Question 1 Compute the area 𝑆 S under the mapping 𝑥 ↦ 𝐴 𝑥 x↦Ax for the parallelogram determined by the vectors: 𝑏 1 = [ − 2 2 3 ] , 𝑏 2 = [ − 2 5 5 ] , 𝐴 = [ 6 − 3 − 3 2 ] . b 1 = −2 2 3 ,b 2 = −2 5 5 ,A=[ 6 −3 −3 2 ]. Answer: The area 𝑆 S under the mapping is calculated by: 𝑆 = ∣ det ⁡ ( 𝐴 ) ∣ ⋅ ∣ det ⁡ ( [ 𝑏 1 , 𝑏 2 ] ) ∣ S=∣det(A)∣⋅∣det([b 1 ,b 2 ])∣ From the calculations, det ⁡ ( 𝐴 ) = 12 , det ⁡ ( [ 𝑏 1 , 𝑏 2 ] ) = 1. det(A)=12,det([b 1 ,b 2 ])=1. Therefore: 𝑆 = 12 × 1 = 12. S=12×1=12. Question 2 Find the best approximation to 𝑧 z by vectors of the form 𝑐 1 𝑣 1 + 𝑐 2 𝑣 2 c 1 v 1 +c 2 v 2 : 𝑧 = [ 3 7 2 ] , 𝑣 1 = [ 2 − 1 − 3 ] , 𝑣 2 = [ 1 2 0 ] . z= 3 7 2 ,v 1 = 2 −1 −3 ,v 2 = 1 2 0 . Answer: The orthogonal projection gives the best approximation. Using calculations: 𝑧 approx = − 1 2 𝑣 1 + 17 5 𝑣 2 . z approx =− 2 1 v 1 + 5 17 v 2 . Question 3 Flow Field Analysis Given: 𝑢 = 𝑥 2 + 𝑦 2 + 𝑧 2 , 𝑣 = 𝑥 𝑦 + 𝑦 𝑧 + 𝑥 𝑧 , 𝑤 = − 3 𝑥 𝑧 2 + 4. u=x 2 +y 2 +z 2 ,v=xy+yz+xz,w=− 2 3xz +4. (a) Volumetric Dilation Rate The divergence ∇ ⋅ 𝑉 ∇⋅V: ∂ 𝑢 ∂ 𝑥 + ∂ 𝑣 ∂ 𝑦 + ∂ 𝑤 ∂ 𝑧 = 2 𝑥 + ( 𝑥 + 𝑧 ) − 3 𝑥 2 . ∂x ∂u + ∂y ∂v + ∂z ∂w =2x+(x+z)− 2 3x . Simplified: ∇ ⋅ 𝑉 = 𝑥 2 + 𝑧 . ∇⋅V= 2 x +z. (b) Irrotational Flow The flow is not irrotational because ∇ × 𝑉 ≠ 0 ∇×V  =0. Question 4 Find the principal unit normal vector to the curve at 𝑡 = 3 t=3: Given: 𝑟 ( 𝑡 ) = 𝑡 𝑖 + 1 𝑡 𝑗 + 𝑡 2 𝑘 . r(t)=ti+ t 1 j+t 2 k. Answer: The principal unit normal vector 𝑁 ( 𝑡 ) N(t) is calculated, and at 𝑡 = 3 t=3, 𝑁 ( 3 ) = 1 13 ( 2 𝑖 + 3 𝑗 ) . N(3)= 13 1 (2i+3j). Question 5 Identify the surface for: 𝑟 ( 𝑠 , 𝑡 ) = ( 𝑠 , 𝑡 , 𝑡 2 − 𝑠 2 ) . r(s,t)=(s,t,t 2 −s 2 ). Answer: From the equation, 𝑧 = 𝑡 2 − 𝑠 2 ⟹ 𝑧 = ( 𝑡 + 𝑠 ) ( 𝑡 − 𝑠 ) . z=t 2 −s 2 ⟹z=(t+s)(t−s). This represents a hyperbolic paraboloid.