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Post by abacus9900 on Jan 6, 2011 18:19:57 GMT 1
Can someone please provide a basic example or two of calculus? I know there are two types, namely, integration and differentiation but what are these?
Thanks.
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Post by eamonnshute on Jan 6, 2011 18:59:03 GMT 1
Calculus is the mathematics of change. For example, If we know the acceleration of a body at any time we can calculate its resulting velocity, and from that we can also calculate the distance travelled. This is integral calculus.
On the other hand, if we know how the distance travelled by an object varies with time then we can calculate its speed at any time, and then its acceleration at any time. This is differential calculus.
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Post by abacus9900 on Jan 6, 2011 19:16:53 GMT 1
Calculus is the mathematics of change. For example, If we know the acceleration of a body at any time we can calculate its resulting velocity, and from that we can also calculate the distance travelled. This is integral calculus. On the other hand, if we know how the distance travelled by an object varies with time then we can calculate its speed at any time, and then its acceleration at any time. This is differential calculus. Thanks Eamonn, that is pretty clear. Could you perhaps give a specific example?
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Post by speakertoanimals on Jan 6, 2011 20:17:59 GMT 1
Distance travelled, speed and acceleration IS a specific example.........
Let's take simplest case, constant acceleration, as when an object falls or rises freely under gravity.
Then we have acceleration is a constant (call it a).
When we integrate this to find the speed as a function of time (acceleration being the rate of change of speed with time), we get the usual formula v = u + at,
where v is the speed at time t, and a is the constant acceleration. Notice that we have had to add another constant (u), which is the speed at the time t=0.
If we integrate again, to get the distance travelled, then we get the other usual formula:
s = ut + (1/2)at^2
where we have set distance s to be zero at t=0 (hence in this case, no extra constant comes in).
If we start from the formula for s and differentiate, we recover the previous two formulae (v=u+at, and acceleration =a).
An easy way to think of this is suppose we draw a graph of the position of our object (just one dimension for the moment!) in terms of height on the y-axis as a function of time on the x-axis. DIfferentiating with respect to time is the slope of this graph (the speed), and differentiating again (the slope of the graph of the slope against time!) is acceleration.
But graphically we can also go the other way. If we have a graph of the speed as a function of time, then the distance travelled between any two times is the AREA under the speed against time graph between those two times.
We hence see that the calculus is just a handy way of doing what we can imagine doing for ANY graph -- finding the slope of the graph (differential calculus), or finding the area under any portion of the graph (the intgeral calculus). All that the fiddly algebra of it all says is -- if this is the formula for the graph, this is how to find the slope, and this is how to find the area.
When it comes to simple graphs, that is about the simplest question you can ask -- slopes and areas. The whole of the calculus (almost) in a simple picture.
Additions to this just have more variables -- for example, rather than a simple graph, we have a landscape, a function that is like a surface, where we have position in x and y, and value of the function in the z direction. Differentiation is then how to find the slope of the surface (think of being on a hill, the slope depends on the direction you look in), whereas integral calculus is area under a cross-section of the hill (imagine cutting through a hill with a straight knife, and thinking of the area of the face you have exposed), or the volume under some section of the hill.
The usefulness should be apparent. For example, if you wanted to build a house on that hill, you'd want to first make a terrace to put it on, and you'd need to know how much earth and rock that would mean digging out. And if you did dig it out, and pile it up at the bottom of the garden, you'd want to know how big the spoil-heap would be, and whether or not it would obscure your view! Capability Brown probably didn't use the calculus though..................
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Post by abacus9900 on Jan 6, 2011 20:57:08 GMT 1
You haven't explained this properly.
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Post by speakertoanimals on Jan 6, 2011 21:25:04 GMT 1
Which part of that did you fail to understand (snip)
Bilious remarks are boring to everyone expect the person who makes them, speaker. Get it under control or quit.
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Post by rsmith7 on Jan 6, 2011 21:38:19 GMT 1
Which part of that did you fail to understand, or is it the plus sign that foxed you....................... If you can't understand a simple, linear equation, with all terms defined, then you'd better give up on trying to understand calculus, and go back to basic arithmetic (or just stop messing about and wasting my time). And you're a lecturer? I sincerely hope you never come into contact with any of my children.
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Post by Progenitor A on Jan 6, 2011 21:46:52 GMT 1
Which part of that did you fail to understand, or is it the plus sign that foxed you....................... If you can't understand a simple, linear equation, with all terms defined, then you'd better give up on trying to understand calculus, and go back to basic arithmetic (or just stop messing about and wasting my time). And you're a lecturer? I sincerely hope you never come into contact with any of my children. ;DQuite easy to spot a fraud isn't it? That's integration! Usual formula Hahaha. Dunno why Newton and Liebniz bothered!
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Post by abacus9900 on Jan 6, 2011 22:06:16 GMT 1
Which part of that did you fail to understand, or is it the plus sign that foxed you....................... If you can't understand a simple, linear equation, with all terms defined, then you'd better give up on trying to understand calculus, and go back to basic arithmetic (or just stop messing about and wasting my time). STA, if you are not prepared to explain things properly to people who may not have had the educational advantages that you have had then I don't really see the value in you posting here. You see, you haven't even defined exactly how acceleration is expressed, so how can you expect people to follow precisely what you have to say? Leibniz and Newton must be turning in their graves!
