Tangent Lines Diagram Along Smooth Curve

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
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I am trying to replicate a diagram and have a minimal example. I don't know how to add additional tangents between the specified points and preserve smoothness.



documentclassarticle
usepackagetikz
usetikzlibrarycalc,intersections

begindocument
begintikzpicture
% Axes
draw [->, name path=x] (-1,0) -- (11,0) node [right] $x$;
draw [->] (0,-1) -- (0,6) node [above] $y$;
% Origin
node at (0,0) [below left] $0$;
% Points
coordinate (start) at (1,-0.8);
coordinate (c1) at (3,3);
coordinate (c2) at (5.5,1.5);
coordinate (c3) at (8,4);
coordinate (end) at (10.5,-0.8);
% show the points
foreach n in start,c1,c2,c3,end fill [black] (n)
circle (2pt) node [below] ;
% join the coordinates
draw [thick,name path=curve] (start) to[out=70,in=180] (c1) to[out=0,in=180]
(c2) to[out=0,in=180] (c3) to[out=0,in=150] (end);
% add tangets and dashed lines
foreach c in 1,2,3
%draw [dashed] let p1=(cc) in (cc) -- (x1,0) node [below] $c_c$;
draw ($(cc)-(0.75,0)$) -- ($(cc)+(0.75,0)$) node [midway,above=4mm]$$; %$f'(c_c)=0$;

% add a and b
path [name intersections=of=x and curve, by=a,b] (a) node [below left] $a$
(b) node [above right] $b$;
endtikzpicture
enddocument


Outputs:



enter image description here
My Goal is create the following:



enter image description here



Feel free to make any changes that make the minimal example approach the goal diagram










share|improve this question





















  • I sense without having a concrete equation for the graph above this will be a challenge. Unless there is another programming tool for TiKz?
    – MathScholar
    Nov 24 at 16:32










  • Marmot mentions tex.stackexchange.com/questions/25928/…
    – MathScholar
    Nov 24 at 16:40















up vote
6
down vote

favorite
2












I am trying to replicate a diagram and have a minimal example. I don't know how to add additional tangents between the specified points and preserve smoothness.



documentclassarticle
usepackagetikz
usetikzlibrarycalc,intersections

begindocument
begintikzpicture
% Axes
draw [->, name path=x] (-1,0) -- (11,0) node [right] $x$;
draw [->] (0,-1) -- (0,6) node [above] $y$;
% Origin
node at (0,0) [below left] $0$;
% Points
coordinate (start) at (1,-0.8);
coordinate (c1) at (3,3);
coordinate (c2) at (5.5,1.5);
coordinate (c3) at (8,4);
coordinate (end) at (10.5,-0.8);
% show the points
foreach n in start,c1,c2,c3,end fill [black] (n)
circle (2pt) node [below] ;
% join the coordinates
draw [thick,name path=curve] (start) to[out=70,in=180] (c1) to[out=0,in=180]
(c2) to[out=0,in=180] (c3) to[out=0,in=150] (end);
% add tangets and dashed lines
foreach c in 1,2,3
%draw [dashed] let p1=(cc) in (cc) -- (x1,0) node [below] $c_c$;
draw ($(cc)-(0.75,0)$) -- ($(cc)+(0.75,0)$) node [midway,above=4mm]$$; %$f'(c_c)=0$;

% add a and b
path [name intersections=of=x and curve, by=a,b] (a) node [below left] $a$
(b) node [above right] $b$;
endtikzpicture
enddocument


Outputs:



enter image description here
My Goal is create the following:



enter image description here



Feel free to make any changes that make the minimal example approach the goal diagram










share|improve this question





















  • I sense without having a concrete equation for the graph above this will be a challenge. Unless there is another programming tool for TiKz?
    – MathScholar
    Nov 24 at 16:32










  • Marmot mentions tex.stackexchange.com/questions/25928/…
    – MathScholar
    Nov 24 at 16:40













up vote
6
down vote

favorite
2









up vote
6
down vote

favorite
2






2





I am trying to replicate a diagram and have a minimal example. I don't know how to add additional tangents between the specified points and preserve smoothness.



documentclassarticle
usepackagetikz
usetikzlibrarycalc,intersections

begindocument
begintikzpicture
% Axes
draw [->, name path=x] (-1,0) -- (11,0) node [right] $x$;
draw [->] (0,-1) -- (0,6) node [above] $y$;
% Origin
node at (0,0) [below left] $0$;
% Points
coordinate (start) at (1,-0.8);
coordinate (c1) at (3,3);
coordinate (c2) at (5.5,1.5);
coordinate (c3) at (8,4);
coordinate (end) at (10.5,-0.8);
% show the points
foreach n in start,c1,c2,c3,end fill [black] (n)
circle (2pt) node [below] ;
% join the coordinates
draw [thick,name path=curve] (start) to[out=70,in=180] (c1) to[out=0,in=180]
(c2) to[out=0,in=180] (c3) to[out=0,in=150] (end);
% add tangets and dashed lines
foreach c in 1,2,3
%draw [dashed] let p1=(cc) in (cc) -- (x1,0) node [below] $c_c$;
draw ($(cc)-(0.75,0)$) -- ($(cc)+(0.75,0)$) node [midway,above=4mm]$$; %$f'(c_c)=0$;

% add a and b
path [name intersections=of=x and curve, by=a,b] (a) node [below left] $a$
(b) node [above right] $b$;
endtikzpicture
enddocument


Outputs:



enter image description here
My Goal is create the following:



enter image description here



Feel free to make any changes that make the minimal example approach the goal diagram










share|improve this question













I am trying to replicate a diagram and have a minimal example. I don't know how to add additional tangents between the specified points and preserve smoothness.



documentclassarticle
usepackagetikz
usetikzlibrarycalc,intersections

begindocument
begintikzpicture
% Axes
draw [->, name path=x] (-1,0) -- (11,0) node [right] $x$;
draw [->] (0,-1) -- (0,6) node [above] $y$;
% Origin
node at (0,0) [below left] $0$;
% Points
coordinate (start) at (1,-0.8);
coordinate (c1) at (3,3);
coordinate (c2) at (5.5,1.5);
coordinate (c3) at (8,4);
coordinate (end) at (10.5,-0.8);
% show the points
foreach n in start,c1,c2,c3,end fill [black] (n)
circle (2pt) node [below] ;
% join the coordinates
draw [thick,name path=curve] (start) to[out=70,in=180] (c1) to[out=0,in=180]
(c2) to[out=0,in=180] (c3) to[out=0,in=150] (end);
% add tangets and dashed lines
foreach c in 1,2,3
%draw [dashed] let p1=(cc) in (cc) -- (x1,0) node [below] $c_c$;
draw ($(cc)-(0.75,0)$) -- ($(cc)+(0.75,0)$) node [midway,above=4mm]$$; %$f'(c_c)=0$;

% add a and b
path [name intersections=of=x and curve, by=a,b] (a) node [below left] $a$
(b) node [above right] $b$;
endtikzpicture
enddocument


Outputs:



enter image description here
My Goal is create the following:



enter image description here



Feel free to make any changes that make the minimal example approach the goal diagram







tikz-pgf intersections






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Nov 24 at 15:53









MathScholar

4658




4658











  • I sense without having a concrete equation for the graph above this will be a challenge. Unless there is another programming tool for TiKz?
    – MathScholar
    Nov 24 at 16:32










  • Marmot mentions tex.stackexchange.com/questions/25928/…
    – MathScholar
    Nov 24 at 16:40

















  • I sense without having a concrete equation for the graph above this will be a challenge. Unless there is another programming tool for TiKz?
    – MathScholar
    Nov 24 at 16:32










  • Marmot mentions tex.stackexchange.com/questions/25928/…
    – MathScholar
    Nov 24 at 16:40
















I sense without having a concrete equation for the graph above this will be a challenge. Unless there is another programming tool for TiKz?
– MathScholar
Nov 24 at 16:32




I sense without having a concrete equation for the graph above this will be a challenge. Unless there is another programming tool for TiKz?
– MathScholar
Nov 24 at 16:32












Marmot mentions tex.stackexchange.com/questions/25928/…
– MathScholar
Nov 24 at 16:40





Marmot mentions tex.stackexchange.com/questions/25928/…
– MathScholar
Nov 24 at 16:40











3 Answers
3






active

oldest

votes

















up vote
7
down vote



accepted










You can use a bit of calculus:



enter image description here



The DrawTangent macro creates two vertical lines around the given x values and computes the intersection of those and the curve. Then, the tangent can be approximated by the secant between those two intersection points.



