Question

A bowl of soup cools according to Newton's Law of Cooling. Its temperature (in *F) at the time of (t) is given bt T(t)=68+144e^-0.04t, where (t) is gven in minutes.
Part I: What was the initial temperature (at the time of t=0) of the soup?
Part II: What is the temperaure of the soup when t=15 minutes?
Part III: To the nearest whole minute at what time (t) is the soups temerature 125 degrees Fahrenheit?

Answer 1

So are given the formula that describes the temperature T as a function of the time t.
All you need to do is substituting t=0 and t=15 for Parts I and II.
For Part II, you have to solve T(t)=125 for t.

Just try it!

Answer 2

I did and I completely failed... Could you possibly give me a step by step instruction?

Answer 3

Part I: set t=0: T(0)=68+144e^0.

What is e^0?

Answer 4

1?

Answer 5

Yes, so T(0)=?

Answer 6

212?

Answer 7

True, again! See, it's not that hard!
Now calculate T(15).

Answer 8

Thank you!

Answer 9

Now i have another question how do i solve e^-0.04(15)?

Answer 10

There is not much to solve here: -0.04*15=0.6, so all you need to do is grab a calculator and type e^0.6

Answer 11

If you multiply that answer by 144 and then add 68, you've found T(15)!

Answer 12

okay thanks:) i am going to have another question will you help me with it too?

Answer 13

how do i solve part 3?

Answer 14

Part 3: Solve T(t)=125 for t, so first write down the equation:
\[68+144e^{-0.04t}=125\]
Remember, you are looking for t, but it is still "up there" somewhere, so we must take a few steps to get there.

First step?

Answer 15

sutract 68?

Answer 16

Go ahead! And then?

Answer 17

divide by 144

Answer 18

Do it! and then?

Answer 19

Thats where im stuck

Answer 20

OK, so you are here now:\[e^{-0.04t}=\frac{ 57 }{ 144 }\]:You can replace this immediately by:\[-0.04t=\ln \frac{ 57 }{ 144 }\]Why?
Because of the definition of the logarithm, which says:\[\log_{a}b=c \Leftrightarrow a^c=b \]In this case, having e as base:\[lna=c \Leftrightarrow e^c=a\]
So whenever you are looking for an exponent (which you are here), you need to come up with a logarithm.

Remember: logarithms are exponents!

E.G: solve \[10^t=25 \Leftrightarrow t=\log_{10}25 =\log 25\]

So go on with \[-0.04t=\ln \frac{ 57 }{ 144 }\]now!

Answer 21

0.396=e^-0.04?

Answer 22

No, we already had -0.04t=ln(57/144)=-0.92676,
so t=-0.92676/-0.04=23.2 min

Answer 23

Just use your calculator fo find ln(57/144) and divide that answer by -0.04

Answer 24

alright. i get it now

Answer 25

\[1000=\frac{ 10000 }{ 5+1245e ^{-0.97t} }\]

how do i solve this?

Answer 26

hello?

Answer 27

I'm back ;)

Divide both sides by 1000 to make it more manageable...