anonymous
  • anonymous
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
js this is a random question
anonymous
  • anonymous
i know it's precal!
anonymous
  • anonymous
oh fun

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anonymous
  • anonymous
don't you hate precalc?
anonymous
  • anonymous
we can do this ready?
anonymous
  • anonymous
you know I do!
anonymous
  • 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
  • anonymous
let me know when you are ready to continue
anonymous
  • anonymous
11.26 & 6.76 ? :)
anonymous
  • 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
  • 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
  • anonymous
before we do it, let me ask if you have any questions about any of the steps so far
anonymous
  • anonymous
where did K come from? :(
anonymous
  • anonymous
lets go slow
anonymous
  • 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
  • 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
  • anonymous
the \(t\) in the formula is the variable time it stays in the formula
anonymous
  • anonymous
Got it! I am making you work so hard sorry :( I just keep on bothering you with all these problems
anonymous
  • 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
  • anonymous
not that I remember :/
anonymous
  • 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
  • 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
  • anonymous
typo there \[6.76=11.16e^{2k}\]
anonymous
  • 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
  • anonymous
\[6.76=11.16e^{2k}\\ .60573=e^{2k}\\ \ln(.60573)=2k\\ \ln(.60573)\div 2=k\]
anonymous
  • anonymous
find that number, that is your \(k\) and put it in \[T=11.16e^{kt}+25\]
anonymous
  • anonymous
11.16? or 11.26? Because I got k=.3001 not sure if it's right
anonymous
  • anonymous
typo on my part
anonymous
  • anonymous
use \(11.26\)
anonymous
  • anonymous
then I got k=.3001 :) ?
anonymous
  • anonymous
ii didn't do it you want me to check?
anonymous
  • anonymous
hmm that is not what i get
anonymous
  • 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
  • anonymous
http://www.wolframalpha.com/input/?i=log%286.76%2F11.26%29%2F2
anonymous
  • anonymous
nevermind, i got -.255 =k is that what you got?
anonymous
  • anonymous
yes
anonymous
  • anonymous
Finally I got it, thank you so much! :)
anonymous
  • anonymous
hope you got the process correct you got another we can try?
anonymous
  • 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
  • anonymous
ok we got the formula for the temp now right? so we can use it
anonymous
  • anonymous
\[T(t)=11.16e^{-.255t}+25\]
anonymous
  • anonymous
damn typos i really mean \[11.16e^{-.255t}+25=37\]
anonymous
  • anonymous
11.26 instead of 11.16 right?
anonymous
  • anonymous
lol jeez
anonymous
  • anonymous
\[11.26e^{-.255t}+25=37\]
anonymous
  • anonymous
I add 11.26 and 25? then divide 37 bu that?
anonymous
  • anonymous
oh dear
anonymous
  • anonymous
no wonder you hate pre calc stuck on algebra how would you solve \[7x+25=37\]?
anonymous
  • anonymous
-25 on both sides then divide by 7?
anonymous
  • anonymous
ok right, not "add 7 and 25"
anonymous
  • anonymous
PreCal will be the death of me
anonymous
  • anonymous
so how would you solve \[11.26e^{-.255t}+25=37\] for \(e^{-.255t}\)
anonymous
  • anonymous
-25 on both sides?
anonymous
  • anonymous
yes, then?
anonymous
  • anonymous
divide by 11.26
anonymous
  • anonymous
yay let me know what you get
anonymous
  • anonymous
\[e ^{-.255t}=1.06\]
anonymous
  • anonymous
yeah or maybe splurge and say \[e^{-.225t}=1.0657\]
anonymous
  • anonymous
now i hope you can find \(t\) still takes two steps
anonymous
  • anonymous
I think I got it :) Thank you so so so SO much!
anonymous
  • anonymous
yw good luck on your test when is it?
anonymous
  • anonymous
Quiz Thursday, test Monday. It will be a miracle if I pass trust me! :(

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