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# Maths 301 key_sem_1_2007_2008

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### Transcript of "Maths 301 key_sem_1_2007_2008"

1. 1. FINAL EXAMINATION <br />Semester I 2007-2008 G<br />MATH-301 (Calculus III) <br />Question 1.<br /> <br />Suppose that each dollar introduced into the economy recirculates as follows: 90% of the <br /> original dollar is spent , then 90% of that \$0.90 is spent , and so on. Find the economic impact (total amount spent) if \$2,000,000 is introduced in to the economy.<br /> Economic Impact = 2,000,000[0.9+ 0.92+ 0.93+ …………]<br /> <br /> =2,000,000 0.91-0.9=2,000,0000.90.1=\$18,000,000 <br /> <br />Use the root test to determine whether the given series converges or diverges or if the test <br /> is inconclusive:n=1∝n55n <br /> Apply root test.----> limn->∞an1/n , Here an= n55n <br />limn->∞ n55n1/n=limn->∞n5/n5n/n = limn->∞n5/nlimn->∞5 = 15 < 1<br /> <br />Therefore the series converges. <br /> <br /> <br />Approximate 01sinx3 dx to four decimal places using:<br />sinx=x- x33!+ x55!- x77!+ ………..<br /> sinx3= x3 - x333!+ x355!- x377!+…<br /> 01sinx3dx= 01x3 - x96+ x15120- x215040+ ………… dx<br /> = x44- x106(10)+ x16120(16) - x225040(22)+ …………01<br /> = 14- 160+ 11920 - 1110,880+ ……………<br /> Since 1110,880=0.000009<0.00005 <br /> 01sinx3dx= 14- 160+ 11920 =0.2339 (Four decimal approximation)<br /> <br />Question 2<br />(a)A basketball player shoots a ball at an angle of 60° from a point 15feet horizontally away<br /> From the center of the basket. The basket is 10ft above the floor and the player releases the <br />X ball from a height of 8 ft. At what speed should the player shoot the ball?. Take the<br /> acceleration due to the gravity as 32 ft/s2<br />Vsin 60 <br />x= x0+ vcos60° t-------->1y= y0+ vsin60° t- g2 t2----->2x0 , y0 =0, 8 and x,y=(15,10) 1 and 2 become15=0+ v2t-->t = 30v and t2 = 900v2 <br />V <br />10ft <br />60°<br />8ftV cos 60 <br />Y <br />(0,0)<br /> <br /> <br /> 10=8+v 32 t -16t2 <br /> 16t2- 32v t+2=0------>3 <br /> Substitution for t and t2 in equation 3 yields<br />16 900v2- 32 v 30v+ 2=0 <br /> 14400v2 = 153 – 2 <br /> v2= 14400153 -2 =600.48 <br /> v= + 600.48 =24. 505 ≅ 24.5 ft/s <br />(b)Find the area of the surface generated by revolving the curve about the x-axis.<br /> x= t2, y=2t, 0 ≤t ≤1<br /> <br /> S= 012π y dxdt2+ dydt2 dt <br /> dxdt=2t and dydt=2 <br /> S= 012π(2t) 2t2+ 22 dt =4π 012t t2+ 1 dt <br /> =4π t2+ 1323201<br />= 8π3 (2)32- (1)32 <br /> ≅ 4.9 π ≅ 15.3 Square Units<br /><ul><li> Sketch the graph of the polar equation : r =6cosθ
2. 2. Y
3. 3. r = 6cosθ
4. 4. 3
5. 5. 36X
