MHB Explore Contour Lines in Geogebra: Drawing and Finding Functions

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The discussion revolves around using Geogebra to draw contour lines for specific functions. For the function f(x,y)=y-x^2-1, the correct approach is to find the intersection of the surface z=f(x,y) with the plane z=3709/2000, rather than plotting y-x^2-1=1374/2000. For the second part, participants suggest using the command "Sequence(IntersectPath(f,z=c), c, 0, 10, 0.4)" to generate contour lines for f(x,y)=cos(xy) and f(x,y)=2/sqrt(x^2+y^2)+2/sqrt((x-1)^2+y^2). The conversation highlights the importance of correctly specifying the functions and commands in Geogebra to achieve the desired contour visualizations. Overall, the discussion provides valuable insights into using Geogebra for contour plotting.
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Hey! :o

I am looking at the following:

1. We have the function $f(x,y)=y-x^2-1$. Write the appropriate commands in Geogebra that draw a contour line with $f(x,y)=\frac{3709}{2000}$.

Could you give me a hint what command we have to use here? Do we just plot $y-x^2-1=\frac{1374}{2000}$ ? (Wondering)
2. Give the graph of the below functions in Geogebra and find the countour lines $f(x,y)=c$ where $c$ is in the interval $[0,10]$ and each contour line has to have distance from the next one $0.4$.
  • $f(x,y)=\cos (xy)$
  • $f(x,y)=\frac{2}{\sqrt{x^2+y^2}}+\frac{2}{\sqrt{(x-1)^2+y^2}}$

Do we use for that the command "Sequence(\cos (xy)=c, c, 0, 10)" ? (Wondering)
 
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mathmari said:
Hey! :o

I am looking at the following:

1. We have the function $f(x,y)=y-x^2-1$. Write the appropriate commands in Geogebra that draw a contour line with $f(x,y)=\frac{3709}{2000}$.

Could you give me a hint what command we have to use here? Do we just plot $y-x^2-1=\frac{1374}{2000}$ ?

Hey mathmari!

A contour line is a curve on a surface $z=f(x,y)$ for a fixed $z$ isn't it? (Wondering)

But if we specify $y-x^2-1=\frac{1374}{2000}$, we do not get such a curve do we?
Instead we are getting a different surface.
Makes sense, because we have effectively specified an equation in x and y without specifying z.
So Geogebra shows a surface that satisfies the equation for x and y, and shows it for any z.
It's a parabolic cylinder. (Worried)

I think that instead we need the intersection of the surface $z=f(x,y)$ and the plane $z=\frac{3709}{2000}$, don't we?
Can we do that? (Wondering)

Btw, should the constant be $\frac{3709}{2000}$ or $\frac{1374}{2000}$? (Nerd)
mathmari said:
2. Give the graph of the below functions in Geogebra and find the countour lines $f(x,y)=c$ where $c$ is in the interval $[0,10]$ and each contour line has to have distance from the next one $0.4$.
  • $f(x,y)=\cos (xy)$
  • $f(x,y)=\frac{2}{\sqrt{x^2+y^2}}+\frac{2}{\sqrt{(x-1)^2+y^2}}$

Do we use for that the command "Sequence(\cos (xy)=c, c, 0, 10)" ?

We can use $\operatorname{Sequence}$ yes.
We'll have to specify an object that actually represents a curve though.
And shouldn't we specify a step size as well? (Wondering)
 
Klaas van Aarsen said:
A contour line is a curve on a surface $z=f(x,y)$ for a fixed $z$ isn't it? (Wondering)

But if we specify $y-x^2-1=\frac{1374}{2000}$, we do not get such a curve do we?
Instead we are getting a different surface.
Makes sense, because we have effectively specified an equation in x and y without specifying z.
So Geogebra shows a surface that satisfies the equation for x and y, and shows it for any z.
It's a parabolic cylinder. (Worried)

I think that instead we need the intersection of the surface $z=f(x,y)$ and the plane $z=\frac{3709}{2000}$, don't we?
Can we do that? (Wondering)

Btw, should the constant be $\frac{3709}{2000}$ or $\frac{1374}{2000}$? (Nerd)

Oh it should be $\frac{3709}{2000}$ and not $\frac{1374}{2000}$, the second one was a typo. (Tmi) So do you mean the following?

View attachment 9598

(Wondering)
 

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mathmari said:
So do you mean the following?

Yep. (Nod)
 
Klaas van Aarsen said:
Yep. (Nod)
Ok, great! So 1. is done.

Let's consider question 2.

Do we maybe use the command "Sequence(Intersect(f,g), c, 0, 10, 0.4)" where $f(x,y)=\cos (xy)$ and $g(x,y)=c$ ? (Wondering)
 
mathmari said:
Let's consider question 2.

Do we maybe use the command "Sequence(Intersect(f,g), c, 0, 10, 0.4)" where $f(x,y)=\cos (xy)$ and $g(x,y)=c$ ?

Well... does it work? (Wondering)

Your f is not actually the intended surface in 3D is it? (Worried)
 
Klaas van Aarsen said:
Well... does it work? (Wondering)

Your f is not actually the intended surface in 3D is it? (Worried)
I tried the following, but something is wrong:

View attachment 9599

What do I have to change? (Wondering)
 

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mathmari said:
I tried the following, but something is wrong:

What do I have to change?

The $g: z=c,\,0\le c\le 10$ doesn't seem to be understood. (Worried)

The $c$ should really be tied to a $\operatorname{Sequence}$.

How about $\operatorname{Sequence}(\operatorname{IntersectPath}(f,z=c), c, 0, 10, 0.4)$? (Wondering)
 
Klaas van Aarsen said:
The $g: z=c,\,0\le c\le 10$ doesn't seem to be understood. (Worried)

The $c$ should really be tied to a $\operatorname{Sequence}$.

How about $\operatorname{Sequence}(\operatorname{IntersectPath}(f,z=c), c, 0, 10, 0.4)$? (Wondering)

Ahh ok! With this command we get:

View attachment 9600 For the other function $f(x,y)=\frac{2}{\sqrt{x^2+y^2}}+\frac{2}{\sqrt{(x-1)^2+y^2}}$ we get:

View attachment 9601

(Malthe)
 

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  • #10
mathmari said:
Ahh ok! With this command we get:

For the other function $f(x,y)=\frac{2}{\sqrt{x^2+y^2}}+\frac{2}{\sqrt{(x-1)^2+y^2}}$ we get:
(Malthe)

Nice! (Happy)
 
  • #11
Klaas van Aarsen said:
Nice! (Happy)

Thank you for your help! (Yes)
 

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