Code:



documentclass[border=2pt]standalone
usepackagetikz
usetikzlibrarycalc,intersections


newcommand*DeltaX0.01
newcommand*DrawTangent[5]%
% #1 = draw options
% #2 = name of curve
% #3 = ymin
% #4 = ymax
% #5 = x value at which tangent is to be drawn

path[name path=Vertical Line Left] (#5-DeltaX,#3) -- (#5-DeltaX,#4);
path[name path=Vertical Line Right] (#5+DeltaX,#3) -- (#5+DeltaX,#4);

path [name intersections=of=Vertical Line Left and #2];
coordinate (X0) at (intersection-1);
path [name intersections=of=Vertical Line Right and #2];
coordinate (X1) at (intersection-1);

draw [shorten <= -1.00cm, shorten >= -1.00cm, #1] (X0) -- (X1);
%

begindocument
begintikzpicture
% Axes
draw [->, name path=x] (-1,0) -- (11,0) node [right] $x$;
draw [->] (0,-1) -- (0,6) node [above] $y$;
% Origin
node at (0,0) [below left] $0$;
% Points
coordinate (start) at (1,-0.8);
coordinate (c1) at (3,3);
coordinate (c2) at (5.5,1.5);
coordinate (c3) at (8,4);
coordinate (end) at (10.5,-0.8);
% show the points
foreach n in start,c1,c2,c3,end fill [black] (n)
circle (2pt) node [below] ;
% join the coordinates
draw [thick,name path=curve] (start) to[out=70,in=180] (c1) to[out=0,in=180]
(c2) to[out=0,in=180] (c3) to[out=0,in=150] (end);
% add tangets and dashed lines
foreach c in 1,2,3
%draw [dashed] let p1=(cc) in (cc) -- (x1,0) node [below] $c_c$;
draw ($(cc)-(0.75,0)$) -- ($(cc)+(0.75,0)$) node [midway,above=4mm]$$; %$f'(c_c)=0$;

% add a and b
path [name intersections=of=x and curve, by=a,b]
(a) node [below left] $a$
(b) node [above right] $b$;

DrawTangent[red, thick]curve-141.5
DrawTangent[orange, thick]curve-143.5

endtikzpicture
enddocument





share|improve this answer




















  • I really like this answer plus the macro. I will allow other some time for others answer as well but this is what I was looking for!
    – MathScholar
    Nov 24 at 17:31











  • Hi Peter if I were to name the curve by draw[domain=.5:3.25,smooth,variable=x,black,<->,thick] plot (x,.5*(x-1.5)*(x-1.5)+1); how would the syntax change>
    – MathScholar
    Nov 27 at 19:52










  • @MathScholar: You need to use name path=xxxx to name the curve (xxxx in this example), and pass that to DrawTangent[<draw options>]xxxx<ymin><ymax>x.
    – Peter Grill
    Nov 27 at 21:09











  • Yes I did that and got a horizontal line. I ask as a question as well so you can see the result and program. I thought this would be more easy to do! Here: tex.stackexchange.com/questions/462070/…
    – MathScholar
    Nov 27 at 21:13











  • Sorry Peter I made a mistake! I got it! I deleted he question
    – MathScholar
    Nov 27 at 21:19


















up vote
6
down vote













A PSTricks solution just for comparison purpose.



enter image description here



documentclass[pstricks]standalone
usepackagepstricks-add,pst-vehicle,tikzducks

defVrput(1,1.5)psscalebox-2 2tikzduck[alien=red]
defF2*cos(x)+x/3
defFpDerive(1,F)

deftangent#1#2#3psplotTangent[linecolor=blue,showpoints]#1.5Fuput[#2](*#1 F)scriptsize$f'#30$
defnormal#1psplotTangent[linecolor=magenta,Derive=-1/Fp,showpoints]#1.2F

pssetplotpoints=200
begindocument
foreach i in -0.8,-0.6,...,10.8
beginpspicture(-2.5,-2)(12.5,6)
psaxes[ticks=none,labels=none,linecolor=gray]->(0,0)(-2,-1.5)(11.5,5)[$x$,0][$y$,90]
psplot[linecolor=red]-1.511F
normal-1.5
tangent-1180>
tangent0.167-90=
tangent1.3340<
tangent2.974-90=
tangent4.71150>
tangent6.449-90=
tangent7.8520<
tangent9.256-90=
tangent10.50>
normal11
psVehicle[vehicle=SelfDefinedVehicle,ownvehicle=V,showSlope=false,frontwheel=SpokesWheelB,backwheel=SpokesWheelA,rB=1.2,rF=0.9,d=3].2iF
endpspicture
enddocument





share|improve this answer


















  • 1




    Many compliments for your answer. +1.
    – Sebastiano
    Nov 24 at 16:30










  • This looks good but I am searching for an answer using using TiKz. Thanks AS!
    – MathScholar
    Nov 24 at 16:34










  • Well, this is certainly nice but of course here you have an analytic expression for the curve. If you have this analytic expression, it is well known how to find the local extremal and tangents and all that, see e.g. here. I thought the challenge here was to do that to curves for which the analytic expression is not known. (This is not to criticize you, just to let you know that there exists quite some literature on this topic on this site.)
    – marmot
    Nov 24 at 19:53










  • @marmot: Analytic expression is no longer as complicated as the previous one.
    – Artificial Stupidity
    Nov 25 at 11:04










  • I presume if the "roller coaster" figure here were done instead with tikz, then the creature riding in the car would be a duck??
    – murray
    Nov 25 at 15:43

















up vote
4
down vote













This is just for fun and not to "steal" the check mark from Peter Grill, who IMHO should get it. Conceptually it is basically the same as Jake's pioneering answer. This answer comes with two styles: tangent at and normal at, which you can pass to the curve, and utilizes decorations.markings rather than intersections. If you want to draw several tangents and/or normals, it is advantageous to use /.list, as in the MWE:



documentclassarticle
usepackagetikz
usetikzlibrarydecorations.markings

begindocument
pgfkeystikz/.cd,
tangent length/.store in=TangentLength,
tangent length=7mm,
normal length/.store in=NormalLength,
normal length=7mm
tikzsettangent/.style=red,thin,normal/.style=blue,thin,
tangent at/.style=postaction=decorate,decoration=markings,
mark=at position #1 with draw[tangent] (-TangentLength,0) -- (TangentLength,0);
fill[tangent] (0,0) circle (2pt);,
normal at/.style=postaction=decorate,decoration=markings,
mark=at position #1 with draw[normal] (0,-NormalLength) -- (0,NormalLength);
fill[normal] (0,0) circle (2pt);,

begintikzpicture
% Axes
draw [-latex] (-1,0) -- (11,0) node [right] $x$;
draw [-latex] (0,-1) -- (0,6) node [above] $y$;
% Origin
node at (0,0) [below left] $0$;
% Points
coordinate (start) at (1,-0.8);
coordinate (c1) at (3,3);
coordinate (c2) at (5.5,1.5);
coordinate (c3) at (8,4);
coordinate (end) at (10.5,-0.8);
% show the points
% foreach n in start,c1,c2,c3,end fill [black] (n)
% circle (2pt) node [below] ;
% join the coordinates
draw [thick,tangent at/.list=0.15,0.3,...,0.75,
normal at/.list=0.25,0.4,...,0.7] (start) to[out=70,in=180] (c1) to[out=0,in=180]
(c2) to[out=0,in=180] (c3) to[out=0,in=150] (end);
endtikzpicture
enddocument


enter image description here



Just if you are wondering: yes, it is possible to draw the tangent at some specific point the curve runs through. One possibility is to use the tricks of this answer, i.e. decompose the path into segments and attach the tangents/normals to the starting or end points of those. If you want to have this spelled out, please ask a new question, this here is really just for fun.



