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

2024
Views:

71

Pages:

4
Author:

banenetta0l

Question 1 Find an equation of the plane tangent to the following surface at the given point: 7 𝑥 𝑦 + 𝑦 𝑧 + 4 𝑥 𝑧 − 48 = 0 ; ( 2 , 2 , 2 ) . 7xy+yz+4xz−48=0;(2,2,2). Answer: Given 𝑓 ( 𝑥 , 𝑦 , 𝑧 ) = 7 𝑥 𝑦 + 𝑦 𝑧 + 4 𝑥 𝑧 − 48 = 0 f(x,y,z)=7xy+yz+4xz−48=0, calculate partial derivatives: 𝑃 𝑥 = 7 𝑦 + 4 𝑧 , 𝑃 𝑦 = 7 𝑥 + 𝑧 , 𝑃 𝑧 = 𝑦 + 4 𝑥 . P x =7y+4z,P y =7x+z,P z =y+4x. At ( 2 , 2 , 2 ) (2,2,2): 𝑃 𝑥 = 7 ( 2 ) + 4 ( 2 ) = 22 , 𝑃 𝑦 = 7 ( 2 ) + 2 = 16 , 𝑃 𝑧 = 2 + 4 ( 2 ) = 10. P x =7(2)+4(2)=22,P y =7(2)+2=16,P z =2+4(2)=10. Tangent plane: 22 ( 𝑥 − 2 ) + 16 ( 𝑦 − 2 ) + 10 ( 𝑧 − 2 ) = 0. 22(x−2)+16(y−2)+10(z−2)=0. Simplify: 22 𝑥 + 16 𝑦 + 10 𝑧 − 96 = 0. 22x+16y+10z−96=0. Question 2 Find a basis for the eigenspace corresponding to the eigenvalue 𝜆 = 3 λ=3 of 𝐴 A: 𝐴 = [ 4 0 2 0 4 2 10 0 3 − 4 17 0 2 − 2 8 3 ] . A= 4 4 3 2 0 2 −4 −2 2 10 17 8 0 0 0 3 . Answer: The eigenvector corresponds to: ( 𝐴 − 3 𝐼 ) 𝑣 = 0. (A−3I)v=0. Solving, the basis for the eigenspace is: Basis vectors: [ 2 − 1 0 0 ] , [ 2 0 − 1 0 ] . Basis vectors: 2 −1 0 0 , 2 0 −1 0 . Question 3 Find the volume of the solid generated by revolving the shaded region about the x-axis. The equation: 4 𝑥 + 3 𝑦 = 24 4x+3y=24. Answer: Rewrite 𝑦 = 8 − 4 3 𝑥 y=8− 3 4 x. The limits are 𝑥 = 0 x=0 to 𝑥 = 6 x=6. Volume by washer method: 𝑉 = 𝜋 ∫ 0 6 [ ( 8 ) 2 − ( 8 − 4 3 𝑥 ) 2 ] 𝑑 𝑥 . V=π∫ 0 6 [(8) 2 −(8− 3 4 x) 2 ]dx. Solve: 𝑉 = 128 𝜋 = 402 cubic units . V=128π=402 cubic units. Question 4 What is 10 ∠ − 30 + 10 ∠ 30 10∠−30+10∠30? Answer in polar form. Answer: As they are on the same facing lines: 10 ∠ − 30 + 10 ∠ 30 = 20 ∠ 0. 10∠−30+10∠30=20∠0. Question 5 Given 𝑤 = 2 ( cos ⁡ 150 + 𝑖 sin ⁡ 150 ) w=2(cos150+isin150) and 𝑧 = − 2 + 𝑖 z=− 2 +i, find the polar form of 𝑧 z. Answer: 𝑟 = ( − 2 ) 2 + 1 2 = 3 , tan ⁡ ( 𝜋 − 𝜃 ) = 1 2 . r= (− 2 ) 2 +1 2 = 3 ,tan(π−θ)= 2 1 . Solve for 𝜃 θ: 𝜃 = 𝜋 − arctan ⁡ ( 1 2 ) . θ=π−arctan( 2 1 ). Polar form: 𝑧 = 3 𝑒 𝑖 ( 𝜋 − arctan ⁡ ( 1 / 2 ) ) . z= 3 e i(π−arctan(1/ 2 )) .