University:
High School
Course:
High School Algebra
Academic year:

2021
Views:

457

Pages:

5
Author:

Alessandro Reed

1. The function v(t) = 12t^2 - 16t is the velocity in m/sec of a particle moving along the x-axis, where t is measured in seconds. Use analytic methods to find the 1 particle's displacement for 0 < t < 5. ∫₀⁵ (12t² - 16t) dt = 4t³ - 8t² |₀⁵ = 4(5)³ - 8(5)² - [4(0)³ - 8(0)²] = 500 - 200 = 300 meters 2. What is the area of the region shown, bounded by the parabola x = y²-2y and the curve x = -y³ + 2y + 4 Find where the graphs intersect y² - 2y = -y³ + 2y + 4 0 = -y³ - y² + 4y + 4 0 = -y²(y + 1) + 4(y + 1) 0 = (y + 1)(-y² + 4) 0 = (y + 1)(2 - y)(2 + y) y = -1, y = 2, y = -2 use these bounds Easier with respects to "y" Top - Bottom ∫[-1, 2] (-y³ + 2y + 4 - (y² - 2y)) dy = ∫[-1, 2] (-y³ - y² + 4y + 4) dy = -¼y⁴ - ⅓y³ + 2y² + 4y |₂₋₁ = (-4 - 8/3 + 8 + 8) - (-¼ + ⅓ + 2 - 4) = 112/12 + 23/12 = 135/12 = 11.25 units² 3. What is the area of the region enclosed by y = sin 2x and y = cos x for -π/2 ≤ x ≤ π/6? A = ∫[-π/2, π/6] (cos x - sin 2x) dx = sin x + ½ cos 2x |_(-π/2)^(π/6) = [sin (π/6) + ½ cos 2(π/6)] - [sin (-π/2) + ½ cos 2(-π/2)] = [½ + ½ (½)] - [-1 + ½ (-1)] = ¾ - (-3/2) = 15/4 units² 4. The base of a solid is the region between the line y = 4 and the parabola y = x². The cross sections of the solid that are perpendicular to the x-axis are semicircles. What is the volume of the solid? Finding the intersections 4 = x² 0 = x² - 4 0 = (x - 2)(x + 2) x = 2, x = -2 but since the graph is symmetric at x=0 one bounds are 0 to 2 Remember that the formula we came up with was (π/8)b² V = ∫₀² (π/8)b² dx = (π/4) ∫₀² (4 - x²)² dx = (π/4) ∫₀² (16 - 8x² + x⁴) dx = (π/4) [16x - (8/3)x³ + (1/5)x⁵]₀² = (π/4) [32 - (64/3) + (32/5)] - (π/4)(0) = 64π/15 units³ 5. A region is bounded by y = √x, y = x - 2, and y = 0. What is the volume of the generated region is rotated around the x-axis? creating a Disk V = π ∫[R² - r²] dx [(f(x) - center) - (g(x) - center)]² dx Split the solid into two parts: 0 ≤ x ≤ 2 2 ≤ x ≤ 4 V = π ∫₀² [ (√x - 0)² - (0 - 0)² ] dx + π ∫₂⁴ [ (√x - 0)² - (x - 2 - 0)² ] dx = π ∫₀² [ x - 0 ] dx + π ∫₂⁴ [ x - (x² - 4x + 4) ] dx = π ∫₀² [ x ] dx + π ∫₂⁴ [ -x² + 5x - 4 ] dx = π [ ½x² ]₀² + π [ -⅓x³ + (5/2)x² - 4x ]₂⁴ = π [(4/2) - 0] + π [ (-64/3 + 40 - 16) - (-8/3 + 4 - 8) ] = 2π + 10π/3 = 16π/3 units³ 6. A certain spring obeys Hooke's Law and requires a force of 10 N to stretch it 8 cm beyond its natural length. How much work would be needed to stretch the spring 12 cm beyond its natural length? Work = Force * Distance Force = k * extension 10 = k * 8 5/4 = k (spring constant) W = ∫₀¹² (5/4)x dx = (5/8)x² |₀¹² = (5/8)(144) - (5/8)(0) = 90 Ncm = 90 J of work needed