ADDENDUM: As for your request in the comments... (and yes, I am using Jake's answer, as I mention above)....



documentclassarticle
usepackagetikz
usetikzlibrarydecorations.markings,intersections

begindocument
pgfkeystikz/.cd,
tangent length/.store in=TangentLength,
tangent length=3.14cm,
normal length/.store in=NormalLength,
normal length=7mm
newcountertangent
newcounternormal
tikzsettangent/.style=red,thin,normal/.style=blue,thin,
tangent at/.style=postaction=decorate,decoration=markings,
mark=at position #1 with stepcountertangent
draw[tangent,name path=tangent-thetangent] (-TangentLength,0) -- (TangentLength,0);
fill[tangent] (0,0) circle (2pt);,
normal at/.style=postaction=decorate,decoration=markings,
mark=at position #1 with stepcounternormal
draw[normal,name path=normal-thenormal] (0,-NormalLength) -- (0,NormalLength);
fill[normal] (0,0) circle (2pt);,

begintikzpicture
% Axes
draw [-latex] (-1,0) -- (11,0) node [right] $x$;
draw [-latex] (0,-1) -- (0,6) node [above] $y$;
% Origin
node at (0,0) [below left] $0$;
% Points
coordinate (start) at (1,-0.8);
coordinate (c1) at (3,3);
coordinate (c2) at (5.5,1.5);
coordinate (c3) at (8,4);
coordinate (end) at (10.5,-0.8);
% show the points
% foreach n in start,c1,c2,c3,end fill [black] (n)
% circle (2pt) node [below] ;
% join the coordinates
draw [name path=curve,thick,tangent at/.list=0.15,0.3,...,0.75,
normal at/.list=0.25,0.4,...,0.7] (start) to[out=70,in=180] (c1) to[out=0,in=180]
(c2) to[out=0,in=180] (c3) to[out=0,in=150] (end);
fill[name intersections=of=curve and tangent-2,total=t,cyan]
(intersection-t) circle (2pt);
endtikzpicture
enddocument


enter image description here






share|improve this answer






















  • this looks fantastic as well and uses methods from tex.stackexchange.com/questions/25928/… I believe. How can I extend the second tangent starting from the left and use the intersection, calc package to show where it intersects the curve further towards the right?
    – MathScholar
    Nov 24 at 20:50











  • @MathScholar I added something. (I guess on the long run one may have to look at the internals of the intersections and fillbetween libraries, which internally compute all sorts of of useful quantities like pgfintersectiongetsolutiontimes, which however did not make it to the manual.)
    – marmot
    Nov 24 at 21:04











  • per your request I will give the check to Peter. Both posts are extremely helpful. Thanks Marmot!
    – MathScholar
    Nov 24 at 22:17










  • @MathScholar Thanks! (I generally do not like if a question that has already an answer that is to the point gets a second answer, which is not to the point or just merely a repetition, like e.g. here. This brings unnecessary tension between the users IMHO, and the last thing I want to do is to step others on the feet.)
    – marmot
    Nov 24 at 22:32










  • @Hi Marmot, if I continue to extend the tangent to the right and fix the left side, there will be another intersection of the curve and tangent. Why does the intersection not get detected by the program?
    – MathScholar
    Nov 24 at 23:46










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3 Answers
3






active

oldest

votes








3 Answers
3






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
7
down vote



accepted










You can use a bit of calculus:



enter image description here



The DrawTangent macro creates two vertical lines around the given x values and computes the intersection of those and the curve. Then, the tangent can be approximated by the secant between those two intersection points.



Code:



documentclass[border=2pt]standalone
usepackagetikz
usetikzlibrarycalc,intersections


newcommand*DeltaX0.01
newcommand*DrawTangent[5]%
% #1 = draw options
% #2 = name of curve
% #3 = ymin
% #4 = ymax
% #5 = x value at which tangent is to be drawn

path[name path=Vertical Line Left] (#5-DeltaX,#3) -- (#5-DeltaX,#4);
path[name path=Vertical Line Right] (#5+DeltaX,#3) -- (#5+DeltaX,#4);

path [name intersections=of=Vertical Line Left and #2];
coordinate (X0) at (intersection-1);
path [name intersections=of=Vertical Line Right and #2];
coordinate (X1) at (intersection-1);

draw [shorten <= -1.00cm, shorten >= -1.00cm, #1] (X0) -- (X1);
%

begindocument
begintikzpicture
% Axes
draw [->, name path=x] (-1,0) -- (11,0) node [right] $x$;
draw [->] (0,-1) -- (0,6) node [above] $y$;
% Origin
node at (0,0) [below left] $0$;
% Points
coordinate (start) at (1,-0.8);
coordinate (c1) at (3,3);
coordinate (c2) at (5.5,1.5);
coordinate (c3) at (8,4);
coordinate (end) at (10.5,-0.8);
% show the points
foreach n in start,c1,c2,c3,end fill [black] (n)
circle (2pt) node [below] ;
% join the coordinates
draw [thick,name path=curve] (start) to[out=70,in=180] (c1) to[out=0,in=180]
(c2) to[out=0,in=180] (c3) to[out=0,in=150] (end);
% add tangets and dashed lines
foreach c in 1,2,3
%draw [dashed] let p1=(cc) in (cc) -- (x1,0) node [below] $c_c$;
draw ($(cc)-(0.75,0)$) -- ($(cc)+(0.75,0)$) node [midway,above=4mm]$$; %$f'(c_c)=0$;

% add a and b
path [name intersections=of=x and curve, by=a,b]
(a) node [below left] $a$
(b) node [above right] $b$;

DrawTangent[red, thick]curve-141.5
DrawTangent[orange, thick]curve-143.5

endtikzpicture
enddocument





share|improve this answer




















  • I really like this answer plus the macro. I will allow other some time for others answer as well but this is what I was looking for!
    – MathScholar
    Nov 24 at 17:31











  • Hi Peter if I were to name the curve by draw[domain=.5:3.25,smooth,variable=x,black,<->,thick] plot (x,.5*(x-1.5)*(x-1.5)+1); how would the syntax change>
    – MathScholar
    Nov 27 at 19:52










  • @MathScholar: You need to use name path=xxxx to name the curve (xxxx in this example), and pass that to DrawTangent[<draw options>]xxxx<ymin><ymax>x.
    – Peter Grill
    Nov 27 at 21:09











  • Yes I did that and got a horizontal line. I ask as a question as well so you can see the result and program. I thought this would be more easy to do! Here: tex.stackexchange.com/questions/462070/…
    – MathScholar
    Nov 27 at 21:13











  • Sorry Peter I made a mistake! I got it! I deleted he question
    – MathScholar
    Nov 27 at 21:19















up vote
7
down vote



accepted










You can use a bit of calculus:



enter image description here



The DrawTangent macro creates two vertical lines around the given x values and computes the intersection of those and the curve. Then, the tangent can be approximated by the secant between those two intersection points.



Code:



documentclass[border=2pt]standalone
usepackagetikz
usetikzlibrarycalc,intersections


newcommand*DeltaX0.01
newcommand*DrawTangent[5]%
% #1 = draw options
% #2 = name of curve
% #3 = ymin
% #4 = ymax
% #5 = x value at which tangent is to be drawn

path[name path=Vertical Line Left] (#5-DeltaX,#3) -- (#5-DeltaX,#4);
path[name path=Vertical Line Right] (#5+DeltaX,#3) -- (#5+DeltaX,#4);

path [name intersections=of=Vertical Line Left and #2];
coordinate (X0) at (intersection-1);
path [name intersections=of=Vertical Line Right and #2];
coordinate (X1) at (intersection-1);

draw [shorten <= -1.00cm, shorten >= -1.00cm, #1] (X0) -- (X1);
%

begindocument
begintikzpicture
% Axes
draw [->, name path=x] (-1,0) -- (11,0) node [right] $x$;
draw [->] (0,-1) -- (0,6) node [above] $y$;
% Origin
node at (0,0) [below left] $0$;
% Points
coordinate (start) at (1,-0.8);
coordinate (c1) at (3,3);
coordinate (c2) at (5.5,1.5);
coordinate (c3) at (8,4);
coordinate (end) at (10.5,-0.8);
% show the points
foreach n in start,c1,c2,c3,end fill [black] (n)
circle (2pt) node [below] ;
% join the coordinates
draw [thick,name path=curve] (start) to[out=70,in=180] (c1) to[out=0,in=180]
(c2) to[out=0,in=180] (c3) to[out=0,in=150] (end);
% add tangets and dashed lines
foreach c in 1,2,3
%draw [dashed] let p1=(cc) in (cc) -- (x1,0) node [below] $c_c$;
draw ($(cc)-(0.75,0)$) -- ($(cc)+(0.75,0)$) node [midway,above=4mm]$$; %$f'(c_c)=0$;

% add a and b
path [name intersections=of=x and curve, by=a,b]
(a) node [below left] $a$
(b) node [above right] $b$;

DrawTangent[red, thick]curve-141.5
DrawTangent[orange, thick]curve-143.5

endtikzpicture
enddocument





share|improve this answer




















  • I really like this answer plus the macro. I will allow other some time for others answer as well but this is what I was looking for!
    – MathScholar
    Nov 24 at 17:31











  • Hi Peter if I were to name the curve by draw[domain=.5:3.25,smooth,variable=x,black,<->,thick] plot (x,.5*(x-1.5)*(x-1.5)+1); how would the syntax change>
    – MathScholar
    Nov 27 at 19:52










  • @MathScholar: You need to use name path=xxxx to name the curve (xxxx in this example), and pass that to DrawTangent[<draw options>]xxxx<ymin><ymax>x.
    – Peter Grill
    Nov 27 at 21:09











  • Yes I did that and got a horizontal line. I ask as a question as well so you can see the result and program. I thought this would be more easy to do! Here: tex.stackexchange.com/questions/462070/…
    – MathScholar
    Nov 27 at 21:13











  • Sorry Peter I made a mistake! I got it! I deleted he question
    – MathScholar
    Nov 27 at 21:19













up vote
7
down vote



accepted







up vote
7
down vote



accepted






You can use a bit of calculus:



enter image description here



The DrawTangent macro creates two vertical lines around the given x values and computes the intersection of those and the curve. Then, the tangent can be approximated by the secant between those two intersection points.



