calculate perimeter of a parable when I change it's shape

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 .ggb-file.

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")

C=(3.48, f(3.48))
D=(-3.48, f(-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 x-axis and cutting this line with your function f.

Now these two points are always on the same height (y-coordinate) 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 can do the deformation of the curve depicted in situation 1

situation 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"=y-coordinate 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 x-axis, e. g. zero and 6.96, so these two points have the same y-coordinate equal zero

f(x)=a*x*(x-6.96)

and now you can play with your slider a and you will see, that your vertex moves up and down and the x-coordinate 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 x-axis, because you determine x-axis-distance and can transfer your result to any given two points with the same y-coordinate 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