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Maths 301 key_sem_1_2009_2010
![FINAL EXAMINATION <br />Semester I , 2009/2010<br />MATH-301 (Calculus III) <br />Question 1. <br />(a)Evaluate the iterated integral.<br />011002x+y+z dy dz dx = 0110x+zy+ y2202 dz dx<br />= 01102x+z+2 dz dx = 01102x+2+2z dz dx <br />= 012x+2z+ 2z2210 dx = 01-2x+2+ 12 dx<br />= 01-2x-3 dx = -2x22- 3x01 = -1-3= -4<br />(b)Sketch the region bounded by the graphs of the following equations and find its area by using double integrals.<br /> ,y = 0, x = 1,x = 2.<br />y = 0x =2x=1y=1x2Area= 1201x2dy dx <br />= 12y01x2 dx = 12x-2 dx<br /> = -1x11= --12- 1 = 12<br /> (c)A solid is bounded by the cone and the plane z = 3. The density at <br />P(x, y, z) is directly proportional to the square of the distance from the origin to P. Find the mass of the solid using integration in cylindrical coordinates.<br />z<br />density= δ=kx2+y2+z2 =kr2+z2mass=m= QδdV = 02π03r3kr2+z2r dz dr dθ<br />Z = 3<br />Z = r<br />y<br />x<br />=k02π03r3r3+rz2dzdrdθ =k02π03r3z+ rz33r3 drdθ<br />=k02π033r3+9r- r4+ r43 drdθ<br />=k02π033r3+9r-43r4 drdθ<br />=k02π3r44+ 9r22-43r5503 dθ <br />=k7292002πdθ = 72920 k θ02π = 72910 πk<br /> <br />(d)Use integration in spherical coordinates to find the centroid of a hemispherical solid Q of radius a = 8.<br />z<br />Mxy= Qz dv , z= ρ cosϕ =02π0π208z ρ2sinϕ dρ dϕ dθρ=a=8 Centroid= 0, 0,z , z= Mxyv <br />y<br />x<br />= 02π0π208 ρ3sinϕcosϕ dρ dϕ dθ<br />=02π0π2ρ4408sinϕcosϕ dϕ dθ = 1024 02π0π2sinϕcosϕ dϕ dθ <br />= 10242 02πsin2 ϕ0π2 dθ= 10242 02πdθ = 10242 θ02π = 1024π<br />z= MxyV = 1024π23 π83 = 1024π1024π3 = 3<br />Question 2.<br />(a)If , prove that f(x, y) is a harmonic function.<br />fx, y =lnx2+ y2<br />fx= 2xx2+ y2 , fxx= 2x2+ y2-2x 2xx2+ y22 = -2x2+2y2x2+ y22 A<br />fy= 2yx2+ y2 , fyy= 2x2+ y2-2y 2yx2+ y22 = 2x2-2y2x2+ y22 (B) <br />From (A) and (B) fxx+ fyy=0 -> f is harmonic.<br />(b)Find an equation for the normal line to the graph of the equation:<br /> - 4 z2 = 2 x2 - 3 y2 - 10 at the point (-1, -2, 3).<br />Fx,y, z=0 -> 2x2-3y2+4z2-10=0 , -1, -2, 3 <br /> x0 y0 z0<br /> ∇F= Fx i+ Fy j+ Fz k = 4x i- 6y j+ 8z k<br />∇F-1, -2, 3= -4 i+ 12 j+ 24 k -> a= -4, 12, 24 <br /> a1 a2 a3<br />Equation of the normal line <br />x=x0+ a1t , y= y0+ a2t , z= z0+ a3t<br /> x= -1-4t , y= -2+12t , z=3+24t<br />(c)The surface of a lake is represented by a region in the xy-plane such that the depth (in feet) under the point (x, y) is . At the point (10, 9), in what direction does the depth remain the same?.<br />fx,y=400-3x2-2y2 , ∇f= fxi+ fyj<br /> ∇f= -6xi- 4yj , ∇f10 , 9= -60i-36j<br />The depth remains the same in a direction orthogonal to ∇f, i.e. in the direction of <br />36i-60j or c36, -60 where c is any real number ≠0.<br />(d)Find the extrema and saddle points, if any, of <br />fx, y= -2x3+ 6xy-2y3<br /> fx=0 -> -6x2+ 6y=0 , fy= 0 ->6x-6y2=0<br /> y= x2 A x= y2 B<br />Substitute from B into A ->y= y22 -> y4-y=0 , yy3-1=0<br /> yy-1y2+y+1=0 -> y=0 or y=1<br />For y=0 ->x=y2=0 -> 0, 0 For y=1 ->x=y2=1 -> 1, 1 are the critical points.<br />Dx, y= fxx∙ fyy- fxy2= -12x-12y- 62<br /> <br />Dx, y= 144xy-36<br /> At 0, 0 ->D=0-36 <0 -> D is negative<br /> So P0, 0, f0,0 is a saddle point.<br />At 1, 1 ->D=14411-36 >0 -> 1, 1 is an extrema.<br />fxx1, 1= -12x1, 1 = -12 <0 ->f is maximum at 1, 1 <br /> fmaximum= -213+ 611-213=2<br />Question 3.<br />,[object Object], Total Distance=x+2x2+ 2x4+ 2x8+ …..<br /> <br /> =x+x+x2+x4+…<br /> Total Distance<br /> =x+x 1+ 12+14+18+ …..x/2x<br />x/4<br />x/8<br />x/16Geometric series with a=1 and r=12 <1<br /> S= a1- r<br />Total Distance=x+xa1-r; Total Distance=x+x11-12=x+2x=3x <br /> 15=3x ->x=153=5m is the initial height.