Are we considering point a line segment?

  • #1

Main Question or Discussion Point

Hello, I have studied difrentiation about some weeks ago and as I was thinking, due to the limiting factor, when dy & dx approch zero, we have actully come up with a point, now how can we find the gradient of a point because a point when magnified is actually a circle, & a circle can have a billion gradients. But we get an exact value after difrentiating so havn't we considered the point as a line segment???

I'll be thankfull for your answer.
:smile:
 

Answers and Replies

  • #2
VietDao29
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mubashirmansoor said:
Hello, I have studied difrentiation about some weeks ago and as I was thinking, due to the limiting factor, when dy & dx approch zero, we have actully come up with a point, now how can we find the gradient of a point because a point when magnified is actually a circle, & a circle can have a billion gradients. But we get an exact value after difrentiating so havn't we considered the point as a line segment???

I'll be thankfull for your answer.
:smile:
No, a point is just... err, a point, not a circle, or a line secment (even if you magnify it). :)
-----------------
Now let's draw a graph of any function y = f(x) on the Catersean plane.
Let's say you want to find f'(x0).
We define a point A(x0, f(x0)), and another point B(x0 + a, f(x0 + a)), a is a real number.
So the line connecting the 2 point A, and B will have the slope of:
[tex]m = \frac{f(x_0 + a) - f(x_0)}{(x_0 + a) - x_0} = \frac{f(x_0 + a) - f(x_0)}{a}[/tex], right?
Now let [tex]a \rightarrow 0[/tex], that means [tex]B \rightarrow A[/tex], right? So the line connecting A, B will approach the tangent line of the graph f(x) at A, right?
So:
[tex]f'(x_0) = \lim_{a \rightarrow 0} \frac{f(x_0 + a) - f(x_0)}{a} = \mbox{Slope of the tangent line at (x_0, f(x_0))}[/tex].
Can you get it? :)
 
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  • #3
matt grime
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you are not fiding the slope *of* a point but the slope *at* a point. which is the nearest I can come to making that make sense.
 
  • #4
Thankyou, But I guess there is still a question mark left for me,

After applying the limiting factor how can we find the tanjent to the point because say, y=x^2 & y+dy=(x+dx)^2 when we say dy & dx approches 0 then we are again talking about the same points of x&y related by the first equation, & that the point of (x,y) is just a point, how can we relate it with the tanjent??

By the way isn't the point a circle with an infintely small radius????

I'll be really thankfull for your kind replies.
 
  • #5
VietDao29
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mubashirmansoor said:
By the way isn't the point a circle with an infintely small radius????
No, a point is just a point, not a circle. When you magnify it, it's still a point.
mubashirmansoor said:
...After applying the limiting factor how can we find the tanjent to the point because say, y=x^2 & y+dy=(x+dx)^2 when we say dy & dx approches 0 then we are again talking about the same points of x&y related by the first equation, & that the point of (x,y) is just a point, how can we relate it with the tanjent??
No, when we take dx -> 0, dx is not 0, it just tends to 0. So we still have 2 separate points, A, and B.
So when dx -> 0, then dy -> 0, too, right? So the line AB will approach the tangent line to f(x) at A, right?
You can look at my #2 post again, and see if you can get it.
Is it all clear now? :)
 
  • #6
So ultimately, should I consider that delta(y)/delta(x) is approximately the gradient, & not exactly. dy/dx ~ delta(y)/delta(x)

Am I considering it correctly? is that what you ment?

Thankyou once again:)
 
  • #7
HallsofIvy
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It's a bit awkward to talk about geometric objects such as points and lines at the same time you are talking about limits! As you say [itex]\frac{\Delta y}{\Delta x}[/itex] for any non-zero [itex]\Delta x[/itex] is only approximately equal to [itex]\frac{dy}{dx}[/itex] which is defined at the limit of such things.
 
  • #8
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VietDao29 said:
No, a point is just a point, not a circle. When you magnify it, it's still a point.
But isn't a circle defined to be the set of points that are equidistant from a fixed point? True, it makes no sense to say a point is a cirlce, but it still fits the definition and equation of a circle of radius 0. I couldn't find reference that stipulated the radius must be positive.
 
  • #9
matt grime
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daveb said:
But isn't a circle defined to be the set of points that are equidistant from a fixed point?
that is one way of thinking about things, but not the only one. as in many parts of maths, what certain things are is a tricky issue. is 0 a natural number? is 0 positive? is it negative? these are the so-called degenerate cases.

However, if you're talking about 'blowing things up' then a point and a circle have strictly different meanings: a circle is locally one dimensional, in this context, i.e. we often preclude circles of radius zero when considering topological properties, or at least we indicate that this case is somehow distinguished...

Anyway, this is all moot, and doesn't alter the fact that the OP's claim that a point 'magnified' is a circle is just not mathematically sound.
 
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  • #10
So how should I define a point?????/
 
  • #11
dav2008
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A dimensionless object whose only property is location.
 
  • #12
matt grime
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Let's get away from assuming things that are unnecessary (such as the idea of a location making sense), a point in some space is just a singleton set. It is better to leave it as an undefined primitive and assume everyone knows what it is, since any attempt to uniquely classify such things is bound to fail since they are context dependent, but evidently that is not possible
 
  • #13
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mubashirmansoor said:
So how should I define a point?????/
Isn’t “point” usually left as an undefined term?
 
  • #14
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Zero is neither positive nor negative (unless you're looking at a limit that approaches zero. But then, it still isn't exactly zero that's positive or negative). Zero is not a natural number, since the natural numbers are the positive integers. Whole numbers are non-negative integers, and thus include zero. That's the reason there is a distinction between non-negative and positive (or non-positive and negative). These aren't what I'd call degenerate cases. Considering a point as a circle or a line segment, yes, those are degenerate cases.

If you consider a point as a circle with radius zero, how much would you have to magnify it to get, say, a unit circle? Similarly, if you consider a point as a line segment of zero length, how much would you have to magnify it to get a unit length line segment? Probably the best way to view dy and dx is that they don't actually reach 0, they just get close enough that we can do some really neat things with them. A bit simplistic, but at least it can be comprehended.
 
  • #15
matt grime
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Now, Nimz, this is where you've got it wrong. Some people do include 0 as a natural number, some people do not. Some people define 0 to be positive and negative and use the phrases strictly positive and strictly negative to exclude it. Others take it to be neither and use the phrase non-negative to mean 0,1,2... It is all a matter of what your conventions are. You have assumed that yours are absolutely correct, and that is not the case.
 
  • #16
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It is true that it is all a matter of convention. However, I have never seen any other convention used besides the one I stated. Human nature being what it is, it isn't surprising that there are several conventions, despite the obvious benefits of having a universal convention. Like meters and feet...

Always more to learn :smile:
 
  • #17
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matt grime said:
Some people define 0 to be positive and negative.
I thought if we take 0 to be an element of the reals itis non positive non negative because R is an ordered field, and 0 is needed to satisfy trichotomy?
 

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