5

I was giving luadraw my first attempt, with extreme hard-working in French documentation(Non-naive English and French language speaker...)😭....

Now I want to draw the tangent line of "x=0.5" in the following example:

\documentclass{standalone}
\usepackage{luadraw}
\begin{document}
\begin{luadraw}{name=implicit_function}
    local g = graph:new{window={-3,3,-3,3},size={10,10}}
    g:Linecap("round")
    local F = function(x,y) return x^2+y^2+x*y-1 end
    g:Dimplicit(F,{draw_options="thick"})
    g:Dgradbox(
        {Z(-2,-2),Z(2,2),1,1},{grid=true,title="\\textbf{Implicit Function Plot}"}
    )
    -- g:Dtangent(F,0.5,3) -- This doesn't work...
    g:Show()
\end{luadraw}
\end{document}

result

But in the documentation, I only found:

  • g:Dtangent(p, t0, long, draw_options)
  • g:DtangentC(f, x0, long, draw_options)

which seemed not easy to handle the implicit funtion like above. Is there exist elegant solution with luadraw?

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1 Answer 1

Reset to default
7

Note: in Lua the comment symbol is -- and not %. There is no (yet) predefined function for tangents to implicit curves, but one can be written:

\documentclass{standalone}
\usepackage{luadraw}
\begin{document}
\begin{luadraw}{name=implicit_function}
    local g = graph:new{window={-3,3,-3,3},size={10,10}}
    g:Linecap("round")
    
    local DtangentI = function(f,x0,y0,len,draw_options) -- f:(x,y()) -> f(x,y)
        -- We assume that f(x0,y0)=0! 
        local h = 1e-6
        local A = Z(x0,y0)
        local a,b = (f(x0+h,y0)-f(x0-h,y0))/(2*h), (f(x0,y0+h)-f(x0,y0-h))/(2*h)
        local v = Z(-b,a)
        if len == nil then 
            g:Dline({A,v},draw_options)
        else
            local u = len*v/cpx.abs(v)/2
            g:Dseg({A-u,A+u},draw_options) -- on renvoie un segment
        end
    end
    
    local F = function(x,y) return x^2+y^2+x*y-1 end
    g:Dimplicit(F,{draw_options="thick"})
    g:Dgradbox(
        {Z(-2,-2),Z(2,2),1,1},{grid=true,title="\\textbf{Implicit Function Plot}"}
    )
    local x0 = 1/math.sqrt(3) 
    local L = solve(function(t) return F(x0,t) end,-2,2)
    for _, y in ipairs(L) do
        DtangentI(F,x0,y,2,"thick,red") -- This work!!
    end
    g:Show()
\end{luadraw}
\end{document}

enter image description here

Note 2: There are actually several questions in your question, because you also want the points of the implicit curve with abscissa 0.5...

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3
  • Sorry, (i)the comment line is the last I added with vscode's "ctrl+/", but LaTeXWorkshop now is still not recognized it, and I copy it here without checking twice, and now edited. (ii) yep, it actually consisted not only-one question like "want the points of the implicit curve with abscissa 0.5...", but I origianally think that luadraw have built-in function to do so(but I didn't find it...) I would keep my question to focus only one point on this site as possible, thanks to your remind! This is a very good learning material for me! Many thanks! Commented Aug 31, 2025 at 11:43
  • I also found the father of DtangentI at here. I hope that, if possible, this could be absorded into powerful luadraw. Thank! Commented Aug 31, 2025 at 12:04
  • 1
    You should draw the gradbox first. Commented Aug 31, 2025 at 15:06

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