pgfplots: How to draw exponential graph with 60° start angle?












4















I want to draw a simplified Michaelis-Menten kinetic (monod-function) to compare it with a linear function.





Minimum Working Example (MWE):



documentclass{standalone}
usepackage{pgfplots}
usepackage{amsmath}

pgfplotsset{compat=1.14, /pgf/declare function={f1(x)=ln(x);}}% <- This is the exponential function which needs to be optimized

begin{document}
begin{tikzpicture}
begin{axis}[
ymin = 0,
xmin = 0,
xmax = 1,
ymax = 0.9,
axis x line = bottom,
axis y line = left,
]

% addplot[no marks, samples=100, draw=blue] {f1(x)};% This is the exponential graph based on the function
addplot[no marks, samples=100, draw=black, thick] coordinates{(0,0) (0.2020725942,0.35)};%
addplot[no marks, samples=100, draw=black, thick] (0.2020725942,0.35) to [out=60,in=180] (0.8,0.7) to [out=0,in=0] (1,0.7);%
draw[draw=black, dashed] (0,0.7) -- node[above] {(y_{text{tot}})} ++(0.8,0.0);%
draw[draw=black, dashed] (0,0.35) -- node[above] {(frac{y_{text{tot}}}{2})} ++(0.2020725942,0) -- (0.2020725942,-0.35);%
end{axis}
end{tikzpicture}
end{document}




Screenshot of the result:



Screenshot of the result





Description of the issue:



How can I replace the current graph with an exponential graph?



Start point of the exponential graph:




  • Start point: x = 0.2020725942,
    y = 0.35,
    angle = 60°,

  • End point: y = ~ 0.7 (of course, wherever the e-function would end)


As soon as I activate the graph with the exponential function, my whole diagram will be distorted. How to implement an exponential graph based on the upper values correctly?










share|improve this question


















  • 3





    This looks like a question of math not of tex/tikz : how should I choose a and b in f(x) = a*exp(x)+b such that f(0.2020725942)=0.35 and f'(0.2020725942)=tan(pi/3) ? If this is the case here is not the right place to ask this question.

    – Kpym
    Apr 3 at 12:23











  • @Kpym: I am sorry, the confusion came because of the mixed axis scalings. NOT because of the function...

    – Dave
    Apr 3 at 13:05
















4















I want to draw a simplified Michaelis-Menten kinetic (monod-function) to compare it with a linear function.





Minimum Working Example (MWE):



documentclass{standalone}
usepackage{pgfplots}
usepackage{amsmath}

pgfplotsset{compat=1.14, /pgf/declare function={f1(x)=ln(x);}}% <- This is the exponential function which needs to be optimized

begin{document}
begin{tikzpicture}
begin{axis}[
ymin = 0,
xmin = 0,
xmax = 1,
ymax = 0.9,
axis x line = bottom,
axis y line = left,
]

% addplot[no marks, samples=100, draw=blue] {f1(x)};% This is the exponential graph based on the function
addplot[no marks, samples=100, draw=black, thick] coordinates{(0,0) (0.2020725942,0.35)};%
addplot[no marks, samples=100, draw=black, thick] (0.2020725942,0.35) to [out=60,in=180] (0.8,0.7) to [out=0,in=0] (1,0.7);%
draw[draw=black, dashed] (0,0.7) -- node[above] {(y_{text{tot}})} ++(0.8,0.0);%
draw[draw=black, dashed] (0,0.35) -- node[above] {(frac{y_{text{tot}}}{2})} ++(0.2020725942,0) -- (0.2020725942,-0.35);%
end{axis}
end{tikzpicture}
end{document}




Screenshot of the result:



Screenshot of the result





Description of the issue:



How can I replace the current graph with an exponential graph?



Start point of the exponential graph:




  • Start point: x = 0.2020725942,
    y = 0.35,
    angle = 60°,

  • End point: y = ~ 0.7 (of course, wherever the e-function would end)


As soon as I activate the graph with the exponential function, my whole diagram will be distorted. How to implement an exponential graph based on the upper values correctly?










share|improve this question


















  • 3





    This looks like a question of math not of tex/tikz : how should I choose a and b in f(x) = a*exp(x)+b such that f(0.2020725942)=0.35 and f'(0.2020725942)=tan(pi/3) ? If this is the case here is not the right place to ask this question.

