental theorem of calculus: °° a: _ 23—2 1 R [a (z—2)2(z—2)d”‘n‘3’3ol2‘”(z—1)+z—lla . R—2 1 1 1 -&5‘;°l(21“(a-1)+"fi ‘(2‘“§+§)l . R—2 1 _2ln(ngrgo—R__l)+0+2ln2—§ (since In 9: is a continuous function) -21 1+2l 2—1—2ln2——1— _ n n 2—. 2. 2. a) Let I = 02* cos2 zdz. Because of the symmetry between cosx and sina: over the interval 21r, we must have also that I 2 f0” sin2 :1: dz. Hence 21f 2x 21:] (coszx+sin2:r)d:c=/ 1dz=21r 0 0 so that I = 1r. Alternatively 2! 21']. I:/ coszzdzzf —(1+cos21:)d;r o o 2 1 1 . 2* 1 1 = §[z + -2-sm2z]o = 5(27r-0) + 1(0 — 0) = 7r In any case the average value is — 1 1 f " 21r — OI ’ 5' n 1 . . 2 b) Write the limit as lim 2 (—) (i,— sin [1 + Then we can "—900 i=1 n n n recognize it as the Riemann sum for the function f(2) = zsin(1 + 1:2) on the interval [0, 1]; so 7| . .2 1 —- ' J_ ._ - 2 L..nli’ng°. 1nzsrn(l+nz).——V/‘o zsm(1+:r)d:c J: Use the subtitution u = 1+ 22; du = 2: dc, or :cdz :: %du; u :1 at a: = 0; andu=2atz= ltofindthat 2 L7” sinudu 1 2 I =[—cosu] =(——cosZ)—(—cosl) cosl—cos2) l 1 w a/ c) The characteristic equation is r2+r+2=0 _1 2_ so that r = W = —% d: = a :tifl. So the general solu— tion is 31(2) : e“ (Acos [31: + Bsin fix) : 6”” (A cos 41: + Bsin 41c) . d) The differential equation is separable so rearrange to get e'V dy = 2: dz. Then integrate to find that ——e‘” = 12:2 — C 2 y:—ln(C-%32). 7rt U e) Let u = 2:3 so that F = / e“ sin dt. Combining the fundamental 0 or theorem of calculus with the chain rule dF dF du _u . 1m 2 H; - mi;- [6 “(a—)l [3* 1- Taking a: = 1, so that u = 1 also, _d_F —dz = (e’1sin (3(1)2) = 2. 3:1 3. a) We can solve the two equations for a: to obtain a: = —2y and a: = 43(6—yz). So the intersection points are where -2y = %(6 — yz), i.e., ——5y = 12 — 2312 or 23,2 -5y—12 =0. Thisfactors as (2y+3)(y—4) = 030 y: —2/3 ory=4. It is now convenient to integrate with respect to y. The ”top”, i.e. rightmost, curve is a: = 3(6 — (as can be checked by substituting a. value of y between —2/3 and 4, and comparing the x—values). So the integral is I = [2/3 (3(6 - 3/2) - (-21.0) dy = [2,3 (45(6 - yz) + 2y)) dy. b) The method of disks gives 3 3 a V =/ 1ry2 dz: 2/ use” dz. 0 o c) It is not possible to solve for a: in terms of y, so we cannot use the method of disks. Instead use cylindrical shells; the height is h = y = ze‘3’, the radius is r = 2: and the thickness is dz. So the volume is 00 CD V = / 21rrh dz = / 21rzze'3" dx. 0 o F'(1) d) Use the formula 8 = f ds where ds = (d:¢:)2 + (dy)2. Here dz = (y -— 1)”2 dy so ds = «(y —1)(dy)2 + (dy)2 = «(y — 1) + 1 dy. Therefore the arc length is 5 s = / ds = / fidy. 1 Alternatively, solve for y: y : 1 + (g3)2/3, on the interval 0 S I S ;3§. so the arc length is 16/3 16/3 2 5:]0 \/1+(y’)2dz=/0 1+ [§(-:—z)'1/3] dz. e) The points of intersection of the curves are where 2:2 : x + 2, that is, 32—2—2: 0. So(z—2)(z+1)= 050:: = 2anda: = —1. To use the method of disks we would have to solve for a: and integrate over two separate intervals, which is inconvenient. Instead use cylindrical shells. Here the variable of integration is r, so the thickness is dx; the height is h = ycop — ybomm = r + 2 — 2:2 (check any point between a: = —1 and :c = 2 to determine which is the top curve); and the radius is the distance from a: to the axis of revolution 2: = 3, that is, r = 3 — 2:, since a: < 3 in the given region. So 2 2 V2] 27rrhdz2/ 21r(3—z)($+2—22) d2). -1 -1 4. Let y denote the depth of the fluid in feet (measured from the bottom of the tank). The area of the hole at the bottom of the sphere is a = «(11—2)2 = 1r/ 144 feetz. If r denotes the radius of a cross-section at height y then, because the tank is a sphere with centre at y = 6, we have (y — 6)2 + r2 = 62, or r2 = 36 — (y — 6)2. A cross-section is a circle of radius r, so its area is A = 1rr2 = 1r(36-—(y—6)2). Now we are ready to plug everything into the well-known differential equation concerning this situation: \/6_4 -174 y (my—flag: “’7 dy — — 2 -— (36 36+ 12y y )dt — _T8— (1211”2 -y3’2)dy= i 18 so integrate to obtain 2 t /2 _ _ 5/2 = __ . 