Finding the Equation of A Graph using Tangents

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To find the quadratic equation that models the cooling of coffee over time, use the three specified points from the provided temperature data: the first point (0, 83), a midpoint between the 7th and 8th points, and the last point (50, 38). The equation should be in the form y = ax² + bx + c. By substituting these points into the equation, you will generate three equations to solve for the coefficients a, b, and c. The discussion emphasizes that the tangents and derivatives are not necessary for this specific problem, focusing instead on the direct substitution method. Following these steps will lead to the desired quadratic model.
healey.cj
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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 separate 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, I am just asking for the path i need to take or an idea of what i need to do...

Thanks everyone,
Chris
 
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Thanks for replying! :-)

yeah, the derivatives you have stated are correct...
 
What the...where did the other post just go?
 
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.
 
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.
 
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.
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
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