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Finding the Equation of A Graph using Tangents

  1. Nov 5, 2006 #1
    Hey everyone,

    We've been covering tangents and derivatives etc in class recently but there is a question on the assignment that we've been given that i don't know how to do.

    The question is:
    "A quadratic equation can be fitted to any three points on a cartesian plane. The model for such an equation in y=ax + bx+ c. By substitution three seperate points (The first, middle and last data points) derive the equation of the Quadratic function that models the data."

    The situation is a cup of hot coffee is left to cool over a 50 minute period and we have been given this table of values:

    mins : Degrees Celcius
    0 :83
    5 :76.5
    8 :70.5
    11:65
    15:61
    18:57.5
    24:52.5
    35: 51
    30:47.5
    34:45
    38:43
    42:41
    45:39.5
    50:38

    From there i have plotted the time vs. temp graph and attached 3 tangents to points time (t) = 0, t = 25 & t = 50.

    at t=0 the Rate of Change or gradient was -1.85Degrees/min
    at t=25, the ROC or Gradient was -0.92degrees/min
    at t=50, the ROC or Gradient was -0.13degrees/min

    I'm not asking for you to do this for me, im just asking for the path i need to take or an idea of what i need to do...

    Thanks everyone,
    Chris
     
  2. jcsd
  3. Nov 5, 2006 #2
    Thanks for replying!!! :-)

    yeah, the derivatives you have stated are correct...
     
  4. Nov 5, 2006 #3
    What the...where did the other post just go?
     
  5. Nov 5, 2006 #4

    verty

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    Homework Helper

    I can only think you should use the derivative of the general formula with 2 gradients to solve for a and b, then peg c not with a gradient but a coordinate.
     
  6. Nov 5, 2006 #5

    verty

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    Homework Helper

    Wait, they specifically say "by substituting 3 separate points, derive the formula", so I don't think you are allowed to use the derivative. So just use 3 points to get 3 equations and hopefully they are easy to solve.
     
  7. Nov 6, 2006 #6

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    How about doing what you were told to do in the problem? Is that to complicated?

    You are told to use a quadratic: y= ax2+ bx+ c. You are also told to use a specific 3 points: "(The first, middle and last data points)". The only problem is that there is NO "middle" point since you are given 14 points. I would be inclined to use halfway between the 7th and 8th points. Put those 3 values in for x and y and solve for the coefficients. I can see no reason to worry about tangents.
     
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