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Math Help - gradient! urgent

  1. #1
    seb
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    gradient! urgent

    Hi someone please help with this one

    a bicycle track can be modelled by the curve with the equation
    y= 2.4x + 3.1x - 2x + 1.5, 0 ≤ x ≤ 5

    where x and y are respectively the horizontal and vertical distances (in km) from the origin
    (i) find the gradient at the start and end of the track
    (i) find when the gradient of the path is zero

    THANKS
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by seb View Post
    Hi someone please help with this one

    a bicycle track can be modelled by the curve with the equation
    y= 2.4x + 3.1x - 2x + 1.5, 0 ≤ x ≤ 5

    where x and y are respectively the horizontal and vertical distances (in km) from the origin
    (i) find the gradient at the start and end of the track
    (i) find when the gradient of the path is zero

    THANKS
    y = 2.4x^3 + 3.1x^2 - 2x + 1.5

    So
    \frac{dy}{dx} = 7.2x^2 + 6.2x - 2

    i) Find the gradient for x = 0 and x = 5.

    ii) Find when \frac{dy}{dx} = 0

    -Dan
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  3. #3
    MHF Contributor
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    Gradient is slope, is dy/dx, is y'.

    y = 2.4x + 3.1x - 2x + 1.5, 0 ≤ x ≤ 5

    y' = 7.2x^2 +6.2x -2

    (i) find the gradient at the start and end of the track

    At the start, where x=0.
    m = y' = -2 -----------------answer
    So the track starts going downward.

    At the end, where x=6.
    m = y' = 7.2(5^2) +6.2(5) -2 = 209 -----------------answer
    So the track ends going upward very steeply.

    (ii) find when the gradient of the path is zero

    So, when y' = 0,
    7.2x^2 +6.2x -2 = 0
    x = {-6.2 +,-sqrt[(6.2)^2 -4(7.2)(-2)]} / (2*7.2)
    x = {-6.2 +,-9.8} / 14.4
    x = 0.25 or -1.1111

    Only the 0.25 is in the domain, so, at x=0.25, the gradient is zero. ------answer.
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