University:
California State University, Northridge
Course:
MATH 262 | Introduction to Linear Algebra
Academic year:

2024
Views:

320

Pages:

4
Author:

Beckett Gray

Question 1 Given the following formula, solve for 𝑟 r. 𝐶 = 2 𝜋 𝑟 C=2πr: A. 𝑟 = 2 𝜋 𝐶 𝜋 r= π 2πC B. 𝑟 = 𝐶 2 r= 2 C C. 𝑟 = 𝐶 2 𝜋 r= 2π C D. 𝑟 = 𝐶 − 2 𝜋 r=C−2π Answer 1 To solve an equation, we use inverse operations. Given: 𝐶 = 2 𝜋 𝑟 C=2πr, to isolate 𝑟 r, divide both sides by 2 𝜋 2π: 𝑟 = 𝐶 2 𝜋 . r= 2π C . Therefore, the correct solution is C. 𝑟 = 𝐶 2 𝜋 r= 2π C . Answer 2 The formula is 𝐶 = 2 𝜋 𝑟 C=2πr. Divide both sides by 2 𝜋 2π: 𝑟 = 𝐶 2 𝜋 . r= 2π C . Answer 3 We need to solve for 𝑟 r: Start with 𝐶 = 2 𝜋 𝑟 C=2πr. Divide by 2 𝜋 2π: 𝑟 = 𝐶 2 𝜋 . r= 2π C . Question 2 Find the measurements of the greatest area possible rectangle that can be enclosed in a circle. Answer We apply the Pythagorean theorem. The diagonal of the rectangle equals the diameter of the circle, 2 𝑟 2r. 𝑥 2 + 𝑦 2 = ( 2 𝑟 ) 2 ⟹ 𝑦 2 = 4 𝑟 2 − 𝑥 2 . x 2 +y 2 =(2r) 2 ⟹y 2 =4r 2 −x 2 . Let 𝐴 A be the area of the rectangle: 𝐴 = 𝑥 𝑦 = 𝑥 4 𝑟 2 − 𝑥 2 . A=xy=x 4r 2 −x 2 . Differentiate 𝐴 A with respect to 𝑥 x: 𝐴 ′ ( 𝑥 ) = 4 𝑟 2 − 𝑥 2 − 𝑥 2 4 𝑟 2 − 𝑥 2 . A ′ (x)= 4r 2 −x 2 − 4r 2 −x 2 x 2 . Solve 𝐴 ′ ( 𝑥 ) = 0 A ′ (x)=0: 4 𝑟 2 − 2 𝑥 2 = 0 ⟹ 𝑥 = 2 𝑟 . 4r 2 −2x 2 =0⟹x= 2 r. Substitute 𝑥 = 2 𝑟 x= 2 r into 𝐴 A: 𝐴 = 2 𝑟 ⋅ 4 𝑟 2 − ( 2 𝑟 ) 2 = 2 𝑟 2 . A= 2 r⋅ 4r 2 −( 2 r) 2 =2r 2 . Result: The greatest area is 2 𝑟 2 2r 2 . Question 3 Find the area of the figure composed of a parallelogram and two semicircles. Answer Parallelogram Area: Base = 32 =32, height = 11 =11: 𝐴 parallelogram = 32 × 11 = 352. A parallelogram =32×11=352. Semicircle Areas: Diameter of each semicircle = 16 =16, so radius = 8 =8. Total area of two semicircles: 𝐴 semicircles = 2 × 1 2 𝜋 𝑟 2 = 𝜋 ( 8 ) 2 = 64 𝜋 ≈ 201.14. A semicircles =2× 2 1 πr 2 =π(8) 2 =64π≈201.14. Total Area: 𝐴 total = 352 + 201.14 = 553.14 square units . A total =352+201.14=553.14square units. Question 4 If the atomic radius of lead is 0.175 nm 0.175nm, find the volume of its unit cell. Answer Given 𝑅 = 0.175 nm R=0.175nm, for FCC (Face-Centered Cubic): 𝑉 𝑐 = 16 𝑅 3 2 . V c =16R 3 2 . Substitute 𝑅 = 0.175 × 1 0 − 9 m R=0.175×10 −9 m: 𝑉 𝑐 = 16 × ( 0.175 × 1 0 − 9 ) 3 × 2 . V c =16×(0.175×10 −9 ) 3 × 2 . Simplify: 𝑉 𝑐 = 1.21 × 1 0 − 28 m 3 . V c =1.21×10 −28 m 3 . Question 5 Difference between a trapezoid and a rhombus: Answer 1 A trapezoid has at least one pair of parallel sides, while a rhombus has two pairs of parallel sides. The sides of a rhombus are all equal, whereas a trapezoid may have sides of different lengths. Answer 2 A rhombus is a special parallelogram where all sides are equal. A trapezoid is a quadrilateral with at least one pair of parallel sides.