Vectors #5 - Questions and Answers

Question 1 Find the curve's unit tangent vector. Also, find the length of the indicated portion of the curve. 𝑟 ( 𝑡 ) = ( 2 cos ⁡ 𝑡 ) 𝑖 + ( 2 sin ⁡ 𝑡 ) 𝑗 + 5 𝑡 𝑘 ,   0 ≤ 𝑡 ≤ 𝜋 r(t)=(2cost)i+(2sint)j+ 5 ​ tk, 0≤t≤π Answer 1 Given: 𝑟 = ( 2 cos ⁡ 𝑡 ) 𝑖 + ( 2 sin ⁡ 𝑡 ) 𝑗 + 5 𝑡 𝑘 r=(2cost)i+(2sint)j+ 5 ​ tk Find 𝑟 ′ ( 𝑡 ) r ′ (t): 𝑟 ′ ( 𝑡 ) = ( − 2 sin ⁡ 𝑡 ) 𝑖 + ( 2 cos ⁡ 𝑡 ) 𝑗 + 5 𝑘 r ′ (t)=(−2sint)i+(2cost)j+ 5 ​ k Magnitude of 𝑟 ′ ( 𝑡 ) r ′ (t): ∣ 𝑟 ′ ( 𝑡 ) ∣ = ( − 2 sin ⁡ 𝑡 ) 2 + ( 2 cos ⁡ 𝑡 ) 2 + ( 5 ) 2 = 4 sin ⁡ 2 𝑡 + 4 cos ⁡ 2 𝑡 + 5 . ∣r ′ (t)∣= (−2sint) 2 +(2cost) 2 +( 5 ​ ) 2 ​ = 4sin 2 t+4cos 2 t+5 ​ . Simplifying: ∣ 𝑟 ′ ( 𝑡 ) ∣ = 4 ( sin ⁡ 2 𝑡 + cos ⁡ 2 𝑡 ) + 5 = 4 + 5 = 3. ∣r ′ (t)∣= 4(sin 2 t+cos 2 t)+5 ​ = 4+5 ​ =3. Unit tangent vector 𝑇 T: 𝑇 = 𝑟 ′ ( 𝑡 ) ∣ 𝑟 ′ ( 𝑡 ) ∣ = − 2 sin ⁡ 𝑡 𝑖 + 2 cos ⁡ 𝑡 𝑗 + 5 𝑘 3 . T= ∣r ′ (t)∣ r ′ (t) ​ = 3 −2sinti+2costj+ 5 ​ k ​ . Thus: 𝑇 = < − 2 3 sin ⁡ 𝑡 , 2 3 cos ⁡ 𝑡 , 5 3 > . T=⟨− 3 2 ​ sint, 3 2 ​ cost, 3 5 ​ ​ ⟩. Length of the curve: 𝐿 = ∫ 0 𝜋 ∣ 𝑟 ′ ( 𝑡 ) ∣   𝑑 𝑡 = ∫ 0 𝜋 3   𝑑 𝑡 = 3 𝑡 ∣ 0 𝜋 = 3 𝜋 . L=∫ 0 π ​ ∣r ′ (t)∣dt=∫ 0 π ​ 3dt=3t ​ 0 π ​ =3π. Result: 𝑇 = < − 2 3 sin ⁡ 𝑡 , 2 3 cos ⁡ 𝑡 , 5 3 > , 𝐿 = 3 𝜋 . T=⟨− 3 2 ​ sint, 3 2 ​ cost, 3 5 ​ ​ ⟩,L=3π. Answer 2 Explanation: To find the unit tangent vector: Differentiate 𝑟 ( 𝑡 ) r(t): 𝑟 ′ ( 𝑡 ) = 𝑑 𝑑 𝑡 ( 2 cos ⁡ 𝑡 ) 𝑖 + 𝑑 𝑑 𝑡 ( 2 sin ⁡ 𝑡 ) 𝑗 + 𝑑 𝑑 𝑡 ( 5 𝑡 ) 𝑘 . r ′ (t)= dt d ​ (2cost)i+ dt d ​ (2sint)j+ dt d ​ ( 5 ​ t)k. 