    – Kpym
    Apr 3 at 12:23











  • @Kpym: I am sorry, the confusion came because of the mixed axis scalings. NOT because of the function...

    – Dave
    Apr 3 at 13:05














4












4








4








I want to draw a simplified Michaelis-Menten kinetic (monod-function) to compare it with a linear function.





Minimum Working Example (MWE):



documentclass{standalone}
usepackage{pgfplots}
usepackage{amsmath}

pgfplotsset{compat=1.14, /pgf/declare function={f1(x)=ln(x);}}% <- This is the exponential function which needs to be optimized

begin{document}
begin{tikzpicture}
begin{axis}[
ymin = 0,
xmin = 0,
xmax = 1,
ymax = 0.9,
axis x line = bottom,
axis y line = left,
]

% addplot[no marks, samples=100, draw=blue] {f1(x)};% This is the exponential graph based on the function
addplot[no marks, samples=100, draw=black, thick] coordinates{(0,0) (0.2020725942,0.35)};%
addplot[no marks, samples=100, draw=black, thick] (0.2020725942,0.35) to [out=60,in=180] (0.8,0.7) to [out=0,in=0] (1,0.7);%
draw[draw=black, dashed] (0,0.7) -- node[above] {(y_{text{tot}})} ++(0.8,0.0);%
draw[draw=black, dashed] (0,0.35) -- node[above] {(frac{y_{text{tot}}}{2})} ++(0.2020725942,0) -- (0.2020725942,-0.35);%
end{axis}
end{tikzpicture}
end{document}




Screenshot of the result:



Screenshot of the result





Description of the issue:



How can I replace the current graph with an exponential graph?



Start point of the exponential graph:




  • Start point: x = 0.2020725942,
    y = 0.35,
    angle = 60°,

  • End point: y = ~ 0.7 (of course, wherever the e-function would end)


As soon as I activate the graph with the exponential function, my whole diagram will be distorted. How to implement an exponential graph based on the upper values correctly?










share|improve this question














I want to draw a simplified Michaelis-Menten kinetic (monod-function) to compare it with a linear function.





Minimum Working Example (MWE):



documentclass{standalone}
usepackage{pgfplots}
usepackage{amsmath}

pgfplotsset{compat=1.14, /pgf/declare function={f1(x)=ln(x);}}% <- This is the exponential function which needs to be optimized

begin{document}
begin{tikzpicture}
begin{axis}[
ymin = 0,
xmin = 0,
xmax = 1,
ymax = 0.9,
axis x line = bottom,
axis y line = left,
]

% addplot[no marks, samples=100, draw=blue] {f1(x)};% This is the exponential graph based on the function
addplot[no marks, samples=100, draw=black, thick] coordinates{(0,0) (0.2020725942,0.35)};%
addplot[no marks, samples=100, draw=black, thick] (0.2020725942,0.35) to [out=60,in=180] (0.8,0.7) to [out=0,in=0] (1,0.7);%
draw[draw=black, dashed] (0,0.7) -- node[above] {(y_{text{tot}})} ++(0.8,0.0);%
draw[draw=black, dashed] (0,0.35) -- node[above] {(frac{y_{text{tot}}}{2})} ++(0.2020725942,0) -- (0.2020725942,-0.35);%
end{axis}
end{tikzpicture}
end{document}




Screenshot of the result:



Screenshot of the result





Description of the issue:



How can I replace the current graph with an exponential graph?



Start point of the exponential graph:




  • Start point: x = 0.2020725942,
    y = 0.35,
    angle = 60°,

  • End point: y = ~ 0.7 (of course, wherever the e-function would end)


As soon as I activate the graph with the exponential function, my whole diagram will be distorted. How to implement an exponential graph based on the upper values correctly?







tikz-pgf pgfplots plot graphs tikz-graphs






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Apr 3 at 11:53









DaveDave

1,236619




1,236619








  • 3





    This looks like a question of math not of tex/tikz : how should I choose a and b in f(x) = a*exp(x)+b such that f(0.2020725942)=0.35 and f'(0.2020725942)=tan(pi/3) ? If this is the case here is not the right place to ask this question.

