# MATH0010-2018-19-mid-solns.pdf - Solutions to MATH0010...

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Solutions to MATH0010 mid-sessional, 2018–2019 1. (a) Rearranging the equation, we find a × y = b × a = - a × b , so a × ( y + b ) = 0 . If a = 0 then any y is a solution. If a 6 = 0 , then y + b must be parallel (or anti-parallel) to a . Therefore y = - b + λ a , for some scalar λ . (b) i. δ ij δ ji a k b k = δ ii a k b k = 3 a · b . ii. δ ij ijk a k b l e l = iik a k b l e l = 0 (repeated index on ). (c) Let the closest points be P (on L 1 ) and Q (on L 2 ). Parametirc forms of the lines are given by r 1 ( s ) = --→ OP 1 + s v 1 and r 2 ( s ) = --→ OP 2 + s v 2 , where P 1 = ( - 1 , 6 , 7), P 2 = (4 , - 3 , - 1), v 1 = 2 i + j + 2 k and v 2 = i + j . The vector -→ PQ is in the same (or opposite) direction to n = v 1 × v 2 = - 2 i + 2 j + k . In fact -→ PQ = n · --→ P 1 P 2 n · n n . Noting that --→ P 1 P 2 = 5 i - 9 j - 8 k , we have -→ PQ = - 10 - 18 - 8 4 + 4 + 1 n = - 4 n = 4(2 i - 2 j - k ) . Therefore the distance between the closest points ( P and Q ) is 4 | 2 i - 2 j - k | = 12. Now let s 0 and t 0 be the parameters such that r 1 ( s 0 ) = -→ OP and r 2 ( t 0 ) = -→ OQ . Then -→ PQ = -→ OQ - -→ OP = r 2 ( t 0 ) - r 1 ( s 0 ) . This gives the equations 2 s 0 - t 0 = - 3 , s 0 - t 0 = - 1 and 2 s 0 = - 4 . Hence s 0 = - 2, t 0 = - 1. Now -→ OP = r 1 ( - 2) and -→ OQ = r 2 ( - 1), so the closest points are P = ( - 5 , 4 , 3) and Q = (3 , - 4 , - 1) . 2. (a) Let f ( x, y, z ) = e yz +1 + sin( y + z ) - x . Then the surface is given by the level set f ( x, y, z ) = 0. Now f ( x, y, z ) = ∂f ∂x i + ∂f ∂y j + ∂f ∂z k = - i + ( z e yz +1 + cos( y + z ) ) j + ( y e yz +1 + cos( y + z ) ) k . Hence a normal vector to the surface at (1 , - 1 , 1) is given by n = f (1 , - 1 , 1) = - i + 2 j . Let P = ( x, y, z ) be an arbitrary point on the tangent plane. We know that P 0 = (1 , - 1 , 1) is a point on the plane. So the equation of the plane is given by n · --→ P 0 P = 0, which gives x - 2 y = 3. 1
(b) Let the level curve be f ( x, y ) = c . We also write the ellipse as the level curve g ( x, y ) = 1, where g ( x, y ) = x 2 + y 2 4 . Suppose that ( x 0 , y 0 ) lies on both curves (i.e., it is an intersection point). The direction of the normal to the level curve f ( x 0 , y 0 ) = c is given by f ( x 0 , y 0 ) = 2( x 0 i - y 0 j ) and the direction of the normal to the ellipse is given by g ( x 0 , y 0 ) = (1 / 2)(4 x 0 i + y 0 j ). We require these vectors to be orthogonal, i.e. f ( x 0 , y 0 ) · ∇ g ( x 0 , y 0 ) = 0. This gives 4 x 2 0 = y 2 0 , i.e. y 0 = ± 2 x 0 . Substituting this into the equation for the ellipse gives 2 x 2 0 = 1. Therefore there are four points where the ellipse intersects some level curve of f at right angles: 1 2 , 2 , 1 2 , - 2 , - 1 2 , 2 , - 1 2 , - 2 . All four of these points lie on the same level curve of f , namely f ( x, y ) = 1 2 2 - 2 2 = - 3 2 . [SKETCH!] 3. (a) Let z = z + i y . Then 1 z + 2 + 3i = 1 ( x + 2) + ( y + 3)i = 1 ( x + 2) + ( y + 3)i ( x + 2) - ( y + 3)i ( x + 2) - ( y + 3)i = ( x + 2) - ( y + 3)i ( x + 2) 2 + ( y + 3) 2 Therefore exp 1 z + 2 + 3i = exp x + 2 ( x + 2) 2 + ( y + 3) 2 - i y + 3 ( x + 2) 2 + ( y + 3) 2 = exp x + 2 ( x + 2) 2 + ( y + 3) 2 exp - i y + 3 ( x + 2) 2 + ( y + 3) 2 .