2. Y

3. r = 6cosθ

4. 3

5. 36X

6. - 3 <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 />