Question 1 Find the arc length of the curve on the given interval. Parametric equations: 𝑥 = 𝑡 , 𝑦 = 3 𝑡 − 1 , Interval: 0 ≤ 𝑡 ≤ 1 x= t ,y=3t−1,Interval: 0≤t≤1 Answer 1 Step 1: Formula for arc length: Arc length = ∫ 𝑎 𝑏 ( 𝑑 𝑥 𝑑 𝑡 ) 2 + ( 𝑑 𝑦 𝑑 𝑡 ) 2 𝑑 𝑡 Arc length=∫ a b ( dt dx ) 2 +( dt dy ) 2 dt Step 2: Compute derivatives: 𝑥 = 𝑡 ⟹ 𝑑 𝑥 𝑑 𝑡 = 1 2 𝑡 x= t ⟹ dt dx = 2 t 1 𝑦 = 3 𝑡 − 1 ⟹ 𝑑 𝑦 𝑑 𝑡 = 3 y=3t−1⟹ dt dy =3 Step 3: Substitute into the arc length formula: Arc length = ∫ 0 1 ( 1 2 𝑡 ) 2 + ( 3 ) 2 𝑑 𝑡 Arc length=∫ 0 1 ( 2 t 1 ) 2 +(3) 2 dt = ∫ 0 1 1 4 𝑡 + 9 𝑑 𝑡 =∫ 0 1 4t 1 +9 dt Step 4: Simplify the integrand: 1 4 𝑡 + 9 = 1 + 36 𝑡 4 𝑡 = 1 + 36 𝑡 4 𝑡 = 1 + 36 𝑡 2 𝑡 4t 1 +9 = 4t 1+36t = 4t 1+36t = 2 t 1+36t Step 5: Substitute and simplify: Arc length = ∫ 0 1 1 + 36 𝑡 2 𝑡 𝑑 𝑡 Arc length=∫ 0 1 2 t 1+36t dt Let 𝑢 = 6 𝑡 u=6 t , so 𝑡 = 𝑢 2 36 t= 36 u 2 and 𝑑 𝑡 = 𝑢 18 𝑑 𝑢 dt= 18 u du. Change of limits: When 𝑡 = 0 t=0, 𝑢 = 0 u=0. When 𝑡 = 1 t=1, 𝑢 = 6 u=6. Substitute: Arc length = 1 2 ∫ 0 6 1 + 𝑢 2 𝑡 ⋅ 𝑢 18 𝑑 𝑢 = 1 6 ∫ 0 6 1 + 𝑢 2 𝑑 𝑢 Arc length= 2 1 ∫ 0 6 t 1+u 2 ⋅ 18 u du= 6 1 ∫ 0 6 1+u 2 du The arc length can be evaluated further numerically or left in integral form. Question 2 The exit velocity of a baseball is 128 ft/s at an angle of 3 0 ∘ 30 ∘ above the horizontal. Find the parametric equations for its motion if the ball starts 4 ft above the ground. Answer 1 Step 1: Resolve velocity into components: 𝑣 𝑥 = 128 cos ⁡ ( 3 0 ∘ ) = 128 ⋅ 3 2 = 64 3 ft/s v x =128cos(30 ∘ )=128⋅ 2 3 =64 3 ft/s 𝑣 𝑦 = 128 sin ⁡ ( 3 0 ∘ ) = 128 ⋅ 1 2 = 64 ft/s v y =128sin(30 ∘ )=128⋅ 2 1 =64ft/s Step 2: Parametric equations for position: Horizontal motion: 𝑥 ( 𝑡 ) = 𝑣 𝑥 𝑡 = 64 3 𝑡 x(t)=v x t=64 3 t Vertical motion: 𝑦 ( 𝑡 ) = 𝑦 0 + 𝑣 𝑦 𝑡 − 1 2 𝑔 𝑡 2 = 4 + 64 𝑡 − 16 𝑡 2 y(t)=y 0 +v y t− 2 1 gt 2 =4+64t−16t 2 (Using 𝑔 = 32 ft/s 2 g=32ft/s 2 ). Final Parametric Equations: 𝑥 ( 𝑡 ) = 64 3 𝑡 , 𝑦 ( 𝑡 ) = 4 + 64 𝑡 − 16 𝑡 2 x(t)=64 3 t,y(t)=4+64t−16t 2 Question 3 Find the equation of the line and parametric equations for the line segment from 𝑃 0 ( − 3 , 5 , 8 ) P 0 (−3,5,8) to 𝑃 1 ( 4 , 2 , − 1 ) P 1 (4,2,−1). Answer 1 Step 1: Direction vector: 𝑑 = ⟨ 4 − ( − 3 ) , 2 − 5 , − 1 − 8 ⟩ = ⟨ 7 , − 3 , − 9 ⟩ d=⟨4−(−3),2−5,−1−8⟩=⟨7,−3,−9⟩ Step 2: Parametric equations: Using 𝑟 ( 𝑡 ) = 𝑟 0 + 𝑡 𝑑 r(t)=r 0 +td: 𝑥 ( 𝑡 ) = − 3 + 7 𝑡 , 𝑦 ( 𝑡 ) = 5 − 3 𝑡 , 𝑧 ( 𝑡 ) = 8 − 9 𝑡 , 𝑡 ∈ [ 0 , 1 ] x(t)=−3+7t,y(t)=5−3t,z(t)=8−9t,t∈[0,1] Final Answer: Parametric equations: 𝑥 ( 𝑡 ) = − 3 + 7 𝑡 , 𝑦 ( 𝑡 ) = 5 − 3 𝑡 , 𝑧 ( 𝑡 ) = 8 − 9 𝑡 , 𝑡 ∈ [ 0 , 1 ] x(t)=−3+7t,y(t)=5−3t,z(t)=8−9t,t∈[0,1]