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pgfplots: How to draw exponential graph with 60° start angle?
plotting two time series with boundsHow do i get the x axis on top but keep a line on the bottomTikZ: Drawing an arc from an intersection to an intersectionHow to prevent rounded and duplicated tick labels in pgfplots with fixed precision?How to hide empty (value 0) ybars with pgfplots?Show mark labels near marks and not centered in ybar interaval graphDrawing rectilinear curves in Tikz, aka an Etch-a-Sketch drawingDisplaying a pgfplots graph above another pgfplots graph in TikZpgfplots: percentage in matrix plotCenter the axes in the coordinate origin
I want to draw a simplified Michaelis-Menten kinetic (monod-function) to compare it with a linear function.
Minimum Working Example (MWE):
documentclassstandalone
usepackagepgfplots
usepackageamsmath
pgfplotssetcompat=1.14, /pgf/declare function=f1(x)=ln(x);% <- This is the exponential function which needs to be optimized
begindocument
begintikzpicture
beginaxis[
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_texttot) ++(0.8,0.0);%
draw[draw=black, dashed] (0,0.35) -- node[above] (fracy_texttot2) ++(0.2020725942,0) -- (0.2020725942,-0.35);%
endaxis
endtikzpicture
enddocument
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):
documentclassstandalone
usepackagepgfplots
usepackageamsmath
pgfplotssetcompat=1.14, /pgf/declare function=f1(x)=ln(x);% <- This is the exponential function which needs to be optimized
begindocument
begintikzpicture
beginaxis[
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_texttot) ++(0.8,0.0);%
draw[draw=black, dashed] (0,0.35) -- node[above] (fracy_texttot2) ++(0.2020725942,0) -- (0.2020725942,-0.35);%
endaxis
endtikzpicture
enddocument
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):
documentclassstandalone
usepackagepgfplots
usepackageamsmath
pgfplotssetcompat=1.14, /pgf/declare function=f1(x)=ln(x);% <- This is the exponential function which needs to be optimized
begindocument
begintikzpicture
beginaxis[
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_texttot) ++(0.8,0.0);%
draw[draw=black, dashed] (0,0.35) -- node[above] (fracy_texttot2) ++(0.2020725942,0) -- (0.2020725942,-0.35);%
endaxis
endtikzpicture
enddocument
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):
documentclassstandalone
usepackagepgfplots
usepackageamsmath
pgfplotssetcompat=1.14, /pgf/declare function=f1(x)=ln(x);% <- This is the exponential function which needs to be optimized
begindocument
begintikzpicture
beginaxis[
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_texttot) ++(0.8,0.0);%
draw[draw=black, dashed] (0,0.35) -- node[above] (fracy_texttot2) ++(0.2020725942,0) -- (0.2020725942,-0.35);%
endaxis
endtikzpicture
enddocument
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
1,190619
1,190619
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.
documentclassamsart
usepackagepgfplots
pgfplotssetcompat=newest
usepackagetikz
begindocument
begintikzpicture
beginaxis[
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_texttot) ++(1,0.0);%
draw[draw=black, dashed] (0,0.35) -- node[above] (fracy_texttot2) ++(0.2020725942,0) -- (0.2020725942,-0.35);%
endaxis
endtikzpicture
enddocument
to get:

Thanks a lot! I am confused: Why doesn't this work withdocumentclassstandalone?
– 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
begindocument
begintikzpicture[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] $y_mathrmtot$;
draw[dashed] (.35*tan(30),0) -- (.35*tan(30),0.35);
draw[dashed] (.35*tan(30),0.35) -- (0,0.35) node[midway,above] $y_mathrmtot/2$;
endtikzpicture
enddocument

@Dave I think this is more of an apt answer.
– Raaja
2 days ago
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
2 days ago
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.
documentclassamsart
usepackagepgfplots
pgfplotssetcompat=newest
usepackagetikz
begindocument
begintikzpicture
beginaxis[
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_texttot) ++(1,0.0);%
draw[draw=black, dashed] (0,0.35) -- node[above] (fracy_texttot2) ++(0.2020725942,0) -- (0.2020725942,-0.35);%
endaxis
endtikzpicture
enddocument
to get:

