What Does 0/dt Mean in Calculus?

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The discussion revolves around the expression 0/dt in calculus, questioning its meaning. It is established that since 0 is a constant, its derivative is always 0, leading to the conclusion that 0/dt equals 0 if dt is a number. However, if dt represents a differential, the expression becomes meaningless. The conversation also hints at the complexity of limits, particularly in cases where both the numerator and denominator approach zero, suggesting that context is crucial for interpretation. Clarification on the original problem involving 0/dt is requested for a more accurate analysis.
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Homework Statement


I am working on a problem and am wondering what 0/dt is.

The Attempt at a Solution


Is it just 0, or does it turn into something with t?
 
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u mean d0/dt ?
the derivative of 0 is 0 no matter what.
 
Hmm, maybe it's a trick question, but logically, how does zero change with time? If your function is zero at all t, then I would assume yes, the answer is zero.

If this is for a lower level calc class, that is probably the answer. If it's higher level, then I am most likely wrong. :)
 
0 is a constant, it does not change. Ever.

So just wait for a math junkie to come in here and say "it depends on what 0 is".
 
If dt is a number, then 0/dt= 0. If dt is a differentianl, then 0/dt is simply meaningless. I wonder if you haven't got a fraction in which the numerator has a limit of 0? If the denominator is going to "dt", a differential, then, essentially, you have a "0/0" situation in which the limit depends on exactly how that limit is taken.

Please tell us what the problem was from which you got "0/dt".
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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