<br />(b)Determine if each of the following series converges or diverges or if the test is inconclusive.<br />(i)<br />an= n!10n -> an+1= n+1!10n+1= n+1n!10n 10<br />L= limn->∞an+1an= limn->∞n+1n!10n 10 ∙10nn! = limn->∞ n+110= ∞<br />So the series diverges.<br />(ii)<br /> an= n2n , L= limn->∞ an1n <br /> L=limn->∞ n2n1n= limn->∞ n1n2nn = limn->∞ n1n2= 12 <1 since limn->∞ n1n=1, <br />So the series converges.<br /> Or limn->∞an+1an= limn->∞n+12n+1∙2nn= limn->∞n+12n 2∙2nn= limn->∞12∙n+1n<br /> = 12limn->∞1+1n= 12 <1, so the series converges.<br />(c)Use <br />to approximate to two decimal places. <br />sinx=x- x33!+ x55!- x77!+ ….. A<br />Substituting sin x from A in<br />01sinxx dx = 01x- x33!+ x55!- x77!+ ….. x12dx<br />=01x12- x523!+ x925!- x1327!+ ….. dx<br />= x3232- x72726+ x112112120- x1521527!+ ……01<br /> I II III<br />= 23- 242+ 0.0015….<br />Since 0.0015 < 0.005, we neglect term III and the remaining terms of the series <br />except I and II.<br /> So 01sinxx dx= 23- 242= 1321 ≅0.619 ≅0.62 to two decimal places.<br />Question 4. <br />y(a)If a rock is thrown from a point 6 ft above the ground with an angle of elevation of 30º and an initial speed of 20 ft/sec. If there is a 8-ft high fence 10 ft in front of you, will the rock sail over the fence?.<br />θ=30°h0=20cos30° =20 32 =103 <br />8’x0=0 v0 =20sin30°=2012=10<br />y0=6'<br />x= x0+ h0t A<br />xt=10'xy= y0+ v0t- 12 gt2 B <br /> g=32 ft/s2<br /> <br />Substituting known values in A and B <br /> C xt =0+103 t and D yt=6+10t-16t2<br />At fence xt=10' so C becomes 10 =103 t->t=13 secs.<br />From D y13fence= 6+1013- 16132=6.44' <8'<br />So the rock will not sail over the fence.<br />(b) Find the points on the curve C at which the tangent line is horizontal.<br />,;t in R.<br />dydx= dydtdxdt= 133t2-1122t-1= t2-1t-1<br /> <br />The tangent line is horizontal at dydx=0 or dydt=0<br /> <br />i.e, at t2- 1=0 -> t2=1 ->t= ±1 <br /> At t = 1 -> x, y= -12, -23 At t = -1 -> x, y= 32, 23 Points at which the tangentline is horizontal. <br />(c)Use integration in polar coordinates to compute the area of the region that is inside the graph of the polar equation r = 3 + 3 cos θ.<br />θ=πθ=0r = 3 + 3 cos θ Area= 12 αβr2 dθ <br /> By symmetry,<br /> A=2∙12 0π3+3cosθ2 dθ= 0π9+18cosθ+9cos2θ dθ =0π 9+18cosθ+921+cos2θdθ<br /> =9θ+18sinθ+ 92θ+94sin2θ0π<br /> = 9π+0+ 92π+0- 0=9π+ 9π2= 27π2 ≅42.41<br />Question 5.<br />(a)A constant force of magnitude 5 N has the same direction as the positive x-axis. If the distance is measured in meters, find the work done if the point of application moves from the point (1, -1, 1) to the point ( 3, 1, -2).<br /> Force=5 1, 0, 0 = 5, 0, 0 <br /> d= 3, 1, -2 – 1, -1, 1 = 2, 2, -3 <br /> Work done= F∙d= 5, 0, 0 ∙ 2, 2, -3 =10 N∙m <br /> (b)Find an equation of the plane through P, Q and R. <br />P( 2, 0, 0), Q ( 0, 2, 0) , R ( 0, 0, 2).<br />PQ= OQ-OP= 0, 2, 0 – 2, 0, 0 = -2, 2, 0 <br />PR= OR-OP= 0, 0, 2 – 2, 0, 0 = -2, 0, 2 <br />PQ × QR= ijk-220-202= i 2002- j -20-22+ k -22-20<br />PQ × PR= i 4-0- j -4 + k 4 =4i+ 4j +4k <br /> Equation of the plane through P, Q, and R<br />a1x-x0+ a2y-y0+ a3z-z0 =0 A<br /> Taking x0, y0, z0= 2, 0,0 , a1, a2, a3 = 4, 4, 4 <br /> So Equation A becomes 4x-2+ 4y-0+ 4z-0=0<br /> or 4x+4y+4z=8<br />(c)Sketch the graph of the following equations in an xyz-coordinate system:<br />(i) <br />z<br /> 1 <br />y<br />x<br />-zzxy(ii) <br />](https://image.slidesharecdn.com/maths301keysem120092010-100517095730-phpapp02/85/Maths-301-key_sem_1_2009_2010-1-320.jpg)
