Finding the Equation of A Graph using Tangents

  • Thread starter healey.cj
  • Start date
  • #1
11
0
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
 

Answers and Replies

  • #2
11
0
Thanks for replying!!! :-)

yeah, the derivatives you have stated are correct...
 
  • #3
11
0
What the...where did the other post just go?
 
  • #4
verty
Homework Helper
2,182
198
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.
 
  • #5
verty
Homework Helper
2,182
198
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.
 
  • #6
HallsofIvy
Science Advisor
Homework Helper
41,847
966
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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