Question

ou have a cup of coffee whose temperature is 200 degrees. You cannot drink it until it reaches 150 degrees. You also have cream which is 40 degrees. The room you are in is 80 degrees. Assuming your coffee cools according to Newton cooling and than when you add you the cream to the coffee the new temperature of the mixture is instantly 11/13 times the temperature of the coffee plus 2/13 times 40 (the temperature of the cream). Note that the cream isn't warming up because you leave in the refrigerator. Assuming the cooling constant for the coffee is k = .6 minutes^-11, when should you add the

Answer 1

continued...when should you add the creamer to ensure you can drink your coffee the soonest

Answer 2

so i started it, newtons law of cooling gives us y' = k ( y-80)

Answer 3

i am stuck
all i know is the temperature of the coffee is \(120e^{-.6t}+80\)

Answer 4

right thats what i got too

Answer 5

the solution is given here, but i dont understand it http://math.stackexchange.com/questions/363066/newton-cooling-coffee

Answer 6

although i have no idea what \(\text{minutes}^{-11}\) means

Answer 7

maybe thats a typo

Answer 8

wow that stack exchange is a lot of words

Answer 9

Just looking at that, too @perl . So, what that answer says is that, using the theory, we want to find a time \(s\) such that, when the cream is poured in, the coffee is immediately drinkable, which is a reasonable reduction based on the problem, as the time per Centigrade/Fahrenheit extends as such \(t\) increases.

Answer 10

@LolWolf right, we want to add cream so that the coffee reaches 150 degrees soonest (so its drinkable)

Answer 11

i think the idea is now to put
\[150=\frac{2}{13}\times 40+\frac{11}{13}(80+120e^{-.6t})\] and solve for \(t\)

Answer 12

Yeah, the simplification, though, comes in the case that it must be \immediately\ drinkable afterwards. So, find a time t, such that, when the cream is added, it becomes immediately below the drinkable-temperature threshold.

Answer 13

but where do we take the derivative, it says to minimize?

Answer 14

The minimization isn't necessary under the assumptions.

Answer 15

the "minimize" is the part we you say intuitively it cools fastest at the beginning, add the cream at the end

Answer 16

oh so we dont need to take the derivative

Answer 17

I mean, if you want to be rigorous, then sure, you can show this, but I don't believe it's necessary for the answer.

Answer 18

I took the derivative of y(T) = 2/13*40 + 11/13(80 + 120 e^(-.6T))

Answer 19

but that didnt work

Answer 20

ok let me solve your equation

Answer 21

after simplifying i get
3/4 = e^(-.6t)
ln(3/4)/(-.6) = t

Answer 22

so at t = .47947 minutes approximately, we add the cream

Answer 23

so the calculus part was getting that equation
y = 120 e^(-.6t) + 80