Finding the Equation of A Graph using Tangents

  • Thread starter Thread starter healey.cj
  • Start date Start date
  • Tags Tags
    Graph
Click For Summary

Homework Help Overview

The discussion revolves around deriving a quadratic equation that models the cooling of coffee over time, based on given temperature data at specific time intervals. The original poster presents a problem involving the use of tangents and derivatives in relation to the quadratic equation.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the use of derivatives and gradients at specific points, with some suggesting the need to derive coefficients from three specific points on the graph. There is also a debate about the interpretation of the "middle" point among the provided data.

Discussion Status

Some participants have confirmed the correctness of the derivatives mentioned, while others are exploring different methods to derive the quadratic equation. There is a recognition of the need to clarify the use of specific points and the implications of the problem's instructions.

Contextual Notes

The problem specifies using three separate points from a set of fourteen data points, leading to questions about how to identify the "middle" point and whether derivatives can be used in this context.

healey.cj
Messages
11
Reaction score
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 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 celsius
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
 
Physics news on Phys.org
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.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
3K
Replies
16
Views
7K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 8 ·
Replies
8
Views
4K
  • · Replies 6 ·
Replies
6
Views
4K
Replies
2
Views
1K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 2 ·
Replies
2
Views
12K
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K