If I have a function given as x =f(y), how can I plot x as independent variable (x axsis) and y dependent variable (y axsis). I also need to control the range of both x and y.
x=y/(0.5+cos3y)

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Walter Roberson
Walter Roberson 2020 年 3 月 5 日
Could you confirm that you need to do this inside a mex routine? Or are you trying to use MATLAB Coder? I am trying to figure out what the Product tag you used implies for this.

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Walter Roberson
Walter Roberson 2020 年 3 月 5 日

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In MATLAB, cos3y would be written as cos(3*y)
In MATLAB, if y is a vector, instead of using / you should use ./ -- so y./something instead of y/something
You have an implicit plot, so you should have a look at fimplicit()

7 件のコメント

Mariam
Mariam 2020 年 3 月 5 日
Let me rephrase my question, I am trying to make a code by matlab for plotting the following equation:
y= (0.5+cos(3*y))*x for both x and y to have the values: form 0 to 1e4. I have never come across with such an equation and I need to know how to solve it numerically.
All the best
Walter Roberson
Walter Roberson 2020 年 3 月 5 日
編集済み: Walter Roberson 2020 年 3 月 5 日
In MATLAB, most of the time you need to express multiplication using the .* operator instead of the * operator.
You have an implicit plot, so you should have a look at fimplicit()
Hint:
If A(x,y) = B(x,y) then A(x,y)-B(x,y) = 0 must be true, and expression being 0 is what fimplicit() draws.
Mariam
Mariam 2020 年 3 月 5 日
I have tried the following:
f=@(x,y) y-(0.5+cos(3.*y)).*x;
fimplicit(f, [0 1e4 0 1e4])
an error message tells me that "undefined function or method 'fimplicit' for input arguments of type 'fuction handle'".
Walter Roberson
Walter Roberson 2020 年 3 月 5 日
Which MATLAB release are you using? fimplicit() was added in R2016b.
Mariam
Mariam 2020 年 3 月 5 日
it is R2009a.
Walter Roberson
Walter Roberson 2020 年 3 月 5 日
In R2009a, you can use
ezplot(f, [0 1e4 0 1e4])
Mariam
Mariam 2020 年 3 月 6 日
Many thanks.

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