Question

A detective finds a murder victim in a room with constant temperature 21◦C. At 5:00am, the body’s temperature was 34.2◦C. One hour later, it was 31.2◦C. Normal body temperature is 37◦C. Assume the body’s temperature, B(t), follows Newton’s Law of Cooling.
(a) Set up and solve the differential equation for B(t).

so far I have the equation B(t) = 16e^(kt)
I was wondering to solve for k
would I set B(1) = 31.2 ?

so I would get

B(1) = 31.2 = 16e^(k(1))
ln(31.2/16) = k

Answer 1

does that make sense? thus,

B(t) = 16e^(ln(31.2/16)(t))

Answer 2

Newton’s Law of Cooling. states that

dT/dt = k(T-T_0)

where T_0 is the temperature of the room

Answer 3

and T is the temperature of the object

Answer 4

Hi Australopithecus :)

How are you dear ?

Answer 5

I'm alright just confused by this question I think I'm going to look for a similar question in the text book as no one seems to be able to assist me I think I may have made a mistake.

Answer 6

Collect data first.

I don't think so it is a too difficult problem. Isn't it?

Answer 7

so B(0) = (T-T_0) = 37-21 = 16
Thus the coefficient 16 is correct

Answer 8

I think my k value is reasonable as well

Answer 9

but perhaps it is wrong because it is the difference between two points

Answer 10

Tell me, what are you trying to find out first?

Answer 11

I am unable to get your concept although your equation for Newton's law of cooling is right.

Answer 12

actually I don't think my k value is reasonable

Answer 13

I'm trying to figure out the time of the murder so when the body's temperature was 34.2, I an then count back from 5:00pm to figure out the time of the murder

Answer 14

\[\Large \frac{dT}{dt}=k(T-T_0)\]
\[\large \int\frac{dT}{(T-T_0)}=\int kdt\]
\[\large T-T_0=Ce^{kt}-------(1)\]
at t=0 i.e. T(t)=T(0)
\[\large T-T_0=C\]
eq(1)=>
\[\large T=(T-T_0)e^{kt}+T_0\]

Answer 15

@Australopithecus kindly bump your question.

So that others could help you out.

Answer 16

I don't know where you are getting the + T_0

dT/dt = k(T - T_0)

dT/(T - T_0) = dtk
=
ln|(T - T_0)| = tk + c
=
T - T_0 = e^(tk + c)
solve for c
Set t=0
16 = e^((0)k + c)
ln(16) = c

Answer 17

look at equation (1) here on the left hand side I have -T_0. So when it moves to right hand side so it will be positive.

Answer 18

the equation shouldn't be set up to = temperature, it is set up to equal time

Answer 19

No. What ever you setup equation equal to. Your first aim is to find the constant k. So it doesn't matter to which equal you setup your equation.

Answer 20

so we are assuming my k value is incorrect because I have a feeling it is

Answer 21

Yes it is incorrect. I didn't calculate. But the procedure you posted is not correct dear.

Answer 22

blah I'm stuck

Answer 23

Now what my friend?

Answer 24

It is so easy. I am sure you are confusing in variables.

Answer 25

I'm sure it is but for some reason I just can't get it

Answer 26

ok I think I might have it, I set up a separate equation to solve for, based on the idea that both share the same rate.

so I have

for the time when it shifts from 34.2 to 31.2 over an hour
K(1) = 31.2 - 21

K(1) = 13.2e^(1)k = 10.2
k = ln(10.2/13.2)


therefore
B(t) = e^((ln(10.2/13.2))t)

Answer 27

does that sound more reasonable?

Answer 28

probably not

Answer 29

sorry I meant B(t) = 16e^((ln(10.2/13.2))t)

Answer 30

since I used a calculator determine the answer is 3:35pm, I know that my k value is still garbage

Answer 31

so
13.2 = 16e^(kt)
34.2 = 21 + 16e^(kt)
10.2 = 16e^(k(t+1))

I solved
kt = ln(13.2/16)
and
k = ln(10.2/16) - ln(13.2/16)

Answer 32

I think I can then just sub k into equation

13.2 = 16e^(kt)
and I get
-3.60
which gives me
5-3.60 = 1:36pm

which is incorrect based on what i calculated with the program I downloaded

Answer 33

I used the method listed on this site to solve the problem

http://mathforum.org/library/drmath/view/62887.html