Modeling with exponential and logarithmic functions help?

HallsofIvy says, t is the number of minutes after death.In summary, we can use the Newton's Law of Cooling equation to solve for the time of death in a given situation. By plugging in the given temperatures and times, we can solve for the negative constant, k, and then use that value to determine the time of death by solving for t in the equation.
  • #1
camp3r101
3
0
Modeling with exponential and logarithmic functions help?

Homework Statement


Use Newton's Lay of Cooling, T = C + (T0 - C)e-kt, to solve this exercise. At 9:00 A.M., a coroner arrived at the home of a person who had died during the night. The temperature of the room was 70 degrees F, and at the time of death the person had a body temperature of 98.6 degrees F. The coroner took the body's temperature at 9:30 A.M., at which time it was 85.6 degrees F, and again at 10:00 A.M., when it was 82.7 degrees F. At what time did the person die?


Homework Equations


T = C + (T0 - C)e-kt
If you do not know what the variable's mean...these are their meanings:
T = temperature of a heated object
C = constant temperature of the surrounding medium (the ambient temp)
T0 = initial temperature of the heated object
k = negative constant associated with the cooling object
t = time (in minutes)

The Attempt at a Solution


I tried solving for k by doing:
Steps (I plugged all the values into their corresponding places):

85.6 = 70 + (98.6 - 70)e-k(30)
15.6 = 28.6e-30k
0.5454545455 = e-30k
ln(0.5454545455) = ln(e-30k)
ln(0.5454545455) = -30k
k = (ln(0.5454545455)/(-30))
k = 0.0202045268


After getting this...i do not know what to do next...of even if I did the process of anything correctly as yet.
 
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  • #2


camp3r101 said:

Homework Statement


Use Newton's Lay of Cooling, T = C + (T0 - C)e-kt, to solve this exercise. At 9:00 A.M., a coroner arrived at the home of a person who had died during the night. The temperature of the room was 70 degrees F, and at the time of death the person had a body temperature of 98.6 degrees F. The coroner took the body's temperature at 9:30 A.M., at which time it was 85.6 degrees F, and again at 10:00 A.M., when it was 82.7 degrees F. At what time did the person die?


Homework Equations


T = C + (T0 - C)e-kt
If you do not know what the variable's mean...these are their meanings:
T = temperature of a heated object
C = constant temperature of the surrounding medium (the ambient temp)
T0 = initial temperature of the heated object
k = negative constant associated with the cooling object
t = time (in minutes)
Excellent! Writing them all out like this is a very good idea- even on your home work paper. Not every one uses the same symbols. One criticism- you say "t= time (in minutes)" but do not say "time from what?" From what you write later, t appears to be the time (in minutes) after 9:00. You should say that.

The Attempt at a Solution


I tried solving for k by doing:
Steps (I plugged all the values into their corresponding places):

85.6 = 70 + (98.6 - 70)e-k(30)
NO. You don't know what the temperature of the body was at 9:00- you are told that " at the time of death the person had a body temperature of 98.6 degrees F" and you don't know when the time of death was. If you had said "t= time (in minutes) after death until the Coroner arrived, at 9:00", you would have
85.6= 70+ (98.6- 70)e-k(t+ 30)
and
82.7= 70+ (98.6- 70)e-k(t+ 60)

That gives you two equations to solve for k and t. Subtracting t minutes from 9:00 gives you the time of death.

15.6 = 28.6e-30k
0.5454545455 = e-30k
ln(0.5454545455) = ln(e-30k)
ln(0.5454545455) = -30k
k = (ln(0.5454545455)/(-30))
k = 0.0202045268


After getting this...i do not know what to do next...of even if I did the process of anything correctly as yet.
 
  • #3


HallsofIvy said:
Excellent! Writing them all out like this is a very good idea- even on your home work paper. Not every one uses the same symbols. One criticism- you say "t= time (in minutes)" but do not say "time from what?" From what you write later, t appears to be the time (in minutes) after 9:00. You should say that.

Oh yeah...sorry I didn't think about that. It helps to clarify those things sometimes. I'll remember.

NO. You don't know what the temperature of the body was at 9:00- you are told that " at the time of death the person had a body temperature of 98.6 degrees F" and you don't know when the time of death was. If you had said "t= time (in minutes) after death until the Coroner arrived, at 9:00", you would have
85.6= 70+ (98.6- 70)e-k(t+ 30)
and
82.7= 70+ (98.6- 70)e-k(t+ 60)

Ohhh I see now haha

That gives you two equations to solve for k and t. Subtracting t minutes from 9:00 gives you the time of death.

