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OK, I've been doing this problem for fun (it's a great problem, BTW!):

Here is the solution:

However, I'm having a problem understanding the solution. How does the first line lead to the second line in the simplification of $F(t + dt)$? It looks like the author has taken $Vdt$ to be zero to get $x/\ell$ but has left in $Vdt$ for the second term in line two. Is that legal?

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BTW, I would like to know what this type of problem is called, and where I can get more practice with problems like these. (I especially like how we're required to come up with the integral from scratch. A lot of books have very routine calculus problems. It's harder to find problems like these.)@Bernhard – Joebevo Mar 4 '12 at 16:13
I don't think there is a special name for this, just the imagination of the teacher. – Bernhard Mar 4 '12 at 20:45

1 Answer 1

up vote 1 down vote accepted

$$ \frac{x+\frac{x}{l} Vdt -u dt}{l+V dt} $$

$$ \frac{x+\frac{x}{l} Vdt}{l+V dt}- \frac{u dt}{l+V dt}$$

$$ \frac{\frac{x}{l}(l+ Vdt)}{l+V dt}- \frac{u dt}{l+V dt}$$

$$ \frac{x}{l}\frac{l+ Vdt}{l+V dt}- \frac{u dt}{l+V dt}$$

$$ \frac{x}{l}- \frac{u dt}{l+V dt}$$

So no assumptions here.

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Thanks. That was simple, shoulda got that! – Joebevo Mar 4 '12 at 12:02

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