Tell me more ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.
up vote 0 down vote favorite

OK, I've been doing this problem for fun (it's a great problem, BTW!): http://www.physics.harvard.edu/academics/undergrad/probweek/prob76.pdf

Here is the solution: http://www.physics.harvard.edu/academics/undergrad/probweek/sol76.pdf

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?

share|improve this question
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

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.

share|improve this answer
Thanks. That was simple, shoulda got that! – Joebevo Mar 4 '12 at 12:02

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.