Code:



documentclass[border=2pt]standalone
usepackagetikz
usetikzlibrarycalc,intersections


newcommand*DeltaX0.01
newcommand*DrawTangent[5]%
% #1 = draw options
% #2 = name of curve
% #3 = ymin
% #4 = ymax
% #5 = x value at which tangent is to be drawn

path[name path=Vertical Line Left] (#5-DeltaX,#3) -- (#5-DeltaX,#4);
path[name path=Vertical Line Right] (#5+DeltaX,#3) -- (#5+DeltaX,#4);

path [name intersections=of=Vertical Line Left and #2];
coordinate (X0) at (intersection-1);
path [name intersections=of=Vertical Line Right and #2];
coordinate (X1) at (intersection-1);

draw [shorten <= -1.00cm, shorten >= -1.00cm, #1] (X0) -- (X1);
%

begindocument
begintikzpicture
% Axes
draw [->, name path=x] (-1,0) -- (11,0) node [right] $x$;
draw [->] (0,-1) -- (0,6) node [above] $y$;
% Origin
node at (0,0) [below left] $0$;
% Points
coordinate (start) at (1,-0.8);
coordinate (c1) at (3,3);
coordinate (c2) at (5.5,1.5);
coordinate (c3) at (8,4);
coordinate (end) at (10.5,-0.8);
% show the points
foreach n in start,c1,c2,c3,end fill [black] (n)
circle (2pt) node [below] ;
% join the coordinates
draw [thick,name path=curve] (start) to[out=70,in=180] (c1) to[out=0,in=180]
(c2) to[out=0,in=180] (c3) to[out=0,in=150] (end);
% add tangets and dashed lines
foreach c in 1,2,3
%draw [dashed] let p1=(cc) in (cc) -- (x1,0) node [below] $c_c$;
draw ($(cc)-(0.75,0)$) -- ($(cc)+(0.75,0)$) node [midway,above=4mm]$$; %$f'(c_c)=0$;

% add a and b
path [name intersections=of=x and curve, by=a,b]
(a) node [below left] $a$
(b) node [above right] $b$;

DrawTangent[red, thick]curve-141.5
DrawTangent[orange, thick]curve-143.5

endtikzpicture
enddocument





share|improve this answer












You can use a bit of calculus:



enter image description here



The DrawTangent macro creates two vertical lines around the given x values and computes the intersection of those and the curve. Then, the tangent can be approximated by the secant between those two intersection points.



Code:



documentclass[border=2pt]standalone
usepackagetikz
usetikzlibrarycalc,intersections


newcommand*DeltaX0.01
newcommand*DrawTangent[5]%
% #1 = draw options
% #2 = name of curve
% #3 = ymin
% #4 = ymax
% #5 = x value at which tangent is to be drawn

path[name path=Vertical Line Left] (#5-DeltaX,#3) -- (#5-DeltaX,#4);
path[name path=Vertical Line Right] (#5+DeltaX,#3) -- (#5+DeltaX,#4);

path [name intersections=of=Vertical Line Left and #2];
coordinate (X0) at (intersection-1);
path [name intersections=of=Vertical Line Right and #2];
coordinate (X1) at (intersection-1);

draw [shorten <= -1.00cm, shorten >= -1.00cm, #1] (X0) -- (X1);
%

begindocument
begintikzpicture
% Axes
draw [->, name path=x] (-1,0) -- (11,0) node [right] $x$;
draw [->] (0,-1) -- (0,6) node [above] $y$;
% Origin
node at (0,0) [below left] $0$;
% Points
coordinate (start) at (1,-0.8);
coordinate (c1) at (3,3);
coordinate (c2) at (5.5,1.5);
coordinate (c3) at (8,4);
coordinate (end) at (10.5,-0.8);
% show the points
foreach n in start,c1,c2,c3,end fill [black] (n)
circle (2pt) node [below] ;
% join the coordinates
draw [thick,name path=curve] (start) to[out=70,in=180] (c1) to[out=0,in=180]
(c2) to[out=0,in=180] (c3) to[out=0,in=150] (end);
% add tangets and dashed lines
foreach c in 1,2,3
%draw [dashed] let p1=(cc) in (cc) -- (x1,0) node [below] $c_c$;
draw ($(cc)-(0.75,0)$) -- ($(cc)+(0.75,0)$) node [midway,above=4mm]$$; %$f'(c_c)=0$;

% add a and b
path [name intersections=of=x and curve, by=a,b]
(a) node [below left] $a$
(b) node [above right] $b$;

DrawTangent[red, thick]curve-141.5
DrawTangent[orange, thick]curve-143.5

endtikzpicture
enddocument






share|improve this answer












share|improve this answer



share|improve this answer










answered Nov 24 at 17:09









Peter Grill

163k24432742




163k24432742











  • I really like this answer plus the macro. I will allow other some time for others answer as well but this is what I was looking for!
    – MathScholar
    Nov 24 at 17:31











  • Hi Peter if I were to name the curve by draw[domain=.5:3.25,smooth,variable=x,black,<->,thick] plot (x,.5*(x-1.5)*(x-1.5)+1); how would the syntax change>
    – MathScholar
    Nov 27 at 19:52










  • @MathScholar: You need to use name path=xxxx to name the curve (xxxx in this example), and pass that to DrawTangent[<draw options>]xxxx<ymin><ymax>x.
    – Peter Grill
    Nov 27 at 21:09











  • Yes I did that and got a horizontal line. I ask as a question as well so you can see the result and program. I thought this would be more easy to do! Here: tex.stackexchange.com/questions/462070/…
    – MathScholar
    Nov 27 at 21:13











  • Sorry Peter I made a mistake! I got it! I deleted he question
    – MathScholar
    Nov 27 at 21:19

















  • I really like this answer plus the macro. I will allow other some time for others answer as well but this is what I was looking for!
    – MathScholar
    Nov 24 at 17:31











  • Hi Peter if I were to name the curve by draw[domain=.5:3.25,smooth,variable=x,black,<->,thick] plot (x,.5*(x-1.5)*(x-1.5)+1); how would the syntax change>
    – MathScholar
    Nov 27 at 19:52










  • @MathScholar: You need to use name path=xxxx to name the curve (xxxx in this example), and pass that to DrawTangent[<draw options>]xxxx<ymin><ymax>x.
    – Peter Grill
    Nov 27 at 21:09











  • Yes I did that and got a horizontal line. I ask as a question as well so you can see the result and program. I thought this would be more easy to do! Here: tex.stackexchange.com/questions/462070/…
    – MathScholar
    Nov 27 at 21:13











  • Sorry Peter I made a mistake! I got it! I deleted he question
    – MathScholar
    Nov 27 at 21:19
















I really like this answer plus the macro. I will allow other some time for others answer as well but this is what I was looking for!
– MathScholar
Nov 24 at 17:31





I really like this answer plus the macro. I will allow other some time for others answer as well but this is what I was looking for!
– MathScholar
Nov 24 at 17:31













Hi Peter if I were to name the curve by draw[domain=.5:3.25,smooth,variable=x,black,<->,thick] plot (x,.5*(x-1.5)*(x-1.5)+1); how would the syntax change>
– MathScholar
Nov 27 at 19:52




Hi Peter if I were to name the curve by draw[domain=.5:3.25,smooth,variable=x,black,<->,thick] plot (x,.5*(x-1.5)*(x-1.5)+1); how would the syntax change>
– MathScholar
Nov 27 at 19:52












@MathScholar: You need to use name path=xxxx to name the curve (xxxx in this example), and pass that to DrawTangent[<draw options>]xxxx<ymin><ymax>x.
– Peter Grill
Nov 27 at 21:09





@MathScholar: You need to use name path=xxxx to name the curve (xxxx in this example), and pass that to DrawTangent[<draw options>]xxxx<ymin><ymax>x.
– Peter Grill
Nov 27 at 21:09













Yes I did that and got a horizontal line. I ask as a question as well so you can see the result and program. I thought this would be more easy to do! Here: tex.stackexchange.com/questions/462070/…
– MathScholar
Nov 27 at 21:13





Yes I did that and got a horizontal line. I ask as a question as well so you can see the result and program. I thought this would be more easy to do! Here: tex.stackexchange.com/questions/462070/…
– MathScholar
Nov 27 at 21:13













Sorry Peter I made a mistake! I got it! I deleted he question
– MathScholar
Nov 27 at 21:19





Sorry Peter I made a mistake! I got it! I deleted he question
– MathScholar
Nov 27 at 21:19











up vote
6
down vote













A PSTricks solution just for comparison purpose.



