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Say for example, delta-P/delta-H. The pressure increases as the depth of the water increases. And you want to find the pressure at a certain depth. I see that most of these problems are solved by differentiation, but I have no real clue why.

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- Thread starter iflabs
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- #1

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Say for example, delta-P/delta-H. The pressure increases as the depth of the water increases. And you want to find the pressure at a certain depth. I see that most of these problems are solved by differentiation, but I have no real clue why.

- #2

ranger

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Is this what you are talking about [tex]\frac{dP}{dH}[/tex]? In that cause you need to separate the variables and have dP on one side and dH on the other. Then you integrate both sides to get rid of dP and dH. Then you will have an equation for P in terms of H. Can you post the entire equation? We do this becuase the given equation is a rate of change (or a derivative). So if we want to go from a derivative to the initial equation we must integrate. Here is another example, if we are give the velocity function for an object. And say we want the position function, we need to integrate the velocity function. Its like working backward.

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- #3

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Just FYI: in most cases, [tex]P \, =\, P_{0}+ \rho g h[/tex]

- #4

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The reason for this is that it frequently happens in science that given some starting conditions and you make a small change it is quite easy to work out pretty exactly what will happen but if you made a large change other factors come into play (the process is non linear) so your results would not be accurate.

To take the example of your pressure versus depth problem it is easy to see that the increase in pressure a little bit deeper in a lake than a given point is related to the change in the depth by the weight of the material involved in the small change. However the total pressure depends on the sum of all the little increases to the surface plus the atmospheric pressure. Assuming that water is essentially incompressible this is reasonably behaved because each of the steps is the same but if it was a compressible gas the amount added would be different for each step because the gas gets denser with each step.

I hope this helps your understanding and appreciation of the need for science to use differential equations

- #5

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I'm not sure the OP is solving ordinary "differential equations", rather than just doing basic calculus/differentiation.

The first thing to understand is that physical effects frequently depend on gradients (eg. heat conduction depends on temperature differences; velocity is the rate that position varies as time progresses).

It sounds like you can accept that, but don't understand why changing [itex]x^4[/itex] to [itex]4x^3[/itex] has anything to do with it. If so, you may need your mathematics teacher to help explain how the particular differentiation technique always has something to do with the rate things change at (or the slopes of hills, whatever). It's really just a tool stolen from mathematics whenever it seems to make things easier for physicists.

The first thing to understand is that physical effects frequently depend on gradients (eg. heat conduction depends on temperature differences; velocity is the rate that position varies as time progresses).

It sounds like you can accept that, but don't understand why changing [itex]x^4[/itex] to [itex]4x^3[/itex] has anything to do with it. If so, you may need your mathematics teacher to help explain how the particular differentiation technique always has something to do with the rate things change at (or the slopes of hills, whatever). It's really just a tool stolen from mathematics whenever it seems to make things easier for physicists.

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- #6

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If something is changing with time or position and you know its value at one point you need something that helps you find it at the other if for example it changes exactly the same amount for every uniform step you take its no problem you just count the number of steps and multiply that by the amount of change by each step. this is a linear relationship. It is called this because the graph of how the value varies with each step is a straight line.

However if the amount it changes with each step changes with each step (that is it is what scientists call an non linear) you have a problem that's where calculus comes in.

Now how you go about solving the problem depends on what you know. The differential equations I was talking about earlier deal with when you can see how things will change in small steps which is the usual scientific way of going about things.

If you know the equation of how the value varies you may need to use differentiation to help you (unfortunately this is a bit arbitrary because teachers tend to start with the statement of the equation and you tend to ask why the answer is probably that someone solved the differential equation in the first place and wrote the answer down for the teacher to use !! :-) )

Now to get back to solving the problem by differentiation. Firstly the basic process of differentiation goes through a lot of little linear steps to show how the slope of the non linear graph produced by the equation changes and comes out with another equation that tells us how the slope changes and allows us to answer the question in one step and not lots of little ones.

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