    – Kpym
    Apr 3 at 12:23











  • @Kpym: I am sorry, the confusion came because of the mixed axis scalings. NOT because of the function...

    – Dave
    Apr 3 at 13:05














  • 3





    This looks like a question of math not of tex/tikz : how should I choose a and b in f(x) = a*exp(x)+b such that f(0.2020725942)=0.35 and f'(0.2020725942)=tan(pi/3) ? If this is the case here is not the right place to ask this question.

    – Kpym
    Apr 3 at 12:23











  • @Kpym: I am sorry, the confusion came because of the mixed axis scalings. NOT because of the function...

    – Dave
    Apr 3 at 13:05








3




3





This looks like a question of math not of tex/tikz : how should I choose a and b in f(x) = a*exp(x)+b such that f(0.2020725942)=0.35 and f'(0.2020725942)=tan(pi/3) ? If this is the case here is not the right place to ask this question.

– Kpym
Apr 3 at 12:23





This looks like a question of math not of tex/tikz : how should I choose a and b in f(x) = a*exp(x)+b such that f(0.2020725942)=0.35 and f'(0.2020725942)=tan(pi/3) ? If this is the case here is not the right place to ask this question.

– Kpym
Apr 3 at 12:23













@Kpym: I am sorry, the confusion came because of the mixed axis scalings. NOT because of the function...

– Dave
Apr 3 at 13:05





@Kpym: I am sorry, the confusion came because of the mixed axis scalings. NOT because of the function...

– Dave
Apr 3 at 13:05










2 Answers
2






active

oldest

votes


















7














One way is via this (note this uses a differnt function than yours). Your MWE is not wrong IMO. However, due to varying domains, your final axis is getting mixed-up.



Nevertheless, you can obtain your desired solution with a summation of two-exponents.



documentclass{amsart}
usepackage{pgfplots}
pgfplotsset{compat=newest}
usepackage{tikz}
begin{document}
begin{tikzpicture}
begin{axis}[
scaled ticks=false,
xmin=0,
xmax=1,
ymin=0,
ymax=1.2,
xlabel=x axis label,
ylabel=y axis label,
axis x line = bottom,
axis y line = left,
]
addplot[domain=0.2:1.2, samples=1000, red, ultra thick,smooth] {(1-e^(-5*x)-exp(-10*x))*0.7};
addplot[no marks, samples=100, draw=black, thick] coordinates{(0,0) (0.2020725942,0.35)};%
draw[draw=black, dashed] (0,0.7) -- node[above] {(y_{text{tot}})} ++(1,0.0);%
draw[draw=black, dashed] (0,0.35) -- node[above] {(frac{y_{text{tot}}}{2})} ++(0.2020725942,0) -- (0.2020725942,-0.35);%
end{axis}
end{tikzpicture}
end{document}


to get:



enter image description here






share|improve this answer
























  • Thanks a lot! I am confused: Why doesn't this work with documentclass{standalone}?

    – Dave
    Apr 3 at 13:04






  • 1





    @Dave in standalone please include amsmath.

    – Raaja
    Apr 3 at 13:07



















7














I would not find a function for that. A curve with exact starting angle (60°) and ending angle (180°) is enough here.



