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

2024
Views:

444

Pages:

4
Author:

Beckett Gray

Question 1 Answer: Given: Small circle radius 𝑟 = 2 𝑚 r=2m Big circle radius 𝑅 = 4 𝑚 R=4m Area of the small circle: 𝐴 small = 𝜋 𝑟 2 = 22 7 ( 2 ) 2 = 88 7 m 2 ≈ 12.57 m 2 . A small =πr 2 = 7 22 (2) 2 = 7 88 m 2 ≈12.57m 2 . Area of the big circle: 𝐴 big = 𝜋 𝑅 2 = 22 7 ( 4 ) 2 = 352 7 m 2 ≈ 50.29 m 2 . A big =πR 2 = 7 22 (4) 2 = 7 352 m 2 ≈50.29m 2 . Area of the shaded region: 𝐴 shaded = 𝐴 big − 𝐴 small = 352 7 − 88 7 = 264 7 m 2 ≈ 37.71 m 2 . A shaded =A big −A small = 7 352 − 7 88 = 7 264 m 2 ≈37.71m 2 . Question 2 Write the area 𝐴 A of a circle as a function of its circumference 𝐶 C: Answer 1: Start with the formulas: 𝐶 = 2 𝜋 𝑟 C=2πr 𝐴 = 𝜋 𝑟 2 A=πr 2 Solve for 𝑟 r in terms of 𝐶 C: 𝑟 = 𝐶 2 𝜋 . r= 2π C . Substitute 𝑟 r into the area equation: 𝐴 = 𝜋 ( 𝐶 2 𝜋 ) 2 . A=π( 2π C ) 2 . Simplify: 𝐴 = 𝐶 2 4 𝜋 . A= 4π C 2 . Answer 2: Using the same formulas 𝐴 = 𝜋 𝑟 2 A=πr 2 and 𝐶 = 2 𝜋 𝑟 C=2πr: Express 𝑟 r in terms of 𝐶 C: 𝐶 2 = 4 ( 𝜋 2 ) 𝑟 2 ⟹ 𝑟 2 = 𝐶 2 4 𝜋 2 . C 2 =4(π 2 )r 2 ⟹r 2 = 4π 2 C 2 . Substitute into the area equation: 𝐴 = 𝜋 𝑟 2 = 𝜋 ( 𝐶 2 4 𝜋 2 ) . A=πr 2 =π( 4π 2 C 2 ). Simplify: 𝐴 = 𝐶 2 4 𝜋 . A= 4π C 2 . Question 3 The minute hand of a clock sweeps out a sector. Find how fast the area of the sector increases. Answer 1: The area of a sector is given by: 𝐴 = 𝜃 2 𝜋 × 𝜋 𝑟 2 = 𝜃 𝑟 2 2 . A= 2π θ ×πr 2 = 2 θr 2 . Differentiate with respect to time 𝑡 t: 𝑑 𝐴 𝑑 𝑡 = 1 2 𝑟 2 𝑑 𝜃 𝑑 𝑡 . dt dA = 2 1 r 2 dt dθ . Given: 𝑟 = 4 in r=4in 𝑑 𝜃 𝑑 𝑡 = 2 𝜋 60 = 𝜋 30 radians/min . dt dθ = 60 2π = 30 π radians/min. Substituting the values: 𝑑 𝐴 𝑑 𝑡 = 1 2 ( 4 ) 2 ( 𝜋 30 ) . dt dA = 2 1 (4) 2 ( 30 π ). Simplify: 𝑑 𝐴 𝑑 𝑡 = 16 𝜋 60 = 4 𝜋 15 in 2 / min . dt dA = 60 16π = 15 4π in 2 /min. Answer 2: Using the same principles: Sector area: 𝐴 = 𝜃 𝑟 2 2 . A= 2 θr 2 . Differentiate: 𝑑 𝐴 𝑑 𝑡 = 1 2 𝑟 2 𝑑 𝜃 𝑑 𝑡 . dt dA = 2 1 r 2 dt dθ . Given: 𝑟 = 4 in , 𝑑 𝜃 𝑑 𝑡 = 𝜋 30 r=4in, dt dθ = 30 π . Substituting: 𝑑 𝐴 𝑑 𝑡 = 1 2 ( 4 ) 2 𝜋 30 . dt dA = 2 1 (4) 2 30 π . Simplify: 𝑑 𝐴 𝑑 𝑡 = 16 𝜋 60 = 4 𝜋 15 in 2 / min . dt dA = 60 16π = 15 4π in 2 /min. Question 4 Find the value of 𝑥 x for two tangents drawn from a point. Answer 2: If two tangents are drawn to a circle from the same external point, they are equal. Given: 𝐴 𝐵 = 𝐴 𝐷 AB=AD. Set up the equation: 2 𝑥 + 7 = 5 𝑥 − 8. 2x+7=5x−8. Solve for 𝑥 x: 2 𝑥 + 7 = 5 𝑥 − 8. 2x+7=5x−8. Combine like terms: 7 + 8 = 5 𝑥 − 2 𝑥 . 7+8=5x−2x. 3 𝑥 = 15 ⟹ 𝑥 = 3. 3x=15⟹x=3. Final Answer: 𝑥 = 3. x=3.