Question 1 At the specified position, determine the vectors 𝑇 T, 𝑁 N, and 𝐵 B. Given: 𝑟 ( 𝑡 ) = ⟨ 𝑡 2 , 𝑡 3 , 𝑡 ⟩ and point ⟨ 4 , − 16 3 , − 2 ⟩ r(t)=⟨t 2 ,t 3 ,t⟩and point ⟨4,− 3 16 ,−2⟩ Answer 1 To determine 𝑇 T, 𝑁 N, and 𝐵 B: Find 𝑡 t: Substituting 𝑡 2 = 4 , 𝑡 3 = − 16 3 , 𝑡 = − 2 t 2 =4,t 3 =− 3 16 ,t=−2. The corresponding parameter is 𝑡 = − 2 t=−2. Find 𝑟 ′ ( 𝑡 ) r ′ (t): 𝑟 ′ ( 𝑡 ) = ⟨ 2 𝑡 , 3 𝑡 2 , 1 ⟩ r ′ (t)=⟨2t,3t 2 ,1⟩ At 𝑡 = − 2 t=−2: 𝑟 ′ ( − 2 ) = ⟨ − 4 , 12 , 1 ⟩ r ′ (−2)=⟨−4,12,1⟩ Magnitude of 𝑟 ′ ( − 2 ) r ′ (−2): ∣ 𝑟 ′ ( − 2 ) ∣ = ( − 4 ) 2 + 1 2 2 + 1 2 = 16 + 144 + 1 = 161 ∣r ′ (−2)∣= (−4) 2 +12 2 +1 2 = 16+144+1 = 161 Unit tangent vector 𝑇 T: 𝑇 = 𝑟 ′ ( − 2 ) ∣ 𝑟 ′ ( − 2 ) ∣ = ⟨ − 4 , 12 , 1 ⟩ 161 T= ∣r ′ (−2)∣ r ′ (−2) = 161 ⟨−4,12,1⟩ Find 𝑟 ′ ′ ( 𝑡 ) r ′′ (t): 𝑟 ′ ′ ( 𝑡 ) = ⟨ 2 , 6 𝑡 , 0 ⟩ r ′′ (t)=⟨2,6t,0⟩ At 𝑡 = − 2 t=−2: 𝑟 ′ ′ ( − 2 ) = ⟨ 2 , − 12 , 0 ⟩ r ′′ (−2)=⟨2,−12,0⟩ Find 𝑁 N: Compute 𝑇 ′ ( 𝑡 ) T ′ (t) and normalize it: 𝑁 = 𝑟 ′ ′ ( − 2 ) ∣ 𝑟 ′ ′ ( − 2 ) ∣ = ⟨ 2 , − 12 , 0 ⟩ 2 2 + ( − 12 ) 2 + 0 2 = ⟨ 2 , − 12 , 0 ⟩ 148 = ⟨ 1 , − 6 , 0 ⟩ 37 N= ∣r ′′ (−2)∣ r ′′ (−2) = 2 2 +(−12) 2 +0 2 ⟨2,−12,0⟩ = 148 ⟨2,−12,0⟩ = 37 ⟨1,−6,0⟩ Find 𝐵 B: 𝐵 = 𝑇 × 𝑁 B=T×N Using 𝑇 = ⟨ − 4 , 12 , 1 ⟩ 161 T= 161 ⟨−4,12,1⟩ and 𝑁 = ⟨ 1 , − 6 , 0 ⟩ 37 N= 37 ⟨1,−6,0⟩ : 𝐵 = 1 161 37 ∣ 𝑖 𝑗 𝑘 − 4 12 1 1 − 6 0 ∣ B= 161 37 1 i −4 1 j 12 −6 k 1 0 Expanding: 𝐵 = 1 161 37 ⟨ ( − 12 ) ( 0 ) − ( 1 ) ( − 6 ) , ( 1 ) ( 1 ) − ( − 4 ) ( 0 ) , ( − 4 ) ( − 6 ) − ( 12 ) ( 1 ) ⟩ B= 161 37 1 ⟨(−12)(0)−(1)(−6),(1)(1)−(−4)(0),(−4)(−6)−(12)(1)⟩ 𝐵 = ⟨ 6 , 1 , − 24 ⟩ 161 37 B= 161 37 ⟨6,1,−24⟩ Question 2 Find ∣ 𝑢 ⋅ 𝑣 ∣ ∣u⋅v∣ and determine whether 𝑢 ⋅ 𝑣 u⋅v points toward or away from the screen. Answer 2 Given ∣ 𝑢 ∣ = 6 , ∣ 𝑣 ∣ = 8 , 𝜃 = 4 5 ∘ ∣u∣=6,∣v∣=8,θ=45 ∘ : ∣ 𝑢 ⋅ 𝑣 ∣ = ∣ 𝑢 ∣ ∣ 𝑣 ∣ ∣ cos ⁡ 𝜃 ∣ = 6 ⋅ 8 ⋅ cos ⁡ ( 4 5 ∘ ) = 6 ⋅ 8 ⋅ 2 2 = 24 2 ∣u⋅v∣=∣u∣∣v∣∣cosθ∣=6⋅8⋅cos(45 ∘ )=6⋅8⋅ 2 2 =24 2 The sign of cos ⁡ ( 4 5 ∘ ) cos(45 ∘ ) is positive, so 𝑢 ⋅ 𝑣 u⋅v points away from the screen. Question 3 Show that 𝑊 W is a vector space or find a spanning set. Given 𝑊 = { [ 4 𝑎 + 3 𝑏 , 0 , 𝑎 + 𝑏 + 𝑐 , 𝑐 − 2 𝑎 ] } W={[4a+3b,0,a+b+c,c−2a]}. Answer 3 Write any vector 𝑤 ∈ 𝑊 w∈W: 𝑤 = 𝑎 [ 4 0 1 − 2 ] + 𝑏 [ 3 0 1 0 ] + 𝑐 [ 0 0 1 1 ] w=a 4 0 1 −2 +b 3 0 1 0 +c 0 0 1 1 Spanning set for 𝑊 W: 𝑆 = { [ 4 0 1 − 2 ] , [ 3 0 1 0 ] , [ 0 0 1 1 ] } S= ⎩ ⎨ ⎧ 4 0 1 −2 , 3 0 1 0 , 0 0 1 1 ⎭ ⎬ ⎫ Question 4 Find a basis for the space spanned by: 𝑣 1 = [ 1 − 2 6 5 0 ] , 𝑣 2 = [ 0 1 − 1 − 3 3 ] , 𝑣 3 = [ 0 1 − 2 3 − 1 ] , 𝑣 4 = [ 1 − 1 − 4 1 0 ] v 1 = 1 −2 6 5 0 ,v 2 = 0 1 −1 −3 3 ,v 3 = 0 1 −2 3 −1 ,v 4 = 1 −1 −4 1 0 Answer 4 Row reduce the matrix formed by 𝑣 1 , 𝑣 2 , 𝑣 3 , 𝑣 4 v 1 ,v 2 ,v 3 ,v 4 : [ 1 0 0 1 − 2 1 1 − 1 6 − 1 − 2 − 4 5 − 3 3 1 0 3 − 1 0 ] 1 −2 6 5 0 0 1 −1 −3 3 0 1 −2 3 −1 1 −1 −4 1 0 After row reduction: [ 1 0 0 1 0 1 0 − 1 0 0 1 − 1 0 0 0 0 0 0 0 0 ] 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 1 −1 −1 0 0 The basis is formed by 𝑣 1 , 𝑣 2 , 𝑣 3 v 1 ,v 2 ,v 3 .