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The purpose of this archive is to organise and centralise the distribution of the exam questions and their solutions. This archive is a non-profit service and it must remain so. Do not let anyone sell and do not buy this archive, or any portion of it. Reproduction or distribution of this archive, or any portion of it, without non-profit purpose may result in severe civil and criminal penalties. 1.) a) Sketch the region R that lies inside the curve r = 2 and outside the curve r = 2 − 2 cos θ. b) Find the area of the region R. Solution: a) The curve r = 2 is a circle of radius 2 with its center located at the origin. To draw the curve r = 2 − 2 cos θ we determine some points that this curve passes through such as: (θ = 0, r = 0), (θ = π/2, r = 2), (θ = π, r = 4), (θ = 3π/2, r = 2), (θ = 2π, r = 0). These two curve intersect when 2 = 2 − cos θ, i.e, at θ = π/2 and θ = 3π/2. So we obtain, theta=pi/2 2                              R −4 −2 o 2 theta=0 −2 b) The area of the region R can be obtained by the integral Z π/2 A = −π/2 Z 1 2 [2 − (2 − 2 cos θ)2 ]dθ = 2 2 π/2 = 2 [2 cos θ − ( −π/2 Z π/2 [2 cos θ − cos2 θ]dθ −π/2 1 + cos 2θ )]dθ = 8 − π 2 Note that since our figure is symmetric with respect to x-axis, this area can also be R π/2 calculated from A=2 0 12 [22 − (2 − 2 cos θ)2 ]dθ. 2.) Find an equation of the plane that passes through the points P1 (0, −3, 2) and P2 (1, 2, 3) and parallel to the line of intersection of the planes 2x + y − z = 1 and x − 2y + z = 7. Solution: ~ 1 =< 2, 1, −1 > is normal to the first plane and N ~ 2 =< 1, −2, 1 > is normal The vector N to the second plane. Therefore the line of intersection of these two planes is parallel to the vector ~1 × N ~ 2 = -i -3j -5 k. V~ = N The vector P1~P2 =< 1, 5, 1 > lies on the required plane. Now, also the vector V~ can be carried onto this plane since this plane is parallel to the line of intersection. Thus, ~ 3 = P1~P2 × V~ = -22 i + 4 j + 2 k N is a normal vector for the required plane. To write down the equation of this plane both of the points P1 and P2 can be used. Using P1 we get −22x + 4y + 2z + 8 = 0 3.) (a) Find an equation of the plane parallel to the plane 3x − y + 2z + 3 = 0 if the point (2,2,-1) is equidistant from both planes. Solution: Since the required plane is parallel to the plane 3x − y + 2z + 3 = 0 , < 3, −1, 2 > is a normal vector for both of them. Therefore the required plane has an equation of the form 3x − y + 2z + D = 0 where D is a real number. Now since the point (2,2,-1) is equidistant from both planes we have |3.2 − 1.2 + 2. − 1 + 3| |3.2 − 1.2 + 2. − 1 + D| p p = . 2 2 2 3 + (−1) + 2 32 + (−1)2 + 22 This equation implies |D+2| = 5 which has two solutions D = 3 and D = −7. The D = 3 solution corresponds to the first plane, therefore the required plane has the equation: 3x − y + 2z − 7 = 0 (b) Consider the straight line through the point (3,2,3) and perpendicular to the plane given by 2x − y + 3z + 1 = 0. Find the coordinates of the point of the intersection of that line and that plane. Solution: The plane has a normal vector < 2, −1, 3 >. Since the required line is perpendicular to the plane, this vector is parallel to the line. Hence, the parametric equation of the line is: x=3+ 2t , y=2-t , z=3+3t , −∞ < t < ∞. To find the value of t at which the line and the plane intersect, we insert the above result into the plane equation: 2(3+2t) - (2-t) + 3(3+3t) +1 =0, which gives t = −1. The intersection point is obtained by putting t = −1 in the parametric equation of the line which gives its coordinates as (1,3,0). (c) Can a vector (whose initial point is at the origin) have direction angles θ1 = π/4, θ2 = 3π/4, θ3 = π/3 where θ1 , θ2 , θ3 are the angles between the vector and x,y,z coordinates respectively? Solution: For a vector V~ whose direction angles are θ1 , θ2 , θ3 we have V~ .i= cos θ1 |V~ | , V~ .j= cos θ2 |V~ | , V~ .k= cos θ3 |V~ | , which leads to, V~ = cos θ1 i + cos θ2 j + cos θ3 k |V~ | Since this is a unit vector, direction cosines should satisfy cos2 θ1 + cos2 θ2 + cos2 θ3 =1 In our problem this is not satisfied since, cos2 (π/4) = cos2 (π/4) = 1/2 and cos2 (π/3) = 1/4. Therefore, such a vector does not exist. (d) Find equations in rectangular and cylindrical coordinates for the surface ρ = 2 sin φ given in spherical coordinates. Solution: We first multiply the equation of this surface by ρ which gives ρ2 = 2ρ sinp φ. To convert 2 2 2 2 this into rectangular coordinates we use ρ = x + y + z , ρ sin φ = x2 + y 2 and obtain p ...
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