Maple Diff.Eq. Help: Pick Correct Value for ktemp

[alane1994]

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?

 

[Dustinsfl]

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]

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.

 

[Dustinsfl]

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]

Stupid question...
How on Earth do I apply that to the problem, and with the given information?

 

[Dustinsfl]

Stupid question...
How on Earth do I apply that to the problem, and with the given information?

Here is a formula for the heat transfer coefficient:

https://en.wikipedia.org/wiki/Heat_transfer_coefficient

 

[alane1994]

Ok,

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

 

[Dustinsfl]

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]

Well...

My textbook 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]

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:

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

The ODE is separable and may be written:

$$\frac{1}{T-M}\,dT=-k\,dt$$

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

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

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

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

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

Hence, we find:

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

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