enter image description here



documentclass[pstricks]standalone
usepackagepstricks-add,pst-vehicle,tikzducks

defVrput(1,1.5)psscalebox-2 2tikzduck[alien=red]
defF2*cos(x)+x/3
defFpDerive(1,F)

deftangent#1#2#3psplotTangent[linecolor=blue,showpoints]#1.5Fuput[#2](*#1 F)scriptsize$f'#30$
defnormal#1psplotTangent[linecolor=magenta,Derive=-1/Fp,showpoints]#1.2F

pssetplotpoints=200
begindocument
foreach i in -0.8,-0.6,...,10.8
beginpspicture(-2.5,-2)(12.5,6)
psaxes[ticks=none,labels=none,linecolor=gray]->(0,0)(-2,-1.5)(11.5,5)[$x$,0][$y$,90]
psplot[linecolor=red]-1.511F
normal-1.5
tangent-1180>
tangent0.167-90=
tangent1.3340<
tangent2.974-90=
tangent4.71150>
tangent6.449-90=
tangent7.8520<
tangent9.256-90=
tangent10.50>
normal11
psVehicle[vehicle=SelfDefinedVehicle,ownvehicle=V,showSlope=false,frontwheel=SpokesWheelB,backwheel=SpokesWheelA,rB=1.2,rF=0.9,d=3].2iF
endpspicture
enddocument





share|improve this answer


















  • 1




    Many compliments for your answer. +1.
    – Sebastiano
    Nov 24 at 16:30










  • This looks good but I am searching for an answer using using TiKz. Thanks AS!
    – MathScholar
    Nov 24 at 16:34










  • Well, this is certainly nice but of course here you have an analytic expression for the curve. If you have this analytic expression, it is well known how to find the local extremal and tangents and all that, see e.g. here. I thought the challenge here was to do that to curves for which the analytic expression is not known. (This is not to criticize you, just to let you know that there exists quite some literature on this topic on this site.)
    – marmot
    Nov 24 at 19:53










  • @marmot: Analytic expression is no longer as complicated as the previous one.
    – Artificial Stupidity
    Nov 25 at 11:04










  • I presume if the "roller coaster" figure here were done instead with tikz, then the creature riding in the car would be a duck??
    – murray
    Nov 25 at 15:43














up vote
6
down vote













A PSTricks solution just for comparison purpose.



enter image description here



documentclass[pstricks]standalone
usepackagepstricks-add,pst-vehicle,tikzducks

defVrput(1,1.5)psscalebox-2 2tikzduck[alien=red]
defF2*cos(x)+x/3
defFpDerive(1,F)

deftangent#1#2#3psplotTangent[linecolor=blue,showpoints]#1.5Fuput[#2](*#1 F)scriptsize$f'#30$
defnormal#1psplotTangent[linecolor=magenta,Derive=-1/Fp,showpoints]#1.2F

pssetplotpoints=200
begindocument
foreach i in -0.8,-0.6,...,10.8
beginpspicture(-2.5,-2)(12.5,6)
psaxes[ticks=none,labels=none,linecolor=gray]->(0,0)(-2,-1.5)(11.5,5)[$x$,0][$y$,90]
psplot[linecolor=red]-1.511F
normal-1.5
tangent-1180>
tangent0.167-90=
tangent1.3340<
tangent2.974-90=
tangent4.71150>
tangent6.449-90=
tangent7.8520<
tangent9.256-90=
tangent10.50>
normal11
psVehicle[vehicle=SelfDefinedVehicle,ownvehicle=V,showSlope=false,frontwheel=SpokesWheelB,backwheel=SpokesWheelA,rB=1.2,rF=0.9,d=3].2iF
endpspicture
enddocument





share|improve this answer


















  • 1




    Many compliments for your answer. +1.
    – Sebastiano
    Nov 24 at 16:30










  • This looks good but I am searching for an answer using using TiKz. Thanks AS!
    – MathScholar
    Nov 24 at 16:34










  • Well, this is certainly nice but of course here you have an analytic expression for the curve. If you have this analytic expression, it is well known how to find the local extremal and tangents and all that, see e.g. here. I thought the challenge here was to do that to curves for which the analytic expression is not known. (This is not to criticize you, just to let you know that there exists quite some literature on this topic on this site.)
    – marmot
    Nov 24 at 19:53










  • @marmot: Analytic expression is no longer as complicated as the previous one.
    – Artificial Stupidity
    Nov 25 at 11:04










  • I presume if the "roller coaster" figure here were done instead with tikz, then the creature riding in the car would be a duck??
    – murray
    Nov 25 at 15:43












up vote
6
down vote










up vote
6
down vote









A PSTricks solution just for comparison purpose.



enter image description here



documentclass[pstricks]standalone
usepackagepstricks-add,pst-vehicle,tikzducks

defVrput(1,1.5)psscalebox-2 2tikzduck[alien=red]
defF2*cos(x)+x/3
defFpDerive(1,F)

deftangent#1#2#3psplotTangent[linecolor=blue,showpoints]#1.5Fuput[#2](*#1 F)scriptsize$f'#30$
defnormal#1psplotTangent[linecolor=magenta,Derive=-1/Fp,showpoints]#1.2F

pssetplotpoints=200
begindocument
foreach i in -0.8,-0.6,...,10.8
beginpspicture(-2.5,-2)(12.5,6)
psaxes[ticks=none,labels=none,linecolor=gray]->(0,0)(-2,-1.5)(11.5,5)[$x$,0][$y$,90]
psplot[linecolor=red]-1.511F
normal-1.5
tangent-1180>
tangent0.167-90=
tangent1.3340<
tangent2.974-90=
tangent4.71150>
tangent6.449-90=
tangent7.8520<
tangent9.256-90=
tangent10.50>
normal11
psVehicle[vehicle=SelfDefinedVehicle,ownvehicle=V,showSlope=false,frontwheel=SpokesWheelB,backwheel=SpokesWheelA,rB=1.2,rF=0.9,d=3].2iF
endpspicture
enddocument





share|improve this answer














A PSTricks solution just for comparison purpose.



enter image description here



documentclass[pstricks]standalone
usepackagepstricks-add,pst-vehicle,tikzducks

defVrput(1,1.5)psscalebox-2 2tikzduck[alien=red]
defF2*cos(x)+x/3
defFpDerive(1,F)

deftangent#1#2#3psplotTangent[linecolor=blue,showpoints]#1.5Fuput[#2](*#1 F)scriptsize$f'#30$
defnormal#1psplotTangent[linecolor=magenta,Derive=-1/Fp,showpoints]#1.2F

pssetplotpoints=200
begindocument
foreach i in -0.8,-0.6,...,10.8
beginpspicture(-2.5,-2)(12.5,6)
psaxes[ticks=none,labels=none,linecolor=gray]->(0,0)(-2,-1.5)(11.5,5)[$x$,0][$y$,90]
psplot[linecolor=red]-1.511F
normal-1.5
tangent-1180>
tangent0.167-90=
tangent1.3340<
tangent2.974-90=
tangent4.71150>
tangent6.449-90=
tangent7.8520<
tangent9.256-90=
tangent10.50>
normal11
psVehicle[vehicle=SelfDefinedVehicle,ownvehicle=V,showSlope=false,frontwheel=SpokesWheelB,backwheel=SpokesWheelA,rB=1.2,rF=0.9,d=3].2iF
endpspicture
enddocument






share|improve this answer














share|improve this answer



share|improve this answer








edited Nov 25 at 10:54

























answered Nov 24 at 16:28









Artificial Stupidity

4,69011035




4,69011035







  • 1




    Many compliments for your answer. +1.
    – Sebastiano
    Nov 24 at 16:30










  • This looks good but I am searching for an answer using using TiKz. Thanks AS!
    – MathScholar
    Nov 24 at 16:34










  • Well, this is certainly nice but of course here you have an analytic expression for the curve. If you have this analytic expression, it is well known how to find the local extremal and tangents and all that, see e.g. here. I thought the challenge here was to do that to curves for which the analytic expression is not known. (This is not to criticize you, just to let you know that there exists quite some literature on this topic on this site.)
    – marmot
    Nov 24 at 19:53










  • @marmot: Analytic expression is no longer as complicated as the previous one.
    – Artificial Stupidity
    Nov 25 at 11:04










  • I presume if the "roller coaster" figure here were done instead with tikz, then the creature riding in the car would be a duck??
    – murray
    Nov 25 at 15:43












  • 1




    Many compliments for your answer. +1.
    – Sebastiano
    Nov 24 at 16:30










  • This looks good but I am searching for an answer using using TiKz. Thanks AS!
    – MathScholar
    Nov 24 at 16:34










