Vegetable Word Problem

[mtiller1]

We were given the following information:

At 11:14am a potato’s temp was X degrees. At 11:24am it was 158 degrees. At 11:34am it was 139 degrees. If the room temp is a steady 75 degrees, what was the temp of the potato at 11:14am? Show this answer, then change the formula you found to base e.

I used this chart:

Time (t) Time (t) Temp f(t)

0 11:14 X

10 11:24 158

20 11:34 139
(The teacher likes us to use “easy” numbers, so I changed the clock time to minutes passed time.)

Using \(\displaystyle f(t) = 75 + Da^t\), I know I have this much:

\(\displaystyle 158 = 75 + Da^t\)
\(\displaystyle 139 = 75 + Da^t\)

where a = exponential base, D = difference in starting temp and room temp (start – room = D)

I’m pretty sure it is a decay graph (although this isn’t part of the solution, I just want to be sure I’m thinking in correct terms) shifted up 75 units to begin.

When I take f(0), I get 76 degrees. I know there’s no way that’s possible. I need another number (D) at zero.

The last thing we did in class was \(\displaystyle log_a b\) = \(\displaystyle \frac{ln b}{ln a}\) = \(\displaystyle \frac{log b}{log a}\).

I am lost at this point. I think I can do the base e conversion once I have the temp of the potato, but I don’t know how to put everything together.

(Note: I tried to get the chart to line up correctly using TeX and Karl's -- I apologize for its appearance.)

Thanks in advance,
Angel

 

[mtiller1]

Okay, I have this much now (by solving for each variable; a, a^10 and D):

a^10 = 64/83 = 0.7710843373

a = 0.9743392282 (by taking the 10th root of each side)

D = 83/0.7710843373 = 107.640625

So the temp of the potato was 182.640625 degrees.

Now I have the formula:

\(\displaystyle 183 = 75 + Da^0\). I need for this formula to be changed to base e. Can anyone help?[/tex]

 

[mtiller1]

Okay, I found an example of how to change to base e, but I'm not sure how to apply it to the values I have.

Here is the example:

f(t) = 50(3^t)
If you think of the P=e^rt formula and apply the log property you get e^rt = (e^r)^t. (So, e^r = 3)

e^r = 3
r = ln3
50e^(1.098612289)t

Thanks,
Angel

 

[rahian2k]

[deleted]

 

[Unco]

G'day, Angel.

Okay, I have this much now (by solving for each variable; a, a^10 and D):

a^10 = 64/83 = 0.7710843373

a = 0.9743392282 (by taking the 10th root of each side)

D = 83/0.7710843373 = 107.640625

So the temp of the potato was 182.640625 degrees.

Now I have the formula:

\(\displaystyle 183 = 75 + Da^0\). I need for this formula to be changed to base e. Can anyone help?[/tex]

You work is great. Just with this last bit, the formula needs to be a general equation for the temperature, T, of the potatio after a time, t.

You had correctly determined D = 107.64..

So it follows from what you have done that the equation is
T = 75 + 107.64a^t, right?

(Remember: D was a constant, and so was the room temperature (of 75deg), so these numbers should appear in the formula.)

Now to change base.

You already know how to change 50(3^t) to 50e^(ln(3))t), so we shouldn't have a problem if 3 was instead a and 50 was 107.64.

T = 75 + 107.64*a^t

T - 75 = 107.64*a^t

Compare the right-hand side with

Ae^(rt) = A(e^r)^t

So e^r = a, just as e^r = 3 before.

Take natural logs of both sides:
r = ln(a).

So your equation is T = 75 + 107.64*e^(ln(a)t), agreed?

Let's check:

107.64*e^(ln(a)t) = (107.64*e^(ln(a)))^t = 107.64a^t because e^ undoes a natural log.