Question

Using Newton’s Law of Cooling:
a) Find the particular equation to describe the cooling of the dead man’s body, if the temperature was 36.26° C upon the arrival of the police and 2 hours later it was 31.76° C. The temperature of the room was a standard 25° C.

Answer 1

js this is a random question

Answer 2

i know it's precal!

Answer 3

oh fun

Answer 4

don't you hate precalc?

Answer 5

we can do this
ready?

Answer 6

you know I do!

Answer 7

first thing to know is you do not work with the numbers 36.26 and 31.76 when making the equation
you work with the difference between the numbers and the room temp
so get out your calculator and compute \[36.26-25\] and \[31.76-25\]

Answer 8

let me know when you are ready to continue

Answer 9

11.26 & 6.76 ? :)

Answer 10

i believe you
now the initial temperature difference was \(11.26\) so you know the equation is going to include \[11.26e^{kt}\] where \(t\) is time

Answer 11

we need \(k\)
you are told that two hours later the temp difference was \(6.76\) so you know \[6.46=11.26e^{2k}\] here i replaced \(t\) by \(2\)
we have to solve this for \(k\)

Answer 12

before we do it, let me ask if you have any questions about any of the steps so far

Answer 13

where did K come from? :(

Answer 14

lets go slow

Answer 15

the formula is going to look like \[A_0e^{kt}+25\] for some \(A_0\) and \(k\)
\(A_0\) you are told, it is the initial difference in temperatures which is \(11.26\)

Answer 16

the \(k\) is what you have to find
it is the rate of decay (since this is decreasing)
sometimes it is the rate of growth, if what you have is increasing

Answer 17

the \(t\) in the formula is the variable time
it stays in the formula

Answer 18

Got it! I am making you work so hard sorry :( I just keep on bothering you with all these problems

Answer 19

no problem, just making sure the set up is clear
before we continue, did you ever do a problem like "at 12 the population of bacteria was 12 and at 2 it was 15, what is the function"?

Answer 20

not that I remember :/

Answer 21

ok fine
usually something like that comes before newton's law of cooling
the only reason why this is more confusing is that you are working not with the actual temperatures, but with the difference between the heated temperature and the room temperature
lets finish

Answer 22

we know that in two hours, the difference in the temperatures is \(6.45\) so we have to solve \[6.45=11.16e^{2k}\] for \(k\)

Answer 23

typo there \[6.76=11.16e^{2k}\]

Answer 24

takes three steps, all require a calculator
a) divide both sides by \(11.16\)
b) rewrite in logarithmic form
c) divide by 2

Answer 25

\[6.76=11.16e^{2k}\\
.60573=e^{2k}\\
\ln(.60573)=2k\\
\ln(.60573)\div 2=k\]

Answer 26

find that number, that is your \(k\) and put it in \[T=11.16e^{kt}+25\]

Answer 27

11.16? or 11.26? Because I got k=.3001 not sure if it's right

Answer 28

typo on my part

Answer 29

use \(11.26\)

Answer 30

then I got k=.3001 :) ?

Answer 31

ii didn't do it
you want me to check?

Answer 32

hmm that is not what i get

Answer 33

also the answer has to be negative since it is decreasing, not increasing
make sure you know how to put this in your calculator correctly

Answer 34

http://www.wolframalpha.com/input/?i=log%286.76%2F11.26%29%2F2

Answer 35

nevermind, i got -.255 =k is that what you got?

Answer 36

yes

Answer 37

Finally I got it, thank you so much! :)

Answer 38

hope you got the process correct
you got another we can try?

Answer 39

The next and last question is "Using your equation, when was the body a normal 37° C, that is, when was the time of death?" not sure if that's the same?

Answer 40

ok we got the formula for the temp now right? so we can use it

Answer 41

\[T(t)=11.16e^{-.255t}+25\]

Answer 42

damn typos
i really mean \[11.16e^{-.255t}+25=37\]

Answer 43

11.26 instead of 11.16 right?

Answer 44

lol jeez

Answer 45

\[11.26e^{-.255t}+25=37\]

Answer 46

I add 11.26 and 25? then divide 37 bu that?

Answer 47

oh dear

Answer 48

no wonder you hate pre calc
stuck on algebra
how would you solve \[7x+25=37\]?

Answer 49

-25 on both sides then divide by 7?

Answer 50

ok right, not "add 7 and 25"

Answer 51

PreCal will be the death of me

Answer 52

so how would you solve \[11.26e^{-.255t}+25=37\] for \(e^{-.255t}\)

Answer 53

-25 on both sides?

Answer 54

yes, then?

Answer 55

divide by 11.26

Answer 56

yay
let me know what you get

Answer 57

\[e ^{-.255t}=1.06\]

Answer 58

yeah or maybe splurge and say \[e^{-.225t}=1.0657\]

Answer 59

now i hope you can find \(t\) still takes two steps

Answer 60

I think I got it :) Thank you so so so SO much!

Answer 61

yw
good luck on your test
when is it?

Answer 62

Quiz Thursday, test Monday. It will be a miracle if I pass trust me! :(