SOLUTIONS MANUAL for Control Systems Engineering 7th Edition by Norman Nise. ISBN 9781118800638

SOLUTIONS MANUAL for Control Systems Engineering 7th Edition by Norman Nise. ISBN 9781118800638 Trapezoid = 31.516 Use the trapezoid rule with n = 4 to approximate the area between the curve f(x) = x^3 -x and the x-axis from x = 3 to x =7ANS-Trapezoid = (1/2)(1)[(3^3 -3) +2(4^3 -4) +2(5^3 -5) +2(6^3 -6) +(7^3 -7)] Trapezoid = 570 Use the trapezoid rule with n = 4 to approximate the area between the curve f(x) = x^2 + 1 and the x-axis from x = 3 to x =7ANSTrapezoid = (1/2)(1)[(3^2 +1) +2(4^2 +1) +2(5^2 +1) +2(6^2 +1) +(7^2 +1)] Trapezoid = 110 Use the trapezoid rule with n = 6 to approximate the area between the curve f(x) = 3x^3 - 4 and the x-axis from x = 0 to x =6ANSTrapezoid = (1/2)(1)[(3(0^3) - 4) + 2(3(1^3) - 4) + 2(3(2^3) - 4) + 2(3(3^3) - 4) + 2(3(4^3) - 4) + 2(3(5^3) - 4) + (3(6^3) - 4)] Trapezoid = 975 Use the trapezoid rule with n = 4 to approximate the area between the curve f(x) = 2x^3 - 1 and the x-axis from x = 2 to x =6ANSTrapezoid = (1/2)(1)[(2(2^3)-1) +2(2(3^3)-1) +2(2(4^3)-1) +(2(5^3)-1) +(2(6^3)-1)] Trapezoid = 527.5 Find the area of the polar equation r = 4cos θANS-A = (1/2) ∫(4cos θ)^2dθ from [0, 2pi] plug into calculator A = 8pi + 8sin(pi) Find the area inside the first curve R = 2 + sin θ and outside the second curve r = 3sin θANS-Find the positions of intersection by setting the equations equal to each other and solving for θ. Find the midpoint Riemann Sum of cos(x^2) with n = 4, from [0, 2]ANS-Mid S4 = (1)(1/2)[cos(.25^2) + cos(.75^2) + cos(1.25^2) + cos(1.75^2) Mid S4 = (1)(1/2)[cos(.625) + cos(.5625) + cos(1.5625) cos(3.0625)]Mid S4 = .824 If the function f is continuous for all real numbers and if f(x) = (x^2-7x +12)/(x -4) when x ≠ 4 then f(4) =ANS-Factor numerator so f(x) = (x-3)(x-4)/(x-4) = x-3 f(4)=4-3 f(4) = 1 If f(x) = (x^2+5) if x < 2, & f(x) = (7x -5) if x ≥ 2 for all real numbers x, which of the following must be true? I. f(x) is continuous everywhere. II. f(x) is differentiable everywhere. III. f(x) has a local minimum at x = 2.ANS-At f(2) both the upper and lower piece of the discontinuity is 9 so the function is continuous everywhere. At f'(2) the upper piece is 4 and lower piece is 7 so f(x) is not differentiable everywhere. Since the slopes of the function on the left and right are both positive the function cannot have a local minimum or maximum at x= 2. Only I is true. For the function f(x) = (ax^3-6x), if x ≤ 1, & f(x) = (bx^2+4), x > 1 to be continuous and differentiable, a = .....ANS2. lim from the left and right are both 8 3. lim f(x) as x approaches 4 is 8 which equals f(4) for the function to be continuous f(1) has to equal f(1): a(1^3) -6(1) = b(1^2) +4 a -6 = b +4 b=a-10 for the functions to be differentiable f'(1) has to equal f'(1): 3a(1^2) -6 = 2b(1) 3a -6 = 2b plug b from the first equation in to find a: 3a -6 = 2(a -10) a = -14 Find k if f(x) = (k) at x = 4 and f(x) = ((x^2 -16)/(x-4))ANS-1. f(4) exists and is equal to 8 k must equal 8 If f(x) is continuous and differentiable and f(x) = (ax^4 +5x) for x ≤ 2, & f(x)= (bx^2 -3) for x > 2 , then b =...ANS-Plug x = 2 into both pieces. f(x) = (16a +10) for x ≤ 2, & (4b -6) for x > 2 They must be equal to be continuous 16a +10 = 4b -6 a=.25b-1 Take the derivative of both pieces of this function and plug in x = 2 f(x) = (32a +5) for x ≤ 2, & f(x) = (4b -3) for x > 2 They must be equal to be differentiable 32a +5 = 4b -3 plug in the first equation to find b 32(.25b-1)+5= 4b-3