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.

- anonymous

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- anonymous

js this is a random question

- anonymous

i know it's precal!

- anonymous

oh fun

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## More answers

- anonymous

don't you hate precalc?

- anonymous

we can do this
ready?

- anonymous

you know I do!

- anonymous

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 \[31.76-25\]

- anonymous

let me know when you are ready to continue

- anonymous

11.26 & 6.76 ? :)

- anonymous

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

- anonymous

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 \(k\)

- anonymous

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

- anonymous

where did K come from? :(

- anonymous

lets go slow

- anonymous

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 \(11.26\)

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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"?

- anonymous

not that I remember :/

- anonymous

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

- anonymous

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 \(k\)

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

typo on my part

- anonymous

use \(11.26\)

- anonymous

then I got k=.3001 :) ?

- anonymous

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

- anonymous

hmm that is not what i get

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

yes

- anonymous

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

- anonymous

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

- anonymous

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?

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

11.26 instead of 11.16 right?

- anonymous

lol jeez

- anonymous

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

- anonymous

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

- anonymous

oh dear

- anonymous

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

- anonymous

-25 on both sides then divide by 7?

- anonymous

ok right, not "add 7 and 25"

- anonymous

PreCal will be the death of me

- anonymous

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

- anonymous

-25 on both sides?

- anonymous

yes, then?

- anonymous

divide by 11.26

- anonymous

yay
let me know what you get

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

yw
good luck on your test
when is it?

- anonymous

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

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