Thanks a lot! I am confused: Why doesn't this work withdocumentclassstandalone?
– 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.
documentclassamsart
usepackagepgfplots
pgfplotssetcompat=newest
usepackagetikz
begindocument
begintikzpicture
beginaxis[
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_texttot) ++(1,0.0);%
draw[draw=black, dashed] (0,0.35) -- node[above] (fracy_texttot2) ++(0.2020725942,0) -- (0.2020725942,-0.35);%
endaxis
endtikzpicture
enddocument
to get:

Thanks a lot! I am confused: Why doesn't this work withdocumentclassstandalone?
– 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.
documentclassamsart
usepackagepgfplots
pgfplotssetcompat=newest
usepackagetikz
begindocument
begintikzpicture
beginaxis[
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_texttot) ++(1,0.0);%
draw[draw=black, dashed] (0,0.35) -- node[above] (fracy_texttot2) ++(0.2020725942,0) -- (0.2020725942,-0.35);%
endaxis
endtikzpicture
enddocument
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.
documentclassamsart
usepackagepgfplots
pgfplotssetcompat=newest
usepackagetikz
begindocument
begintikzpicture
beginaxis[
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_texttot) ++(1,0.0);%
draw[draw=black, dashed] (0,0.35) -- node[above] (fracy_texttot2) ++(0.2020725942,0) -- (0.2020725942,-0.35);%
endaxis
endtikzpicture
enddocument
to get:

answered Apr 3 at 12:22
RaajaRaaja
5,31421644
5,31421644
Thanks a lot! I am confused: Why doesn't this work withdocumentclassstandalone?
– 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 withdocumentclassstandalone?
– 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
documentclassstandalone?– Dave
Apr 3 at 13:04
Thanks a lot! I am confused: Why doesn't this work with
documentclassstandalone?– 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
begindocument
begintikzpicture[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] $y_mathrmtot$;
draw[dashed] (.35*tan(30),0) -- (.35*tan(30),0.35);
draw[dashed] (.35*tan(30),0.35) -- (0,0.35) node[midway,above] $y_mathrmtot/2$;
endtikzpicture
enddocument

@Dave I think this is more of an apt answer.
– Raaja
2 days ago
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
2 days ago
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
begindocument
begintikzpicture[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] $y_mathrmtot$;
draw[dashed] (.35*tan(30),0) -- (.35*tan(30),0.35);
draw[dashed] (.35*tan(30),0.35) -- (0,0.35) node[midway,above] $y_mathrmtot/2$;
endtikzpicture
enddocument

@Dave I think this is more of an apt answer.
– Raaja
2 days ago
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
2 days ago
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
begindocument
begintikzpicture[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] $y_mathrmtot$;
draw[dashed] (.35*tan(30),0) -- (.35*tan(30),0.35);
draw[dashed] (.35*tan(30),0.35) -- (0,0.35) node[midway,above] $y_mathrmtot/2$;
endtikzpicture
enddocument

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
begindocument
begintikzpicture[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] $y_mathrmtot$;
draw[dashed] (.35*tan(30),0) -- (.35*tan(30),0.35);
draw[dashed] (.35*tan(30),0.35) -- (0,0.35) node[midway,above] $y_mathrmtot/2$;
endtikzpicture
enddocument

edited Apr 3 at 16:19
answered Apr 3 at 13:23
JouleVJouleV
11.1k22560
11.1k22560
@Dave I think this is more of an apt answer.
– Raaja
2 days ago
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
2 days ago
add a comment |
@Dave I think this is more of an apt answer.
– Raaja
2 days ago
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
2 days ago
@Dave I think this is more of an apt answer.
– Raaja
2 days ago
@Dave I think this is more of an apt answer.
– Raaja
2 days ago
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
2 days ago
@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
2 days ago
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