Question

At 2:00 P.M., a thermometer reading 80 degrees F is taken outside, where the are temperature is 20 degrees F. At 2:03 P.M., the temperature reading yielded by the thermometer is 42 degrees F. Later, the thermometer is brought inside, where the are is at 80 degrees F. At 2:10 P.M., the reading is 71 degrees F. When was the thermometer brought indoors?

Answer 1

that's degrees Fahrenheit :)

Answer 2

i guess we need the rate of change of the thermometer

Answer 3

if this is like a newton's law of cooling problem we work with the differences in the temperature. the inital difference in the thermometer and the ouside temperature is 60 degrees, and 3 minutes later it is 22 degrees so the temperature can be modelled by
\[60\times (\frac{22}{60})^{\frac{t}{3}}\]

Answer 4

but my guess is you are supposed to write an exponential equation that looks like
\[A_0e^{rt}\] yes?

Answer 5

that is perhaps start with
\[22=60e^{3r}\] and solve for r

Answer 6

we get
\[\frac{11}{30}=e^{3r}\]
\[3r=\ln(\frac{11}{30})\]
\[r=\frac{1}{3}\ln(\frac{11}{30})\]

Answer 7

my book says \[du/dt = -k \Delta u\] where u is the change in temperature.

Answer 8

i get r = -.3344 to four decimal places. then

Answer 9

yeah that is because they want you to think this is a differential equation, but it is simply an exponential model
integrate both sides and you get
\[u=A_0e^{-kt}\] and i think we just found k

Answer 10

in other words we know
\[-k=.3344\]

Answer 11

what's \[A _{o}e ^{rt}\]?

Answer 12

i used r (for rate) instead of -k

Answer 13

do you know what newton's law of cooling says?

Answer 14

change in the difference of the ambient temperature and the temperature of heated (or cooled) object decays exponentially.

Answer 15

the time rate of change of temperature du/dt is proportional to the temperature difference delta u.. yes?

Answer 16

yes that is essentially the same thing i said, but you are working with the DIFFERENCES in the temperatures

Answer 17

so initial difference is 80 - 20 = 60

Answer 18

that is
\[A_0\] the initial differnce in temperature, that is, the difference at time t = 0

Answer 19

can you show me how did you do it step by step? I want to know how you came up with the answer. and yes, you answered it right! :)

Answer 20

ok sure, but that is not your final answer, that is just
\[-k=-.3344\] or
\[k=.3344\] you still have to find the time the thermometer was brought in.
we can do that second

Answer 21

what's k btw? XD

Answer 22

ok lets start at the beginning

Answer 23

the initial temperature difference is 60 right?

Answer 24

okay. i'm really sorry for asking simple questions. it's just that i'm really clueless. :S

Answer 25

yes

Answer 26

and after 3 minutes the temperature differnce is 22 yes?

Answer 27

*difference

Answer 28

yes...

Answer 29

ok now forget about differntial equations for a second

Answer 30

okay...^^

Answer 31

i know that the temperature difference decays exponentially, that is, decays proportional to the current amount

Answer 32

okay. then?

Answer 33

so i can write an equation for the temperature difference as
\[A(t)=A_0e^{-kt}\] where t is time ,
\[A_0\] is the initial temperature difference (when t = 0) and k is the rate of change. that is what you are looking for, k

Answer 34

we are given two pieces of information. when t = 0 the temp difference is 60
when t = 3 the temperature diffence is 22
this is like saying i have an exponential function
\[A(t)=A_0e^{-kt}\] and i know two points, (0,60) and (3,22)

Answer 35

so i know what
\[A_0\] is right away, when t = 0 we know
\[A(0)=A_0e^{-k\times 0}=A_0=60\] more simply put,
\[A_0\] is the initial value, which is 60

Answer 36

ohh.. okay. i'm slowly getting it. :D

Answer 37

lets make sure it is clear what we are after. we want k

Answer 38

like finding the slope of a line given two points, only in this case the function is exponential not linear, and k is the rate of change

Answer 39

now we also know one other point , namely (3,22) because we know after 3 minutes the temperature differnence is 22 degrees.

Answer 40

so we take our as yes unfinished function
\[A(t)=60e^{-kt}\] replace t by 3 and set the result equal to 22

Answer 41

we get
\[A(3)=22=60e^{-k\times 3}\]
\[22=60e^{-3k}\] and solve this equation for k

Answer 42

so we take our as yes unfinished function <~ what? XD

Answer 43

we don't yet know k
that is what we need

Answer 44

like if you have a line and you know the y intercept is (0,60) so you know it looks like
\[f(x)=mx+60\] but you don't yet know m
that is what i mean by "as yet unfinished" we still need k

Answer 45

so our last job is to find k, which we do by solving
\[22=60e^{-3k}\] is it clear where i got that equation from?

Answer 46

setting up that equation is really the whole problem.
solving it is easy enough

Answer 47

yep! super clear! thank you so much! I'm gonna practice how you got that equation now and try to solve other problems. :)

Answer 48

so now we solve via
\[\frac{22}{60}=e^{-3k}\]
\[-3k=\ln(\frac{11}{30})\]
\[k=-\frac{1}{3}\ln(\frac{11}{30})=.3344\] and our equation is now
\[A(t)=60e^{-.3344t}\]

Answer 49

hmm. okay.

Answer 50

sorry forgot to finish
we now bring the thermometer back into the room. at 2:10 the temperature differnce is 9 so we now have
\[A(t)=9e^{-.3344t}\] where again A(t) is the temperature difference, but now t is time starting at 2:10

Answer 51

we want to know when we brought it back in to the room. at that time the temperature differnce is 80 - 42 = 38 so we have to solve
\[38=9e^{-.3344t}\] for t
this time we get
\[\frac{38}{9}=e^{-.3344t}\]
\[t=\frac{\ln(\frac{38}{9})}{-.3344}\]
\[t=-4.3\] so 4.3 minutes before 2:10