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| View Poll Results: Do you understand change? | |||
| Yes I understand change |
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4 | 17.39% |
| Yes I understand rate of change |
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1 | 4.35% |
| Yes I understand differences in rate of change |
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17 | 73.91% |
| I don't get it at all |
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1 | 4.35% |
| Voters: 23. You may not vote on this poll | |||
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#1
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The more I think about it, rate of change is a tricky concept. When you're moving at 60 mph, your position is changing but the rate at which it is changing is constant. A body falling in a vacuum, say 1,000 miles away from earth, is changing speed, accelerating at a steady rate.
Best I can recall, calculus has two main functions: to find the slope of any line tangent to an X - Y axis graph at any point on it -- the derivative; and to find the area between a non - linear graph and, say, the X axis -- the integral. Imagine a graph of the function Y = sinX + 2. Since the sin of any angle will always be between one and negative one, this function will be a sin wave completely above the X axis. With calculus, ideally you could find the area between the graph and the X axis for some portion of the graph. I say ideally because I can't recall if it only works with polynomials, where you have a finite number of constants and variables which are combined using only addition, subtraction, multiplication, and non-negative whole number exponents (raising to a power). This from Wikipedia, better than I could have said it. The sin function is not one of those. Not sure how you could use the ability to find the area between a curved line and the X axis, or between two curved lines, dlineated by straight lines on either end. I understand it opens all sorts of boxes though. More to the point, derivatives find the slope of a line tangent to the graph at any point on the graph and give a value for the rate of change of that graph at that point. Again from Wikipedia: A derivative is an instantaneous rate of change: it is calculated at a specific instant rather than as an average over time. And again, I'm not sure exactly what that operation is used for but I imagine physicists and aerospace engineers know quite well.
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Te futueo et caballum tuum 1986 300SDL, 362K 1984 300D, 138K Last edited by cmac2012; 01-09-2007 at 05:41 AM. |
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#2
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Of all the tools available to answering a question concerning rates of change, calculus answers it directly, quickly and as precisely as needs be. B |
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#3
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I can see how it would be really useful. The hard part I imagine, is representing real phemomena with a mathematical equation, such that the answer you get from the pure math has validity.
In some fields of endeavor, that could be relatively easy. In others, not so much, but I'm talking through my hat somewhat. Unusual for me.
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Te futueo et caballum tuum 1986 300SDL, 362K 1984 300D, 138K |
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#4
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Quote:
let's say you want to model the atmosphere over time. the atmosphere is 3-d and curved parallel to the Earth's curvature. It is also dynamic -- density changes with temperature and chemistry and so forth. You have lots of ground data and you want to INTEGRATE in over time and through space. let's ay you want to DERIVE a particular climatic condition from that data set at a particular time. B |
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#5
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I was looking at some calculus on-line to refresh my memory. Oh mah God, half or more looks like Greek to me now. I do remember getting a real kick out of that program I wrote, Newton's method for finding zero roots. Before the advent of computers, Good Lord, the application of that method to just one polynomial, let's say X to the fourth power on down, would take weeks of computation by hand. Basically, you start with a number far to the left of the Y axis, plug it in, and reduce the number by regular increments until the sign of the result changes, meaning you just crossed the X axis. I would multiply the result of each trial and check to see if it was less than 0, meaning you had just entered a zone where the previous result was either positive or negative and the new result was the opposite sign, yielding a negative product. You then do a derivative of the graph at the point of the most recent result, the sign changing figure, and use that line as a pointer to take you back to the X axis at a point close to where the graph crossed, and repeat the process until you get as close to zero as you'd like. With my program, I could do equations starting with X to the tenth power in seconds. Oh man, what Newton or Copernicus would have given to be in my shoes. Poor guys had to do lenghty cube root calculations and the like by hand.
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Te futueo et caballum tuum 1986 300SDL, 362K 1984 300D, 138K |
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