  • Well, this is certainly nice but of course here you have an analytic expression for the curve. If you have this analytic expression, it is well known how to find the local extremal and tangents and all that, see e.g. here. I thought the challenge here was to do that to curves for which the analytic expression is not known. (This is not to criticize you, just to let you know that there exists quite some literature on this topic on this site.)
    – marmot
    Nov 24 at 19:53










  • @marmot: Analytic expression is no longer as complicated as the previous one.
    – Artificial Stupidity
    Nov 25 at 11:04










  • I presume if the "roller coaster" figure here were done instead with tikz, then the creature riding in the car would be a duck??
    – murray
    Nov 25 at 15:43







1




1




Many compliments for your answer. +1.
– Sebastiano
Nov 24 at 16:30




Many compliments for your answer. +1.
– Sebastiano
Nov 24 at 16:30












This looks good but I am searching for an answer using using TiKz. Thanks AS!
– MathScholar
Nov 24 at 16:34




This looks good but I am searching for an answer using using TiKz. Thanks AS!
– MathScholar
Nov 24 at 16:34












Well, this is certainly nice but of course here you have an analytic expression for the curve. If you have this analytic expression, it is well known how to find the local extremal and tangents and all that, see e.g. here. I thought the challenge here was to do that to curves for which the analytic expression is not known. (This is not to criticize you, just to let you know that there exists quite some literature on this topic on this site.)
– marmot
Nov 24 at 19:53




Well, this is certainly nice but of course here you have an analytic expression for the curve. If you have this analytic expression, it is well known how to find the local extremal and tangents and all that, see e.g. here. I thought the challenge here was to do that to curves for which the analytic expression is not known. (This is not to criticize you, just to let you know that there exists quite some literature on this topic on this site.)
– marmot
Nov 24 at 19:53












@marmot: Analytic expression is no longer as complicated as the previous one.
– Artificial Stupidity
Nov 25 at 11:04




@marmot: Analytic expression is no longer as complicated as the previous one.
– Artificial Stupidity
Nov 25 at 11:04












I presume if the "roller coaster" figure here were done instead with tikz, then the creature riding in the car would be a duck??
– murray
Nov 25 at 15:43




I presume if the "roller coaster" figure here were done instead with tikz, then the creature riding in the car would be a duck??
– murray
Nov 25 at 15:43










up vote
4
down vote













This is just for fun and not to "steal" the check mark from Peter Grill, who IMHO should get it. Conceptually it is basically the same as Jake's pioneering answer. This answer comes with two styles: tangent at and normal at, which you can pass to the curve, and utilizes decorations.markings rather than intersections. If you want to draw several tangents and/or normals, it is advantageous to use /.list, as in the MWE:



documentclassarticle
usepackagetikz
usetikzlibrarydecorations.markings

begindocument
pgfkeystikz/.cd,
tangent length/.store in=TangentLength,
tangent length=7mm,
normal length/.store in=NormalLength,
normal length=7mm
tikzsettangent/.style=red,thin,normal/.style=blue,thin,
tangent at/.style=postaction=decorate,decoration=markings,
mark=at position #1 with draw[tangent] (-TangentLength,0) -- (TangentLength,0);
fill[tangent] (0,0) circle (2pt);,
normal at/.style=postaction=decorate,decoration=markings,
mark=at position #1 with draw[normal] (0,-NormalLength) -- (0,NormalLength);
fill[normal] (0,0) circle (2pt);,

begintikzpicture
% Axes
draw [-latex] (-1,0) -- (11,0) node [right] $x$;
draw [-latex] (0,-1) -- (0,6) node [above] $y$;
% Origin
node at (0,0) [below left] $0$;
% Points
coordinate (start) at (1,-0.8);
coordinate (c1) at (3,3);
coordinate (c2) at (5.5,1.5);
coordinate (c3) at (8,4);
coordinate (end) at (10.5,-0.8);
% show the points
% foreach n in start,c1,c2,c3,end fill [black] (n)
% circle (2pt) node [below] ;
% join the coordinates
draw [thick,tangent at/.list=0.15,0.3,...,0.75,
normal at/.list=0.25,0.4,...,0.7] (start) to[out=70,in=180] (c1) to[out=0,in=180]
(c2) to[out=0,in=180] (c3) to[out=0,in=150] (end);
endtikzpicture
enddocument


enter image description here



Just if you are wondering: yes, it is possible to draw the tangent at some specific point the curve runs through. One possibility is to use the tricks of this answer, i.e. decompose the path into segments and attach the tangents/normals to the starting or end points of those. If you want to have this spelled out, please ask a new question, this here is really just for fun.



ADDENDUM: As for your request in the comments... (and yes, I am using Jake's answer, as I mention above)....



documentclassarticle
usepackagetikz
usetikzlibrarydecorations.markings,intersections

begindocument
pgfkeystikz/.cd,
tangent length/.store in=TangentLength,
tangent length=3.14cm,
normal length/.store in=NormalLength,
normal length=7mm
newcountertangent
newcounternormal
tikzsettangent/.style=red,thin,normal/.style=blue,thin,
tangent at/.style=postaction=decorate,decoration=markings,
mark=at position #1 with stepcountertangent
draw[tangent,name path=tangent-thetangent] (-TangentLength,0) -- (TangentLength,0);
fill[tangent] (0,0) circle (2pt);,
normal at/.style=postaction=decorate,decoration=markings,
mark=at position #1 with stepcounternormal
draw[normal,name path=normal-thenormal] (0,-NormalLength) -- (0,NormalLength);
fill[normal] (0,0) circle (2pt);,

begintikzpicture
% Axes
draw [-latex] (-1,0) -- (11,0) node [right] $x$;
draw [-latex] (0,-1) -- (0,6) node [above] $y$;
% Origin
node at (0,0) [below left] $0$;
% Points
coordinate (start) at (1,-0.8);
coordinate (c1) at (3,3);
coordinate (c2) at (5.5,1.5);
coordinate (c3) at (8,4);
coordinate (end) at (10.5,-0.8);
% show the points
% foreach n in start,c1,c2,c3,end fill [black] (n)
% circle (2pt) node [below] ;
% join the coordinates
draw [name path=curve,thick,tangent at/.list=0.15,0.3,...,0.75,
normal at/.list=0.25,0.4,...,0.7] (start) to[out=70,in=180] (c1) to[out=0,in=180]
(c2) to[out=0,in=180] (c3) to[out=0,in=150] (end);
fill[name intersections=of=curve and tangent-2,total=t,cyan]
(intersection-t) circle (2pt);
endtikzpicture
enddocument


enter image description here






share|improve this answer






















  • this looks fantastic as well and uses methods from tex.stackexchange.com/questions/25928/… I believe. How can I extend the second tangent starting from the left and use the intersection, calc package to show where it intersects the curve further towards the right?
    – MathScholar
    Nov 24 at 20:50











  • @MathScholar I added something. (I guess on the long run one may have to look at the internals of the intersections and fillbetween libraries, which internally compute all sorts of of useful quantities like pgfintersectiongetsolutiontimes, which however did not make it to the manual.)
    – marmot
    Nov 24 at 21:04











  • per your request I will give the check to Peter. Both posts are extremely helpful. Thanks Marmot!
    – MathScholar
    Nov 24 at 22:17










  • @MathScholar Thanks! (I generally do not like if a question that has already an answer that is to the point gets a second answer, which is not to the point or just merely a repetition, like e.g. here. This brings unnecessary tension between the users IMHO, and the last thing I want to do is to step others on the feet.)
    – marmot
    Nov 24 at 22:32










  • @Hi Marmot, if I continue to extend the tangent to the right and fix the left side, there will be another intersection of the curve and tangent. Why does the intersection not get detected by the program?
    – MathScholar
    Nov 24 at 23:46














up vote
4
down vote













This is just for fun and not to "steal" the check mark from Peter Grill, who IMHO should get it. Conceptually it is basically the same as Jake's pioneering answer. This answer comes with two styles: tangent at and normal at, which you can pass to the curve, and utilizes decorations.markings rather than intersections. If you want to draw several tangents and/or normals, it is advantageous to use /.list, as in the MWE:



documentclassarticle
usepackagetikz
usetikzlibrarydecorations.markings

begindocument
pgfkeystikz/.cd,
tangent length/.store in=TangentLength,
tangent length=7mm,
normal length/.store in=NormalLength,
normal length=7mm
tikzsettangent/.style=red,thin,normal/.style=blue,thin,
tangent at/.style=postaction=decorate,decoration=markings,
mark=at position #1 with draw[tangent] (-TangentLength,0) -- (TangentLength,0);
fill[tangent] (0,0) circle (2pt);,
normal at/.style=postaction=decorate,decoration=markings,
mark=at position #1 with draw[normal] (0,-NormalLength) -- (0,NormalLength);
fill[normal] (0,0) circle (2pt);,

begintikzpicture
% Axes
draw [-latex] (-1,0) -- (11,0) node [right] $x$;
draw [-latex] (0,-1) -- (0,6) node [above] $y$;
% Origin
node at (0,0) [below left] $0$;
% Points
coordinate (start) at (1,-0.8);
coordinate (c1) at (3,3);
coordinate (c2) at (5.5,1.5);
coordinate (c3) at (8,4);
coordinate (end) at (10.5,-0.8);
% show the points
% foreach n in start,c1,c2,c3,end fill [black] (n)
% circle (2pt) node [below] ;
% join the coordinates
draw [thick,tangent at/.list=0.15,0.3,...,0.75,
normal at/.list=0.25,0.4,...,0.7] (start) to[out=70,in=180] (c1) to[out=0,in=180]
(c2) to[out=0,in=180] (c3) to[out=0,in=150] (end);
endtikzpicture
enddocument


enter image description here



Just if you are wondering: yes, it is possible to draw the tangent at some specific point the curve runs through. One possibility is to use the tricks of this answer, i.e. decompose the path into segments and attach the tangents/normals to the starting or end points of those. If you want to have this spelled out, please ask a new question, this here is really just for fun.



