Engineering Mathematics Miscellaneous


Engineering Mathematics Miscellaneous

Engineering Mathematics

  1. A calculator has accuracy up to 8 digits after decimal place. The value of

    when evaluated using this calculator by trapezoidal method with 8 equal intervals, to 5 significant digits is









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    h =
    2π - 0
    =
    π
    84

    y0 = sin (0) = 0
    y1 = sin
    π
    = 0.7071
    4

    y2 = sin
    π
    = 1
    2

    y3 = sin
    = 0.7071
    4

    y4 = sin (π) = 0
    y5 = sin
    = -0.7071
    4

    y6 = sin
    = -1
    4

    y7 = sin
    = -0.7071
    4

    y8 = sin
    = 0
    4

    Trapezoidal rule
    h
    [ y0 + yn + 2(y1 + y2 + ... + yn - 1) ]
    2


    [(0 + 0) + 2 (0.7071 + 1 + .7071 + -0.7071 - 0.7071)] = 0

    Correct Option: A

    h =
    2π - 0
    =
    π
    84

    y0 = sin (0) = 0
    y1 = sin
    π
    = 0.7071
    4

    y2 = sin
    π
    = 1
    2

    y3 = sin
    = 0.7071
    4

    y4 = sin (π) = 0
    y5 = sin
    = -0.7071
    4

    y6 = sin
    = -1
    4

    y7 = sin
    = -0.7071
    4

    y8 = sin
    = 0
    4

    Trapezoidal rule
    h
    [ y0 + yn + 2(y1 + y2 + ... + yn - 1) ]
    2


    [(0 + 0) + 2 (0.7071 + 1 + .7071 + -0.7071 - 0.7071)] = 0


  1. The accuracy of Simpson's rule quadrature for a step size h is









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    NA

    Correct Option: C

    NA



  1. An explicit forward Euler method is used to numerically integrate the differential equation
    dy
    = y
    dt

    using a time step of 0.1. With the initial condition y(0) = 1, the value of y(10) computed by this method is ______ (correct to two decimal places).









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    General formula
    yn + 1 = yn + hf (tn , yn)
    For n = 0, y1 = y0 + hf (t0, y0)
    = y0 + hy0
    = 1 + 0.1 (1)
    y1 = 1.1
    For n = 1, y2 = y1 + hf (t1, y1)
    = y1 + hy1
    = 1.1 + 0.1 (1.1)
    y2 = 1.21
    For n = 2, y3 = y2 + hf (t2, y2)
    = y2 + hy2
    = 1.21 + 0.1 × 1.21
    y3 = 1.331
    For n = 3, y4 = y3 + hf (t3, y3)
    = y3 + hy3
    = 1.331 + 0.1 × 1.331
    y4 = 1.4641
    For n = 4, y5 = y4 + hf (t4, y4)
    = y4 + hy4
    = 1.4641 + 0.1 × (1.4641)
    y5 = 1.61051
    For n = 5, y6 = y5 + hf (t5, y5)
    = y5 + hy5
    = 1.61051 + 0.1 × 1.61051
    y6 = 1.771561
    For n = 6, y7 = y6 + hf (t6, y6)
    = y6 + hy6
    = 1.771561 + 0.1 × 1.771561 = 1.9487
    For n = 7, y8 = y7 + hf (t7, y7)
    = y7 + hy7
    = 1.9487 + 0.1 × (1.9487)
    y8 = 2.14357
    For n = 8, y9 = y8 + hf (t8, y8)
    = y8 + hy8
    = 2.14357 + 0.1 × 2.14357
    y9 = 2.3579
    For n = 9, y10 = y9 + hf (t9, y9)
    = y9 + hy9
    = 2.3579 + 0.1 × (2.3579)
    y10 = 2.5937

    Correct Option: C

    General formula
    yn + 1 = yn + hf (tn , yn)
    For n = 0, y1 = y0 + hf (t0, y0)
    = y0 + hy0
    = 1 + 0.1 (1)
    y1 = 1.1
    For n = 1, y2 = y1 + hf (t1, y1)
    = y1 + hy1
    = 1.1 + 0.1 (1.1)
    y2 = 1.21
    For n = 2, y3 = y2 + hf (t2, y2)
    = y2 + hy2
    = 1.21 + 0.1 × 1.21
    y3 = 1.331
    For n = 3, y4 = y3 + hf (t3, y3)
    = y3 + hy3
    = 1.331 + 0.1 × 1.331
    y4 = 1.4641
    For n = 4, y5 = y4 + hf (t4, y4)
    = y4 + hy4
    = 1.4641 + 0.1 × (1.4641)
    y5 = 1.61051
    For n = 5, y6 = y5 + hf (t5, y5)
    = y5 + hy5
    = 1.61051 + 0.1 × 1.61051
    y6 = 1.771561
    For n = 6, y7 = y6 + hf (t6, y6)
    = y6 + hy6
    = 1.771561 + 0.1 × 1.771561 = 1.9487
    For n = 7, y8 = y7 + hf (t7, y7)
    = y7 + hy7
    = 1.9487 + 0.1 × (1.9487)
    y8 = 2.14357
    For n = 8, y9 = y8 + hf (t8, y8)
    = y8 + hy8
    = 2.14357 + 0.1 × 2.14357
    y9 = 2.3579
    For n = 9, y10 = y9 + hf (t9, y9)
    = y9 + hy9
    = 2.3579 + 0.1 × (2.3579)
    y10 = 2.5937


