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Math Help - Velocity and Normal line problems

  1. #1
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    Velocity and Normal line problems

    1.Suppose that "f" is an objects position at time "x" in seconds.

    x is greater than or equal to 0 AND is less than or equal to 3


    Position function:
    f(x) = 2x^3 - 6x^2 + 4x + 5

    Velocity:

    f'(x) = 6x^2 - 12x + 4

    Acceleration:

    f''(x) = 12x - 12

    I'm asked to find when the object is stationary. Now correct me if I'm wrong but that should be when the slopes of the position function = 0 or rather the x-intercepts of the velocity function.

    However i'm having trouble finding the x-intercepts

    2. Find the equation of a normal line (perpendicular to a tangent line) of
    y = \sqrt{x} + 1/\sqrt[3]{x} at x = 1


    What I did was solve for y which is y=2

    So I have a point now (1,2)

    Then I find the derivative of the function which is
    y' = (1/2)(x^(-1/2)) + (-1/3)(x^(-4/3))

    I then plug in 1 which is y' = 1/6 at x =1

    I use the point slope equation

    y - 2 = 1/6(x-1)
    y = 1/6x -1/6 + 2
    y = 1/6x + 11/6

    Since the equation of a normal line is the negative reciprocal:
    y = -1/(1/6x + 11/6)

    Is this right or not I just want to confirm

    Thanks in advance
    Last edited by GameTheory; June 9th 2010 at 01:15 AM.
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  2. #2
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    Quote Originally Posted by GameTheory View Post
    1.Suppose that "f" is an objects position at time "x" in seconds.

    x is greater than or equal to 0 AND is less than or equal to 3


    Position function:
    f(x) = 2x^3 - 6x^2 + 4x + 5

    Velocity:

    f'(x) = 6x^2 - 12x + 4

    Acceleration:

    f"(x) = 12x - 12

    I'm asked to find when the object is stationary. Now correct me if I'm wrong but that should be when the slopes of the position function = 0 or rather the x-intercepts of the velocity function.

    However i'm having trouble finding the x-intercepts
    Your derivatives are correct. The object is stationary when the velocity f'(x) = 0.

    So 6x^2 - 12x + 4 = 0

    x^2 - 2x + \frac{2}{3} = 0

    x^2 - 2x + (-1)^2 - (-1)^2 + \frac{2}{3} = 0

    (x - 1)^2 - 1 + \frac{2}{3} = 0

    (x - 1)^2 - \frac{1}{3} = 0

    (x - 1)^2 = \frac{1}{3}

    x - 1 = \pm \sqrt{\frac{1}{3}}

    x - \frac{3}{3} = \frac{\pm \sqrt{3}}{3}

    x = \frac{3 \pm \sqrt{3}}{3}.


    So the object is stationary when x = \frac{3 - \sqrt{3}}{3} and when x = \frac{3 + \sqrt{3}}{3}.
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  3. #3
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    Quote Originally Posted by GameTheory View Post
    2. Find the equation of a normal line (perpendicular to a tangent line) of
    y = \sqrt{x} + 1/\sqrt[3]{x} at x = 1


    What I did was solve for y which is y=2

    So I have a point now (1,2)

    Then I find the derivative of the function which is
    y' = (1/2)(x^(-1/2)) + (-1/3)(x^(-4/3))

    I then plug in 1 which is y' = 1/6 at x =1

    I use the point slope equation

    y - 2 = 1/6(x-1)
    y = 1/6x -1/6 + 2
    y = 1/6x + 11/6

    Since the equation of a normal line is the negative reciprocal:
    y = -1/(1/6x + 11/6)

    Is this right or not I just want to confirm

    Thanks in advance
    Your evaluation of y is correct. So (1, 2) lies on the normal line.

    Now you need the gradient.

    If m_T is the gradient of line tangent to the curve at (x, y) = (1, 2),

    then m_N = -\frac{1}{m_T}.


    So to find m_T...

    y = \sqrt{x} + \frac{1}{\sqrt[3]{x}}

     = x^{\frac{1}{2}} + x^{-\frac{1}{3}}


    \frac{dy}{dx} = \frac{1}{2}x^{-\frac{1}{2}} - \frac{1}{3}x^{-\frac{4}{3}}

     = \frac{1}{2\sqrt{x}} - \frac{1}{3\sqrt[3]{x^4}}


    m_T = \frac{dy}{dx}|_{x = 1}

     = \frac{1}{2\sqrt{1}} - \frac{1}{3\sqrt{1^4}}

     = \frac{1}{2} - \frac{1}{3}

     = \frac{1}{6}.


    Therefore m_N = -\frac{1}{m_T}

     = -\frac{1}{\frac{1}{6}}

     = -6.


    So now you have the gradient of the normal line.


    The normal line is of the form

    y = mx + c

    Substituting (x, y) = (1, 2) and m = -6 gives

    2 = -6(1) + c

    2 = -6 + c

    c = 8.



    Therefore, the equation of the normal line is

    y = -6x + 8.
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    Thanks alot, I realize now where I made my mistake in no.2.
    However I would like to know what this process is:
     (-1)^2 - (-1)^2

    of

    <br /> <br />
x^2 - 2x + (-1)^2 - (-1)^2 + \frac{2}{3} = 0<br />

    I don't understand how you just plugged in those -1s
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  5. #5
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    Quote Originally Posted by GameTheory View Post
    Thanks alot, I realize now where I made my mistake in no.2.
    However I would like to know what this process is:
     (-1)^2 - (-1)^2

    of

    <br /> <br />
x^2 - 2x + (-1)^2 - (-1)^2 + \frac{2}{3} = 0<br />

    I don't understand how you just plugged in those -1s
    Completing the square.
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  6. #6
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    1.Suppose that "f" is an objects position at time "x" in seconds.

    x is greater than or equal to 0 AND is less than or equal to 3


    Position function:
    f(x) = 2x^3 - 6x^2 + 4x + 5

    Velocity:

    f'(x) = 6x^2 - 12x + 4

    Acceleration:

    f''(x) = 12x - 12

    I am also asked to find when the object is moving the fastest. How would I go about in doing this?
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  7. #7
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    Quote Originally Posted by GameTheory View Post
    1.Suppose that "f" is an objects position at time "x" in seconds.

    x is greater than or equal to 0 AND is less than or equal to 3


    Position function:
    f(x) = 2x^3 - 6x^2 + 4x + 5

    Velocity:

    f'(x) = 6x^2 - 12x + 4

    Acceleration:

    f''(x) = 12x - 12

    I am also asked to find when the object is moving the fastest. How would I go about in doing this?
    The object is moving the fastest when the velocity is maximised.

    You have a velocity function f'(x). How would you find the point where it is maximised?
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  8. #8
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    x interval is between 0 and 3. So I suppose f'(3) = maximized velocity?

    so f'(3) = 22 so the object is moving fastest at x = 3?
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