array of $x$ and $y$ in an implicit plot in Matlab
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If I use implicit plot of a relation (say $xcdot y+sin(x+y)=0$ ) and obtain a plot. How do I get the list/array of the values of $x$ and $y$ plotted?
matlab implicit-function
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add a comment |
$begingroup$
If I use implicit plot of a relation (say $xcdot y+sin(x+y)=0$ ) and obtain a plot. How do I get the list/array of the values of $x$ and $y$ plotted?
matlab implicit-function
$endgroup$
add a comment |
$begingroup$
If I use implicit plot of a relation (say $xcdot y+sin(x+y)=0$ ) and obtain a plot. How do I get the list/array of the values of $x$ and $y$ plotted?
matlab implicit-function
$endgroup$
If I use implicit plot of a relation (say $xcdot y+sin(x+y)=0$ ) and obtain a plot. How do I get the list/array of the values of $x$ and $y$ plotted?
matlab implicit-function
matlab implicit-function
edited Jan 17 at 11:51
Larry
2,39131129
2,39131129
asked Jan 17 at 11:34
RockRock
324
324
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1 Answer
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How accurate do you need the values to be? To get a 2D array of points satisfying an equation, you could just make a grid with boundaries $a_x,b_x$ and $a_y,b_y$ and steps $h_x,h_y$, then substitute every point into the equation and assign some small tolerance $t$ such that if $$x_n y_m+sin(x_n+y_m) < t$$ then the point approximately satisfies the equation. Then you make a loop through all the points and obtain an array corresponding to an approximate implicit plot.
$endgroup$
$begingroup$
I will check it. Thanks
$endgroup$
– Rock
Jan 17 at 13:19
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
How accurate do you need the values to be? To get a 2D array of points satisfying an equation, you could just make a grid with boundaries $a_x,b_x$ and $a_y,b_y$ and steps $h_x,h_y$, then substitute every point into the equation and assign some small tolerance $t$ such that if $$x_n y_m+sin(x_n+y_m) < t$$ then the point approximately satisfies the equation. Then you make a loop through all the points and obtain an array corresponding to an approximate implicit plot.
$endgroup$
$begingroup$
I will check it. Thanks
$endgroup$
– Rock
Jan 17 at 13:19
add a comment |
$begingroup$
How accurate do you need the values to be? To get a 2D array of points satisfying an equation, you could just make a grid with boundaries $a_x,b_x$ and $a_y,b_y$ and steps $h_x,h_y$, then substitute every point into the equation and assign some small tolerance $t$ such that if $$x_n y_m+sin(x_n+y_m) < t$$ then the point approximately satisfies the equation. Then you make a loop through all the points and obtain an array corresponding to an approximate implicit plot.
$endgroup$
$begingroup$
I will check it. Thanks
$endgroup$
– Rock
Jan 17 at 13:19
add a comment |
$begingroup$
How accurate do you need the values to be? To get a 2D array of points satisfying an equation, you could just make a grid with boundaries $a_x,b_x$ and $a_y,b_y$ and steps $h_x,h_y$, then substitute every point into the equation and assign some small tolerance $t$ such that if $$x_n y_m+sin(x_n+y_m) < t$$ then the point approximately satisfies the equation. Then you make a loop through all the points and obtain an array corresponding to an approximate implicit plot.
$endgroup$
How accurate do you need the values to be? To get a 2D array of points satisfying an equation, you could just make a grid with boundaries $a_x,b_x$ and $a_y,b_y$ and steps $h_x,h_y$, then substitute every point into the equation and assign some small tolerance $t$ such that if $$x_n y_m+sin(x_n+y_m) < t$$ then the point approximately satisfies the equation. Then you make a loop through all the points and obtain an array corresponding to an approximate implicit plot.
answered Jan 17 at 11:40
Yuriy SYuriy S
15.8k433118
15.8k433118
$begingroup$
I will check it. Thanks
$endgroup$
– Rock
Jan 17 at 13:19
add a comment |
$begingroup$
I will check it. Thanks
$endgroup$
– Rock
Jan 17 at 13:19
$begingroup$
I will check it. Thanks
$endgroup$
– Rock
Jan 17 at 13:19
$begingroup$
I will check it. Thanks
$endgroup$
– Rock
Jan 17 at 13:19
add a comment |
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