One remark 

Hello everybody.

Instead of calculating the k value from the boundary conditions as I did below, we should take into consideration that the rate constant k represents the fractional loss of mass per unit lenght. But from the problem we know the donkey eats 1 kg of wheat per km, that is, 1 kg/km.

This means that

K = 1

so that y(x) = Yo e^ - k ( x/s ) becomes

y(x) = Yo e^ - (x/s)

This is the general equation we were searching.

As x = s when the donkey arrives at the mill we have finally

y(s) = Yo / e

This is the particular solution of the problem.


Analytical result: 3678,794 kg (approximate result)

Computer result: 3647,427 kg (exact result)

Difference: 31,367 kg

In a short time I'll come back at this problem to answer to the following question:

Why the discrepance above?


Manuel C. Sousa

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