- Thread starter Indranil
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First, please tell us why you are asking this question. It is a very odd one, as "dv/dx" normally does not mean anything like what it has to mean in this question. Where did this come from?How to calculate the value of f if 'f = A (dv/dx)'

if dx is 1 cm, A is 1cm^2 and dv is 1cm/s. Kindly show me the methods.

Second, please show us your attempt, so we can see what help you need. If, as you require, we take dv and dx as variables with given values, the evaluation is very easy. The only hard part I see might be to determine the units in the answer, and even that should not be too hard if you think of division by dx as multiplication by its reciprocal. The more you show of your own thinking, the more effectively we can help.

If dx is 1 cm, A is 1cm^2 and dv is 1cm/s.First, please tell us why you are asking this question. It is a very odd one, as "dv/dx" normally does not mean anything like what it has to mean in this question. Where did this come from?

Second, please show us your attempt, so we can see what help you need. If, as you require, we take dv and dx as variables with given values, the evaluation is very easy. The only hard part I see might be to determine the units in the answer, and even that should not be too hard if you think of division by dx as multiplication by its reciprocal. The more you show of your own thinking, the more effectively we can help.

putting the value of A, dv,and dx f = A (dv/dx) = 1cm² [(1cm/s) / (1cm)] = I don't understand how to solve that portion. Could you help me, please?

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1cm² [(1cm/s) / (1cm)] = 1cm² [(1/s)]= 1cmIf dx is 1 cm, A is 1cm^2 and dv is 1cm/s.

putting the value of A, dv,and dx f = A (dv/dx) = 1cm² [(1cm/s) / (1cm)] = I don't understand how to solve that portion. Could you help me, please?

a/b/c = a/(bc)

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But I am inclined to think, since, as I said, "dv/dx" usually means the "derivative of v with respect to x", that there is

Thank you very much1cm² [(1cm/s) / (1cm)] = 1cm² [(1/s)]= 1cm^{2}/s

a/b/c = a/(bc)

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The way I suggested to make it easier is to change the division to a multiplication:If dx is 1 cm, A is 1cm^2 and dv is 1cm/s.

putting the value of A, dv,and dx f = A (dv/dx) = 1cm² [(1cm/s) / (1cm)] = I don't understand how to solve that portion. Could you help me, please?

1cm² [(1cm/s) / (1cm)] = 1cm² (1cm/s) (1/(1cm)) = 1cm² (1/s) (1/1) = 1 cm²/s

But as two of us have now said, we don't trust that the problem really is as you said. Please quote the entire original problem.

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I would say rather that "a/b/c" is1cm² [(1cm/s) / (1cm)] = 1cm² [(1/s)]= 1cm^{2}/s

a/b/c = a/(bc)

Thanks a lotThe way I suggested to make it easier is to change the division to a multiplication:

1cm² [(1cm/s) / (1cm)] = 1cm² (1cm/s) (1/(1cm)) = 1cm² (1/s) (1/1) = 1 cm²/s

But as two of us have now said, we don't trust that the problem really is as you said. Please quote the entire original problem.

Then what would be the right way, could you suggest, please?I would say rather that "a/b/c" isambiguousit could mean either (a/b)/c= a/bcora/(b/c)= ac/b.

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I would not call it ambiguous; it is standard to evaluate a/b/c from left to right, though some people might read it otherwise. (An expression that is more definitely ambiguous is a/bc, for which texts actually teach different interpretations. So Halls actually erred in not writing that as a/(bc).)Then what would be the right way, could you suggest, please?

But the way to avoid ambiguity is to use parentheses, just as was shown -- and just as