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Determine the dimensions of dP/dx.

Where P is a force and x represents distance

So dimensions of P would be ML/T^2

Dimensions of dP/dx would be (ML/T^2)' = M/T^2

Is that right?

- Thread starter pyroknife
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- #1

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Determine the dimensions of dP/dx.

Where P is a force and x represents distance

So dimensions of P would be ML/T^2

Dimensions of dP/dx would be (ML/T^2)' = M/T^2

Is that right?

- #2

TSny

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Yes, that look's correct.

[Although, I've never seen anyone take a derivative of a dimensional expression as you did! Dimensionally, you can think of a derivative as just a fraction. So, the dimensions of dP/dx is just the dimensions of the numerator (P) divided by the dimensions of the denominator (x). Think of dP as essentially the same as ΔP which is a change in pressure and therefore has the dimensions of pressure. Similarly for dx.]

[Although, I've never seen anyone take a derivative of a dimensional expression as you did! Dimensionally, you can think of a derivative as just a fraction. So, the dimensions of dP/dx is just the dimensions of the numerator (P) divided by the dimensions of the denominator (x). Think of dP as essentially the same as ΔP which is a change in pressure and therefore has the dimensions of pressure. Similarly for dx.]

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- #3

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I posted in the introductory forum an extension of this question https://www.physicsforums.com/showthread.php?p=4050555#post4050555.

This one really confused me, I've never seen anything like it.

- #4

Chestermiller

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chet

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No he stated that P was force.

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TSny

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Ugh, I mean't to say force instead of pressure. Sorry, Chestermiller and pyroknife.Yes, that look's correct.

[Although, I've never seen anyone take a derivative of a dimensional expression as you did! Dimensionally, you can think of a derivative as just a fraction. So, the dimensions of dP/dx is just the dimensions of the numerator (P) divided by the dimensions of the denominator (x). Think of dP as essentially the same as ΔP which is a change in pressure and therefore has the dimensions of pressure. Similarly for dx.]

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