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

2024
Views:

423

Pages:

4
Author:

banenetta0l

Question 1 Find the maximum volume of a rectangular box inscribed in a sphere of radius 𝑟 r. Answer: The volume of a box is given by 𝑉 = 𝑙 × 𝑤 × ℎ V=l×w×h. The box must satisfy the sphere equation 𝑥 2 + 𝑦 2 + 𝑧 2 = 𝑟 2 x 2 +y 2 +z 2 =r 2 . Substituting 𝑧 = 𝑟 2 − 𝑥 2 − 𝑦 2 z= r 2 −x 2 −y 2 , the volume becomes: 𝑉 = 8 𝑥 𝑦 𝑧 = 8 𝑥 𝑦 𝑟 2 − 𝑥 2 − 𝑦 2 V=8xyz=8xy r 2 −x 2 −y 2 Optimization: Solving partial derivatives for 𝑉 V, the solution gives: 𝑥 = 𝑦 = 𝑧 = 𝑟 3 x=y=z= 3 r Maximum volume: 𝑉 = 8 × ( 𝑟 3 ) 3 = 8 𝑟 3 3 3 V=8×( 3 r ) 3 = 3 3 8r 3 Question 2 Assume 𝑇 T is a linear transformation. Find the standard matrix of 𝑇 T: 𝑇 ( 𝑒 1 ) = ( 7 , 1 , 7 , 1 ) , 𝑇 ( 𝑒 2 ) = ( − 8 , 5 , 0 , 0 ) T(e 1 )=(7,1,7,1),T(e 2 )=(−8,5,0,0) Answer: Matrix 𝑀 M is created with columns 𝑇 ( 𝑒 1 ) T(e 1 ) and 𝑇 ( 𝑒 2 ) T(e 2 ): 𝑀 = [ 7 − 8 1 5 7 0 1 0 ] M= 7 1 7 1 −8 5 0 0 Question 3 Find equivalent polar coordinates for ( 0 , − 3 ) (0,−3). Answer: Convert rectangular to polar using 𝑟 = 𝑥 2 + 𝑦 2 r= x 2 +y 2 , 𝜃 = tan ⁡ − 1 ( 𝑦 / 𝑥 ) θ=tan −1 (y/x). Calculations: 𝑟 = 3 , 𝜃 = 27 0 ∘ r=3,θ=270 ∘ Equivalent polar coordinates: ( 3 , 27 0 ∘ ) (3,270 ∘ ). Question 4 Convert 𝑥 2 + 𝑦 2 + 8 𝑥 = 0 x 2 +y 2 +8x=0 to polar form. Answer: Substituting 𝑥 = 𝑟 cos ⁡ 𝜃 , 𝑦 = 𝑟 sin ⁡ 𝜃 x=rcosθ,y=rsinθ: 𝑟 2 + 8 𝑟 cos ⁡ 𝜃 = 0 r 2 +8rcosθ=0 Factorize: 𝑟 ( 𝑟 + 8 cos ⁡ 𝜃 ) = 0 r(r+8cosθ)=0 Polar form: 𝑟 = − 8 cos ⁡ 𝜃 r=−8cosθ Question 5 Graph the solution of the system of nonlinear inequalities: 𝑥 2 + 𝑦 2 ≥ 9 , 𝑥 2 + 𝑦 2 ≤ 25 x 2 +y 2 ≥9,x 2 +y 2 ≤25 Answer: The solution is the region between two circles: Inner circle: 𝑥 2 + 𝑦 2 = 9 x 2 +y 2 =9 (radius = 3). Outer circle: 𝑥 2 + 𝑦 2 = 25 x 2 +y 2 =25 (radius = 5). Shade the region between these circles.