ADDENDUM: As for your request in the comments... (and yes, I am using Jake's answer, as I mention above)....



documentclassarticle
usepackagetikz
usetikzlibrarydecorations.markings,intersections

begindocument
pgfkeystikz/.cd,
tangent length/.store in=TangentLength,
tangent length=3.14cm,
normal length/.store in=NormalLength,
normal length=7mm
newcountertangent
newcounternormal
tikzsettangent/.style=red,thin,normal/.style=blue,thin,
tangent at/.style=postaction=decorate,decoration=markings,
mark=at position #1 with stepcountertangent
draw[tangent,name path=tangent-thetangent] (-TangentLength,0) -- (TangentLength,0);
fill[tangent] (0,0) circle (2pt);,
normal at/.style=postaction=decorate,decoration=markings,
mark=at position #1 with stepcounternormal
draw[normal,name path=normal-thenormal] (0,-NormalLength) -- (0,NormalLength);
fill[normal] (0,0) circle (2pt);,

begintikzpicture
% Axes
draw [-latex] (-1,0) -- (11,0) node [right] $x$;
draw [-latex] (0,-1) -- (0,6) node [above] $y$;
% Origin
node at (0,0) [below left] $0$;
% Points
coordinate (start) at (1,-0.8);
coordinate (c1) at (3,3);
coordinate (c2) at (5.5,1.5);
coordinate (c3) at (8,4);
coordinate (end) at (10.5,-0.8);
% show the points
% foreach n in start,c1,c2,c3,end fill [black] (n)
% circle (2pt) node [below] ;
% join the coordinates
draw [name path=curve,thick,tangent at/.list=0.15,0.3,...,0.75,
normal at/.list=0.25,0.4,...,0.7] (start) to[out=70,in=180] (c1) to[out=0,in=180]
(c2) to[out=0,in=180] (c3) to[out=0,in=150] (end);
fill[name intersections=of=curve and tangent-2,total=t,cyan]
(intersection-t) circle (2pt);
endtikzpicture
enddocument


enter image description here






share|improve this answer






















  • this looks fantastic as well and uses methods from tex.stackexchange.com/questions/25928/… I believe. How can I extend the second tangent starting from the left and use the intersection, calc package to show where it intersects the curve further towards the right?
    – MathScholar
    Nov 24 at 20:50











  • @MathScholar I added something. (I guess on the long run one may have to look at the internals of the intersections and fillbetween libraries, which internally compute all sorts of of useful quantities like pgfintersectiongetsolutiontimes, which however did not make it to the manual.)
    – marmot
    Nov 24 at 21:04











  • per your request I will give the check to Peter. Both posts are extremely helpful. Thanks Marmot!
    – MathScholar
    Nov 24 at 22:17










  • @MathScholar Thanks! (I generally do not like if a question that has already an answer that is to the point gets a second answer, which is not to the point or just merely a repetition, like e.g. here. This brings unnecessary tension between the users IMHO, and the last thing I want to do is to step others on the feet.)
    – marmot
    Nov 24 at 22:32










  • @Hi Marmot, if I continue to extend the tangent to the right and fix the left side, there will be another intersection of the curve and tangent. Why does the intersection not get detected by the program?
    – MathScholar
    Nov 24 at 23:46












up vote
4
down vote










up vote
4
down vote









This is just for fun and not to "steal" the check mark from Peter Grill, who IMHO should get it. Conceptually it is basically the same as Jake's pioneering answer. This answer comes with two styles: tangent at and normal at, which you can pass to the curve, and utilizes decorations.markings rather than intersections. If you want to draw several tangents and/or normals, it is advantageous to use /.list, as in the MWE:



documentclassarticle
usepackagetikz
usetikzlibrarydecorations.markings

begindocument
pgfkeystikz/.cd,
tangent length/.store in=TangentLength,
tangent length=7mm,
normal length/.store in=NormalLength,
normal length=7mm
tikzsettangent/.style=red,thin,normal/.style=blue,thin,
tangent at/.style=postaction=decorate,decoration=markings,
mark=at position #1 with draw[tangent] (-TangentLength,0) -- (TangentLength,0);
fill[tangent] (0,0) circle (2pt);,
normal at/.style=postaction=decorate,decoration=markings,
mark=at position #1 with draw[normal] (0,-NormalLength) -- (0,NormalLength);
fill[normal] (0,0) circle (2pt);,

begintikzpicture
% Axes
draw [-latex] (-1,0) -- (11,0) node [right] $x$;
draw [-latex] (0,-1) -- (0,6) node [above] $y$;
% Origin
node at (0,0) [below left] $0$;
% Points
coordinate (start) at (1,-0.8);
coordinate (c1) at (3,3);
coordinate (c2) at (5.5,1.5);
coordinate (c3) at (8,4);
coordinate (end) at (10.5,-0.8);
% show the points
% foreach n in start,c1,c2,c3,end fill [black] (n)
% circle (2pt) node [below] ;
% join the coordinates
draw [thick,tangent at/.list=0.15,0.3,...,0.75,
normal at/.list=0.25,0.4,...,0.7] (start) to[out=70,in=180] (c1) to[out=0,in=180]
(c2) to[out=0,in=180] (c3) to[out=0,in=150] (end);
endtikzpicture
enddocument


enter image description here



Just if you are wondering: yes, it is possible to draw the tangent at some specific point the curve runs through. One possibility is to use the tricks of this answer, i.e. decompose the path into segments and attach the tangents/normals to the starting or end points of those. If you want to have this spelled out, please ask a new question, this here is really just for fun.



ADDENDUM: As for your request in the comments... (and yes, I am using Jake's answer, as I mention above)....



documentclassarticle
usepackagetikz
usetikzlibrarydecorations.markings,intersections

begindocument
pgfkeystikz/.cd,
tangent length/.store in=TangentLength,
tangent length=3.14cm,
normal length/.store in=NormalLength,
normal length=7mm
newcountertangent
newcounternormal
tikzsettangent/.style=red,thin,normal/.style=blue,thin,
tangent at/.style=postaction=decorate,decoration=markings,
mark=at position #1 with stepcountertangent
draw[tangent,name path=tangent-thetangent] (-TangentLength,0) -- (TangentLength,0);
fill[tangent] (0,0) circle (2pt);,
normal at/.style=postaction=decorate,decoration=markings,
mark=at position #1 with stepcounternormal
draw[normal,name path=normal-thenormal] (0,-NormalLength) -- (0,NormalLength);
fill[normal] (0,0) circle (2pt);,

begintikzpicture
% Axes
draw [-latex] (-1,0) -- (11,0) node [right] $x$;
draw [-latex] (0,-1) -- (0,6) node [above] $y$;
% Origin
node at (0,0) [below left] $0$;
% Points
coordinate (start) at (1,-0.8);
coordinate (c1) at (3,3);
coordinate (c2) at (5.5,1.5);
coordinate (c3) at (8,4);
coordinate (end) at (10.5,-0.8);
% show the points
% foreach n in start,c1,c2,c3,end fill [black] (n)
% circle (2pt) node [below] ;
% join the coordinates
draw [name path=curve,thick,tangent at/.list=0.15,0.3,...,0.75,
normal at/.list=0.25,0.4,...,0.7] (start) to[out=70,in=180] (c1) to[out=0,in=180]
(c2) to[out=0,in=180] (c3) to[out=0,in=150] (end);
fill[name intersections=of=curve and tangent-2,total=t,cyan]
(intersection-t) circle (2pt);
endtikzpicture
enddocument


enter image description here






share|improve this answer














This is just for fun and not to "steal" the check mark from Peter Grill, who IMHO should get it. Conceptually it is basically the same as Jake's pioneering answer. This answer comes with two styles: tangent at and normal at, which you can pass to the curve, and utilizes decorations.markings rather than intersections. If you want to draw several tangents and/or normals, it is advantageous to use /.list, as in the MWE:



