How could I graph z=sqrt(x^2-y^2) using Pgftools? I believe I've all suggestions on the internet but to no...
1
I have tried graphing it with the parametric notation (i.e. ({x},{y},{sort(x^2-y^2)}) and as seen on my code below but nothing renders the image shown on graphing calculators.
begin{tikzpicture}[scale=1.5]
begin{axis}[
grid=major,
3d box=complete,
enlargelimits=false,
colormap/cool,
xlabel=$x$,
ylabel=$y$,
zlabel=$z$,
zlabel style = {sloped like x axis}
]
addplot3 [
surf,
shader=faceted,
samples=50,
z buffer=sort,
] {sqrt((x)^2-(y)^2)};
end{axis}
end{tikzpicture}
tikz-pgf pgfplots plot 3d tikz-3dplot
asked Feb 28 at 2:28
Daniel HernándezDaniel Hernández
152
add a comment |
1
I have tried graphing it with the parametric notation (i.e. ({x},{y},{sort(x^2-y^2)}) and as seen on my code below but nothing renders the image shown on graphing calculators.
begin{tikzpicture}[scale=1.5]
begin{axis}[
grid=major,
3d box=complete,
enlargelimits=false,
colormap/cool,
xlabel=$x$,
ylabel=$y$,
zlabel=$z$,
zlabel style = {sloped like x axis}
]
addplot3 [
surf,
shader=faceted,
samples=50,
z buffer=sort,
] {sqrt((x)^2-(y)^2)};
end{axis}
end{tikzpicture}
tikz-pgf pgfplots plot 3d tikz-3dplot
asked Feb 28 at 2:28
Daniel HernándezDaniel Hernández
152
add a comment |
1
1
1
I have tried graphing it with the parametric notation (i.e. ({x},{y},{sort(x^2-y^2)}) and as seen on my code below but nothing renders the image shown on graphing calculators.
begin{tikzpicture}[scale=1.5]
begin{axis}[
grid=major,
3d box=complete,
enlargelimits=false,
colormap/cool,
xlabel=$x$,
ylabel=$y$,
zlabel=$z$,
zlabel style = {sloped like x axis}
]
addplot3 [
surf,
shader=faceted,
samples=50,
z buffer=sort,
] {sqrt((x)^2-(y)^2)};
end{axis}
end{tikzpicture}
tikz-pgf pgfplots plot 3d tikz-3dplot
asked Feb 28 at 2:28
Daniel HernándezDaniel Hernández
152
I have tried graphing it with the parametric notation (i.e. ({x},{y},{sort(x^2-y^2)}) and as seen on my code below but nothing renders the image shown on graphing calculators.
begin{tikzpicture}[scale=1.5]
begin{axis}[
grid=major,
3d box=complete,
enlargelimits=false,
colormap/cool,
xlabel=$x$,
ylabel=$y$,
zlabel=$z$,
zlabel style = {sloped like x axis}
]
addplot3 [
surf,
shader=faceted,
samples=50,
z buffer=sort,
] {sqrt((x)^2-(y)^2)};
end{axis}
end{tikzpicture}
tikz-pgf pgfplots plot 3d tikz-3dplot
tikz-pgf pgfplots plot 3d tikz-3dplot
asked Feb 28 at 2:28
Daniel HernándezDaniel Hernández
152
asked Feb 28 at 2:28
Daniel HernándezDaniel Hernández
152
asked Feb 28 at 2:28
Daniel HernándezDaniel Hernández
152
asked Feb 28 at 2:28
Daniel HernándezDaniel Hernández
152
asked Feb 28 at 2:28
Daniel HernándezDaniel Hernández
152
152
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
3
You did almost everything right, pgfplots is just not very good at taking the square root of a negative number. So if you adjust the domains a bit you get
documentclass[tikz,border=3.14mm]{standalone}
usepackage{tikz,pgfplots}
pgfplotsset{compat=1.16}
begin{document}
begin{tikzpicture}[scale=1.5]
begin{axis}[
grid=major,
3d box=complete,
enlargelimits=false,
colormap/cool,
xlabel=$x$,
ylabel=$y$,
zlabel=$z$,
zlabel style = {sloped like x axis}
]
addplot3 [domain=2:4,domain y=-2:2,
surf,
shader=faceted,
samples=50,
z buffer=sort,
] {sqrt(x^2-y^2)};
end{axis}
end{tikzpicture}
end{document}
As for your request in the comments, one can write the function as sqrt(u*v), where u=x+y and v=x-y. Then x=(u+v)/2 and y=(u-v)/2. Since the variables are just placeholders in a parametric plot, we can rename u and v to x and y, respectively, and arrive at
documentclass[tikz,border=3.14mm]{standalone}
usepackage{pgfplots}
pgfplotsset{compat=1.16}
begin{document}
begin{tikzpicture}[scale=1.5]
begin{axis}[xmax=2,xmin=-2,
grid=major,
3d box=complete,
enlargelimits=false,
colormap/cool,
xlabel=$x$,
ylabel=$y$,
zlabel=$z$,
zlabel style = {sloped like x axis}
]
addplot3 [domain=0:4,domain y=0:4,
surf,
shader=faceted,
samples=50,
z buffer=sort,
] ({(x+y)/2},{(x-y)/2},{sqrt(x*y)});
addplot3 [domain=0:4,domain y=0:4,
surf,
shader=faceted,
samples=50,
z buffer=sort,
] ({-(x+y)/2},{-(x-y)/2},{sqrt(x*y)});
end{axis}
end{tikzpicture}
end{document}
answered Feb 28 at 3:35
marmotmarmot
106k5129243
Thanks though that fixed most of it. Though yesterday I forgot to post the image of how I would like my graph to look like. How can I make the graph look more like this?
