Question

help with a calc problem! medal!

Answer 1

i can show you how to do this with no calculus at all if you like
otherwise let someone else do it

Answer 2

I see 4 and 5, but not 3

Answer 3

good point

Answer 4

You can solve this equation using separation of variables.

Answer 5

can you work this out with me please

Answer 6

\[
\frac{dT}{dt} = k(T-38)\implies \frac{dT/dt}{k(T-38)} = 1
\]The integrate both sides with respect to \(t\).
Remember that \(dT = \frac{dT}{dt}dt\), so:\[
\int \frac{dT}{k(T-38)}=\int dt
\]

Answer 7

Do you follow so far?

Answer 8

yes i got it so far!

Answer 9

Can you integrate the left side?

Answer 10

uhh yeah i can try

Answer 11

You can try? It seems like a fairly easy integral

Answer 12

is it dT/k(T-38)x

Answer 13

No, you are integrating with respect to \(T\).

Answer 14

\(k\) is constant with respect to \(T\).

Answer 15

Dtx?

Answer 16

Dtx? What?

Answer 17

There is no \(x\).

Answer 18

Okay let's do this again...

Answer 19

wow this is a lot of work ...

Answer 20

\[
\int\frac{dT/dt}{k(T-38)}dt = \frac 1k\int \frac{dT/dt}{T-38}dt = \frac 1k\int \frac{1}{T-38}dT
\]

Answer 21

This is just making it more clear what is going on here. We can make it more simple if we make a substitution. \(u=T-38\) results in \(du=dT\). Our integral becomes: \[
\frac 1k\int \frac 1u du
\]

Answer 22

Can you integrate this?

Answer 23

is the answer \(51\) degrees?

Answer 24

all my answers has an x

Answer 25

to the nearest degree, what is the temp after another 30 minutes? that cannot have an x in it, that is a number, that you can get pretty quickly with no calc at all

Answer 26

then i dont know how to get to the next step

Answer 27

it cools from 75 to 60 degrees in a 38 degree fridge
the difference in the beer and the fridge goes form \(75-38=37\) to \(60-38=22\) degrees
the ratio is \(\frac{22}{37}\) so in another 30 minutes it will cool by a factor of \(\frac{22}{37}\) to
\[22\times \frac{22}{37}=13.08\] add that to the fridge temp and the beer is now
\[13+38=51\] approximately

Answer 28

did you get 51 @wio ?

Answer 29

....

Answer 30

Did you not learn how to integrate anything?

Answer 31

If you assume it is an exponential function, then you can use the properties of exponential functions to find the answer.

Answer 32

But I'm pretty sure they want you to find the actual function of \(T(t)\).

Answer 33

ln(u)/k +C

Answer 34

Okay, good. Now how about \(\int dt\), what does that become?

Answer 35

wait that was actually right?

Answer 36

Yes

Answer 37

t+C

Answer 38

Yes

Answer 39

Now, remember that \(u=T-38\).

Answer 40

Solve for \(T\).

Answer 41

T=u+38

Answer 42

No, wait

Answer 43

You just solved the integrals:\[
\int \frac{dT}{k(T-38)}=\int dt
\]

Answer 44

And the result was \[
f(T)= g(t)
\]That is, an expression of \(T\) on the left and an expression of \(t\) on the right.
Now you solve for \(T\).

Answer 45

ok now im confused im gonna start a new question

Answer 46

Hold on

Answer 47

When you integrated... okay?

Answer 48

\[
\int \frac{dT}{k(T-38)}=\int dt
\]Became: \[
\frac1k\ln(T-38)+C_1=t+C_2
\]

Answer 49

Now you solve for \(T\). Do you know any algebra?

Answer 50

@Michele_Laino can you please continue this???

Answer 51

I rewrite the last equation, as below:
\[\ln(T-38)=kt+c\]
where c is an arbitrary constant.
Now I apply the definition of logarithm, and I get:
\[T-38=c _{1}e ^{kt}\]
where c_1 is another arbitrary constant which depends on c.
FInally, I can write:
\[T=c _{1}e ^{kt}+38\]

Answer 52

thats it?

Answer 53

It is the solution

Answer 54

I rewrite the solution as below:
\[T(t)=c _{1}e ^{kt}+38\]
which expresses the temperature T as function of time t.
Now that solution is generic, since depends on a constant c_1.
So in order to find our solution, I have to apply the initial condition.

Answer 55

okk

Answer 56

Then I write T(0), using your initial data, namely at t=0 T=75:

Answer 57

So I write:
\[75=c _{1}e ^{k*0}+38=c _{1}+38\]
since e^(k*0)=e^0=1

Answer 58

Now we knw the value of c_1, which is:
\[c _{1}=75-38=37\]
and now I can rewrite your solution without arbitrary constant, since it is:
\[T(t)=37*e ^{kt}+38\]

Answer 59

now, we insert your data, namely T=60 at t=30:
\[T(30)=60=37*e ^{k*30}+38\]
please solve for k

Answer 60

wait whats e im confused, can you just show me the steps this is why i never understand it

Answer 61

ok!

Answer 62

we have:
\[37*e ^{k*30}=60-38=22\]

Answer 63

so:
\[e ^{k*30}=\frac{ 22 }{ 37 }\]

Answer 64

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

Answer 65

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

Answer 66

that is the value of your constant k

Answer 67

is it clear, please?

Answer 68

yes! very good explanation!

Answer 69

thanks!

Answer 70

Now, in order to answer to question c, we have to apply our formula, with our value for k, and set T=55, namely:

Answer 71

what did you get for exact value of k?

Answer 72

\[T(t)=55=37*e ^{kt}+38\]
where k=\[=\frac{ 1 }{ 30 }\ln \left( \frac{ 22 }{ 37 } \right)\]

Answer 73

I got the value of k, using our solution and inserting t=30 and T=60

Answer 74

now, we got this condition:
\[55=37*e ^{kt}+38\]
please solve for t

Answer 75

17/37= e^kt

Answer 76

ok! So we can write:
\[kt=\ln \left( \frac{ 17 }{ 37 } \right)\]

Answer 77

now, insert the value of k

Answer 78

hint:
\[t=\frac{ 1 }{ k }\ln \left( \frac{ 17 }{ 37 } \right)\]

Answer 79

im confused, can you do the next step?

Answer 80

ok!

Answer 81

\[t=30\frac{ \ln(17/37) }{ \ln(22/37) }\]
that's your answer

Answer 82

that is my final answer?

Answer 83

yes!

Answer 84

no, t is in minutes, we have answered to question 5

Answer 85

55 is your numerical dat, and t is your answer

Answer 86

we haven't answered to question 4, yet!

Answer 87

in order to answer to question 4, we have to apply our formula for T8t) and insert t=30+30=60 minutes,namely:

Answer 88

\[T(60)=37*e ^{k*60}+38\]
where k is:
\[k=\frac{ 1 }{ 30 }\ln \left( \frac{ 22 }{ 37 } \right)\]

Answer 89

its funny, cause you were completely wrong lol