University:
University of Michigan
Course:
MATH 116 | Calculus II
Academic year:

2024
Views:

84

Pages:

3
Author:

banenetta0l

Question 1 Find parametric equations for the lines in The line through the origin parallel to the vector 2 𝑗 + 𝑘 2j+k Answer: The vector equation is 𝑟 = ( 0 ) + 𝑡 ( 2 𝑗 + 𝑘 ) r=(0)+t(2j+k). Parametric equations are: 𝑥 = 0 , 𝑦 = 2 𝑡 , 𝑧 = 𝑡 . x=0,y=2t,z=t. Question 2 Find both the parametric and vector equations of the line through ( 0 , 0 , 1 ) (0,0,1) in the direction of 𝑣 = ( 4 , 7 , 0 ) v=(4,7,0). Answer: Vector equation: 𝑟 = ( 0 , 0 , 1 ) + 𝑡 ( 4 , 7 , 0 ) . r=(0,0,1)+t(4,7,0). Parametric equations: 𝑥 = 4 𝑡 , 𝑦 = 7 𝑡 , 𝑧 = 1. x=4t,y=7t,z=1. Question 3 Determine the area of the region below the parametric curve 𝑥 ( 𝑡 ) = 𝑡 4 + 5 𝑡 − 40 , 𝑦 ( 𝑡 ) = 2 𝑡 + 5 x(t)=t 4 +5t−40,y(t)=2t+5 over 𝑡 ∈ [ − 2 , 5 ] t∈[−2,5]. Answer: The area under the parametric curve is: 𝐴 = ∫ − 2 5 𝑔 ( 𝑡 ) 𝑓 ′ ( 𝑡 ) 𝑑 𝑡 . A=∫ −2 5 g(t)f ′ (t)dt. Given 𝑓 ( 𝑡 ) = 𝑥 ( 𝑡 ) , 𝑔 ( 𝑡 ) = 𝑦 ( 𝑡 ) f(t)=x(t),g(t)=y(t): 𝐴 = ∫ − 2 5 ( 2 𝑡 + 5 ) ( 4 𝑡 3 + 5 ) 𝑑 𝑡 . A=∫ −2 5 (2t+5)(4t 3 +5)dt. Simplify and solve: 𝐴 = 2892.167 units 2 . A=2892.167units 2 . Question 4 Find the parametric equation of the line through 𝑎 = [ 3 , − 4 ] a=[3,−4] parallel to 𝑏 = [ − 7 , 8 ] b=[−7,8]. Answer: Vector equation: 𝑥 = 𝑎 + 𝑡 𝑏 = [ 3 , − 4 ] + 𝑡 [ − 7 , 8 ] . x=a+tb=[3,−4]+t[−7,8]. Parametric equations: 𝑥 1 = 3 − 7 𝑡 , 𝑥 2 = − 4 + 8 𝑡 . x 1 =3−7t,x 2 =−4+8t.