Geometrical interpretation of vector derivative

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Let f and g be real differentiable functions.
Let v be a vectorial function of a real variable t, such that:

v(t) = ( f(t), g(t) )

I know that by definition v'(t) = ( f'(t), g'(t) )

In many books they give a geomtrical interpretation of this. They draw the curve (the values of f(t) in the x axis, the values of g(t) in y axis). They take two points in the curve (say v(t0) and v(t0 + h) ) and they draw the positions vectors of both.

They say: if you take smaller "h" then the vector v(t0 + h) - v(t0) gets nearer and nearer the tangent line that touches the curve in v(t0).

So they assume that if you take smallers "h" then the point v(t0 + h) will be nearer to v(t0) in the curve. Which is the basis of this assumption?

Thanks.
 

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D H
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The assumption is that the two functions that define the curve are continuous and differentiable.
 
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Lets take 0 < h1 < h2

Does the continuuty of functions f and g guarantees that, in the curve, between v(t0) and v(t0 + h2) you wil necesarily find v(t0 + h1) ?
 
HallsofIvy
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Yes, that is the "intermediate value" property of continuous functions.
 
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Does the IVT applies to a vectorial function?

Remember thar v is a vectorial function. v(t) = ( f(t), g(t) ).

Does the continuity of f and g implies that we can apply the IVT to the vectorial function v?

Please an answer.
 
Hurkyl
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What do you mean by "between" here? I think if you can give a definition of the phrase:

B is between A and C

for three points A, B, and C on a curve, then we could give a clear answer you would accept. (And you could probably do it yourself!)
 
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Let p be a real function of real variable.

Let A = (x1, p(x1))
B = ( x2, p(x2))
C = (x3, p(x3))

To say that, in the graphic of this function p, point B is between point A and point C, means that x2 is between x1 and x3.
 
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Am I okey? Can someone answer me? How to use this previous fact to prove the point I wish to prove?
 
Hurkyl
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So, in the special case of a graph, you wish to prove that if b is between a and c, then (b, p(b)) is between (a, p(a)) and (c, p(c)), correct?
 
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OK.
We have v(t) = ( f(t), g(t) ), where f and g are real functions of one real variable, continuous and differentiable.

Let t, h2 be fixed numbers.
Suppose we have f(t) < f(t + h2). Let "c" be / f(t) < c < f(t + h2).

The function f is continuous in [t, t+h2] so by the IVT there exists a number "d" in [t, t+h2] such that f(d) = c.

So there exist a number h1 such that 0 < h1 < h2 and such that d = t + h1.
Replacing: f(t + h1) = c, so f(t) < f(t+h1) < f(t+h2).

So in the curve the point ( f(t+h1), g(t+h1) ) is "between" ( f(t), g(t) ) and ( f(t+h2),g(t+h2) ).

In other words, v(t+h1) is between v(t) and v(t + h2).

But I only have proved that for that h2 I have an h1. I have not prove that if "h" is ANY number < h2 we will have v(t +h) between v(t) and v(t + h2)...
 
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Hurkyl, or Hallsoftivy, or anyone, can you help?
 
Hurkyl
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Well, you've only defined what "between" means in the case where your curve is a graph, so you can't really prove anything in the general case. Why not work through the problem for graphs first, before moving onto the general case (where we still need to define "between")?
 
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Hurkyl
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Oh, on a side note, I just now noticed that nobody explicitly stated that

[tex]\lim_{h \rightarrow 0} \vec{v}(t + h) = \vec{v}(t).[/tex]

Maybe that's what you really wanted to know all along?
 
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Oh, on a side note, I just now noticed that nobody explicitly stated that

[tex]\lim_{h \rightarrow 0} \vec{v}(t + h) = \vec{v}(t).[/tex]

Maybe that's what you really wanted to know all along?
From an analytical point of view, I understand this equation. It simply follows from: 1) the definition of continuity for a vectorial function, 2) the fact that the component real functions f and g are continuous.

What I am trying to understand is why, if we draw the vectorial function in a plane, your equation is also "confirmated" geometrically. But maybe I am asking some sort of tautology and I fail to see it like that, or maybe I am handling a definition as if it was a theorem.

In any case, thanks, Hurkyl.
 

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