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Math Help - Differentiation Help!

  1. #1
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    Differentiation Help!

    Can someone help me on the following questions cos im just so stuck!

    1.Find the equations of the tangent and normal to the curve xy=4 at the point where x=2

    2.Given that y=ax^2+bx-a^2 has a gradient -4 at (-2,-13), find the possible values for a and b.

    3.The radius, r cm of a circle at time 't's is given by r=(t^3)/3 -2t, t >&= 0
    Find the rate at which the radius is changing when t=1 and when t=2.

    Any help will be much appreciated! Thanks in advance
    Last edited by Confuzzled?; November 11th 2005 at 10:59 AM.
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  2. #2
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    Quote Originally Posted by Confuzzled?
    1.Find the equations of the tangent and normal to the curve xy=4 at the point where x=2
    I'll start with the first one. In order to find the tangent line and the normal line, you need a point and a slope. The point you can find by plugging in your x value, and how would you find the slope? How can you rewrite xy=4 into something easier to differentiate?
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  3. #3
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    Quote Originally Posted by Confuzzled?
    2.Given that y=ax^2+bx-a^2 has a gradient -4 at (-2,-13), find the possible values for a and b.

    3.The radius, r cm of a circle at time 't's is given by r=(t^3)/3 -2t, t >&= 0
    For the second one, I don't know why a function of one variable makes a reference to gradient, as one normally sees that with functions of three variables. But, I guess you could define the gradient for f(x) as:

    (1) \nabla{f(x)}=\frac{\partial{f}}{\partial{x}}\hat{x  }

    and work from there. It's really just the derivative for this function.

    And for the third one, I don't even see a question being asked. Inforrmation is provided, what are they looking for?
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  4. #4
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    Gradient is a synonym for "slope of the tangent at the
    point in question" in this context.

    RonL
    Last edited by CaptainBlack; November 11th 2005 at 10:20 AM. Reason: clarification
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  5. #5
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    Quote Originally Posted by Confuzzled?
    Can someone help me on the following questions cos im just so stuck!

    1.Find the equations of the tangent and normal to the curve xy=4 at the point where x=2

    2.Given that y=ax^2+bx-a^2 has a gradient -4 at (-2,-13), find the possible values for a and b.

    3.The radius, r cm of a circle at time 't's is given by r=(t^3)/3 -2t, t >&= 0

    Any help will be much appreciated! Thanks in advance
    sorry guys i was kinda in a rush. Anyways the actual question for no. 3 was

    3. Find the rate at which the radius is changing when t=1 and when t=2.

    And Jameson, I converted xy=4 into y=4x^-1 (is that right?) I worked with that yet I didnt get the right answer in the end (the answers are at the back)
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  6. #6
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    oh and for question 2 I think have been able to attempt a similar question where I had to form a simultaneous equation to find a and b but I wasnt sure how I could do this for that question.
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  7. #7
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    "1.Find the equations of the tangent and normal to the curve xy=4 at the point where x=2 "

    Slope of tangent line is dy/dx.
    Slope of normal line is the negative reciprocal of dy/dx.

    xy = 4
    Differentiate both sides with respect to x,
    x[dy/dx] +y[dx/dx] = 0
    dy/dx = -y/x

    When x=2:
    xy = 4
    2y = 4
    y = 2 also.
    So, dy/dx = -y/x = -2/2 = -1 -----slope of tangent line.
    And so, slope of normal line = -1/(-1) = 1 ----------------***

    Then, for equation of tangent line at point (2,2), using the point-slope form,
    (y-y1) = m(x-x1)
    y-2 = -1(x-2) --------answer.
    Or, in its slope-intercept form or y=mx+b,
    y -2 = -x +2
    y = -x +4 -------answer.

    The normal line at (2,2):
    y-2 = 1(x-2) -----point-slope form, answer.
    Or, y = x -------slope-intercept form, answer.

    -----------------------
    " 2.Given that y=ax^2+bx-a^2 has a gradient -4 at (-2,-13), find the possible values for a and b. "

    gradient is dy/dx, so,
    y = ax^2 +bx -a^2 -------(i)
    dy/dx = 2ax +b = -4
    At point (-2,-13),
    2a(-2) +b = -4
    -4a +b = -4
    b = 4a -4 ------(1)

    Two unknowns, a&b, we need another equation.
    y = ax^2 +bx -a^2 -------(i)
    At point (-2,-13),
    -13 = a(-2)^2 +b(-2) -a^2
    -13 = 4a -2b -a^2 ----------(2)
    Substitute into (2) the b from (1)
    -13 = 4a -2(4a -4) -a^2
    -13 = 4a -8a +8 -a^2
    a^2 +4a -21 = 0
    (a+7)(a-3) = 0
    a = -7 or 3 -------------------answer.

    When a= -7,
    b = 4a -4 = 4(-7) -4 = -32
    When a = 3,
    b = 4(3) -4 = 8
    Hence, b = -32 or 8 -----------------answer.

    ------------------
    " 3.The radius, r cm of a circle at time 't's is given by r=(t^3)/3 -2t, t >&= 0 "

    If I can understand that, I can answer it.
    Last edited by ticbol; November 11th 2005 at 10:36 AM.
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  8. #8
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    Ah, so there is the correction for #3.

    I am using a mac Powerbook now because my Dell laptop was damaged by ligtning storm a few days ago. I am new at Mac OS so I am really very slow in answering now. (I don't even know how to use the Clipboard!).
    The "Quote" feature above doesn't seem to work with Mac.

    " 3. Find the rate at which the radius is changing when t=1 and when t=2 "
    " 3.The radius, r cm of a circle at time 't's is given by r=(t^3)/3 -2t, t >&= 0 "

    r = (t^3)/3 -2t
    Differentiate both sides with respect to time t,
    dr/dt = t^2 -2 ------rate of change of the radius.
    So,
    When t=1, dr/dt = (1)^2 -2 = -1 -------answer.
    When t=2, dr/dt = (2)^2 -2 = 2 -------answer.
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