Are you sure this is calculus? You want exponential decay to room temp?
Or is it algebra and you want a linear function?
In a murder investigation, the temperature of the corpse was 32.5 C at 1:30pm and 30.3 C an hour later. Normal body temperature is 37.0 C and the temperature of the surrounding was 20.0 C. When did the murder take place?
PLEASE SHOW STEP BY STEP
5 answers
I am going to assume it is algebra first
take t = 0 at 1:30
T = Ti - k t
Ti = 32.5
so
T = 32.5 - k t
30.3 = 32.5 - k (1 hour)
k = 2.2
so
T = 32.5 - 2.2 t
37 = 32.5 - 2.2 t
t = - 2.04 call it 2 hours
so by that linear model 11:30 am
now I will work on the more realistic calculus version
take t = 0 at 1:30
T = Ti - k t
Ti = 32.5
so
T = 32.5 - k t
30.3 = 32.5 - k (1 hour)
k = 2.2
so
T = 32.5 - 2.2 t
37 = 32.5 - 2.2 t
t = - 2.04 call it 2 hours
so by that linear model 11:30 am
now I will work on the more realistic calculus version
rate of change of temp proportional to temp above room temp which is 20
to make it easy on the arithmetic define
T' = real T - 20
dT'/dt = k (T')
dT/T' = k dt
ln T' = k t
T' = C e^kt
let's call t = 0 at 1:30
T = 32.5 so T' = 12.5
12.5 = C e^0 = C
so
T' = 12.5 e^kt
now when t = 1 hour T = 30.3 so T' = 10.3
10.3 = 12.5 e^k
ln .824 = k
k = - .194
so
T' = 12.5 e^-.194 t
so what was t when T= 37 so T'= 17 ?
17 = 12.5 e^-.194 t
1.36 = e^-.194 t
.307 = -.194 t
t = - 1.87 hours
so 1.87 hours before 1:30 pm
11:38 am
to make it easy on the arithmetic define
T' = real T - 20
dT'/dt = k (T')
dT/T' = k dt
ln T' = k t
T' = C e^kt
let's call t = 0 at 1:30
T = 32.5 so T' = 12.5
12.5 = C e^0 = C
so
T' = 12.5 e^kt
now when t = 1 hour T = 30.3 so T' = 10.3
10.3 = 12.5 e^k
ln .824 = k
k = - .194
so
T' = 12.5 e^-.194 t
so what was t when T= 37 so T'= 17 ?
17 = 12.5 e^-.194 t
1.36 = e^-.194 t
.307 = -.194 t
t = - 1.87 hours
so 1.87 hours before 1:30 pm
11:38 am
According to Newton's Law of Cooling
T(t) = roomtemp + (37 - 20)e^(-kt) , where t is the time in hours and k is a constant
so we get two equations:
32.5 = 20 + 17e^(-kt) ---> 12.5 = 17e^(-kt)
and
30.3 = 20 + 17e^(-k(t+1)) ---> 10.3 = 17e^(-kt - k)
divide them:
115/103 = e^k , remember to divide powers of the same base, you keep the base and subtract the exponents.
k = ln115 - ln103
then in 12.5 =17e^(-kt)
125/170 = e^(-kt)
-kt = ln125 - ln170
t = (ln125 - ln170)/-k
= (ln125 - ln170)/(ln103 - ln115) , .... only now will I do any actual calculation
= 2.79 hrs
so death occurred 2.79 hrs or 2 hrs and 47.4 minutes before 1:30 pm
I will leave it up to you to figure out that time.
T(t) = roomtemp + (37 - 20)e^(-kt) , where t is the time in hours and k is a constant
so we get two equations:
32.5 = 20 + 17e^(-kt) ---> 12.5 = 17e^(-kt)
and
30.3 = 20 + 17e^(-k(t+1)) ---> 10.3 = 17e^(-kt - k)
divide them:
115/103 = e^k , remember to divide powers of the same base, you keep the base and subtract the exponents.
k = ln115 - ln103
then in 12.5 =17e^(-kt)
125/170 = e^(-kt)
-kt = ln125 - ln170
t = (ln125 - ln170)/-k
= (ln125 - ln170)/(ln103 - ln115) , .... only now will I do any actual calculation
= 2.79 hrs
so death occurred 2.79 hrs or 2 hrs and 47.4 minutes before 1:30 pm
I will leave it up to you to figure out that time.
One of us made an arithmetic mistake. It is up to Tanya to get it right :)