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Math Help - In need of verification

  1. #1
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    In need of verification

    Last years midterm
    Q1B:
    find derivative with respect to x
    y = (4x+3)^1/2 / 2x
    u = (4x+3)
    u' = 4

    T'B-TB'/B^2
    1/2(4x+3)-^1/2 * 4
    2(4x+3)^-1/2

    (2(4x+3)-^1/2 * 2x - (4x+3)^1/2 * 2) / (4x^2)
    =
    (4x(4x+3)^-1/2 - 2(4x+3)^1/2) / (4x^2)

    (however, this derivative is in respect to y. Can someone still tell me if the derivative is correct at least, thanks)

    Q2:
    Find equation of line normal to
    y = 2x^3 - 3x - 1 at x = 1
    y - y1 = m(x - x1)
    y' = m
    y' = 6x^2 - 3
    y' = 6(1)^2 - 3 = 36 - 3 = 33
    y = 2(1)^3 - 3(1) - 1 = 8 - 3 - 1 = 4
    x = 1, y = 4
    y - 4 = 33(x - 1)
    y - 4 = 33x - 33
    y - 37 = 33x

    Q4:
    A rectangular area divided into two areas, wall inside parallel to 2 opposite sides
    Find max dimensions for display if total fencing available is 240 m.

    (sketch: 4 sides with one partitioning wall)

    2x + 3y
    240 = 2x + 3y
    240 - 2x = 3y
    80 - 2/3x = y

    A = x * y
    A = x * (80 - 2/3x)
    A = 80x - 2/3x^2
    A' = 80 - 4/3x

    x = 60 m
    y = 40 m

    Q6:
    Evaluate integrals:
    A:
    (x3 - x^1/2) / x
    (x^3)/x - (x^1/2)/x
    x^2 - x-^1/2
    (1/3x^3) - (2x^1/2)

    B:
    8x(2x^2-5)^4
    (2x^2-5)(2x^2-5)(2x^2-5)(2x^2-5)
    4x^4 - 10x^2 - 10x^2 + 25
    4x^4 - 20x^2 + 25
    integrate
    16x^8 - 80x^6 + 100x^4 - 80x^2 + 400x^4 + 500 + 100x^4 - 500x^2 + 625
    16x^8 - 80x^6 + 500x^4 - 580x^2 + 1125
    simplify
    16/9x^9 - 80/7x^7 + 100x^5 + 580/3x^3 + 1125x

    Q7:
    Find area under curve
    y = 1/(x + 1)^1/2
    Between x = 3 and x = 8
    (x + 1)^-1/2
    2(x+1)^1/2
    x = 3: 2(3+1)^1/2: y = 4
    x = 8: 2(8+1)^1/2: y = 6

    4 - 6 = -2

    Thanks for looking at it.
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  2. #2
    MHF Contributor
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    Re: In need of verification

    Hey togo.

    Q7 is almost right: you did the subtraction the wrong way around (it should be 6-4 = 2).

    Q6 is correct.

    Q4 is correct.

    Q2 is not right since 6(1)^2 - 3 = 6 - 3 = 3 instead of 33 - 3 = 33. Also the equation of a normal is m1*m2 = -1 which means the gradient of the normal is -1/3. Try doing this one again.

    Q1B is correct.
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  3. #3
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    Re: In need of verification

    Q 1B is correct but please note that it is derivative with respect to x and not y as mentioned at the end of the question
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  4. #4
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    Re: In need of verification

    Thanks, wonderful to know where work is needed.
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