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Post by abacus9900 on Jan 7, 2011 12:00:53 GMT 1
Which part of that did you fail to understand, or is it the plus sign that foxed you....................... STA, as per usual, I have not had a decent answer to my question. Frankly, I can't see why people bother to use science messageboards if all you get is either a perfunctory answer, an answer that goes completely over one's head, or simply given a link to a site that deals with the material (something I could do on my own anyway). There are good messageboards I have visited (based in America I think) that do often seem to provide in-depth answers to posters' questions but I am afraid that this one and also your home one do not compare favourably with them. It's surprising that those individuals who do seem to possess specialist knowledge seem rather loathed to make a proper effort in sharing it with others, perhaps assuming that the questioners already have attained a certain level of education, thereby saving the 'specialist' the effort to provide proper explanations. It really is a pity that forums such as this are not being used to their full potential because the Internet can be a wonderful resource if only people would realize it and prepared to take a little time to share their knowledge.
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Post by speakertoanimals on Jan 7, 2011 12:55:02 GMT 1
I assumed that people of a general level of intelligence (snip)
More bile. You have been warned, speaker.
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Post by abacus9900 on Jan 7, 2011 13:11:14 GMT 1
I assumed that people of a general level of intelligence would know what acceleration was (ditto speed and distance and time). Or do you really need those explaining to you as well? If so, you're dimmer than even I thought you were....................... Of course I and others here know what acceleration is but it is the mathematical formulation of it that needs to be clearly conveyed and how it is combined in an equation to compute velocity and so on. Again, this is another example of a thread that seems to have descended into petty squabbling instead of being used as an opportunity to exchange knowledge. What a pity.
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Post by speakertoanimals on Jan 7, 2011 14:39:50 GMT 1
v = u +at
I've already said it, WHY do you keep claiming I haven't.................
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Post by Progenitor A on Jan 7, 2011 15:58:47 GMT 1
Hi Abacus
You may find this short lesson helpful on differentiation and how the equations of velocity acceleration and distance are calculated using the calculus (where integration is the function that existed before differentiation). You may also find it unhelpful, but do refard ir as a start and we can move on from there.
Differential Calculus We are all accustomed , for example, to calculating average speed. It is s the distance travelled, divided by t the time taken to travel that distance:
v= Delta s/Delta t
If the distance travelled is constant with time, then the speed is constant
Similarly for acceleration, , if the speed increases in equal increments for equal periods of time then the acceleration is constant
a=Delta v/Delta t
We do not require differential calculus
But if the distance s (of the velocity v) is varying with equal time increments, how do we calculate the instantaneous speed at any time? (Similarly if the velocity v increases non-linearly with equal time intervals, how do we measure the instantaneous acceleration)
These were, in essence , the problems that Newton and Leibniz were addressing when they invented their differential calculus.
If we do not have the differential calculus, an approximation to the instantaneous speed (or acceleration) is arrived at by taking the slope of the graph at various points that we are interested in. This is accomplished by drawing a tangent to the curve at the point of interest and the slope is given by Delta y/Delta x. But once again this local slope gives only an approximation to the instantantous rate of change, and the process is very tedious
If we know the mathematical function that describes a term such as changing acceleration, then Leibniz and Newton give us a shorthand method of calculating the true instantaneous rate of change at any point on the curve that describes the mathematical function
Thus if we have the mathematical function
y= 5x2 + 7 …………………(i)
Newton tells us to add a small increment dy to y. This causes a small increment to x So we have y + dy=5(x + dx)2 +7…………..(ii)
Now (ii) minus (i) gives us
[y + dy=5(x + dx)2 +7] –[ y= 5x2 + 7] dy =5(x + dx)2 – 5x2 dy =5[(x + dx)2 – x2] expand ing out the term in the inner brackets gives us:
dy =5(2x + dx) dx
and dy/dx = 5(2x + dx)
Now if we make dx approach 0 dy/dx = 5(2x + 0)
dy/dx=10x
This is called differentiation from 1st principles
In general if (for the simplest cases) y= xn
dy/dx=nxn-1
Now let’s have a look at rate of change of distance s
s=kt This equation tells us that distance is increasing uniformly as a function of time
s=kt1 and using the general result above
ds/dt=1 x kt(1-1) = k = v (velocity)….the speed is constant (and is simply the slope of the graph)
if velocity is increasing as a function of time we have acceleration
a=dv/dt in other words acceleration (as a function of time) is the rate of change of velocity
Now what function will, before it is differentiated, give us a?
Try this: at=v dv/dt= at1-1 =a.......!
Now what we have done here, finding the form that a differentiated function had before it was differentiated is called Integration. The instantaneous velocity is given by v= at But that assumes that the initial velocity was zero, if it isn’t we must add the initial velocity
v= at + u whre u is th einitial velocity
If we wish to find the distance travelled during the acceleration phase: v=ds/dt ds/dt=at+u
Again what function gives us (at +u) before it is differentiated?
And we find (at2/2 + ut) fits the bill (we have integrated the function)
s = at2/2 + ut
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Post by abacus9900 on Jan 7, 2011 16:21:36 GMT 1
Thank you naymissus, although I will have to study what you have told me, at least you have taken the trouble to provide a proper account of what calculus does and I can come back and ask you about points I need help on.
Thanks again.
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