Question

@phi

Answer 1

did you solve the differential equation ?

Answer 2

no

Answer 3

does this make sense:
\[ \frac{dT}{dt}= k(T-38) \\ \frac{dT}{(T-38)} = k \ dt \]

Answer 4

oh yeah i did that for number 4 also

Answer 5

integrate both sides, what do you get ?

Answer 6

we only need C2 (it's arbitrary at this point)
I would write it as
\[ \ln(T-38) = k \ t + C \]
make each side the exponent of e
\[ e^{\ln(T-38) } = e^{ k \ t + C } \\ T-38 = A e^{k \ t}
\]
where e^C is arbitrary, so rename it A

Answer 7

or
\[ T = A e^{k\ t} +38 \]
now we solve for A and k.
at t=0 , T=75:
\[ 75 = A e^0 +38 \]
and e^0 is 1

Answer 8

what do we get for A ?

Answer 9

A= 37

Answer 10

now we have
\[ T = 37 e^{k\ t} +38 \]

now we use at t=30, T= 60
so
\[ 60 = 37 e^{30k}+38 \]can you solve for k ?

Answer 11

22=37e^30k , what what is e?

Answer 12

e is the base for natural logarithms 2.7318281828...
now divide both sides by 37
\[ \frac{22}{27}= e^{30k} \]
take the "natural log" of both sides. the natural log of e^x is x (the natural log "grabs" the exponent)
\[ \ln\left( \frac{22}{27}\right) = \ln\left( e^{30k} \right) \\
\ln\left( \frac{22}{27}\right) = 30 k
\]

Answer 13

*that should be 37 not 27

Answer 14

\[ \ln\left( \frac{22}{37}\right) = 30 k \\
k= \frac{1}{30} \ln\left( \frac{22}{37}\right)
\]

Answer 15

and then divide 30 to get k alone?

Answer 16

ohh got it!

Answer 17

yes. we will need a calculator to figure out the exact number.

Answer 18

you can type into the google search window
ln(22/37)/30=

Answer 19

once you get k, you have the complete equation and can answer parts 4 and 5
for part 4. set t=60 (I think that is the time they want)
for part 5, set T= 55 and solve for t. to solve we will have to use ln

Answer 20

\[ T = 37 e^{-0.017329\ t} +38 \]

Answer 21

@DanJS i dont get what to do next here

Answer 22

@DanJS

Answer 23

i just did this on the other thread i thought. lol

Answer 24

yeah but i got everything you said, phi answered it, it just didnt give me the notification hahaha

Answer 25

i just dont get what to do next to get my answer for 4 and 5

Answer 26

separable DE, then get

T(t) = 38 + Ce^(kt)

need to use points to get C and k

Answer 27

cause he just went offline randomly

Answer 28

i already got k, it is 1/30ln(22/37)

Answer 29

An additional 30 minutes.... what time would that be ?

Answer 30

\[T(t) = 38 + 37e ^{-.017329*t}\]

I just copied down the constants from above, assuming they are right

Answer 31

So they gave you out to T(30) = 60

An ADDITIONAL 30 minutes would be T(60) = ....

Answer 32

These are the 2 easy questions, you already found the solution to the differential equation. lol

Answer 33

for 5) How long to cool to 55?

T(t) = 55 find t

Answer 34

wait what is number 4? im so lost

Answer 35

from now on im only putting up one question haha

Answer 36

For 4, they want you to just add 30 minutes on to the last time point they gave you

T(30) = 60
30 more minutes is

T(60) = 38 + 37e^(-0.017392*60)

Answer 37

Just evaluate T(60) for number 4

Answer 38

For 5) They tell you the temp is 55 and want to know the time, solve for t here

T(t) = 55 = 38 + 37e^(-0.017392t)

Answer 39

51 degrees for number 4?

Answer 40

umm, let me see

Answer 41

yeah , i got about 51 too

Answer 42

yay(:

Answer 43

here is the graph for this problem function

Answer 44

Notice
T(0) = 75,
T(60) = 51.03
It approaches 38 degrees as time goes on to infinity

Answer 45

That is temperature on the Y axis, and Time on the X axis

Answer 46

omg theres someone going on with openstudy, the lag is crzy

Answer 47

okay so i got number 4, that is finally done(: now onto number 5, what is the next step??

Answer 48

For 5) They tell you the temp is 55 and want to know the time, solve for t here
T(t) = 55 = 38 + 37e^(-0.017392t)

Answer 49

Here is the graph, with a horizontal line at y=55 degrees, it intersects the curve at time 44.72 minutes

Answer 50

17=37e^(-.017392t)
17/37=e^(-.017392t)

Answer 51

im stuck

Answer 52

That is just a plot of the T(t) function, the red curve is the temperature at time t

Answer 53

just drew a line at y = 55 degrees, the corresponding time is x = 44.72 minutes

Answer 54

so my answe is 44.72 minutes and thats it?

Answer 55

yes

Answer 56

sorry this site is a piece of crap server

Answer 57

thank you!!