documentclassarticle
usepackagetikz
usetikzlibrarydecorations.markings

begindocument
pgfkeystikz/.cd,
tangent length/.store in=TangentLength,
tangent length=7mm,
normal length/.store in=NormalLength,
normal length=7mm
tikzsettangent/.style=red,thin,normal/.style=blue,thin,
tangent at/.style=postaction=decorate,decoration=markings,
mark=at position #1 with draw[tangent] (-TangentLength,0) -- (TangentLength,0);
fill[tangent] (0,0) circle (2pt);,
normal at/.style=postaction=decorate,decoration=markings,
mark=at position #1 with draw[normal] (0,-NormalLength) -- (0,NormalLength);
fill[normal] (0,0) circle (2pt);,

begintikzpicture
% Axes
draw [-latex] (-1,0) -- (11,0) node [right] $x$;
draw [-latex] (0,-1) -- (0,6) node [above] $y$;
% Origin
node at (0,0) [below left] $0$;
% Points
coordinate (start) at (1,-0.8);
coordinate (c1) at (3,3);
coordinate (c2) at (5.5,1.5);
coordinate (c3) at (8,4);
coordinate (end) at (10.5,-0.8);
% show the points
% foreach n in start,c1,c2,c3,end fill [black] (n)
% circle (2pt) node [below] ;
% join the coordinates
draw [thick,tangent at/.list=0.15,0.3,...,0.75,
normal at/.list=0.25,0.4,...,0.7] (start) to[out=70,in=180] (c1) to[out=0,in=180]
(c2) to[out=0,in=180] (c3) to[out=0,in=150] (end);
endtikzpicture
enddocument


enter image description here



Just if you are wondering: yes, it is possible to draw the tangent at some specific point the curve runs through. One possibility is to use the tricks of this answer, i.e. decompose the path into segments and attach the tangents/normals to the starting or end points of those. If you want to have this spelled out, please ask a new question, this here is really just for fun.



ADDENDUM: As for your request in the comments... (and yes, I am using Jake's answer, as I mention above)....



documentclassarticle
usepackagetikz
usetikzlibrarydecorations.markings,intersections

begindocument
pgfkeystikz/.cd,
tangent length/.store in=TangentLength,
tangent length=3.14cm,
normal length/.store in=NormalLength,
normal length=7mm
newcountertangent
newcounternormal
tikzsettangent/.style=red,thin,normal/.style=blue,thin,
tangent at/.style=postaction=decorate,decoration=markings,
mark=at position #1 with stepcountertangent
draw[tangent,name path=tangent-thetangent] (-TangentLength,0) -- (TangentLength,0);
fill[tangent] (0,0) circle (2pt);,
normal at/.style=postaction=decorate,decoration=markings,
mark=at position #1 with stepcounternormal
draw[normal,name path=normal-thenormal] (0,-NormalLength) -- (0,NormalLength);
fill[normal] (0,0) circle (2pt);,

begintikzpicture
% Axes
draw [-latex] (-1,0) -- (11,0) node [right] $x$;
draw [-latex] (0,-1) -- (0,6) node [above] $y$;
% Origin
node at (0,0) [below left] $0$;
% Points
coordinate (start) at (1,-0.8);
coordinate (c1) at (3,3);
coordinate (c2) at (5.5,1.5);
coordinate (c3) at (8,4);
coordinate (end) at (10.5,-0.8);
% show the points
% foreach n in start,c1,c2,c3,end fill [black] (n)
% circle (2pt) node [below] ;
% join the coordinates
draw [name path=curve,thick,tangent at/.list=0.15,0.3,...,0.75,
normal at/.list=0.25,0.4,...,0.7] (start) to[out=70,in=180] (c1) to[out=0,in=180]
(c2) to[out=0,in=180] (c3) to[out=0,in=150] (end);
fill[name intersections=of=curve and tangent-2,total=t,cyan]
(intersection-t) circle (2pt);
endtikzpicture
enddocument


enter image description here







share|improve this answer














share|improve this answer



share|improve this answer








edited Nov 24 at 20:59

























answered Nov 24 at 19:49









marmot

79.8k490169




79.8k490169











  • this looks fantastic as well and uses methods from tex.stackexchange.com/questions/25928/… I believe. How can I extend the second tangent starting from the left and use the intersection, calc package to show where it intersects the curve further towards the right?
    – MathScholar
    Nov 24 at 20:50











  • @MathScholar I added something. (I guess on the long run one may have to look at the internals of the intersections and fillbetween libraries, which internally compute all sorts of of useful quantities like pgfintersectiongetsolutiontimes, which however did not make it to the manual.)
    – marmot
    Nov 24 at 21:04











  • per your request I will give the check to Peter. Both posts are extremely helpful. Thanks Marmot!
    – MathScholar
    Nov 24 at 22:17










  • @MathScholar Thanks! (I generally do not like if a question that has already an answer that is to the point gets a second answer, which is not to the point or just merely a repetition, like e.g. here. This brings unnecessary tension between the users IMHO, and the last thing I want to do is to step others on the feet.)
    – marmot
    Nov 24 at 22:32










  • @Hi Marmot, if I continue to extend the tangent to the right and fix the left side, there will be another intersection of the curve and tangent. Why does the intersection not get detected by the program?
    – MathScholar
    Nov 24 at 23:46
















  • this looks fantastic as well and uses methods from tex.stackexchange.com/questions/25928/… I believe. How can I extend the second tangent starting from the left and use the intersection, calc package to show where it intersects the curve further towards the right?
    – MathScholar
    Nov 24 at 20:50











  • @MathScholar I added something. (I guess on the long run one may have to look at the internals of the intersections and fillbetween libraries, which internally compute all sorts of of useful quantities like pgfintersectiongetsolutiontimes, which however did not make it to the manual.)
    – marmot
    Nov 24 at 21:04











  • per your request I will give the check to Peter. Both posts are extremely helpful. Thanks Marmot!
    – MathScholar
    Nov 24 at 22:17










  • @MathScholar Thanks! (I generally do not like if a question that has already an answer that is to the point gets a second answer, which is not to the point or just merely a repetition, like e.g. here. This brings unnecessary tension between the users IMHO, and the last thing I want to do is to step others on the feet.)
    – marmot
    Nov 24 at 22:32










  • @Hi Marmot, if I continue to extend the tangent to the right and fix the left side, there will be another intersection of the curve and tangent. Why does the intersection not get detected by the program?
    – MathScholar
    Nov 24 at 23:46















this looks fantastic as well and uses methods from tex.stackexchange.com/questions/25928/… I believe. How can I extend the second tangent starting from the left and use the intersection, calc package to show where it intersects the curve further towards the right?
– MathScholar
Nov 24 at 20:50





this looks fantastic as well and uses methods from tex.stackexchange.com/questions/25928/… I believe. How can I extend the second tangent starting from the left and use the intersection, calc package to show where it intersects the curve further towards the right?
– MathScholar
Nov 24 at 20:50













@MathScholar I added something. (I guess on the long run one may have to look at the internals of the intersections and fillbetween libraries, which internally compute all sorts of of useful quantities like pgfintersectiongetsolutiontimes, which however did not make it to the manual.)
– marmot
Nov 24 at 21:04





@MathScholar I added something. (I guess on the long run one may have to look at the internals of the intersections and fillbetween libraries, which internally compute all sorts of of useful quantities like pgfintersectiongetsolutiontimes, which however did not make it to the manual.)
– marmot
Nov 24 at 21:04













per your request I will give the check to Peter. Both posts are extremely helpful. Thanks Marmot!
– MathScholar
Nov 24 at 22:17




per your request I will give the check to Peter. Both posts are extremely helpful. Thanks Marmot!
– MathScholar
Nov 24 at 22:17












@MathScholar Thanks! (I generally do not like if a question that has already an answer that is to the point gets a second answer, which is not to the point or just merely a repetition, like e.g. here. This brings unnecessary tension between the users IMHO, and the last thing I want to do is to step others on the feet.)
– marmot
Nov 24 at 22:32




@MathScholar Thanks! (I generally do not like if a question that has already an answer that is to the point gets a second answer, which is not to the point or just merely a repetition, like e.g. here. This brings unnecessary tension between the users IMHO, and the last thing I want to do is to step others on the feet.)
– marmot
Nov 24 at 22:32












@Hi Marmot, if I continue to extend the tangent to the right and fix the left side, there will be another intersection of the curve and tangent. Why does the intersection not get detected by the program?
– MathScholar
Nov 24 at 23:46




@Hi Marmot, if I continue to extend the tangent to the right and fix the left side, there will be another intersection of the curve and tangent. Why does the intersection not get detected by the program?
– MathScholar
Nov 24 at 23:46

















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