And also, why don't you simply use tan function in TikZ? 0.2020725942 ≈ 0.35 × tan(30°), but certainly if you type {.35*tan(30)} it is more accurate than 0.2020725942.



documentclass[tikz]{standalone}
begin{document}
begin{tikzpicture}[scale=8,>=stealth]
draw[<->] (1,0) -- (0,0) -- (0,.9);
draw[thick] (0,0) -- ({.35*tan(30)},0.35) coordinate (a);
draw[thick] (a) to[out=60,in=180] (0.8,0.7) -- (1,0.7);
foreach i in {0,0.2,0.4,0.6,0.8} {
draw (i,.01) -- (i,-.01) node[below] {$i$};
draw (.01,i) -- (-.01,i) node[left] {$i$};
}
draw (1,.01) -- (1,-.01) node[below] {$1$};
draw[dashed] (0.8,0.7) -- (0,0.7) node[midway,above] {$y_mathrm{tot}$};
draw[dashed] ({.35*tan(30)},0) -- ({.35*tan(30)},0.35);
draw[dashed] ({.35*tan(30)},0.35) -- (0,0.35) node[midway,above] {$y_mathrm{tot}/2$};
end{tikzpicture}
end{document}


enter image description here






share|improve this answer


























  • @Dave I think this is more of an apt answer.

    – Raaja
    Apr 4 at 4:31






  • 1





    @JouleV: Thanks a lot for your answer! Yes, I have plotted the line as a simple draw with start angle before posting my request. However, the previous line was too uniform - the Michaelis-Menten-kinetic should have a exponential arc instead of a uniform arc. :-( But your idea of posting lengths calculated by angles is just great! Thanks a lot!

    – Dave
    Apr 4 at 15:51












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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









7














One way is via this (note this uses a differnt function than yours). Your MWE is not wrong IMO. However, due to varying domains, your final axis is getting mixed-up.



Nevertheless, you can obtain your desired solution with a summation of two-exponents.



documentclass{amsart}
usepackage{pgfplots}
pgfplotsset{compat=newest}
usepackage{tikz}
begin{document}
begin{tikzpicture}
begin{axis}[
scaled ticks=false,
xmin=0,
xmax=1,
ymin=0,
ymax=1.2,
xlabel=x axis label,
ylabel=y axis label,
axis x line = bottom,
axis y line = left,
]
addplot[domain=0.2:1.2, samples=1000, red, ultra thick,smooth] {(1-e^(-5*x)-exp(-10*x))*0.7};
addplot[no marks, samples=100, draw=black, thick] coordinates{(0,0) (0.2020725942,0.35)};%
draw[draw=black, dashed] (0,0.7) -- node[above] {(y_{text{tot}})} ++(1,0.0);%
draw[draw=black, dashed] (0,0.35) -- node[above] {(frac{y_{text{tot}}}{2})} ++(0.2020725942,0) -- (0.2020725942,-0.35);%
end{axis}
end{tikzpicture}
end{document}


to get:



enter image description here






share|improve this answer
























  • Thanks a lot! I am confused: Why doesn't this work with documentclass{standalone}?

    – Dave
    Apr 3 at 13:04






  • 1





    @Dave in standalone please include amsmath.

    – Raaja
    Apr 3 at 13:07
















7














One way is via this (note this uses a differnt function than yours). Your MWE is not wrong IMO. However, due to varying domains, your final axis is getting mixed-up.



Nevertheless, you can obtain your desired solution with a summation of two-exponents.



documentclass{amsart}
usepackage{pgfplots}
pgfplotsset{compat=newest}
usepackage{tikz}
begin{document}
begin{tikzpicture}
begin{axis}[
scaled ticks=false,
xmin=0,
xmax=1,
ymin=0,
ymax=1.2,
xlabel=x axis label,
ylabel=y axis label,
axis x line = bottom,
axis y line = left,
]
addplot[domain=0.2:1.2, samples=1000, red, ultra thick,smooth] {(1-e^(-5*x)-exp(-10*x))*0.7};
addplot[no marks, samples=100, draw=black, thick] coordinates{(0,0) (0.2020725942,0.35)};%
draw[draw=black, dashed] (0,0.7) -- node[above] {(y_{text{tot}})} ++(1,0.0);%
draw[draw=black, dashed] (0,0.35) -- node[above] {(frac{y_{text{tot}}}{2})} ++(0.2020725942,0) -- (0.2020725942,-0.35);%
end{axis}
end{tikzpicture}
end{document}


to get:



enter image description here






share|improve this answer
























  • Thanks a lot! I am confused: Why doesn't this work with documentclass{standalone}?

