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Question

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inkyvoyd · Answer

stop roasting people

inkyvoyd · Answer

On a more serious note, this is a separable differential equation. Do you know how to solve it?

bmk614 · Answer

First you need to take the integral of both sides.

inkyvoyd · Answer

Well, I think you'll want to rewrite it in the form f(T) dT/dt=g(t) first

inkyvoyd · Answer

Let's keep going. we can call e^c some new constant C, correct? and in this case we can take the abs. value out, so just go ahead and solve for T(t)

inkyvoyd · Answer

you'll wind up with a system of two equations in two variables for your initial value problem. Let me know what you get.

irishboy123 · Answer

$\large  |T−A|=e^c∗e^{−kt} $
which leads to next step $\large  e^c = \alpha$
so
$\large  |T−A|=\alpha e^{−kt} $


ah!

@surjithayer

greatlife44 · Answer

$$\frac{ dT }{ dt } = -k(T-A)$$

$$\int\limits \frac{ dT }{ (T-A) } = \int\limits -k*dT$$

$$\ln|T-A| = -kt+C$$

$$e^{\ln|T-A|} = e^{-kt}+e^{C}$$

$$|T-A| = e^{c}+e^{-kt}$$

$$|T-A| = \alpha*e^{-kt}$$

greatlife44 · Answer

This is what i'm thinking 

$$|T-A| = \alpha*e^{0}$$

so i'm guessing that at t = 0 

$$68^{o}F$$

$$|T-A| = \alpha $$ 


$$\alpha*e^{-kt} = |T-A|$$

$$e^{-kt} =  \frac{ |T-A|  }{ \alpha }$$

$$\ln(e^{-kt}) =  \ln(\frac{ |T-A|  }{ \alpha })$$

$$k = -\ln(\frac{ |T-A|  }{ \alpha })*t^{-1}$$ 

$$k = -\ln(\frac{ |68-25|  }{ 68 })*(5)^{-1}$$

$$k = 9.17*10^{-2}$$

greatlife44 · Answer

$$t = \ln (\frac{ |T-A| }{ \alpha })*-k^{-1}$$

$$\ln(\frac{ 68-20 }{ 68 })*-9.17*10^{-2} = 4~hours$$

greatlife44 · Answer

@chris215

greatlife44 · Answer

this is just my guess because i'm assuming at t =0 the original temperature is 68 F

chris215 · Answer

hmm 4 isn't an answer choice but thanks for your help!!! I'll just ask my teacher about this one

greatlife44 · Answer

@chris215 when you get the answer can you tell me lol?