Which values of k ant t do I use to find the time of death?
Do I use both; do I add both t values together and then subtract from 9:00 ??
 
  • #4


HallsofIvy said:
If you had said "t= time (in minutes) after death until the Coroner arrived, at 9:00", you would have
85.6= 70+ (98.6- 70)e-k(t+ 30)
and
82.7= 70+ (98.6- 70)e-k(t+ 60)

camp3r101: while what HallsofIvy said is correct, I wouldn't do it this way, because you defined T0 as the initial temperature of the object... that is to say, the temperature at t = 0. I would define t as simply the number of minutes after death. So the equations I would write are
[tex]85.6 = 70 + (98.6 - 70)e^{-kt}[/tex]
for 09:30am, and
[tex]82.7 = 70 + (98.6 - 70)e^{-k(t + 30)}[/tex]
for 10:00am.

camp3r101 said:
Which values of k ant t do I use to find the time of death?
Do I use both; do I add both t values together and then subtract from 9:00 ??
No. (The hints I'm going to give may be too much, so apologies to the mods in advance.)

Take my 1st equation and simplify it a bit:
[tex]15.6 = 28.6e^{-kt}[/tex]

Take my 2nd equation and simplify:
[tex]12.7 = 28.6e^{-k(t + 30)}[/tex]
[tex]12.7 = 28.6e^{-kt - 30k}[/tex]

Use property of exponents:
[tex]12.7 = \frac{28.6e^{-kt}}{e^{30k}}[/tex]

Notice the numerator? That equals 15.6, from the first equation, so we substitute:
[tex]12.7 = \frac{15.6}{e^{30k}}[/tex]

Now I'll shut up. Solve for k, and plug this in back into
[tex]15.6 = 28.6e^{-kt}[/tex]
and solve for t. Remember that this t value corresponds to 09:30am, so subtract to find the time of death.
 
Last edited:
  • #5


eumyang said:
camp3r101: while what HallsofIvy said is correct, I wouldn't do it this way, because you defined T0 as the initial temperature of the object... that is to say, the temperature at t = 0. I would define t as simply the number of minutes after death. So the equations I would write are
[tex]85.6 = 70 + (98.6 - 70)e^{-kt}[/tex]
for 09:30am, and
[tex]82.7 = 70 + (98.6 - 70)e^{-k(t + 30)}[/tex]
for 10:00am.


No. (The hints I'm going to give may be too much, so apologies to the mods in advance.)

Take my 1st equation and simplify it a bit:
[tex]15.6 = 28.6e^{-kt}[/tex]

Take my 2nd equation and simplify:
[tex]12.7 = 28.6e^{-k(t + 30)}[/tex]
[tex]12.7 = 28.6e^{-kt - 30k}[/tex]

Use property of exponents:
[tex]12.7 = \frac{28.6e^{-kt}}{e^{30k}}[/tex]

Notice the numerator? That equals 15.6, from the first equation, so we substitute:
[tex]12.7 = \frac{15.6}{e^{30k}}[/tex]

Now I'll shut up. Solve for k, and plug this in back into
[tex]15.6 = 28.6e^{-kt}[/tex]
and solve for t. Remember that this t value corresponds to 09:30am, so subtract to find the time of death.

DANNGGG! You guys are soo much help! Much thanks to you!
I see what steps I didn't take. Thanks again! :)
 

FAQ: Modeling with exponential and logarithmic functions help?

1. What is the purpose of using exponential and logarithmic functions in modeling?

Exponential and logarithmic functions are often used in modeling to represent patterns of growth or decay. They can also be used to describe relationships between two variables that change at a constant rate.

2. How do you determine which function to use in a modeling situation?

The type of function used in a modeling situation depends on the specific scenario and the relationship between the variables. Exponential functions are typically used for growth or decay situations, while logarithmic functions are used for inverse relationships.

3. What are some real-life applications of exponential and logarithmic functions?

Exponential and logarithmic functions are commonly used in fields such as finance, biology, and physics. They can be used to model population growth, compound interest, radioactive decay, and more.

4. Can you give an example of how to solve a problem using exponential or logarithmic functions?

Sure! Let's say we want to model the growth of a bacteria population that doubles every hour. We can use the exponential function P(t) = P0(2t), where P0 is the initial population and t is the time in hours. If we start with 100 bacteria, after 3 hours we would have P(3) = 100(23) = 800 bacteria.

5. Are there other types of functions that can be used for modeling?

Yes, there are many different types of functions that can be used for modeling, such as linear, quadratic, and trigonometric functions. The choice of function depends on the specific situation and the type of relationship between the variables being modeled.

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