– Daniel Hernández
Feb 28 at 13:34
@DanielHernández I added a graph of this sort.
– marmot
Feb 28 at 15:22
©marmot Thanks! That's what I needed. Pgfplots still is kind of confusing for me.
– Daniel Hernández
Feb 28 at 17:52
@DanielHernández You are welcome! (The issue here is more a parametrization question than a pgfplots question, I think. But you're right, pgfplots is a nontrivial packages with lots of functionality that one keeps discovering.)
– marmot
Feb 28 at 18:03
add a comment |
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1 Answer
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1 Answer
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3
You did almost everything right, pgfplots is just not very good at taking the square root of a negative number. So if you adjust the domains a bit you get
documentclass[tikz,border=3.14mm]{standalone}
usepackage{tikz,pgfplots}
pgfplotsset{compat=1.16}
begin{document}
begin{tikzpicture}[scale=1.5]
begin{axis}[
grid=major,
3d box=complete,
enlargelimits=false,
colormap/cool,
xlabel=$x$,
ylabel=$y$,
zlabel=$z$,
zlabel style = {sloped like x axis}
]
addplot3 [domain=2:4,domain y=-2:2,
surf,
shader=faceted,
samples=50,
z buffer=sort,
] {sqrt(x^2-y^2)};
end{axis}
end{tikzpicture}
end{document}
As for your request in the comments, one can write the function as sqrt(u*v), where u=x+y and v=x-y. Then x=(u+v)/2 and y=(u-v)/2. Since the variables are just placeholders in a parametric plot, we can rename u and v to x and y, respectively, and arrive at
documentclass[tikz,border=3.14mm]{standalone}
usepackage{pgfplots}
pgfplotsset{compat=1.16}
begin{document}
begin{tikzpicture}[scale=1.5]
begin{axis}[xmax=2,xmin=-2,
grid=major,
3d box=complete,
enlargelimits=false,
colormap/cool,
xlabel=$x$,
ylabel=$y$,
zlabel=$z$,
zlabel style = {sloped like x axis}
]
addplot3 [domain=0:4,domain y=0:4,
surf,
shader=faceted,
samples=50,
z buffer=sort,
] ({(x+y)/2},{(x-y)/2},{sqrt(x*y)});
addplot3 [domain=0:4,domain y=0:4,
surf,
shader=faceted,
samples=50,
z buffer=sort,
] ({-(x+y)/2},{-(x-y)/2},{sqrt(x*y)});
end{axis}
end{tikzpicture}
end{document}
answered Feb 28 at 3:35
marmotmarmot
106k5129243
Thanks though that fixed most of it. Though yesterday I forgot to post the image of how I would like my graph to look like. How can I make the graph look more like this?
– Daniel Hernández
Feb 28 at 13:34
@DanielHernández I added a graph of this sort.
– marmot
Feb 28 at 15:22
©marmot Thanks! That's what I needed. Pgfplots still is kind of confusing for me.
– Daniel Hernández
Feb 28 at 17:52
@DanielHernández You are welcome! (The issue here is more a parametrization question than a pgfplots question, I think. But you're right, pgfplots is a nontrivial packages with lots of functionality that one keeps discovering.)
