Question 1 Find two different sets of parametric equations for the rectangular equation: 𝑦 = 𝑥 2 − 2 y=x 2 −2 Answer 1 Let 𝑥 = 𝑡 x=t: 𝑦 = 𝑡 2 − 2 y=t 2 −2 First set of parametric equations: 𝑥 = 𝑡 , 𝑦 = 𝑡 2 − 2 x=t,y=t 2 −2 Let 𝑥 = 𝑡 + 2 x=t+2: 𝑦 = ( 𝑡 + 2 ) 2 − 2 = 𝑡 2 + 4 𝑡 + 4 − 2 = 𝑡 2 + 4 𝑡 + 2 y=(t+2) 2 −2=t 2 +4t+4−2=t 2 +4t+2 Second set of parametric equations: 𝑥 = 𝑡 + 2 , 𝑦 = 𝑡 2 + 4 𝑡 + 2 x=t+2,y=t 2 +4t+2 Question 2 Find two different parametric equations for the rectangular equation: 𝑦 = − 2 𝑥 + 4 y=−2x+4 Answer 1 Let 𝑥 = 𝑡 x=t: 𝑦 = − 2 𝑡 + 4 y=−2t+4 First set of parametric equations: 𝑥 = 𝑡 , 𝑦 = − 2 𝑡 + 4 x=t,y=−2t+4 Let 𝑥 = 𝑡 + 1 x=t+1: 𝑦 = − 2 ( 𝑡 + 1 ) + 4 = − 2 𝑡 − 2 + 4 = − 2 𝑡 + 2 y=−2(t+1)+4=−2t−2+4=−2t+2 Second set of parametric equations: 𝑥 = 𝑡 + 1 , 𝑦 = − 2 𝑡 + 2 x=t+1,y=−2t+2 Question 3 Verify the Sum and Product Rules for derivatives of vector-valued functions. Answer 1 Sum Rule: ( 𝐹 ( 𝑡 ) + 𝐺 ( 𝑡 ) ) ′ = 𝐹 ′ ( 𝑡 ) + 𝐺 ′ ( 𝑡 ) (F(t)+G(t)) ′ =F ′ (t)+G ′ (t) Let 𝐹 ( 𝑡 ) = ⟨ 𝑡 2 , 5 𝑡 ⟩ F(t)=⟨t 2 ,5t⟩ and 𝐺 ( 𝑡 ) = ⟨ sin ⁡ 𝑡 , cos ⁡ 𝑡 ⟩ G(t)=⟨sint,cost⟩. LHS: ( 𝐹 ( 𝑡 ) + 𝐺 ( 𝑡 ) ) ′ = ( ⟨ 𝑡 2 + sin ⁡ 𝑡 , 5 𝑡 + cos ⁡ 𝑡 ⟩ ) ′ = ⟨ 2 𝑡 + cos ⁡ 𝑡 , 5 − sin ⁡ 𝑡 ⟩ (F(t)+G(t)) ′ =(⟨t 2 +sint,5t+cost⟩) ′ =⟨2t+cost,5−sint⟩ RHS: 𝐹 ′ ( 𝑡 ) + 𝐺 ′ ( 𝑡 ) = ⟨ 2 𝑡 , 5 ⟩ + ⟨ cos ⁡ 𝑡 , − sin ⁡ 𝑡 ⟩ = ⟨ 2 𝑡 + cos ⁡ 𝑡 , 5 − sin ⁡ 𝑡 ⟩ F ′ (t)+G ′ (t)=⟨2t,5⟩+⟨cost,−sint⟩=⟨2t+cost,5−sint⟩ LHS = RHS, so the Sum Rule is verified. Product Rule: ( 𝑓 ( 𝑡 ) 𝐺 ( 𝑡 ) ) ′ = 𝑓 ′ ( 𝑡 ) 𝐺 ( 𝑡 ) + 𝑓 ( 𝑡 ) 𝐺 ′ ( 𝑡 ) (f(t)G(t)) ′ =f ′ (t)G(t)+f(t)G ′ (t) Let 𝑓 ( 𝑡 ) = 𝑡 2 f(t)=t 2 , 𝐺 ( 𝑡 ) = ⟨ 𝑡 4 , ln ⁡ 𝑡 ⟩ G(t)=⟨t 4 ,lnt⟩. LHS: ( 𝑓 ( 𝑡 ) 𝐺 ( 𝑡 ) ) ′ = ( 𝑡 2 ⋅ ⟨ 𝑡 4 , ln ⁡ 𝑡 ⟩ ) ′ = ⟨ 𝑡 6 , 𝑡 2 ln ⁡ 𝑡 ⟩ ′ = ⟨ 6 𝑡 5 , 2 𝑡 ln ⁡ 𝑡 + 𝑡 ⟩ (f(t)G(t)) ′ =(t 2 ⋅⟨t 4 ,lnt⟩) ′ =⟨t 6 ,t 2 lnt⟩ ′ =⟨6t 5 ,2tlnt+t⟩ RHS: 𝑓 ′ ( 𝑡 ) 𝐺 ( 𝑡 ) + 𝑓 ( 𝑡 ) 𝐺 ′ ( 𝑡 ) = 2 𝑡 ⋅ ⟨ 𝑡 4 , ln ⁡ 𝑡 ⟩ + 𝑡 2 ⋅ ⟨ 4 𝑡 3 , 1 𝑡 ⟩ = ⟨ 2 𝑡 5 + 4 𝑡 5 , 2 𝑡 ln ⁡ 𝑡 + 𝑡 ⟩ f ′ (t)G(t)+f(t)G ′ (t)=2t⋅⟨t 4 ,lnt⟩+t 2 ⋅⟨4t 3 , t 1 ⟩=⟨2t 5 +4t 5 ,2tlnt+t⟩ LHS = RHS, so the Product Rule is verified. Question 4 Vector-valued functions and continuity: Given 𝑟 ( 𝑡 ) = ⟨ 4 𝑡 + 2 , sin ⁡ ( 𝜋 𝑡 ) , 1 𝑡 3 ⟩ r(t)=⟨4t+2,sin(πt), t 3 1 ⟩, find lim ⁡ 𝑡 → − 1 𝑟 ( 𝑡 ) lim t→−1 r(t). Answer 1 Step 1: Take limits component-wise: lim ⁡ 𝑡 → − 1 𝑟 ( 𝑡 ) = ⟨ lim ⁡ 𝑡 → − 1 ( 4 𝑡 + 2 ) , lim ⁡ 𝑡 → − 1 sin ⁡ ( 𝜋 𝑡 ) , lim ⁡ 𝑡 → − 1 1 𝑡 3 ⟩ t→−1 lim r(t)=⟨ t→−1 lim (4t+2), t→−1 lim sin(πt), t→−1 lim t 3 1 ⟩ Step 2: Evaluate each limit: lim ⁡ 𝑡 → − 1 ( 4 𝑡 + 2 ) = 4 ( − 1 ) + 2 = − 4 + 2 = − 2 t→−1 lim (4t+2)=4(−1)+2=−4+2=−2 lim ⁡ 𝑡 → − 1 sin ⁡ ( 𝜋 𝑡 ) = sin ⁡ ( − 𝜋 ) = 0 t→−1 lim sin(πt)=sin(−π)=0 lim ⁡ 𝑡 → − 1 1 𝑡 3 = 1 ( − 1 ) 3 = − 1 t→−1 lim t 3 1 = (−1) 3 1 =−1 Final Answer: lim ⁡ 𝑡 → − 1 𝑟 ( 𝑡 ) = ⟨ − 2 , 0 , − 1 ⟩ t→−1 lim r(t)=⟨−2,0,−1⟩ Question 5 Find parametric and symmetric equations for the line passing through the origin and the point ( 9 , 3 , − 1 ) (9,3,−1). Answer 1 Parametric equations: The direction vector is: 𝑑 = ⟨ 9 − 0 , 3 − 0 , − 1 − 0 ⟩ = ⟨ 9 , 3 , − 1 ⟩ d=⟨9−0,3−0,−1−0⟩=⟨9,3,−1⟩ Using the parametric form 𝑟 ( 𝑡 ) = 𝑟 0 + 𝑡 𝑑 r(t)=r 0 +td: 𝑥 = 9 𝑡 , 𝑦 = 3 𝑡 , 𝑧 = − 𝑡 x=9t,y=3t,z=−t Symmetric equations: Eliminate 𝑡 t: 𝑡 = 𝑥 9 , 𝑡 = 𝑦 3 , 𝑡 = 𝑧 − 1 t= 9 x ,t= 3 y ,t= −1 z 𝑥 9 = 𝑦 3 = 𝑧 − 1 9 x = 3 y = −1 z Final Answer: Parametric equations: 𝑥 = 9 𝑡 , 𝑦 = 3 𝑡 , 𝑧 = − 𝑡 x=9t,y=3t,z=−t Symmetric equations: 𝑥 9 = 𝑦 3 = 𝑧 − 1 9 x = 3 y = −1 z