An oddly worded question: A yam is put in an oven...

[Sophie]

Question: A yam is put in an oven maintained at a constant temperature of 250C. Suppose that after 30 minutes, the temperature of the yam is 150C and is increasing at a rate of 3C/Min. If the Temperature of the yam t minutes after it is put in the oven is modeled by T(t)=250-ae^bt
Find a and b.



\(\displaystyle \L\\\begin{array}{l}
150 + 3t = Temperature(after30\min s) \\
T(35) = 165 \\
T(40) = 180 \\
\\
165 = 250 - ae^{ - b35} \\
85 = ae^{ - b35} \\
180 = 250 - ae^{ - b40} \\
70 = ae^{ - b40} \\
\\
a = \frac{{85}}{{ae^{ - b35} }} \\
70 = \frac{{85}}{{ae^{ - b35} }}e^{ - b40} \\
\frac{{85}}{{70}} = \frac{{e^{ - b35} }}{{e^{ - b40} }} \\
\ln \frac{{17}}{{14}} = \ln e^{ - b35} - \ln e^{ - b40} \\
\ln \frac{{17}}{{14}} = b40 - b35 \\
\frac{{0.194}}{5} = b \\
b = 0.0388312 \\
\end{array}\)

I put b into


\(\displaystyle \L\\\begin{array}{l}
a = \frac{{85}}{{e^{ - 0.0388312*35} }} \\
a = 330.8759 \\
\end{array}\)

I then put a and b into model to get

a=330.8759
b= 0.0388312

\(\displaystyle T(t) = 250 - 330.8759e^{ - 0.0388312*t}\)

I put 30 into the model and got 146.79, which is close to 150

I am sure this is incorrect as all the other questions require limits and derivatives of some sort, but I do not know how to use those in this question.

Some direction would be appreciated, thanks Sophie

 

[skeeter]

\(\displaystyle \L T(t)=250-ae^{bt}\)

\(\displaystyle \L \frac{dT}{dt} = -abe^{bt}\)

at t = 30, T = 150 ...

\(\displaystyle \L 150=250-ae^{30b}\)

\(\displaystyle \L ae^{30b} = 100\)

at t = 30, dT/dt = 3

\(\displaystyle \L 3 = -abe^{30b}\)

\(\displaystyle \L 3 = -100b\)

\(\displaystyle \L b = -.03\)

\(\displaystyle \L ae^{30(-.03)} = 100\)

\(\displaystyle \L a = 100e^{.9}\)

so ...

\(\displaystyle \L T(t) = 250 - 100e^{.9 - .03t}\)

 

[galactus]

I will try this. If I get off on a tangent, I trust someone will come along and give their input.

Take the derivative of the given formula(that's the rate of change which we know is 3 at 30 minutes):

\(\displaystyle \L\\\frac{d}{dt}[250-ae^{bt}]={-}abe^{bt}\)

\(\displaystyle \L\\{-}abe^{30b}=3\)........[1]


\(\displaystyle \L\\250-ae^{bt}\)

\(\displaystyle \L\\250-ae^{30b}=150\)............[2]


Solve [1] for a and sub into [2]:

\(\displaystyle \L\\a=\frac{100}{e^{30b}}\)

Which gives with [2]: \(\displaystyle \L\\\frac{-100}{e^{30b}}be^{30b}=3\)

Solving for b gives: \(\displaystyle \L\\b=\frac{-3}{100}\)

And \(\displaystyle a=100e^{\frac{9}{100}}\)

This gives a formula of:

\(\displaystyle \L\\250-100e^{\frac{9}{10}-\frac{3t}{100}}\)

If you plug in 30 you get 150 as a check.

What do you think?. Seem viable?.

I reckon it woud depend on the size of the yam to determine when it's done.

250 C is 482 F. That's too hot to cook a potato. It would more than likely burn on the outside before it was done on the inside.

EDIT: Looks like I am onto something. Skeeter concurs. As does Soroban.

 

[soroban]

Re: An oddly worded question

Hello, Sophie!

This is not an easy one . . .

A yam is put in an oven maintained at a constant temperature of 250°C.
Suppose that after 30 minutes, the temperature of the yam is 150°C
. . and is increasing at a rate of 3°C/min.
If the Temperature of the yam \(\displaystyle t\) minutes after it is put in the oven
is modeled by: \(\displaystyle \:T(t)\:=\:250\,-\,a\cdot e^{^{bt}}\), find \(\displaystyle a\) and \(\displaystyle b.\)

We are told that, when \(\displaystyle t\,=\,30,\;T \,=\,150\)

. . We have: \(\displaystyle \:250\,-\,a\cdot e^{^{30b}}\:=\:150\;\;\Rightarrow\;\;a\cdot e^{^{30b}}\:=\:100\;\) [1]

We are told that, when \(\displaystyle t\,=\,30,\:\frac{dT}{dt}\,=\,3\)
. . Since \(\displaystyle \frac{dT}{dt}\:=\:-ab\cdot e^{bt}\)
. . we have: \(\displaystyle \:-ab\cdot e^{30b}\,=\,3\;\;\Rightarrow\;\;a\cdot e^{30b} \:=\:-\frac{3}{b}\;\) [2]

Equate [1] and [2]:\(\displaystyle \:100 \:=\:-\frac{3}{b}\;\;\Rightarrow\;\;\L\fbox{b\,=\,-0.03}\)

Substitute into [1]: \(\displaystyle \:a\cdot e^{^{30(-0.03)}} \:=\:100\;\;\Rightarrow\;\;\L\fbox{a\:=\:100e^{^{0.9}}}\)

Hmmm, too slow again . . .

 

[galactus]

Yer not too slow Soroban. The Skeetster beat me while I was typing mine up.
At least we all agree.

 

[Sophie]

galactus, soroban and skeeter

Thanks very much for the input, I knew the question was easer than I made it. I am also greatful that you thought it was tough too, sometimes I just think I am of with the Fairys.

Thanks Sophie