    – Dave
    Apr 3 at 13:04






  • 1





    @Dave in standalone please include amsmath.

    – Raaja
    Apr 3 at 13:07














7












7








7







One way is via this (note this uses a differnt function than yours). Your MWE is not wrong IMO. However, due to varying domains, your final axis is getting mixed-up.



Nevertheless, you can obtain your desired solution with a summation of two-exponents.



documentclass{amsart}
usepackage{pgfplots}
pgfplotsset{compat=newest}
usepackage{tikz}
begin{document}
begin{tikzpicture}
begin{axis}[
scaled ticks=false,
xmin=0,
xmax=1,
ymin=0,
ymax=1.2,
xlabel=x axis label,
ylabel=y axis label,
axis x line = bottom,
axis y line = left,
]
addplot[domain=0.2:1.2, samples=1000, red, ultra thick,smooth] {(1-e^(-5*x)-exp(-10*x))*0.7};
addplot[no marks, samples=100, draw=black, thick] coordinates{(0,0) (0.2020725942,0.35)};%
draw[draw=black, dashed] (0,0.7) -- node[above] {(y_{text{tot}})} ++(1,0.0);%
draw[draw=black, dashed] (0,0.35) -- node[above] {(frac{y_{text{tot}}}{2})} ++(0.2020725942,0) -- (0.2020725942,-0.35);%
end{axis}
end{tikzpicture}
end{document}


to get:



enter image description here






share|improve this answer













One way is via this (note this uses a differnt function than yours). Your MWE is not wrong IMO. However, due to varying domains, your final axis is getting mixed-up.



Nevertheless, you can obtain your desired solution with a summation of two-exponents.



documentclass{amsart}
usepackage{pgfplots}
pgfplotsset{compat=newest}
usepackage{tikz}
begin{document}
begin{tikzpicture}
begin{axis}[
scaled ticks=false,
xmin=0,
xmax=1,
ymin=0,
ymax=1.2,
xlabel=x axis label,
ylabel=y axis label,
axis x line = bottom,
axis y line = left,
]
addplot[domain=0.2:1.2, samples=1000, red, ultra thick,smooth] {(1-e^(-5*x)-exp(-10*x))*0.7};
addplot[no marks, samples=100, draw=black, thick] coordinates{(0,0) (0.2020725942,0.35)};%
draw[draw=black, dashed] (0,0.7) -- node[above] {(y_{text{tot}})} ++(1,0.0);%
draw[draw=black, dashed] (0,0.35) -- node[above] {(frac{y_{text{tot}}}{2})} ++(0.2020725942,0) -- (0.2020725942,-0.35);%
end{axis}
end{tikzpicture}
end{document}


to get:



enter image description here







share|improve this answer












share|improve this answer



share|improve this answer










answered Apr 3 at 12:22









RaajaRaaja

5,32421644




5,32421644













  • Thanks a lot! I am confused: Why doesn't this work with documentclass{standalone}?

    – Dave
    Apr 3 at 13:04






  • 1





    @Dave in standalone please include amsmath.

    – Raaja
    Apr 3 at 13:07



















  • Thanks a lot! I am confused: Why doesn't this work with documentclass{standalone}?

    – Dave
    Apr 3 at 13:04






  • 1





    @Dave in standalone please include amsmath.

    – Raaja
    Apr 3 at 13:07

















Thanks a lot! I am confused: Why doesn't this work with documentclass{standalone}?

– Dave
Apr 3 at 13:04





Thanks a lot! I am confused: Why doesn't this work with documentclass{standalone}?

– Dave
Apr 3 at 13:04




1




1





@Dave in standalone please include amsmath.

– Raaja
Apr 3 at 13:07





@Dave in standalone please include amsmath.

– Raaja
Apr 3 at 13:07











7














I would not find a function for that. A curve with exact starting angle (60°) and ending angle (180°) is enough here.