  1. Gauss seidel method is used to solve the following equations (as per the given order):
    x1 + 2x2 + 3x3 = 1
    2x1 + 3x2 + x3 = 1
    3x1 + 2x2 + x3 = 1
    Assuming initial guess as x1 = x2 = x3 = 0, the value of x3 after the first iteration is ______.









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    The equations are
    x1 + 2x2 + 3x3 = 5
    2x1 + 3x2 + x3 = 1
    3x1 + 2x2 + x3 = 3
    By pivoting we get

    x1 =
    3 - 2x2 - x3
    .....(1)
    3

    x2 =
    1 - 2x1 - x3
    .....(2)
    3

    x3 =
    5 - x1 - 2x2
    .....(3)
    3

    When we put x2 = 0, x3 = 0 in equaiton (1), x1 = 1
    put x1 = 1, x3 = 0 in equation (3), x2 = – .333
    put x1 = 1, x2 = – .333 in equation (3), x3 = 1.555

    Correct Option: A

    The equations are
    x1 + 2x2 + 3x3 = 5
    2x1 + 3x2 + x3 = 1
    3x1 + 2x2 + x3 = 3
    By pivoting we get

    x1 =
    3 - 2x2 - x3
    .....(1)
    3

    x2 =
    1 - 2x1 - x3
    .....(2)
    3

    x3 =
    5 - x1 - 2x2
    .....(3)
    3

    When we put x2 = 0, x3 = 0 in equaiton (1), x1 = 1
    put x1 = 1, x3 = 0 in equation (3), x2 = – .333
    put x1 = 1, x2 = – .333 in equation (3), x3 = 1.555



  1. Consider an ordinary differential equation
    dw
    = 0 - iz
    dz

    If x = x0 at t = 0 , the increment in x calculated using Runge-Kutta fourth order multi-step method with a step size of ∆t = 0.2 is









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    Given ,
    dx
    = 4t + 4
    dt

    x = x0 at t = 0
    n = 0.2
    Calculate x(0.2) value
    K1 = f(t0, x0) = f(0, x 0) = 4
    K2 = f t0 +
    h
    , x0 +
    1
    K1h
    22

    = f(0 + 0.1, x0 + 0.4)
    = f(0.1, x0 + 0.4) = 4(0.1) + 4 = 4.4
    K3 = f x0 +
    h
    , x0 +
    1
    K2h
    22

    = f[t0 + 0.1, x0 + (2.2)(0.2)]
    = f(0.1, x0 + 0.44) = 4(0.1) + 4 = 4.4
    K4 = f(t0 + h, x0 + K3 h)
    = f(0 + 0.2, x0 + 0.88)
    = f(0.2, x0 + 0.88)
    = 4(0.2) + 4 = 4.8
    x(0.2) = x1 = x0 +
    h
    (K1 + 2K2 + 2K3 + K4)
    6

    = x0 +
    0.2
    [ 4 + 2(4.4) + 2(4.4) + (4.8) ]
    6

    = x0 +
    0.2
    (4 + 8.8 + 8.8 + 4.8)
    6

    = x0 + 0.88
    Increment as x = x 1 – x0 = x0 + 0.88 – x0 = 0.88

    Correct Option: D

    Given ,
    dx
    = 4t + 4
    dt

    x = x0 at t = 0
    n = 0.2
    Calculate x(0.2) value
    K1 = f(t0, x0) = f(0, x 0) = 4
    K2 = f t0 +
    h
    , x0 +
    1
    K1h
    22

    = f(0 + 0.1, x0 + 0.4)
    = f(0.1, x0 + 0.4) = 4(0.1) + 4 = 4.4
    K3 = f x0 +
    h
    , x0 +
    1
    K2h
    22

    = f[t0 + 0.1, x0 + (2.2)(0.2)]
    = f(0.1, x0 + 0.44) = 4(0.1) + 4 = 4.4
    K4 = f(t0 + h, x0 + K3 h)
    = f(0 + 0.2, x0 + 0.88)
    = f(0.2, x0 + 0.88)
    = 4(0.2) + 4 = 4.8
    x(0.2) = x1 = x0 +
    h
    (K1 + 2K2 + 2K3 + K4)
    6

    = x0 +
    0.2
    [ 4 + 2(4.4) + 2(4.4) + (4.8) ]
    6

    = x0 +
    0.2
    (4 + 8.8 + 8.8 + 4.8)
    6

    = x0 + 0.88
    Increment as x = x 1 – x0 = x0 + 0.88 – x0 = 0.88