5
$\begingroup$

I want to add some kind of arrows (or some kind of function) on the x-axis (as depicted). if sin(x) is higher than bx, then arrow should point to the right, in the interval between its two intersection. A left arrow, if bx is higher than sin(x). Arrows should vary with change of paramters.

adding arrows

func[x_, b_] := b*x;
Manipulate[
 Plot[{a*Sin[x], func[x, b]}, {x, 0, 2 Pi}],
 {a, 1, 10}, {b, 0.5, 10}]
Improve this question
$\endgroup$
2
  • $\begingroup$ Can you be more clear as to what you mean when Sin[x] greater than bx? For a=1 and b=0.5, at $x=1$ Sin[x]>bx at $x=4$ Sin[x]<bx. At the intersection Sin[x]==bx. At which point of x are you evaluating at? $\endgroup$ – I should change my Username Jul 7 '17 at 9:12
  • $\begingroup$ if i lower b or raise a, then the interval where sin[x] is greater than bx gets larger. Then the left arrow pointing to the right should get longer aswell (until sin[x] is lower than bx). There is no specific point of x where i am evaluating at. The arrow should vary with the parameters a and b. $\endgroup$ – Manuel Fülling Jul 7 '17 at 9:44
6
$\begingroup$

Update 2: Dealing with question in the comments:

My original code is much longer and with multiple functions. I don't use the command Show in it. So if i add /. Line->Arrow all functions get an arrow. Is there a possibility to use /. Line ->Arrow separately on functions?

You can put the /. Line->Arrow piece after Plot[...], and specify Arrowheads[0] for the lines that you want to be rendered as Lines as part of the PlotStyle setting. For example:

Manipulate[Plot[{a*Sin[x], func[x, b], 
     ConditionalExpression[0, func[x, b] <= a Sin[x]], 
     ConditionalExpression[0, func[x, b] >= a Sin[x]]}, {x, 0, 2 Pi}, 
    PlotStyle -> {Directive[Thick, Arrowheads[0], ColorData[54, "ColorList"][[1]]], 
      Directive[Thick, Arrowheads[0], ColorData[54, "ColorList"][[2]]], 
      Directive[Red, Thick, Arrowheads[{0, .05}]], 
      Directive[Blue, Thick, Arrowheads[{-.05, 0}]]}] /. Line -> Arrow,
   {a, 1, 10}, {b, 0.5, 10}]

enter image description here

Note: In version 9, you don't have to use /. Line->Arrow trick. You can use functions to specify the PlotStyle and use ({Directive[Blue, Thick, Arrowheads[{-.05, 0}], Arrow@@#}&) in place ofDirective[Blue, Thick, Arrowheads[{-.05, 0}]`. But this method no longer works in newer versions.

Update: If you have Version 10 or newer versions, you can also use NumberLinePlot as Epilog:

Manipulate[Plot[{a*Sin[x], func[x, b]}, {x, 0, 2 Pi}, 
    Epilog -> First[NumberLinePlot[{a Sin[x] <= func[x, b], 
       a Sin[x] > func[x, b]}, {x, 0, 2 Pi},
      PlotStyle -> {Directive[Red, Thick, Arrowheads[{-.05, 0}]], 
         Directive[Blue, Thick, Arrowheads[{0, .05}]]}, 
      Spacings -> 0] /. Line -> Arrow]],
   {a, 1, 10}, {b, 0.5, 10}]

enter image description here

Original answer:

Manipulate[Show[Plot[{a*Sin[x], func[x, b]}, {x, 0, 2 Pi}], 
  Plot[{ConditionalExpression[0, func[x, b] <= a Sin[x]], 
        ConditionalExpression[0, func[x, b] >= a Sin[x]]}, {x, 0, 2 Pi}, 
    PlotStyle -> {Directive[Red, Thick, Arrowheads[{0, .05}]], 
      Directive[Blue, Thick, Arrowheads[{-.05, 0}]]}] /. Line -> Arrow], 
 {a, 1, 10}, {b, 0.5, 10}]

enter image description here

You can also use a combination of Epilog and ParametricPlot with Mesh options:

Manipulate[Plot[{a*Sin[x], func[x, b]}, {x, 0, 2 Pi}, 
  Epilog -> First[ParametricPlot[{x, 0}, {x, 0, 2 Pi}, 
      MeshFunctions -> {a Sin[#] - func[#, b] &}, Mesh -> {{0}}, 
      MeshShading -> {Directive[Red, Thick, Arrowheads[{-.05, 0}]], 
        Directive[Blue, Thick, Arrowheads[{0, .05}]]}] /. Line -> Arrow]], 
  {a, 1, 10}, {b, 0.5, 10}]
Improve this answer
$\endgroup$
10
  • $\begingroup$ Can you explain a bit of how the ConditionalExpression[] is working? It returns 0 when func[x,b]<=a Sin[x] and 0 when func[x,b]>=a Sin[x]. Since the 2nd argument to ConditionalExpression[] is the cond., I can't seem to be able to wrap my head around how you are getting the coordinates. $\endgroup$ – I should change my Username Jul 7 '17 at 10:49
  • 1
    $\begingroup$ @IshouldchangemyUsername, ConditionalExpression[0,condition] gives 0 if the condition is satisfied and Nothing otherwise. The Nothing portion is not plotted so we get, in the current case, a line from {0,0} to {0,x*} where x* is the maximum point where condition is satisfied. BTW, ConditionalExpression[0,condition] gives the same result as Piecewise[{{0,condition}}, Nothing] (which may be more intuitive). Hope this helps. $\endgroup$ – kglr Jul 7 '17 at 10:57
  • $\begingroup$ ConditionalExpression[0, func[x, b] <= a Sin[x] is a condition where i only compare two functions with each other. What if i need to compare func[x,b] with two functions (for example with asin(x) and acos(x)? ConditionalExpression[0, func[x, b] <= {a Sin[x], a Cos[x]} does not work $\endgroup$ – Manuel Fülling Jul 7 '17 at 11:37
  • 1
    $\begingroup$ @Manuel, condition in ConditionalExpression[0,condition] should be a Boolean function that evaluates to True or False. So, you can use And @@ Thread[func[x, b] <= {a Sin[x], a Cos[x]}] or Or @@ Thread[func[x, b] <= {a Sin[x], a Cos[x]}] as condition in ConditionalExpression[0,condition] $\endgroup$ – kglr Jul 7 '17 at 11:59
  • $\begingroup$ if i add /. Line -> Arrow after my closing Plot Bracket ] , all my function get an arrow. Where do i add /. Line -> Arrow and do i need the command Show[] for it? $\endgroup$ – Manuel Fülling Jul 7 '17 at 14:17
3
$\begingroup$

Another solution using NSolve to find the intersection point and Graphics to draw the arrows:

func[x_, b_] := b*x;
Manipulate[
 With[{xIntersection = 
    NSolve[{a*Sin[x] == func[x, b], x > 0}, x, Reals][[1, 1, 2]]},
  Show[
   Plot[{a*Sin[x], func[x, b]},
    {x, 0, 2 Pi},
    PlotRange -> {{0, 2 Pi}, {-6, 6}}
    ],
   Graphics[{
     Point@{xIntersection, func[xIntersection, b]},
     Arrowheads@0.03,
     Red, Arrow@{{0, 0}, {xIntersection, 0}},
     Blue, Arrow@{{2 Pi, 0}, {xIntersection, 0}}
     }]
   ]
  ],
 {a, 1, 10}, {b, 0.5, 10}
 ]

enter image description here

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.