– marmot
Feb 28 at 18:03
add a comment |
3
You did almost everything right, pgfplots is just not very good at taking the square root of a negative number. So if you adjust the domains a bit you get
documentclass[tikz,border=3.14mm]{standalone}
usepackage{tikz,pgfplots}
pgfplotsset{compat=1.16}
begin{document}
begin{tikzpicture}[scale=1.5]
begin{axis}[
grid=major,
3d box=complete,
enlargelimits=false,
colormap/cool,
xlabel=$x$,
ylabel=$y$,
zlabel=$z$,
zlabel style = {sloped like x axis}
]
addplot3 [domain=2:4,domain y=-2:2,
surf,
shader=faceted,
samples=50,
z buffer=sort,
] {sqrt(x^2-y^2)};
end{axis}
end{tikzpicture}
end{document}
As for your request in the comments, one can write the function as sqrt(u*v), where u=x+y and v=x-y. Then x=(u+v)/2 and y=(u-v)/2. Since the variables are just placeholders in a parametric plot, we can rename u and v to x and y, respectively, and arrive at
documentclass[tikz,border=3.14mm]{standalone}
usepackage{pgfplots}
pgfplotsset{compat=1.16}
begin{document}
begin{tikzpicture}[scale=1.5]
begin{axis}[xmax=2,xmin=-2,
grid=major,
3d box=complete,
enlargelimits=false,
colormap/cool,
xlabel=$x$,
ylabel=$y$,
zlabel=$z$,
zlabel style = {sloped like x axis}
]
addplot3 [domain=0:4,domain y=0:4,
surf,
shader=faceted,
samples=50,
z buffer=sort,
] ({(x+y)/2},{(x-y)/2},{sqrt(x*y)});
addplot3 [domain=0:4,domain y=0:4,
surf,
shader=faceted,
samples=50,
z buffer=sort,
] ({-(x+y)/2},{-(x-y)/2},{sqrt(x*y)});
end{axis}
end{tikzpicture}
end{document}
answered Feb 28 at 3:35
marmotmarmot
106k5129243
Thanks though that fixed most of it. Though yesterday I forgot to post the image of how I would like my graph to look like. How can I make the graph look more like this?
– Daniel Hernández
Feb 28 at 13:34
@DanielHernández I added a graph of this sort.
– marmot
Feb 28 at 15:22
©marmot Thanks! That's what I needed. Pgfplots still is kind of confusing for me.
– Daniel Hernández
Feb 28 at 17:52
@DanielHernández You are welcome! (The issue here is more a parametrization question than a pgfplots question, I think. But you're right, pgfplots is a nontrivial packages with lots of functionality that one keeps discovering.)
– marmot
Feb 28 at 18:03
add a comment |
3
3
3
You did almost everything right, pgfplots is just not very good at taking the square root of a negative number. So if you adjust the domains a bit you get
documentclass[tikz,border=3.14mm]{standalone}
usepackage{tikz,pgfplots}
pgfplotsset{compat=1.16}
begin{document}
begin{tikzpicture}[scale=1.5]
begin{axis}[
grid=major,
3d box=complete,
enlargelimits=false,
colormap/cool,
xlabel=$x$,
ylabel=$y$,
zlabel=$z$,
zlabel style = {sloped like x axis}
]
addplot3 [domain=2:4,domain y=-2:2,
surf,
shader=faceted,
samples=50,
z buffer=sort,
] {sqrt(x^2-y^2)};
end{axis}
end{tikzpicture}
end{document}
As for your request in the comments, one can write the function as sqrt(u*v), where u=x+y and v=x-y. Then x=(u+v)/2 and y=(u-v)/2. Since the variables are just placeholders in a parametric plot, we can rename u and v to x and y, respectively, and arrive at
documentclass[tikz,border=3.14mm]{standalone}
usepackage{pgfplots}
pgfplotsset{compat=1.16}
begin{document}
begin{tikzpicture}[scale=1.5]
begin{axis}[xmax=2,xmin=-2,
grid=major,
3d box=complete,
enlargelimits=false,
colormap/cool,
xlabel=$x$,
ylabel=$y$,
zlabel=$z$,
zlabel style = {sloped like x axis}
]
addplot3 [domain=0:4,domain y=0:4,
surf,
shader=faceted,
samples=50,
z buffer=sort,
] ({(x+y)/2},{(x-y)/2},{sqrt(x*y)});
addplot3 [domain=0:4,domain y=0:4,
surf,
shader=faceted,
samples=50,
z buffer=sort,
] ({-(x+y)/2},{-(x-y)/2},{sqrt(x*y)});
end{axis}
end{tikzpicture}
end{document}
answered Feb 28 at 3:35
marmotmarmot
106k5129243
You did almost everything right, pgfplots is just not very good at taking the square root of a negative number. So if you adjust the domains a bit you get
documentclass[tikz,border=3.14mm]{standalone}
usepackage{tikz,pgfplots}
pgfplotsset{compat=1.16}
begin{document}
begin{tikzpicture}[scale=1.5]
begin{axis}[
grid=major,
