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Gradient of the secant

  1. Jan 6, 2010 #1
    1. The problem statement, all variables and given/known data

    There are two parts to this problem.

    On the curve 2x^2-5 lie two points P and Q. Let the abscissa of P be "x" and the abscissa of Q be "x+h". No numerical coordinates are given.

    a) State the coordinates of P and Q.

    b) Using these points find the gradient of the secant PQ


    2. Relevant equations



    3. The attempt at a solution

    For the first part of the question Im assuming that the coordinates would be P (x, f(x)) and Q (x+h, f(x+h))

    Now when it comes to finding the gradient of the secant - Can use a table to find a set of coordinates and then plug in the m = y2 - y1 / x2 - x1 ? Not should why the question would ask to use the above coordinates when the gradient of the secant is a different equation?

    Could use some guidence on how I attack this problem. Many thanks.
     
  2. jcsd
  3. Jan 6, 2010 #2

    Dick

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    Just evaluate the coordinates of P and Q in terms of x and h and using the explicit function f(x)=2x^2-5. Find a formula for the gradient of the secant. Nothing else to do.
     
  4. Jan 6, 2010 #3
    Okay - correct me here if necessary

    f(x) = 2x^2 - 5
    f(x+h) = 2(x+h)^2 - 5
    = 2x^2+4xh+2h^2 - 5

    f(x+h) - f(x) = 4xh+2h^2

    f(x+h)-f(x) / h = 4x + 2h

    f'(x) lim = 4x+2h
    approach 0 = 4x gradient of the secant is 4x

    Can you please take a moment to check my calculation to confirm my understanding.
     
  5. Jan 6, 2010 #4

    Dick

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    The calculation is fine. But what you have found is that the gradient of the secant is 4x+2h. Yes, if you take the limit h->0 you get 4x, but that's not the gradient of the secant. That's the gradient of the TANGENT at x. You are jumping ahead.
     
  6. Jan 6, 2010 #5
    Many thanks. So, 4x+2h is the gradient of the secant and I should leave the statement there?
     
  7. Jan 6, 2010 #6

    Dick

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    Sure.
     
  8. Jan 6, 2010 #7
    Many thanks Dick - this is greatly appreciated.

    Cheers
     
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