# Petroleum engineering maths help

• HW.
In summary: I am not going to keep discussing this with you, because we have no way of knowing what you are thinking, and so I can't follow your argument. The person who wrote the text made a mistake. He or she meant 'velocity' rather than 'distance.'
HW.
Hi guys. I am having trouble with my petroleum engineering course. I have attached the example that's giving me problems, start at 1.1.1 limits and just read through to the second pic where it states that the graph at 3 sec shows a height of -96.522ft and at 2 and 4 secs the height is at -48.261ft and -241.305. Am I missing something here? I have no clue where these numbers came from. Theres a formula on pic 1 but I can't seem to get the same answers and also from the graph the heights are not coinciding with time. This is a renowned handbook that pro's use so it's not a mistake. Any help appreciated!

#### Attachments

• pic 1.jpg
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• Pic 2.jpg
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Well, I'm going to bet that, even though it is a renowned handbook, there's a mistake. I don't see where the -96 is coming from. It looks more like -150. The -48.261 and -241.305 coming from reading where the tangent line intersects the t = 2 and t = 4 lines, but the six-figure accuracy is ridiculous. You'd be hard-pressed to get two figures from these graphs.

I think the only mistake is in the sentence:

In Fig. 1.1 at the time of 3 seconds, the distance is -96.522 ft

If should really state:

In Fig. 1.1 at the time of 3 seconds, the slope is -96.522 ft/s

Then, they continue to state how the can calculate the slope of the curve at 3 seconds by using the 2- and 4-second points and doing (y2-y1)/(x2-x1) ...but these x's and y's are those of the straight line, not the curve.

As you know, the curve is represented by y = - gt2/2

And the slope of the curve at any time is simply its derivative, and so, the equation of the straight line is y = -2gt/2 = -gt

And so, having this last equation and substituting gravity, y = -32.174t and so, it is not ridiculous to say that at 3 seconds, y = -96.522 ...nobody said that they read the number from the graph...it came from the formula.

gsal said:
I think the only mistake is in the sentence:

In Fig. 1.1 at the time of 3 seconds, the distance is -96.522 ft

If should really state:

In Fig. 1.1 at the time of 3 seconds, the slope is -96.522 ft/s

Then, they continue to state how the can calculate the slope of the curve at 3 seconds by using the 2- and 4-second points and doing (y2-y1)/(x2-x1) ...but these x's and y's are those of the straight line, not the curve.

As you know, the curve is represented by y = - gt2/2

And the slope of the curve at any time is simply its derivative, and so, the equation of the straight line is y = -2gt/2 = -gt

And so, having this last equation and substituting gravity, y = -32.174t and so, it is not ridiculous to say that at 3 seconds, y = -96.522 ...nobody said that they read the number from the graph...it came from the formula.

First of all, you're awesome. Secondly, shouldn't dy/dt=-32.174t as apposed to y? so actually velocity=-96.522ft/s and not distance?

gsal said:
I think the only mistake is in the sentence:

In Fig. 1.1 at the time of 3 seconds, the distance is -96.522 ft

If should really state:

In Fig. 1.1 at the time of 3 seconds, the slope is -96.522 ft/s
Ah! Yes, I'm sure you're right.

Then, they continue to state how the can calculate the slope of the curve at 3 seconds by using the 2- and 4-second points and doing (y2-y1)/(x2-x1) ...but these x's and y's are those of the straight line, not the curve.

As you know, the curve is represented by y = - gt2/2

And the slope of the curve at any time is simply its derivative, and so, the equation of the straight line is y = -2gt/2 = -gt
Here you lose me. First, the equation of a line is y = m t + y0. If the slope of the line is -gt, then at t=3 m = -3g, and the equation of the tangent is y = (-3g)t + y0. Second, you can't assume y0 = 0. It is obviously about 48.

And so, having this last equation and substituting gravity, y = -32.174t and so, it is not ridiculous to say that at 3 seconds, y = -96.522 ...nobody said that they read the number from the graph...it came from the formula.
Oh, come on. y at 3 sec is not -96.522, or anything close to it. Whether it came from the formula or a graph, the statement that y is -96.522 is WRONG. I think your first idea was right: they said "distance" when they meant "slope".

To HW: You are kind of right, I am badly re-using the y...the equation is more like dy/dt = -32.174t...but for the purposes of having it in a form more familiar when ploting in the x,y cartesian plane ( more like t,y, here), I simply referred to that equation as being y = -32.174t ...we could also replace time, t, with x, if you want...and write y=-32.174x...

To pmsrw3: There are two equations at play, here, the one for distance at any time during free falling [ y = -gt2/2] and its corresponding derivative which indicates its slope at any given time, as well [y = -gt] ...AGAIN, noticed that these two equations can be treated as two different equation and hence, I am re-using the 'y' ...but, of course, one should not forget that they are actually related...

Anyway, for as long as HW understood...

gsal said:
To pmsrw3: There are two equations at play, here, the one for distance at any time during free falling [ y = -gt2/2] and its corresponding derivative which indicates its slope at any given time, as well [y = -gt]
No, y = -gt is not the derivative equation that indicates the slope at any given time. The equation that indicates the slope at any given time is dy/dt = -gt. This is not a trivial difference.

## What is Petroleum Engineering?

Petroleum engineering is a field of engineering that focuses on the exploration, extraction, and production of oil and natural gas. It involves using various techniques and technologies to find and extract these resources from the earth's subsurface.

## What is the importance of math in Petroleum Engineering?

Math is a crucial component of petroleum engineering as it is used to solve complex problems and make accurate calculations related to drilling, production, and reservoir engineering. It is also used to analyze data and make informed decisions about the efficiency and profitability of oil and gas operations.

## What are the key mathematical concepts used in Petroleum Engineering?

The key mathematical concepts used in petroleum engineering include calculus, linear algebra, differential equations, statistics, and numerical analysis. These concepts are applied to solve problems related to drilling, reservoir characterization, and production optimization.

## Are there any specific software used for Petroleum Engineering math calculations?

Yes, there are several software programs used in the field of petroleum engineering for mathematical calculations, such as MATLAB, Petrel, Eclipse, and CMG. These programs use algorithms and mathematical models to simulate and analyze various aspects of oil and gas operations.

## How can I improve my mathematical skills for Petroleum Engineering?

To improve your mathematical skills for petroleum engineering, it is important to have a strong foundation in calculus, linear algebra, and differential equations. Additionally, practicing problem-solving and utilizing software programs can help you gain a better understanding of how math is applied in the field of petroleum engineering.

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