3d box=complete,
enlargelimits=false,
colormap/cool,
xlabel=$x$,
ylabel=$y$,
zlabel=$z$,
zlabel style = {sloped like x axis}
]
addplot3 [domain=2:4,domain y=-2:2,
surf,
shader=faceted,
samples=50,
z buffer=sort,
] {sqrt(x^2-y^2)};
end{axis}
end{tikzpicture}
end{document}
As for your request in the comments, one can write the function as sqrt(u*v), where u=x+y and v=x-y. Then x=(u+v)/2 and y=(u-v)/2. Since the variables are just placeholders in a parametric plot, we can rename u and v to x and y, respectively, and arrive at
documentclass[tikz,border=3.14mm]{standalone}
usepackage{pgfplots}
pgfplotsset{compat=1.16}
begin{document}
begin{tikzpicture}[scale=1.5]
begin{axis}[xmax=2,xmin=-2,
grid=major,
3d box=complete,
enlargelimits=false,
colormap/cool,
xlabel=$x$,
ylabel=$y$,
zlabel=$z$,
zlabel style = {sloped like x axis}
]
addplot3 [domain=0:4,domain y=0:4,
surf,
shader=faceted,
samples=50,
z buffer=sort,
] ({(x+y)/2},{(x-y)/2},{sqrt(x*y)});
addplot3 [domain=0:4,domain y=0:4,
surf,
shader=faceted,
samples=50,
z buffer=sort,
] ({-(x+y)/2},{-(x-y)/2},{sqrt(x*y)});
end{axis}
end{tikzpicture}
end{document}
answered Feb 28 at 3:35
marmotmarmot
106k5129243
edited Feb 28 at 14:49
answered Feb 28 at 3:35
marmotmarmot
106k5129243
answered Feb 28 at 3:35
marmotmarmot
106k5129243
answered Feb 28 at 3:35
marmotmarmot
106k5129243
106k5129243
Thanks though that fixed most of it. Though yesterday I forgot to post the image of how I would like my graph to look like. How can I make the graph look more like this?
– Daniel Hernández
Feb 28 at 13:34
@DanielHernández I added a graph of this sort.
– marmot
Feb 28 at 15:22
©marmot Thanks! That's what I needed. Pgfplots still is kind of confusing for me.
– Daniel Hernández
Feb 28 at 17:52
@DanielHernández You are welcome! (The issue here is more a parametrization question than a pgfplots question, I think. But you're right, pgfplots is a nontrivial packages with lots of functionality that one keeps discovering.)
– marmot
Feb 28 at 18:03
add a comment |
Thanks though that fixed most of it. Though yesterday I forgot to post the image of how I would like my graph to look like. How can I make the graph look more like this?
– Daniel Hernández
Feb 28 at 13:34
@DanielHernández I added a graph of this sort.
– marmot
Feb 28 at 15:22
©marmot Thanks! That's what I needed. Pgfplots still is kind of confusing for me.
– Daniel Hernández
Feb 28 at 17:52
@DanielHernández You are welcome! (The issue here is more a parametrization question than a pgfplots question, I think. But you're right, pgfplots is a nontrivial packages with lots of functionality that one keeps discovering.)
– marmot
Feb 28 at 18:03
Thanks though that fixed most of it. Though yesterday I forgot to post the image of how I would like my graph to look like. How can I make the graph look more like this?
– Daniel Hernández
Feb 28 at 13:34
Thanks though that fixed most of it. Though yesterday I forgot to post the image of how I would like my graph to look like. How can I make the graph look more like this?
– Daniel Hernández
Feb 28 at 13:34
@DanielHernández I added a graph of this sort.
– marmot
Feb 28 at 15:22
@DanielHernández I added a graph of this sort.
– marmot
Feb 28 at 15:22
©marmot Thanks! That's what I needed. Pgfplots still is kind of confusing for me.
– Daniel Hernández
Feb 28 at 17:52
©marmot Thanks! That's what I needed. Pgfplots still is kind of confusing for me.
– Daniel Hernández
Feb 28 at 17:52
@DanielHernández You are welcome! (The issue here is more a parametrization question than a pgfplots question, I think. But you're right, pgfplots is a nontrivial packages with lots of functionality that one keeps discovering.)
– marmot
Feb 28 at 18:03
@DanielHernández You are welcome! (The issue here is more a parametrization question than a pgfplots question, I think. But you're right, pgfplots is a nontrivial packages with lots of functionality that one keeps discovering.)
– marmot
Feb 28 at 18:03
add a comment |
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