6. 6. - 3</li></ul> <br />Question 3.<br />(a)Find the unit vector that has the same direction as the vector a = <-1, 3, -2><br /> The unit vector of a = a a = <-1 , 3 , -2>-12+32+ -22<br /> = <-1 , 3 , -2>1 + 9 + 4= <-1 , 3 , -2>14 <br />P(b)Find the distance from point P to the line through Q and R :<br /> P(-1, 3, 0), Q(0, 3, -3), R(5, 0, 1)<br />d <br />Distanced = QR × QPQR Q <br />R <br /> QP= <-1, 3, 0>- <0, 3,-3>= <-1, 0 , 3><br />QR= <5, 0, 1> - <0, 3, -3> = <5, -3, 4><br />QR x QP= ijk5-34-103=i -3403-j 54-13+k 5-3-10<br /> = i -9-0-j 15+4+ k0-3= -9i-19j-3k<br />Distance = QR × QPQR = -92+-192+-3252+ -32+42 = 45150 ≅3.003 ≅3.0<br />(c)Sketch the graph of y= x2+ z2 in an xyz – coordinate system. <br />Z <br />Y <br />X <br />Question 4. <br /> <br />(a)Test if fx,y= x2+ xy- y2 is harmonic.<br /> <br /> fx=2x+y ----> fxx=2<br /> fy =x-2y----> fyy=-2<br /> fxx+ fyy=2-2=0 --->f is harmonic<br />r (b) The radius r and altitude h of a right circular cylinder are decreasing at rates of 0.03 cm/min and 0.04 cm/min, respectively. At what rate is the curved surface area changing at the time when r =3cm and h = 4cm<br />S = Curved surface area = 2πrh <br />h Rate of change= dsdt= ∂s∂r drdt+ ∂s∂h dhdt <br /> =2π h-0.03+ (2πr)(-0.04)<br /> <br /> =2π[4-0.03+(3)(-0.04) ]<br /> = -0.48π= -1.5 cm2/min<br />(c) Find the directional derivative of fx,y= tan-1yx at the point (1,-1) in the direction of the vector a = 3i -4j<br />.<br />The directional derivative of f = ∇f . u= Du f(x, y)<br /> ∇f = fxi+ fyj = - yx21+yx2 i+ 1x1+ yx2 j <br /> <br /> = -yx2+y2 i+ xx2+y2j<br /> u= a a= 3i- 4j32 + -42= 35 i- 45 j<br /> <br />Du fx, y= ∇f . u= -3y5( x2+y2) i- 4x5(x2+y2)j = -3y-4x5(x2+y2) <br /> Du fx, y1, -1)= -3-1-4(1)512+ -12 = -110= -0.1<br />Question 5<br /> <br />(a)Sketch the region bounded by the graphs of the following equations and find its area by using only one double integral: y= - x2 , y=2, x=0, x=2<br /> <br />y<br />R<br />y =2<br />x=2x =0 <br />x<br />Y = -x2<br />Area= RdA = 02-x22dy dx =02y-x22 dx<br /> <br />=022+ x2dx= 2x+ x3302<br /> =4+ 83-0+0= 203 =6.66 Square Units<br /> (b)Find area of the region R that lies outside the circle r =3 and inside the circle r = 6cosθ<br />Y <br />θ=π3 <br />r=3r=6cosθ<br />θ=0<br />X <br />36<br />R <br />θ=- π3<br />Intersection points: 3=6cosθ ---> cosθ= 12 , θ=±π3<br />Area= RdA= Rr dr dθ<br />Using symmetry , Area=20π336cosθr dr dθ <br />Area= 20π3 r2236cosθdθ =0π3(36cosθ2-9) dθ <br /> = 0π3 3621+cos2θdθ- 0π39 dθ Rule: cos2θ= 1+cos2θ2 <br /> = 18 θ+ sin2θ20π3-9θ0π3 <br /> =18 π3+322 -0-3π=3π+ 923 ≅17.2 Square Units <br />(c)Evaluate the following triple integral: 010-2023x2z+ 2xy2dz dx dy<br />: 010-2023x2z+ 2xy2dz dx dy= 010-232 x2z2+ 2xy2z02 dx dy<br /> <br /> =010-26x2+4xy2 dx dy <br /> =016x33+ 42 x2 y20-2 dy <br /> = 01-16+8y2 dy <br />= -16y+ 83 y301<br /> = -16+ 83 = - 403 ≅ -13.3<br />