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Draw the secants of a curve (line?)

  1. Jul 28, 2009 #1
    1. The problem statement, all variables and given/known data
    An automobile's distance is given by d = 30t (t in seconds, d in metres):

    "Draw the secants from the point (2, 60) to each of the points on the graph ending at t = 3, 2.5, 2.1, 1.01"

    "Determine the equations of each of the secants above."


    2. Relevant equations



    3. The attempt at a solution

    To my understanding, d = 30t is a line.
    According to the text, "A line that intersects a curve in two places is called a secant".
    How can a line have any secants?
     
    Last edited: Jul 28, 2009
  2. jcsd
  3. Jul 28, 2009 #2

    Mark44

    Staff: Mentor

    A curve in its most liberal interpretation can also include a straight line.
     
  4. Jul 28, 2009 #3

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    How can a point end? :confused:
     
  5. Jul 28, 2009 #4
    Uh I'm not sure, but I'm guessing it's asking me to draw the secant from (2, 60) to these points (3, d(3)), (2.5, d(2.5)), etc.

    Though I still don't see the point since they're all gonna overlap anyway lol.
     
  6. Jul 28, 2009 #5

    ideasrule

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    Perhaps the point is to show that for a straight line, all secant lines have the same slope.
     
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