And also, why don't you simply use tan function in TikZ? 0.2020725942 ≈ 0.35 × tan(30°), but certainly if you type {.35*tan(30)} it is more accurate than 0.2020725942.



documentclass[tikz]{standalone}
begin{document}
begin{tikzpicture}[scale=8,>=stealth]
draw[<->] (1,0) -- (0,0) -- (0,.9);
draw[thick] (0,0) -- ({.35*tan(30)},0.35) coordinate (a);
draw[thick] (a) to[out=60,in=180] (0.8,0.7) -- (1,0.7);
foreach i in {0,0.2,0.4,0.6,0.8} {
draw (i,.01) -- (i,-.01) node[below] {$i$};
draw (.01,i) -- (-.01,i) node[left] {$i$};
}
draw (1,.01) -- (1,-.01) node[below] {$1$};
draw[dashed] (0.8,0.7) -- (0,0.7) node[midway,above] {$y_mathrm{tot}$};
draw[dashed] ({.35*tan(30)},0) -- ({.35*tan(30)},0.35);
draw[dashed] ({.35*tan(30)},0.35) -- (0,0.35) node[midway,above] {$y_mathrm{tot}/2$};
end{tikzpicture}
end{document}


enter image description here






share|improve this answer


























  • @Dave I think this is more of an apt answer.

    – Raaja
    Apr 4 at 4:31






  • 1





    @JouleV: Thanks a lot for your answer! Yes, I have plotted the line as a simple draw with start angle before posting my request. However, the previous line was too uniform - the Michaelis-Menten-kinetic should have a exponential arc instead of a uniform arc. :-( But your idea of posting lengths calculated by angles is just great! Thanks a lot!

    – Dave
    Apr 4 at 15:51
















7














I would not find a function for that. A curve with exact starting angle (60°) and ending angle (180°) is enough here.



And also, why don't you simply use tan function in TikZ? 0.2020725942 ≈ 0.35 × tan(30°), but certainly if you type {.35*tan(30)} it is more accurate than 0.2020725942.



documentclass[tikz]{standalone}
begin{document}
begin{tikzpicture}[scale=8,>=stealth]
draw[<->] (1,0) -- (0,0) -- (0,.9);
draw[thick] (0,0) -- ({.35*tan(30)},0.35) coordinate (a);
draw[thick] (a) to[out=60,in=180] (0.8,0.7) -- (1,0.7);
foreach i in {0,0.2,0.4,0.6,0.8} {
draw (i,.01) -- (i,-.01) node[below] {$i$};
draw (.01,i) -- (-.01,i) node[left] {$i$};
}
draw (1,.01) -- (1,-.01) node[below] {$1$};
draw[dashed] (0.8,0.7) -- (0,0.7) node[midway,above] {$y_mathrm{tot}$};
draw[dashed] ({.35*tan(30)},0) -- ({.35*tan(30)},0.35);
draw[dashed] ({.35*tan(30)},0.35) -- (0,0.35) node[midway,above] {$y_mathrm{tot}/2$};
end{tikzpicture}
end{document}


enter image description here






share|improve this answer


























  • @Dave I think this is more of an apt answer.

    – Raaja
    Apr 4 at 4:31






  • 1





    @JouleV: Thanks a lot for your answer! Yes, I have plotted the line as a simple draw with start angle before posting my request. However, the previous line was too uniform - the Michaelis-Menten-kinetic should have a exponential arc instead of a uniform arc. :-( But your idea of posting lengths calculated by angles is just great! Thanks a lot!

    – Dave
    Apr 4 at 15:51














7












7








7







I would not find a function for that. A curve with exact starting angle (60°) and ending angle (180°) is enough here.



