Question

Did I do this correctly? pls help.

Answer 1

I came here because I saw Will Smith. But yeah bring it on. What's the problem?

Answer 2

* 8.256821718 hours before 10:23 pm?

Answer 3

is t supposed to be hours?

Answer 4

Please wait. My net is kinda slow. It hasnt finished downloading the image yet.

Answer 5

oh ok :)

Answer 6

And I also apologize, you might have to zoom in a bit.

Answer 7

That's what's in the image. :)

Answer 8

Okay done with the reading.

I think the unit of time doesn't matter when you do it in the start.
If you take 1 hour and solve you get an end result in hours in the second case.

Answer 9

But keep in mind all standard formulas use time in seconds.

Answer 10

ohh ok. Did you see my process? and my results?

Answer 11

First case looks okay.
What about the second one ? Why 68-98.6? It's just to make sure.

Answer 12

Oh God, not sorry that was a typing error, it's supposed to be 98.6-68,

Answer 13

no sorry *

Answer 14

Ah that's better XD

Answer 15

Your approach is right :) I can't check the numerical answer as I have to take dinner now. Let me know if you have any more problems!

Answer 16

I'm just not sure since I got such a long decimal number lol, I was expecting like an exact number or so lol. >.<

Answer 17

@Data_LG2

Answer 18

@freckles

Answer 19

80 = 68 + (80-68)
78 = 68 + (80-68)e^(-r)

-r = log(5/6)

98.6 = 68 + (80-68)e^(-rt)

30.6 = 12 e^(-rt)

log(30.6/12) = -rt

t = log(30.6/12)/log(5/6) = -5.134

Answer 20

our initial body temp is 80 when we started the whole process .... we determine the rate of change after an hour, and then determine the value of t that gets us back into 98.6

Answer 21

in other words, the only thing we change in the setup is, t and compare it to the temperature of the body at any value of t

Answer 22

So I was doing it wrong?

Answer 23

I thought the initial temperature was 98, so I used that to find t in the 2nd problem.

Answer 24

how did you get log?

Answer 25

*copy and paste*


So I figured I had to look for r first right? so
78=68+(80−68)e−r1

t=1 because one hour passed?

then I got r = 0.025317808

now that I know r I did this:

80=68+(68−98.6)e−(whatIgotforr)t


So I got 8.256821718,

So 8.256821718 passed after the person died?

Answer 26

the first 2 temperature readings give us a way to determine the rate of change (r)

80 = 68 + (80-68)e^(-r0) , at t=0

78 = 68 + (80-68)e^(-r1), after an hour, the temperature is 78

the first setup is not really needed, but it proves that the formula works when t=0, our temperature is initially determined to be 80

the second setup solves for -r, subtract, divide, log

10 = 12 e^(-r)

log(5/6) = -r
-------------------------

we know have a value for the rate of change ... we do not know WHEN the body was 98.6 degrees, but we do know WHEN the body was 80 degrees, and this is what has calibrated our calculations to determine the rate of change.

all that is left to do is determine how far back the body was at 98.6, given that our initial time is t=0, and a rate of -log(5/6)
--------------------------

98.6 = 68 + (80-68)e^(-rt)

solve for t

Answer 27

80 = 68 + (98.6-68)e^(-rt) might be useful as well i spose

12 = 30.6 e^(-rt)

t = log(12/30.6)/log(5/6) = 5.134 hours to get down to 80 degrees

Answer 28

then again, to determine the hours that have passed when the body is 78 degrees

78 - 68 = (98.6-68) e^(-rt)

log(10/30.6)/log(5/6) = t = 6.134 as expectd since it was an hour later

Answer 29

also, in higher math classes there is only one log function, the natural log. log = ln in my responses

Answer 30

78 = 68 + (80-68)e^(-r1)

the second reading of course

Answer 31

Ohh I see what I was doing wrong lol, it was that I added 68 and (t_0-t_e) lol

Answer 32

:)

Answer 33

So the person died around 5:37 p.m.?