Successfully reported this slideshow.
We use your LinkedIn profile and activity data to personalize ads and to show you more relevant ads. You can change your ad preferences anytime.
FINAL EXAMINATION <br />Semester I 2007-2008 G<br />MATH-301 (Calculus III) <br />Question 1.<br />         <br />Suppose ...
Y
r = 6cosθ
3
36X
- 3</li></ul>           <br />Question 3.<br />(a)Find the unit vector that has the same direction as the vector   a = <-1...
Maths 301 key_sem_1_2007_2008
Maths 301 key_sem_1_2007_2008
Upcoming SlideShare
Loading in …5
×

Maths 301 key_sem_1_2007_2008

365 views

Published on

Published in: Education, Technology
  • Be the first to comment

  • Be the first to like this

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 />

×