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Maple Diff.Eq. Help

alane1994

Active member
Oct 16, 2012
126
Maple Diff.Eq.

Not really sure where to place this, as it crosses a couple lines. However, much of the confusion arises from the programming.

Here is what I have.

My code,



My problem,



My confusion is from what does it mean for Pick correct value for ktemp in the last line of code that I have and in the recommended codes?
 
Last edited:

dwsmith

Well-known member
Feb 1, 2012
1,673
I think you are supposed to use the values that correspond to the problem you are solving. For instance, take the line that says \(T(0) = ???\), that would want the initial condition for the desired problem. So I think the questions marks are related to what the problem statement contains.
 

alane1994

Active member
Oct 16, 2012
126
Yeah, I got those lines.

T(0)= 40

My concern is with line 7 of recommended code.
And the notated line after (13) in my code.
 
Last edited:

dwsmith

Well-known member
Feb 1, 2012
1,673
Yeah, I got those lines.

T(0)= 40

My concern is with line 7 of recommended code.
And notated line (13) in my code.
\(k = hA\) where \(h\) is the heat transfer coefficient and \(A\) is the heat transfer surface area.
 

alane1994

Active member
Oct 16, 2012
126
Stupid question...
How on earth do I apply that to the problem, and with the given information?
 

dwsmith

Well-known member
Feb 1, 2012
1,673

alane1994

Active member
Oct 16, 2012
126
Ok,

Now... how would I find the surface area of my object? I see nothing that would indicate lengths, widths, heights, or area.
 

dwsmith

Well-known member
Feb 1, 2012
1,673
Ok,

Now... how would I find the surface area of my object? I see nothing that would indicate lengths, widths, heights, or area.
I suppose you could Google what the surface area of a gallon jug or half gallon. It may even be in the back of your book.
 

alane1994

Active member
Oct 16, 2012
126
Well...

My text book and my Maple Lab Book are two different things.
And I can't imagine that they would give you a problem that you don't have all the necessary information in the question. And hence, where some of my confusion of this problem comes from.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Newton's law of Cooling states that the time rate of change of the temperature $T$ of an object is proportional to the difference between the ambient temperature $M$ and the temperature of the object. Stated mathematically, this is:

\(\displaystyle \frac{dT}{dt}=-k(T-M)\) where \(\displaystyle T(0)=T_0,\,0<k\in\mathbb{R}\) and \(\displaystyle T>M\).

The ODE is separable and may be written:

\(\displaystyle \frac{1}{T-M}\,dT=-k\,dt\)

Integrating, using the boundaries, and dummy variables of integration, we find:

\(\displaystyle \int_{T_0}^{T(t)}\frac{1}{u-M}\,du=-k\int_0^t v\,dv\)

(1) \(\displaystyle \ln\left(\frac{T(t)-M}{T_0-M} \right)=-kt\)

If we know another point \(\displaystyle \left(t_1,T_1 \right)\) we may now find $k$:

\(\displaystyle -k=\frac{1}{t_1}\ln\left(\frac{T_1-M}{T_0-M} \right)\)

Hence, we find:

(2) \(\displaystyle T(t)=\left(T_0-M \right)\left(\frac{T_1-M}{T_0-M} \right)^{\frac{t}{t_1}}+M\)

(3) \(\displaystyle t=\frac{t_1\ln\left(\frac{T(t)-M}{T_0-M} \right)}{\ln\left(\frac{T_1-M}{T_0-M} \right)}\)