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- Dec 19th 2009, 03:09 PMJaydeeExponential Cooling Curve Problem
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- Dec 19th 2009, 07:00 PMmr fantastic
Your model is wrong. The model to use is $\displaystyle T = a (b)^t + 20$ (other models, such as $\displaystyle T = a e^{kt} + 20$, could also be used).

Substitute T = 95 at t = 0: 95 = a + 20 => a = 75.

Update the model (just like Ronnie): $\displaystyle T = 75 (b)^t + 20$.

Now substitute T = 90 when t = 2 and solve for b. - Dec 19th 2009, 07:22 PMJaydee
Ah, I see what I did wrong.

So ...

T = 75(b)^t + 20

90 = 75(b)^2 + 20

90 - 20 = 75(b)^2

70 / 75 = b^2

±√70/75 = b

b = ±0.97 ? - Dec 20th 2009, 02:49 AMmr fantastic
- Dec 20th 2009, 10:14 AMJaydee
- Dec 21st 2009, 02:28 AMmr fantastic
You don't need any special knowledge of exponential curves - just some mathematical common sense. Let's say you use the negative value of b.

Substitute some value of t into the rule. Do you get values of T that make sense? What happens if you substitute t = 1/2, for example ...? What values of T do you get for odd values of t ....?

Use some mathematical common sense. - Dec 21st 2009, 12:54 PMJaydee
Oh, I guess -0.97 wouldn't make sense. It seems as though the temperture fluctuates too wildly.

At 2 hours I came up with: T = 90.6

Then at 3 hours it went to: T = -48.5

Then at 4: T = 86.4

That doesn't make a heck of a lot of sense. Especially when a half hour yields a 'domain error' on my calculator.

So b has to be positive, then...

(And I apologize again. I'm pretty much devoid of mathematical common sense. It's never made any sense to me...)