## theEric 2 years ago Hi! Any help with this ODE word problem will be appreciated! I'll simplify it to the important data: We are keeping track of the temperature of coffee. Well, we have the temperature at two times and want a third. I would like to be able to conceptualize what's going on, if anyone can help! 7:30am $$200^\circ\text F$$ 7:45am $$120^\circ\text F$$ 8:00am $$?^\circ\text F$$ The rate of change of heat is a simplified version of Newton's Law of Cooling, that $$\dfrac{dT}{dt}=k(T-T_r)$$ where $$k$$ is a proportionality constant and $$T_r$$ is the room temperature ($$70^\circ\text F$$).

1. ganeshie8

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2. theEric

hM?

3. theEric

Hm?*

4. theEric

I figure that I probably need to find out what $$k$$ is...

5. ganeshie8

@sauravshakya

6. theEric

Can I use $$\Large{\int\limits_{7.5}^{7.75}}\normalsize\dfrac{dT}{dt}dt=\Large{\int\limits_{7.5}^{7.75}}\normalsize k(T-T_r)\ dt=200-120$$? I'll think on that...

7. theEric

I haven't done anything like it in class, that I recall.

8. theEric

But I don't have $$T(t)$$ to do integration...

9. Saeeddiscover

You can say nothing about 8:00 AM here. Because We don't know how the temperature function is changing over time.

10. theEric

That's what I was thinking as well... But we know the rate-of-change function, two points on the $$T$$ vs $$t$$ curve, which is one solution to the rate-of-change function...

11. theEric

I know that initial value problems require one point...

12. phi

I think I would solve the equation then use the given data to find the constants $\frac{dT}{dt}= k(T -T_r) \\ \frac{dT}{(T -T_r)}= k \ dt \\\ln(T -T_r)= k \ t + C$ where t is time in minutes from 7:30 make each side the exponent of e $T- T_r= A e^{k\ t}$ now use 7:30 t=0 T= 200 to find A and 7:45 t=15 to find k

13. theEric

Thank you very much! I'm working on digesting it now.

14. theEric

$$A$$ is an arbitrary constant like $$C$$, right?

15. phi

yes A = e^C we don't care about C per se.

16. theEric

This is somewhat related, @phi , but do we always need to look at things like $$t=0$$ for initial value problems? Or can we use non-zero values for the independent variable?

17. theEric

Thank you!

18. phi

you can use non zero values of t. But obviously t=0 makes it especially easy to find A

19. theEric

Right, thank you very much! :)

20. theEric

In case it is of any viewer's concern: $$A=130^\circ\text F$$ $$k=-0.0637\ [s^{-1}]$$ and the coffee will be about $$89.2^\circ\text F$$ at 8:00am. :) The problem was about a professor with an 8am class, who never drinks coffee unless it is at $$90^\circ\text F$$ or cooler. Good news for him - he can drink the coffee in class! I might show off and solve for the time $$t$$ at which he can take his first sip. :P But not until I get the real problems done. Thank you!