𝑟 ′ ( 𝑡 ) = ( − 2 sin ⁡ 𝑡 ) 𝑖 + ( 2 cos ⁡ 𝑡 ) 𝑗 + 5 𝑘 . r ′ (t)=(−2sint)i+(2cost)j+ 5 ​ k. Find ∣ 𝑟 ′ ( 𝑡 ) ∣ ∣r ′ (t)∣: ∣ 𝑟 ′ ( 𝑡 ) ∣ = ( − 2 sin ⁡ 𝑡 ) 2 + ( 2 cos ⁡ 𝑡 ) 2 + ( 5 ) 2 . ∣r ′ (t)∣= (−2sint) 2 +(2cost) 2 +( 5 ​ ) 2 ​ . Simplify: ∣ 𝑟 ′ ( 𝑡 ) ∣ = 4 sin ⁡ 2 𝑡 + 4 cos ⁡ 2 𝑡 + 5 = 4 + 5 = 3. ∣r ′ (t)∣= 4sin 2 t+4cos 2 t+5 ​ = 4+5 ​ =3. Unit tangent vector 𝑇 T: 𝑇 ( 𝑡 ) = 𝑟 ′ ( 𝑡 ) ∣ 𝑟 ′ ( 𝑡 ) ∣ = − 2 sin ⁡ 𝑡 𝑖 + 2 cos ⁡ 𝑡 𝑗 + 5 𝑘 3 . T(t)= ∣r ′ (t)∣ r ′ (t) ​ = 3 −2sinti+2costj+ 5 ​ k ​ . Length of the curve: 𝐿 = ∫ 0 𝜋 ∣ 𝑟 ′ ( 𝑡 ) ∣   𝑑 𝑡 = ∫ 0 𝜋 3   𝑑 𝑡 = 3 𝜋 . L=∫ 0 π ​ ∣r ′ (t)∣dt=∫ 0 π ​ 3dt=3π. Answer 3 Steps: Differentiate 𝑟 ( 𝑡 ) r(t): 𝑟 ′ ( 𝑡 ) = ( − 2 sin ⁡ 𝑡 ) 𝑖 + ( 2 cos ⁡ 𝑡 ) 𝑗 + 5 𝑘 . r ′ (t)=(−2sint)i+(2cost)j+ 5 ​ k. Find ∣ 𝑟 ′ ( 𝑡 ) ∣ ∣r ′ (t)∣: ∣ 𝑟 ′ ( 𝑡 ) ∣ = ( − 2 sin ⁡ 𝑡 ) 2 + ( 2 cos ⁡ 𝑡 ) 2 + ( 5 ) 2 . ∣r ′ (t)∣= (−2sint) 2 +(2cost) 2 +( 5 ​ ) 2 ​ . Simplify: ∣ 𝑟 ′ ( 𝑡 ) ∣ = 3. ∣r ′ (t)∣=3. Unit tangent vector 𝑇 T: 𝑇 ( 𝑡 ) = 𝑟 ′ ( 𝑡 ) ∣ 𝑟 ′ ( 𝑡 ) ∣ . T(t)= ∣r ′ (t)∣ r ′ (t) ​ . Length of the curve: 𝐿 = ∫ 0 𝜋 ∣ 𝑟 ′ ( 𝑡 ) ∣   𝑑 𝑡 = 3 𝜋 . L=∫ 0 π ​ ∣r ′ (t)∣dt=3π. Answer 4 Differentiate 𝑟 ( 𝑡 ) r(t): 𝑟 ′ ( 𝑡 ) = ( − 2 sin ⁡ 𝑡 ) 𝑖 + ( 2 cos ⁡ 𝑡 ) 𝑗 + 5 𝑘 . r ′ (t)=(−2sint)i+(2cost)j+ 5 ​ k. Find ∣ 𝑟 ′ ( 𝑡 ) ∣ ∣r ′ (t)∣: ∣ 𝑟 ′ ( 𝑡 ) ∣ = 4 sin ⁡ 2 𝑡 + 4 cos ⁡ 2 𝑡 + 5 = 3. ∣r ′ (t)∣= 4sin 2 t+4cos 2 t+5 ​ =3. Unit tangent vector 𝑇 T: 𝑇 ( 𝑡 ) = 𝑟 ′ ( 𝑡 ) ∣ 𝑟 ′ ( 𝑡 ) ∣ = − 2 sin ⁡ 𝑡 𝑖 + 2 cos ⁡ 𝑡 𝑗 + 5 𝑘 3 . T(t)= ∣r ′ (t)∣ r ′ (t) ​ = 3 −2sinti+2costj+ 5 ​ k ​ . Length of the curve: 𝐿 = ∫ 0 𝜋 ∣ 𝑟 ′ ( 𝑡 ) ∣   𝑑 𝑡 = 3 𝜋 . L=∫ 0 π ​ ∣r ′ (t)∣dt=3π. Question 2 Find the vector, not with determinants, but by using properties of cross products. ( 𝑖 + 𝑗 ) × ( 𝑖 − 𝑗 ) (i+j)×(i−j) Answer 1 Use the distributive property: ( 𝑖 + 𝑗 ) × ( 𝑖 − 𝑗 ) = 𝑖 × 𝑖 − 𝑖 × 𝑗 + 𝑗 × 𝑖 − 𝑗 × 𝑗 . (i+j)×(i−j)=i×i−i×j+j×i−j×j. 𝑖 × 𝑖 = 0 i×i=0 𝑗 × 𝑗 = 0 j×j=0 𝑖 × 𝑗 = 𝑘 i×j=k 𝑗 × 𝑖 = − 𝑘 j×i=−k. Substitute back: ( 𝑖 + 𝑗 ) × ( 𝑖 − 𝑗 ) = 0 − 𝑘 + ( − 𝑘 ) + 0. (i+j)×(i−j)=0−k+(−k)+0. Simplify: ( 𝑖 + 𝑗 ) × ( 𝑖 − 𝑗 ) = − 2 𝑘 . (i+j)×(i−j)=−2k. Answer 2 Simplify directly: ( 𝑖 + 𝑗 ) × ( 𝑖 − 𝑗 ) = ( 𝑖 × 𝑖 ) − ( 𝑖 × 𝑗 ) + ( 𝑗 × 𝑖 ) − ( 𝑗 × 𝑗 ) . (i+j)×(i−j)=(i×i)−(i×j)+(j×i)−(j×j). 𝑖 × 𝑖 = 0 i×i=0, 𝑗 × 𝑗 = 0 j×j=0, 𝑖 × 𝑗 = 𝑘 i×j=k, 𝑗 × 𝑖 = − 𝑘 j×i=−k. Thus: ( 𝑖 + 𝑗 ) × ( 𝑖 − 𝑗 ) = − 2 𝑘 . (i+j)×(i−j)=−2k. Answer 3 Cross Product Approach: 𝑏 = [ − 1 1 2 ] , 𝑐 = [ 1 − 3 8 ] . b= ​ −1 1 2 ​ ​ ,c= ​ 1 −3 8 ​ ​ . Calculate 𝑏 × 𝑐 b×c using the determinant: 𝑏 × 𝑐 = ∣ 𝑖 𝑗 𝑘 − 1 1 2 1 − 3 8 ∣ . b×c= ​ i −1 1 ​ j 1 −3 ​ k 2 8 ​ ​ . Question 4 Find the vector and parametric equations for the line segment connecting 𝑃 ( 0 , − 1 , 1 ) P(0,−1,1) and 𝑄 ( 1 2 , 1 3 , 1 4 ) Q( 2 1 ​ , 3 1 ​ , 4 1 ​ ). Answer 1 Vector equation: 𝑟 ( 𝑡 ) = 𝑃 + 𝑡 ( 𝑄 − 𝑃 ) . r(t)=P+t(Q−P). Substitute the values: 𝑟 ( 𝑡 ) = ( 0 , − 1 , 1 ) + 𝑡 ( 1 2 , 1 3 , 1 4 ) . r(t)=(0,−1,1)+t( 2 1 ​ , 3 1 ​ , 4 1 ​ ). Simplify: 𝑥 = 𝑡 2 , 𝑦 = − 1 + 𝑡 3 , 𝑧 = 1 + 𝑡 4 . x= 2 t ​ ,y=−1+ 3 t ​ ,z=1+ 4 t ​ . Parametric equations: 𝑥 = 𝑡 2 , 𝑦 = − 1 + 𝑡 3 , 𝑧 = 1 + 𝑡 4 , 𝑡 ∈ [ 0 , 1 ] . x= 2 t ​ ,y=−1+ 3 t ​ ,z=1+ 4 t ​ ,t∈[0,1].