Rate of Change Problem: Finding Derivative of Temperature Function T(t)

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Homework Help Overview

The discussion revolves around finding the derivative of a temperature function, T(t), which describes the temperature of food in a fridge over time. The function is defined as T(t) = 15 - 3t + 4/(t - 1) for the interval 0 ≤ t ≤ 5, and the goal is to determine the rate of change of temperature after one hour.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the need to find the derivative of the temperature function to determine the rate of change. There is a focus on the application of the quotient rule in differentiation. Some participants express concern about the function's behavior at t = 1, questioning the continuity and implications for the derivative at that point.

Discussion Status

The conversation is ongoing, with participants exploring the implications of the function's behavior near t = 1. There is recognition of a potential issue with the function's continuity, and some participants suggest considering values slightly greater than 1 for evaluation.

Contextual Notes

Participants note the peculiar nature of the temperature function, particularly its behavior suggesting an increase in temperature when placed in a refrigerator, prompting further examination of the problem's setup.

Incog
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Homework Statement



Suppose that t hours after a piece of food is put in the fridge its temperature (in Celsius) is

T(t) = 15 - 3t + [tex]\frac{4}{t - 1}[/tex]

where 0 [tex]\leq[/tex] t [tex]\leq[/tex] 5.

Find the rate of change of temperature after one hour.


The Attempt at a Solution



Since it's asking for rate of change, I'm guessing I have to find the derivative of the equation with respect to t.


T(t) = 15 - 3t + [tex]\frac{4}{t - 1}[/tex]

T`(t) = 0 - 3 + [tex]\frac{0(t - 1) - 1(4)}{(t-1)^{2}}[/tex] (Quotient Rule)

T`(t) = -3 + [tex]\frac{0 - 4}{(t-1)^{2}}[/tex]

T`(t) = -3 + [tex]\frac{-4}{(t-1)^{2}}[/tex]

T`(t) = -3 - [tex]\frac{4}{(t-1)^{2}}[/tex]


Would I just plug in 1 after this?
 
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Incog said:

Homework Statement



Suppose that t hours after a piece of food is put in the fridge its temperature (in Celsius) is

T(t) = 15 - 3t + [tex]\frac{4}{t - 1}[/tex]

where 0 [tex]\leq[/tex] t [tex]\leq[/tex] 5.

Find the rate of change of temperature after one hour.


The Attempt at a Solution



Since it's asking for rate of change, I'm guessing I have to find the derivative of the equation with respect to t.
Don't guess! The derivative of a function is its rate of change!


T(t) = 15 - 3t + [tex]\frac{4}{t - 1}[/tex]

T`(t) = 0 - 3 + [tex]\frac{0(t - 1) - 1(4)}{(t-1)^{2}}[/tex] (Quotient Rule)

T`(t) = -3 + [tex]\frac{0 - 4}{(t-1)^{2}}[/tex]

T`(t) = -3 + [tex]\frac{-4}{(t-1)^{2}}[/tex]

T`(t) = -3 - [tex]\frac{4}{(t-1)^{2}}[/tex]


Would I just plug in 1 after this?
That's what you would like to do- but this function has serious problem at t= 1. Do you remember that, in order to have a derivative at a point, the function must be continuous there? Are you sure you have copied the problem correctly? That's a very strange temperature function! Isn't it peculiar that the temperature of the food goes up when it is put in the refridgerator?
 
Last edited by a moderator:
Yes, I checked and checked again and that is the equation.

What if I were to plug in a value slightly greater than 1? Would that give me the rate of change after one hour?
 
Last edited:
Well, I just don't know what to say about a refrigerator where the temperature goes to infinity in one hour!
 

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