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Tangets and lines and things

  1. May 13, 2008 #1
    damn annoying question. I'm basically new to this stuff, simple HS maths to you guys.

    f(x) = X/2 -4/x^2 + 1

    thats the line.

    dy/dx is 1/2 + 8/x^3 that was first question.


    2nd = the normal at point (2/1) on the curve above cuts the x and y axes at A and B respectively. Calculate the length of AB leaving answer in simplified surd form.

    ok.

    gradient of tangent at that point is 1/2 + 8/16 = 3/2

    normal is -2/3.

    ok, now from here i'm not sure where to go or what to do =( halp =)
     
  2. jcsd
  3. May 13, 2008 #2

    dx

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    The gradient is 1/2 + 1.

    Now, do you know what the general equation of a line is? And what are the two pieces of information that you have about the required line?
     
  4. May 13, 2008 #3
    y = mx + c

    y = X/2 -4/x^2 + 1

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

    1 = 3/2 - 4/4 + 1 -wrrroong


    I have no idea what i'm doing you see. that's the problem =(
     
  5. May 13, 2008 #4

    dx

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    Ok. I'll give you a hint. If you have

    1. a point on the line
    2. its slope

    You'll be able to determine what the line is.
     
  6. May 13, 2008 #5
    I don't know what to do!
     
  7. May 13, 2008 #6
    Yes you do.

    You know the equation for a straight line, and you know there are three points, one of which is the intersection with the curve (which you have), the other two will pop out when you fit the line to the curve, with the slope you also already have.
     
  8. May 13, 2008 #7
    .... i dont know what to do with the straight line or these points, i dont know where to put the gradient because the line doesnt look like mx + c, i dont know if i have to change some of the X into the number 2 because of the coordinates..


    i seriously dont know what the hell im doing.
     
  9. May 13, 2008 #8
    DeanBH,
    the slope-intercept form of a line equation (y = mx + c) is only useful if you already have the y - intercept (the 'c' in the equation). If you have a point that is not the y - intercept, which is frequently the case in calculus problems, then the point - slope form of the line equation is much more handy. Try to find it in your text. It will contain the terms x1 & y1 - substitute the coordinates of the point you know for these.

    BTW - because it can be used with any point, the point - slope equation is worth commiting to memory. Even if the point you know is the y - intercept, since x = 0, it reduces to the slope - intercept form.
     
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