Expert: Paul Klarreich - 10/4/2008

Question
QUESTION: I have a question on transformation of functions, it goes like this.   
A hot brick is removed from a kiln and set on the floor to cool. Let t be time in minutes after the brick was removed. The difference, D(t), between the brick's temperature, initially 350F, and room temperature, 70F, decays exponentially over time at a rate of 3 per minute. The brick's temperature, H(t), is a transformation of D(t).  Find a formula for H(t).   
This is what i was thinking it is but I am not sure:  
H(t)= 280e^-0.03t  
Thank You

ANSWER: Questioner:   Robert
Category:  Calculus
Private:  No
 
Subject:  Transformations of Functions
Question:  I have a question on transformation of functions, it goes like this.   
A hot brick is removed from a kiln and set on the floor to cool. Let t be time in minutes after the brick was removed. The difference, D(t), between the brick's temperature, initially 350F, and room temperature, 70F, decays exponentially over time at a rate of 3 per minute. The brick's temperature, H(t), is a transformation of D(t).  Find a formula for H(t).   
This is what i was thinking it is but I am not sure:  
H(t)= 280e^-0.03t  
Thank You
..................................................
Hi, Robert,

I think you are missing a few details.  It is an exponential function, but your analysis, as Spock would say, is flawed.

D(t) is indeed an exponential function.  [We will worry about H(t) later.]  It is given by:

D(t) = D0 e^(-kt).

and your D0 and k must be computed from facts in the example.  But your sentence: "The difference, D(t),... decays exponentially ... at a rate of 3 per minute." is not correct.  D(t) decays at a VARIABLE rate.  So you must say something like "INITIALLY at a rate of 3 [degrees] per minute".

Then you can compute a value of k, based on D'(t), which, I think, will go as follows:

[keep in mind that  e^(0) = 1.]

First, D0 = 280, the difference at  t = 0, so:

D(t) = 280 e^(-kt)

Now D'(t) = - 280k e^(-kt)

The phrase "INITIALLY at a rate of 3 [degrees] per minute" means D'(0) = - 3.  [It's cooling, right?  So that is a decrease of 3 degrees per min.]
- 3 = - 280k, and  k = 3/280.

Ok, then  D(t) = 280 e^(-3/280 t), and  3/280 is not 0.03, more like 0.0107 or so.

Finally,  H(t) = D(t) + 70.


---------- FOLLOW-UP ----------

QUESTION: Hi thank you for your response, but i think i left something out.  I think spell check removed the percent sign after the 3 on the sentence you were refering to.  It should read: The difference, D(t), between the brick's temperature, initially 350F, and room temperature, 70F, decays exponentially over time at a rate of 3[percent] per minute.
This changes everything right?  
Thank you again.

Answer
ANSWER: Questioner:   Robert
Category:  Calculus
Private:  No

Subject:  Transformations of Functions
Question:  I have a question on transformation of functions, it goes like this.   
A hot brick is removed from a kiln and set on the floor to cool. Let t be time in minutes after the brick was removed. The difference, D(t), between the brick's temperature, initially 350F, and room temperature, 70F, decays exponentially over time at a rate of 3 per minute. The brick's temperature, H(t), is a transformation of D(t).  Find a formula for H(t).   
This is what i was thinking it is but I am not sure:  
H(t)= 280e^-0.03t  
Thank You
..................................................
Hi, Robert,

I think you are missing a few details.  It is an exponential function, but your analysis, as Spock would say, is flawed.

D(t) is indeed an exponential function.  [We will worry about H(t) later.]  It is given by:

D(t) = D0 e^(-kt).

and your D0 and k must be computed from facts in the example.  But your sentence: "The difference, D(t),... decays exponentially ... at a rate of 3 per minute." is not correct.  D(t) decays at a VARIABLE rate.  So you must say something like "INITIALLY at a rate of 3 [degrees] per minute".

Then you can compute a value of k, based on D'(t), which, I think, will go as follows:

[keep in mind that  e^(0) = 1.]

First, D0 = 280, the difference at  t = 0, so:

D(t) = 280 e^(-kt)

Now D'(t) = - 280k e^(-kt)

The phrase "INITIALLY at a rate of 3 [degrees] per minute" means D'(0) = - 3.  [It's cooling, right?  So that is a decrease of 3 degrees per min.]
- 3 = - 280k, and  k = 3/280.

Ok, then  D(t) = 280 e^(-3/280 t), and  3/280 is not 0.03, more like 0.0107 or so.

Finally,  H(t) = D(t) + 70.


---------- FOLLOW-UP ----------

QUESTION: Hi thank you for your response, but i think i left something out.  I think spell check removed the percent sign after the 3 on the sentence you were referring to.  It should read: The difference, D(t), between the brick's temperature, initially 350F, and room temperature, 70F, decays exponentially over time at a rate of 3[percent] per minute.
This changes everything right?  
Thank you again.
.................................

Hi, again, Robert,

Yes, it does make a difference, but you still have to state that it decays exponentially over time at an INITIAL rate of 3[percent] per minute.  If it always decayed at 3%/minute it would become zero in 33.3 (or so) minutes; that would not be exponential decay at all, but linear.

As to the example, now you would write:

D'(t) = - 280k e^(-kt)

D'(0) = -0.03(280) = 280k
and  k = 0.03.

D(t) = 280 e^(-0.03t), etc.