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calculate perimeter of a parable when I change it's shape
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Hi,
what I want to do is to draw an arch form and define it's perimeter. Then I want to change its shape like in the pictures (red: initial arch, blue: deformation) and measure how the perimeter changes with deformation in different parts of the arch (lateral expansion, anterior expansion...)
Any help would be really appreciated
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Your "schermata" :smile: (Italian?) doesn't look like showing an arc of a parabola, but rather the union of smaller arcs or a polyline.
In any case, you could use the integral formula for the calculation of the length of an arc of a curve.
There are many, depending on how your function is defined (functional notation, parametric curve). Just check them out e.g. in Wikipedia or on a Calculus book.
GeoGebra shows automatically only the perimeter of a circle or an ellipse (using the Perimeter command) or the length of a circle arc, when this is created using the related tools or commands.
Moreover, in some cases you can use the Length command to calculate the length of a part of a curve/function. See https://wiki.geogebra.org/e...
yes, Italian :) thank you very much for your help, I'm trying to use "lenght command" but I got a "slider". Where am I wrong?
usa lenght(f,C,D)
f está ya definida
I've already tried, it doesen't work , I got a "slider" :(
My reply inside the screenshot. Cheers
Hi zazi!
Hmm, please post your .ggbfile.
As you can see  for me it works and so the question is  Why not for you?
Kind Regards
mire2
Length not Lenght
thanks to everyone...using mire2 ggb file I noticed that when I move the A slider the parable "expands" but the C and D points move down. That's not what I want. I want to get the trasformation depicted in the first images I uploaded...they seems like a polyline because of the bad drawing but actually they are a curve. Any suggestion to do what I want?
I'm very inexperienced with this program please be patient
Hi zazi!
The reason, why C and D move that way you described is the definition of them (you can see by looking at "properties")
So if you change f by drawing a slider, the ycoordinate change, but the xcoordinate is always 3.48.
If you want two points at the same "height" you have to define them in a other way, e. g. drawing a parallel line to the xaxis and cutting this line with your function f.
Now these two points are always on the same height (ycoordinate) and they move left and right by drawing a slider to stay on the graph of function f (see attached file)
Hope that helps you and kind regards
mire2
mire2 , thank you very much!!! I still need your help because now I can do the deformation of the curve depicted in situation 1. How can I do also the deformation of situation 2 and 3? see pictures attached
I think that y=a x^2+b x+c through 3 points is unique
I think situation1 is not possiblesituation 1 is the one depicted by mire2, I need to do also situation 2 and 3 but I don't know how to do
Hi!
I'm not sure that I interpret the pictures correct, but situation looks for me like fixing two points at the same "height"=ycoordinate and the vertex of the parabola moves up or down.
Hm, I think one possibility to realize this situation in a relative easy way is to have a short look here:
https://en.wikipedia.org/wi...
The factored form give us two fixed points on the xaxis, e. g. zero and 6.96, so these two points have the same ycoordinate equal zero
and now you can play with your slider a and you will see, that your vertex moves up and down and the xcoordinate will always be in the middle of 0 and 6.96, 3.48.
So it is helpful to change the form of your quadratic function and it is indeed no restriction to fix two points on the xaxis, because you determine xaxisdistance and can transfer your result to any given two points with the same ycoordinate in an easy way.
Hmm, I can only guess what you mean by situation 3  perhaps parabolas with the same vertex?
Hope that helps you a little bit and kind regards
mire2
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