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Contour plots?
Answered
Is it possible that we can define a function, say f(x,y)=x^3+y^3, and then draw its contour plot f(x,y)=k, say k from 1 to 5?
What if we want to change the definition of f later on?
How do we do that?
Regards,
Pegasus Roe
 GeoGebra
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Hi,
if you just want to draw, maybe you can use slider for k, and TraceOn command on function.
Something like this (you need to turn on animation)
https://ggbm.at/549647
I hope that can be useful.
Hi
Quickely like that but must be finished later
https://ggbm.at/549649
Daniel
hello
x^3+y^3=k is possible with implicitcurve because it is a polynomial expression, but another like x^2+ exp(x) it is more hard
contour plots are implicit functions really
i could try with this methods
http://www.geogebra.org/for...
saludos
Hi,
Thanks for your replies, but maybe I didn't make my point clear enough,
I'm sorry about that, let me try again. :)
The situation is this: now I have a function, say f(x,y)=x^3+sin(y) (or anything else), and a slider k,
and a textfield linked to function f.
I want to draw the curve f(x,y) = k, and then I will drag the slider k to observe its movement,
and then I still want to change the definition of function f through that textfield assigned to it,
and observe the curve's movement again.
How can I achieve that?
Regards,
Pegasus Roe
This feature (f(x, y) = k) would be very handy in mutivariable calculus!
Hi, I think this is possible to add for polynomials, but I'm afraid once sin() or sqrt() are involved you'll have to use the SolveOde command. E.g. for function sin(x)+sin(y) create a point A and type
SolveODE[cos(x),cos(y),x(A),y(A),2,0.1]
http://tube.geogebra.org/ma...
hello
for more difficult contours you can use the "ghost" method of rafael losada
saludos
Hi, kondr
Can you show me how to do that for polynomials? Thanks.
P.S. I tried to link to your demo file:
http://www.geogebratube.org...
but nothing showed up, except for the point A and an error message showing that
something's wrong with the operation "1*e(x,y)" :?
Regards,
Pegasus Roe
Hi Pegasus Roe,
what I meant was that we can add this feature into 4.0 for polynomials (now there is even a ticket http://www.geogebra.org/tra...), but with more complicated functions one will have to find workarounds. My demo (which now works, sorry for the bug with 1*e(x,y)) shows that it's not hard to algorithmically draw contours for a "nice" nonpolynomial function that go through given point. Unfortunatelly it's harder to draw contours with given zvalue. But maybe in 5.0 or 6.0 ...
Cheers,
Zbynek
Hello,
I use an implicit features that are in GG today. I want to propose to this problem a method based on Taylor series.
In the example pegasusroe:
f(x,y)=x^3+sin(y) >y^3+sin(x)=k
sin(x)= 0.09983 + 0.995 (x  0.1)  0.09983(x  0.1)² / 2!  0.995(x  0.1)³ / 3! + 0.09983(x  0.1)⁴ / 4! + 0.995(x  0.1)⁵ / 5!  0.09983(x  0.1)⁶ / 6!  0.995(x  0.1)⁷ / 7! With grad 7!!!
Unfortunately, in an implicit function (yet?) you cannot insert the Taylor series, it should be done by hand. The graph shows that in the interval (3.3) approximation of the sinus with Taylor is good. For another intervals is necessary to changed slider a.
As you can see the implicit curve b: & the exact solution p (x) on the interval (3.3) is very well suited.
P.s. I had been asked, or is or may be in an implicit function opportunity to record the explicit function. E.g b: x ^ 2 + f (x, y) * y = 5, where f (x, y) = x + y.
This wil l be very useful!
Roman Chijner
https://ggbm.at/549939
If you have f(x,y)=x^3+y^3, you can try f(x,y)>1, f(x,y)>2, etc. (syntax f>3 is automatically rewritten as f(x,y)>3). Extending this from inequalities to equations is planned.
For single variable functions you can try e.g. TaylorPolynomial[sin(x),0,7]>0.5.
Hi, just to play with javascript
I'd hope that axes don't desapear but how ?
Michel
https://ggbm.at/549941
This will work in the next release (together with ImplicitCurve[<f(x,y)>] )
[polynomials only]
Thanks, works !!!
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