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

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
add a comment |
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:

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
3
This looks like a question of math not of tex/tikz : how should I chooseaandbinf(x) = a*exp(x)+bsuch thatf(0.2020725942)=0.35andf'(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
add a comment |
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:

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
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:

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
tikz-pgf pgfplots plot graphs tikz-graphs
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 chooseaandbinf(x) = a*exp(x)+bsuch thatf(0.2020725942)=0.35andf'(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
add a comment |
3
This looks like a question of math not of tex/tikz : how should I chooseaandbinf(x) = a*exp(x)+bsuch thatf(0.2020725942)=0.35andf'(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
add a comment |
2 Answers
2
active
oldest
votes
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:

Thanks a lot! I am confused: Why doesn't this work withdocumentclass{standalone}?
– Dave
Apr 3 at 13:04
1
@Dave instandaloneplease includeamsmath.
– Raaja
Apr 3 at 13:07
add a comment |
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}

@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
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
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:

Thanks a lot! I am confused: Why doesn't this work withdocumentclass{standalone}?
– Dave
Apr 3 at 13:04
1
@Dave instandaloneplease includeamsmath.
– Raaja
Apr 3 at 13:07
add a comment |
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:

Thanks a lot! I am confused: Why doesn't this work withdocumentclass{standalone}?
– Dave
Apr 3 at 13:04
1
@Dave instandaloneplease includeamsmath.
– Raaja
Apr 3 at 13:07
add a comment |
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:

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:

answered Apr 3 at 12:22
RaajaRaaja
5,32421644
5,32421644
Thanks a lot! I am confused: Why doesn't this work withdocumentclass{standalone}?
– Dave
Apr 3 at 13:04
1
@Dave instandaloneplease includeamsmath.
– Raaja
Apr 3 at 13:07
add a comment |
Thanks a lot! I am confused: Why doesn't this work withdocumentclass{standalone}?
– Dave
Apr 3 at 13:04
1
@Dave instandaloneplease includeamsmath.
– 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
add a comment |
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}

@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
add a comment |
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}

@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
add a comment |
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}

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}

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
add a comment |
@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
add a comment |
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3
This looks like a question of math not of tex/tikz : how should I choose
aandbinf(x) = a*exp(x)+bsuch thatf(0.2020725942)=0.35andf'(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