And also, why don't you simply use tan function in TikZ? 0.2020725942 ≈ 0.35 × tan(30°), but certainly if you type {.35*tan(30)} it is more accurate than 0.2020725942.



documentclass[tikz]{standalone}
begin{document}
begin{tikzpicture}[scale=8,>=stealth]
draw[<->] (1,0) -- (0,0) -- (0,.9);
draw[thick] (0,0) -- ({.35*tan(30)},0.35) coordinate (a);
draw[thick] (a) to[out=60,in=180] (0.8,0.7) -- (1,0.7);
foreach i in {0,0.2,0.4,0.6,0.8} {
draw (i,.01) -- (i,-.01) node[below] {$i$};
draw (.01,i) -- (-.01,i) node[left] {$i$};
}
draw (1,.01) -- (1,-.01) node[below] {$1$};
draw[dashed] (0.8,0.7) -- (0,0.7) node[midway,above] {$y_mathrm{tot}$};
draw[dashed] ({.35*tan(30)},0) -- ({.35*tan(30)},0.35);
draw[dashed] ({.35*tan(30)},0.35) -- (0,0.35) node[midway,above] {$y_mathrm{tot}/2$};
end{tikzpicture}
end{document}


enter image description here






share|improve this answer















I would not find a function for that. A curve with exact starting angle (60°) and ending angle (180°) is enough here.



And also, why don't you simply use tan function in TikZ? 0.2020725942 ≈ 0.35 × tan(30°), but certainly if you type {.35*tan(30)} it is more accurate than 0.2020725942.



documentclass[tikz]{standalone}
begin{document}
begin{tikzpicture}[scale=8,>=stealth]
draw[<->] (1,0) -- (0,0) -- (0,.9);
draw[thick] (0,0) -- ({.35*tan(30)},0.35) coordinate (a);
draw[thick] (a) to[out=60,in=180] (0.8,0.7) -- (1,0.7);
foreach i in {0,0.2,0.4,0.6,0.8} {
draw (i,.01) -- (i,-.01) node[below] {$i$};
draw (.01,i) -- (-.01,i) node[left] {$i$};
}
draw (1,.01) -- (1,-.01) node[below] {$1$};
draw[dashed] (0.8,0.7) -- (0,0.7) node[midway,above] {$y_mathrm{tot}$};
draw[dashed] ({.35*tan(30)},0) -- ({.35*tan(30)},0.35);
draw[dashed] ({.35*tan(30)},0.35) -- (0,0.35) node[midway,above] {$y_mathrm{tot}/2$};
end{tikzpicture}
end{document}


enter image description here







share|improve this answer














share|improve this answer



share|improve this answer








edited Apr 3 at 16:19

























answered Apr 3 at 13:23









JouleVJouleV

12.3k22663




12.3k22663













  • @Dave I think this is more of an apt answer.

    – Raaja
    Apr 4 at 4:31






  • 1





    @JouleV: Thanks a lot for your answer! Yes, I have plotted the line as a simple draw with start angle before posting my request. However, the previous line was too uniform - the Michaelis-Menten-kinetic should have a exponential arc instead of a uniform arc. :-( But your idea of posting lengths calculated by angles is just great! Thanks a lot!

    – Dave
    Apr 4 at 15:51



















  • @Dave I think this is more of an apt answer.

    – Raaja
    Apr 4 at 4:31






  • 1





    @JouleV: Thanks a lot for your answer! Yes, I have plotted the line as a simple draw with start angle before posting my request. However, the previous line was too uniform - the Michaelis-Menten-kinetic should have a exponential arc instead of a uniform arc. :-( But your idea of posting lengths calculated by angles is just great! Thanks a lot!

    – Dave
    Apr 4 at 15:51

















@Dave I think this is more of an apt answer.

– Raaja
Apr 4 at 4:31





@Dave I think this is more of an apt answer.

– Raaja
Apr 4 at 4:31




1




1





@JouleV: Thanks a lot for your answer! Yes, I have plotted the line as a simple draw with start angle before posting my request. However, the previous line was too uniform - the Michaelis-Menten-kinetic should have a exponential arc instead of a uniform arc. :-( But your idea of posting lengths calculated by angles is just great! Thanks a lot!

– Dave
Apr 4 at 15:51





@JouleV: Thanks a lot for your answer! Yes, I have plotted the line as a simple draw with start angle before posting my request. However, the previous line was too uniform - the Michaelis-Menten-kinetic should have a exponential arc instead of a uniform arc. :-( But your idea of posting lengths calculated by angles is just great! Thanks a lot!

– Dave
Apr 4 at 15:51


















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