81/3 5:1 18 +0 (1) Now use the initial conditions to find C. At t = 0, the tank is full, so y :: diameter = 12. This gives 8(12)3/2 — §(12)5/2 = o + c which is c = 123/203 — g - 12) = Egfi. We wish to find t for which y = 0. So substitute y = 0 into (1): t 6912\/§ , so t = 180 = 5 = 2394.48. 5. a) Let I = f sin” 2dr. The form of the equation to be proved resembles integration by parts, so we will try this. The presence of the term sin"'1 2: cos 2: suggests that we use u = sin"‘1x and dv : sins-dz; consequently we have du = (n — 1) sin"’2 a:(coszdx) and v = —cos 1:. This leads to I = —sin"’1 zcosa: — — 1) sin"“2zcosx][— cosz] dz = —sin"’1 :rcosa: + (n — 1)‘/‘sin""2 .1:cos2 3 dz. Now the given equation does not have a cos2 a: in it; so this suggests that we eliminate this term using Pythagoras, cos2 a: = 1 — sin2 9:. Then I = —sin"'1 zcosa: + (n —1)/sin"‘2 2(1 — sin2 1:) d2: = — sin"‘1 zcosz + (n —1)‘/(sin"‘2 a: — sin" 2:) dz = —sin"'1 zcosx + (n - 1) [(sin"'2zd:c — (n — 1)I Thus n1 2 —sin"‘1 zcosz + (n — 1) [(sin”"2xdz 1 .n_ —srn zcosa: n—l . _ I=—-———+ (sm" 2:cda: n n which is the required formula. b) Using the result of a), 7/2 x/Z “7/2 / sinszdxz—l[si 7xcos:c] +Z/ sinezda: o 8 o 8 0 l2 «[2 =[0—0]+§ {—é-[sinszcosz]; +2] sin4zd2] o _7 5 1.3 r/23'/2.2 _§[[0—0]+6[—Z[sm :|:cos:¢:](J +4—/0 sm zdz 7 5 3 1 . #2 1 m . 0 _ g.E[[0—0]+Z [—§[smzcosz]o +5]0 sm ads! 7 6. We will use the trapezoidal error estimate M(1—0)3 _ M 127.2 ‘ 12112 lTn-IIS where |f”(:c)| g M on the interval [0,1]. To compute f’(:c), let u = :62, so that u f = / sin x/t dt. By the chain rule and the fundamental theorem of calculus o f’(a:) = g; 2 £975“ = (sin J17)(2:c) = 2::sinz. Then f”(1:)= 2sinz + 2::cosx. For 0 g a: _<_ 1, Isinzl S sinl and [cost] S 1 (since sin: and cost are increasing and decreasing functions respectively over this interval). Using the triangle inequality If"(z)| = |2sin$+22cosz| g 2 |sinz| +2 Izl [cos 1:] g 2sin1+2(1)(1) = 23in1+2 = M To acheive five decimal places of accuracy, a reasonable requirement on the error ITfl — I I is that |Tn — I | 3 0.000001. (Note that, depending on the actual digits of I, this may not be good enough. For instance if I 2 0.123459 then I +0.000001 has a 6 in the 5th decimal place, not a 5. This situation is infrequent and need not concern us.) So solve for n in M 12112 3 0.000001 2 sin 1 + 2 < n2 12(0.000001) - ’2sin1 + 2 < n 0.000012 — 553.996 3 n So we should take n = 554. 7. a) Since f(x) is to be a probability density, solve for k in fan f(:c) dz = 1. Thus 20 1 = / k(8000 — 2:3) dz 0 20 = I: [80002: — £24] = k [(160, 000 — £160, 000) — (0 — 0)] = 120, 0001: 0 _ 1 — 120, 000. b) The probability is given by sok l 5 _ 3 120,000/0 (8000 a: )da: 1 = 120i000 [8000” _ 234i: 1 1 _ 120,000 [(40,000 — Z625) — (0 — 0)] 85 . _ fl _ 0.332031. P(XS5)=/05f(z)dz= c) The mean is 20 1 20 4 fl=A zf(z)dz=120,000 0 (SOOOz—x )dz 1 = 120, 000 1 = 120,000 = 8. [40003:2 -— $35130 [(1, 600, 000 — 640, 000) — (0 — 0)] ...
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Math 101 Dec 00 Solutions - Math 101 December 2000...

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