# Maple Diff.Eq. Help

#### alane1994

##### Active member
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?

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#### dwsmith

##### Well-known member
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
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.

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#### dwsmith

##### Well-known member
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
Stupid question...
How on earth do I apply that to the problem, and with the given information?

#### alane1994

##### Active member
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
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
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

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