Question 1 Find the points on the surface 𝑦 2 = 9 + 𝑥 𝑧 y 2 =9+xz that are closest to the origin. Answer 1 Let ( 𝑥 , 𝑦 , 𝑧 ) (x,y,z) be a point on the surface 𝑦 2 = 9 + 𝑥 𝑧 y 2 =9+xz. The distance of this point from the origin is given by: 𝑑 = 𝑥 2 + 𝑦 2 + 𝑧 2 . d= x 2 +y 2 +z 2 . To minimize the distance, it is simpler to minimize the function 𝑑 2 d 2 : 𝑑 2 = 𝑥 2 + 𝑦 2 + 𝑧 2 . d 2 =x 2 +y 2 +z 2 . Using the constraint 𝑦 2 = 9 + 𝑥 𝑧 y 2 =9+xz, we substitute 𝑦 2 y 2 into 𝑑 2 d 2 : 𝑑 2 = 𝑥 2 + ( 9 + 𝑥 𝑧 ) + 𝑧 2 . d 2 =x 2 +(9+xz)+z 2 . 𝑑 2 = 𝑥 2 + 9 + 𝑥 𝑧 + 𝑧 2 . d 2 =x 2 +9+xz+z 2 . To minimize, compute partial derivatives: 𝑓 𝑥 = 2 𝑥 + 𝑧 , 𝑓 𝑧 = 2 𝑧 + 𝑥 . f x =2x+z,f z =2z+x. Setting 𝑓 𝑥 = 0 f x =0 and 𝑓 𝑧 = 0 f z =0: 2 𝑥 + 𝑧 = 0 (1) . 2x+z=0(1). 2 𝑧 + 𝑥 = 0 (2) . 2z+x=0(2). From (1): 𝑧 = − 2 𝑥 z=−2x. Substitute 𝑧 = − 2 𝑥 z=−2x into (2): 2 ( − 2 𝑥 ) + 𝑥 = 0. 2(−2x)+x=0. − 4 𝑥 + 𝑥 = 0 ⟹ 𝑥 = 0. −4x+x=0⟹x=0. From 𝑧 = − 2 𝑥 z=−2x: 𝑧 = 0 z=0. Substitute 𝑥 = 0 x=0, 𝑧 = 0 z=0 into the constraint 𝑦 2 = 9 + 𝑥 𝑧 y 2 =9+xz: 𝑦 2 = 9 ⟹ 𝑦 = ± 3. y 2 =9⟹y=±3. Hence, the points are: ( 0 , 3 , 0 ) , ( 0 , − 3 , 0 ) . (0,3,0),(0,−3,0). Question 2 Determine the dimension of the subspace spanned by the given vectors [ 1 0 2 ] , [ 3 1 − 1 ] , [ 9 4 − 3 ] . 1 0 2 , 3 1 −1 , 9 4 −3 . Answer 1 Form a matrix with the given vectors as columns: 𝐴 = [ 1 3 9 0 1 4 2 − 1 − 3 ] . A= 1 0 2 3 1 −1 9 4 −3 . Perform row reduction: [ 1 3 9 0 1 4 2 − 1 − 3 ] ∼ [ 1 3 9 0 1 4 0 − 7 − 21 ] ∼ [ 1 3 9 0 1 4 0 0 0 ] . 1 0 2 3 1 −1 9 4 −3 ∼ 1 0 0 3 1 −7 9 4 −21 ∼ 1 0 0 3 1 0 9 4 0 . There are 2 pivot columns. Thus, the dimension of the subspace is: Dimension = 2. Dimension=2. Question 3 Is the velocity field 𝑉 = 4 𝑡 𝑖 − 2 𝑡 2 𝑥 𝑦 𝑗 + 4 𝑥 𝑧 𝑘 V=4ti−2t 2 xyj+4xzk steady or unsteady? Is it two- or three-dimensional? At ( − 1 , 1 , 0 ) (−1,1,0), compute the acceleration vector. Answer 1 Steady or Unsteady: Since 𝑉 V explicitly depends on 𝑡 t, the velocity field is unsteady. Two- or Three-Dimensional: The field involves all three coordinates 𝑥 , 𝑦 , 𝑧 x,y,z, so it is three-dimensional. Acceleration Vector: 𝑎 = ∂ 𝑉 ∂ 𝑡 + ( 𝑉 ⋅ ∇ ) 𝑉 . a= ∂t ∂V +(V⋅∇)V. Compute components of 𝑉 V: 𝑢 = 4 𝑡 , 𝑣 = − 2 𝑡 2 𝑥 𝑦 , 𝑤 = 4 𝑥 𝑧 . u=4t,v=−2t 2 xy,w=4xz. Partial derivatives with respect to 𝑡 t: ∂ 𝑢 ∂ 𝑡 = 4 , ∂ 𝑣 ∂ 𝑡 = − 4 𝑡 𝑥 𝑦 , ∂ 𝑤 ∂ 𝑡 = 0. ∂t ∂u =4, ∂t ∂v =−4txy, ∂t ∂w =0. At ( − 1 , 1 , 0 ) (−1,1,0): 𝑎 = ⟨ 4 , − 4 ( − 1 ) ( 1 ) ( 0 ) , 0 ⟩ = ⟨ 4 , 0 , 0 ⟩ . a=⟨4,−4(−1)(1)(0),0⟩=⟨4,0,0⟩. Question 4 Find two unit vectors orthogonal to both ( 3 , 2 , 1 ) (3,2,1) and ( − 1 , 1 , 0 ) (−1,1,0). Answer 1 Compute cross product: 𝑣 = 𝑎 × 𝑏 = ∣ 𝑖 𝑗 𝑘 3 2 1 − 1 1 0 ∣ = 𝑖 ∣ 2 1 1 0 ∣ − 𝑗 ∣ 3 1 − 1 0 ∣ + 𝑘 ∣ 3 2 − 1 1 ∣ . v=a×b= i 3 −1 j 2 1 k 1 0 =i 2 1 1 0 −j 3 −1 1 0 +k 3 −1 2 1 . 𝑣 = 𝑖 ( 2 ⋅ 0 − 1 ⋅ 1 ) − 𝑗 ( 3 ⋅ 0 − ( − 1 ) ⋅ 1 ) + 𝑘 ( 3 ⋅ 1 − ( − 1 ) ⋅ 2 ) . v=i(2⋅0−1⋅1)−j(3⋅0−(−1)⋅1)+k(3⋅1−(−1)⋅2). 𝑣 = ⟨ − 1 , − 1 , 5 ⟩ . v=⟨−1,−1,5⟩. Normalize 𝑣 v: ∣ 𝑣 ∣ = ( − 1 ) 2 + ( − 1 ) 2 + 5 2 = 27 . ∣v∣= (−1) 2 +(−1) 2 +5 2 = 27 . Unit vector: 𝑢 1 = 𝑣 ∣ 𝑣 ∣ = ⟨ − 1 , − 1 , 5 ⟩ 27 . u 1 = ∣v∣ v = 27 ⟨−1,−1,5⟩ . Find another orthogonal unit vector by taking cross product of 𝑣 v with 𝑎 a: 𝑢 2 = 𝑣 × 𝑎 ∣ 𝑣 × 𝑎 ∣ . u 2 = ∣v×a∣ v×a . Question 5 Find direction numbers for the line of intersection of the planes 𝑥 + 𝑦 + 𝑧 = 1 x+y+z=1 and 𝑥 + 𝑧 = 0 x+z=0. Answer 1 Normal vectors: 𝑁 1 = ⟨ 1 , 1 , 1 ⟩ , 𝑁 2 = ⟨ 1 , 0 , 1 ⟩ . N 1 =⟨1,1,1⟩,N 2 =⟨1,0,1⟩. Cross product 𝑑 = 𝑁 1 × 𝑁 2 d=N 1 ×N 2 : 𝑑 = ∣ 𝑖 𝑗 𝑘 1 1 1 1 0 1 ∣ = ⟨ 1 , 0 , − 1 ⟩ . d= i 1 1 j 1 0 k 1 1 =⟨1,0,−1⟩. Direction numbers: 𝑑 = ( − 1 , 1 , 1 ) . d=(−1,1,1).