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-