Application of derivative help (1 Viewer)

[blazer78]

completely lost..

The thing that screws me most is the fact that it doesn't start from time=0 , but 30 minutes after t=0. =/

Any help is greatly appreciated.

 

[abcd9146]

i dont know the answer, my class hasn't done this stuff yet, but are these questions from a textbook, if so what textbook?

 

[SoulSearcher]

Here you have the temperature of the water as 20 degrees Celcius and you have 2 equations that describe the temperature of the body in the water so

(a) 20 + Ae^30k = 33
therefore Ae^30k = 13 ... (1)
20 + Ae^90k = 31
therefore Ae^90k = 11 ... (2)

[ (2) / (1) ] Ae^90k / Ae^30k = 11 / 13
e^60k = 11 / 13
60k = ln (11/13)
k = -0.00278 to 5 dec. pl.

so sub in value of k into (1)
Ae^-0.0834 = 13
A = 14.13 to 2 dec. pl.

so for initial temp when body was thrown into water,
20 + 14.13 = 34.13 degrees Celcius

(b) now let the temperature of the body when it was thrown into the water equal T
34.13 = 30 + 7e^-0.106t
4.13 = 7e^-0.106t
0.59 = e^-0.106t
-0.106t = ln 0.59
t = ln 0.59 / -0.106
t = 4.97 hours = 5 hours to nearest hour. Therefore the person was murdered 5 hours before it was dumped into the water, which is 11am

 

[blazer78]

@abcd, they don't come from a textbook.

@soul, thanks!