If f = A (dv/dx), then how to calculate the value of f?

[Indranil]

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.

 

[Dr.Peterson]

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.

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.

 

[Indranil]

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.

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?

 

[Steven G]

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 [(1/s)]= 1cm²/s

a/b/c = a/(bc)

 

[HallsofIvy]

What is confusing us is that "dv/dx" usually means that v is some function of the variable x and dv/dx is the derivative of v with respect to x. But here you are telling us that dv and dx are simply quantities and dv/dx is a simple division. 1 "cm per second" divided by "1 cm" is just "1 per second" (Whatever that means. These are your units.) With A also equal to 1 square centimeter then "Adv/dx" is "1 square centimeter per second".

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 more to this than you have said. Are you given v as a function of x? If so, what function?

 

[Indranil]

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

a/b/c = a/(bc)

Thank you very much

 

[Dr.Peterson]

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?

The 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.

 

[HallsofIvy]

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

a/b/c = a/(bc)

I would say rather that "a/b/c" is ambiguous it could mean either (a/b)/c= a/bc or a/(b/c)= ac/b.

 

[Indranil]

The 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.

Thanks a lot

 

[Indranil]

I would say rather that "a/b/c" is ambiguous it could mean either (a/b)/c= a/bc or a/(b/c)= ac/b.

Then what would be the right way, could you suggest, please?

 

[Dr.Peterson]

Then what would be the right way, could you suggest, please?

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).)

But the way to avoid ambiguity is to use parentheses, just as was shown -- and just as you did, in fact: 1cm² [(1cm/s) / (1cm)]. The comment was not made to you, but to Jomo.

 

[Steven G]

Please note that post was in Beginning Algebra so I suspect that dV/dt is not what we usually think of when we see it.