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

2024
Views:

207

Pages:

3
Author:

Beckett Gray

Question 1 Circle P is dilated to form circle P'. Answer Circle P is dilated to form circle P'. This means circle P will be similar to circle P' due to dilation. The scale factor determines if it gets bigger or smaller. Option (A) is correct because the ratio of the diameter to circumference remains 1 : 𝜋 1:π for both circles. Question 2 Write an equation to solve the following problems involving circles: a) The circumference of Fabians Answer (a) Circumference 𝐶 = 125.6 mm C=125.6mm: Use the formula: 𝐶 = 2 𝜋 𝑟 . C=2πr. Solve for 𝑟 r: 125.6 = 2 × 3.1428 × 𝑟 ( 𝜋 = 3.1428 ) . 125.6=2×3.1428×r(π=3.1428). 𝑟 = 125.6 6.2856 ≈ 19.982 mm . r= 6.2856 125.6 ≈19.982mm. Diameter: 𝑑 = 2 𝑟 = 2 × 19.982 = 39.964 mm . d=2r=2×19.982=39.964mm. (b) Circumference 𝐶 = 102.6 feet C=102.6feet: Solve for 𝑟 r: 102.6 = 2 × 3.1428 × 𝑟 . 102.6=2×3.1428×r. 𝑟 = 102.6 6.2856 ≈ 16.324 feet . r= 6.2856 102.6 ≈16.324feet. Diameter: 𝑑 = 2 𝑟 = 2 × 16.323 = 32.646 feet . d=2r=2×16.323=32.646feet. (c) Area 𝐴 = 21.98 yd 2 A=21.98yd 2 : Use the formula 𝐶 = 2 𝜋 𝑟 C=2πr. Solve for 𝑟 r: 21.98 = 2 × 3.1428 × 𝑟 . 21.98=2×3.1428×r. 𝑟 ≈ 3.454 yd . r≈3.454yd. Area: 𝐴 = 𝜋 𝑟 2 = 3.1428 × ( 3.454 ) 2 ≈ 3749.39 yd 2 . A=πr 2 =3.1428×(3.454) 2 ≈3749.39yd 2 . Question 3 Which area corresponds to circle B's diameter given circle A's area and diameter? Answer Given: 𝑑 𝐴 = 20 inches , 𝐴 𝐴 = 300 inches 2 d A =20inches,A A =300inches 2 . 𝑑 𝐵 = 60 inches d B =60inches. The area 𝐴 A of a circle is proportional to the square of its diameter: 𝐴 𝐵 𝐴 𝐴 = ( 𝑑 𝐵 𝑑 𝐴 ) 2 . A A A B =( d A d B ) 2 . Substitute the values: 𝐴 𝐵 300 = ( 60 20 ) 2 . 300 A B =( 20 60 ) 2 . 𝐴 𝐵 300 = 3 2 = 9. 300 A B =3 2 =9. 𝐴 𝐵 = 300 × 9 = 2700 inches 2 . A B =300×9=2700inches 2 . The correct answer is 2700 inches 2 inches 2 . Question 4 Find the equation for the surface obtained by rotating 𝑦 = 𝑥 2 y=x 2 about the y-axis. Answer When the curve 𝑦 = 𝑥 2 y=x 2 is rotated about the y-axis: Each trace parallel to the 𝑥 𝑧 xz-plane forms a circle with radius 𝑥 x. Since 𝑥 2 = 𝑦 x 2 =y, the surface equation becomes: 𝑥 2 + 𝑧 2 = 𝑦 . x 2 +z 2 =y. Adding the rotation: 𝑥 2 + 𝑧 2 = 𝑦 2 . x 2 +z 2 =y 2 . Question 5 Find the relative error in the surface area of a sphere when 𝐶 = 84 cm C=84cm with a possible error of 0.5 cm 0.5cm. Answer Circumference Formula: 𝐶 = 2 𝜋 𝑟 ⟹ 𝑟 = 𝐶 2 𝜋 = 84 2 𝜋 ≈ 13.37 cm . C=2πr⟹r= 2π C = 2π 84 ≈13.37cm. Surface Area: 𝑆 = 4 𝜋 𝑟 2 . S=4πr 2 . Differentiate 𝑆 S: 𝑑 𝑆 𝑑 𝑟 = 8 𝜋 𝑟 . dr dS =8πr. Error in Surface Area: Δ 𝑆 = 8 𝜋 𝑟 Δ 𝑟 . ΔS=8πrΔr. Substitute 𝑟 = 13.37 r=13.37 and Δ 𝑟 = 0.5 Δr=0.5: Δ 𝑆 = 8 𝜋 ( 13.37 ) ( 0.5 ) ≈ 27 cm 2 . ΔS=8π(13.37)(0.5)≈27cm 2 . Relative Error: Δ 𝑆 𝑆 = 27 4 𝜋 ( 13.37 ) 2 ≈ 0.012 ( 1.2 % ) . S ΔS = 4π(13.37) 2 27 ≈0.012(1.2%). Final Relative Error: 0.012 0.012 or 1.2%.