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

Incog
Messages
17
Reaction score
0

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 + \frac{4}{t - 1}

where 0 \leq t \leq 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 + \frac{4}{t - 1}

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

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

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

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


Would I just plug in 1 after this?
 
Physics news on Phys.org
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 + \frac{4}{t - 1}

where 0 \leq t \leq 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 + \frac{4}{t - 1}

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

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

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

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


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!
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top