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MTH 279
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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I'm having difficulty with your process on one solution to a 2/9 question and I think you did the same thing in class. In the following you say that since b-a= 0 means that a=-b. I put ## around the section I'm referring to. This leads to a problem later on. In the next # section the integral is wrong because the neg disappears. It was (1 / L) / P + (-1 / L ) / (L - P) = k dt . The second term is (-1 / L ) / (L - P) and when you integrate that using u sub with u= L-P then du= -dP the negatives cancel making the term positive but you have it neg still so that the two ln terms can be divided but if the integral is in fact pos then they cannot be divided so that goes all the way back to say that a = b and not a=-b. Does this make sense? It bothered me working it and comparing to your solution in the notes because it just doesn't seem right this way.
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We rewrite 1 / (P ( L - P ) ), using partial fractions:
1 / (P ( L - P ) ) = a / P + b / (L - P) for appropriate values of A and B.
a / P + b / (L - P) = ( a ( L - P ) + b P ) / (P ( L - P )) = (a * L + ( b - a ) * P )
1 / (P ( L - P ) ) = (a * L + ( b - a ) * P ) / (P ( L - P ) ).
The denominators of both sides are equal, so their numerators must be equal. So
a * L + (b - a) * P = 1.
This must be so for all values of P. Thus
a * L = 1 and
##
(b - a) = 0.
We conclude that a = 1 / L and b = - 1 / L.
##
Thus
1 / (P ( L - P ) ) = (1 / L) / P + (-1 / L ) / (L - P).
Our equation becomes
##
(1 / L) / P + (-1 / L ) / (L - P) = k dt
Integrating both sides we get
1 / L * ln | P | - 1 /L * ln | L - P | = k t + c
ln | P | - ln | L - P | = k L t + c
ln (| P | / | L - P |) = k L t + c
##
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You are right. My final result is correct, but I was careless on two steps. The second error was in integrating 1 / (L - P), which should be - ln | L - P |. With my former error in sign, which would have resulted in solution 1 / L * ln | P | + 1 /L * ln | L - P | , which would of course be a diaster. Correcting both errors we end up with the correct integral 1 / L * ln | P | - 1 /L * ln | L - P | :
The corrected steps:
(b - a) = 0 so a = b
We conclude that a = 1 / L and b = 1 / L.
##
Thus
1 / (P ( L - P ) ) = (1 / L) / P + (1 / L ) / (L - P).
Our equation becomes
##
(1 / L) / P + (1 / L ) / (L - P) = k dt
Integrating both sides we get
1 / L * ln | P | - 1 /L * ln | L - P | = k t + c
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