6
$\begingroup$

Please help. Trying to find area between three curves, e^-x, x = 2, y = 1. Can't find out how to plot x = 2. Don't want to use Epilogue unless it can shade the area enclosed by the three curves.

$\endgroup$
4
  • 2
    $\begingroup$ stackoverflow.com/questions/2897277/… $\endgroup$
    – Igor Rivin
    Sep 12, 2014 at 1:15
  • $\begingroup$ gridlines and epilog are the answers to that question - if I'm not mistaken both are done after the initial graphics rendering, meaning that the filling command will continue shading past the point the lines are drawn. I need a solution that will allow me to actually plot the equation x = 2 $\endgroup$
    – Cory
    Sep 12, 2014 at 1:19
  • $\begingroup$ A related question is Calling Correct Function for Plotting DiracDelta. Not the same because the plot has value 0 away from the spike. $\endgroup$
    – Jens
    Sep 12, 2014 at 5:50
  • 1
    $\begingroup$ Couldn't you use the HeavisideTheta function? Just multiply HeavisideTheta(x-2) by something that is infinite fpr the purposes of your application (i.e. bigger than your visible y-max). One such number could be 100. $\endgroup$ Sep 12, 2014 at 8:08

8 Answers 8

12
$\begingroup$

you can also try:

PolarPlot[2/Cos[t], {t, 0, Pi/4}]

or

ContourPlot[x == 2, {x, 0, 4 Pi}, {y, 0, 4 Pi}]

If you want to find the area using other method, I would suggest to use Area and ImplicitRegion in V10 as follows:

r = ImplicitRegion[y >= Exp[-x] && x <= 2 && y <= 1, {x, y}];
Area[r]

(*(1 + E^2)/E^2*)

for shading issue you may find this interesting:

Show[{ContourPlot[{x == 2, y == Exp[-x], y == 1}, {x, -1, 3}, {y, 0, 
    2}], RegionPlot[r, ColorFunction -> "Rainbow"]}]      (*@ Rahul Narain*)

or

Show[{Plot[{Exp[-x], 1}, {x, -2, 3}], 
  PolarPlot[2/Cos[t], {t, 0, 2 \[Pi]}], 
  RegionPlot[r, ColorFunction -> "Rainbow"]}]

enter image description here

$\endgroup$
3
  • $\begingroup$ Used the advice above. Thank you. $\endgroup$
    – Cory
    Sep 12, 2014 at 2:24
  • $\begingroup$ You could combine both the Plot and the PolarPlot into a single ContourPlot in your last example. In any case I feel the PolarPlot form is a little opaque while ContourPlot[x == 2, ...] is perfectly clear. $\endgroup$
    – user484
    Sep 12, 2014 at 2:50
  • $\begingroup$ @RahulNarain, thanks for the advice. in polar, you know x=rCos[t] or r=x/Cos[t], for this case x=2 and r=2/Cos[t]. $\endgroup$ Sep 12, 2014 at 2:57
8
$\begingroup$

Its my understanding that you want to insist on using Plot for this problem. Then how about defining a function that has a vertical jump at x=2 and otherwise exceeds the required PlotRange so that its remaining parts won't show up?

Plot[100 Sign[x - 2], {x, -3, 3}, ExclusionsStyle -> Red, 
 PlotRange -> {-1, 1}]

plot

$\endgroup$
3
  • $\begingroup$ I was about to suggest $MaxMachineNumber might be a good choice for the coefficient 100, but apparently not. (+1) $\endgroup$
    – Michael E2
    Sep 12, 2014 at 2:17
  • $\begingroup$ This works well enough. I'm really surprised there is no plain way to plot a vertical line. Thank you. $\endgroup$
    – Cory
    Sep 12, 2014 at 2:23
  • $\begingroup$ Plot[9!(x-2),{x,-3,3},PlotRange->{-3,3}] $\endgroup$
    – matrix89
    Sep 12, 2014 at 16:23
7
$\begingroup$

The new V10 region functionality is rather suited to implementing your description of the problem in a direct way:

reg = ImplicitRegion[y < 1 && y > E^-x && x < 2, {x, y}];
Show[BoundaryDiscretizeRegion[reg, {{0, 2}, {E^-2, 1}}], Axes -> True,
  AxesOrigin -> {0, 0}, AspectRatio -> 1/GoldenRatio]

Mathematica graphics

Also for finding the area:

RegionMeasure[reg]
(* (1 + E^2)/E^2 *)
$\endgroup$
1
  • $\begingroup$ There are, except that the OP has pooh-poohed all of them before. $\endgroup$
    – Igor Rivin
    Sep 12, 2014 at 3:19
2
$\begingroup$
Show[
 RegionPlot[y > E^-x && y < 1 && x < 2,
  {x, -1, 3}, {y, 0, 1.5}],
 Plot[{
   Tooltip[E^-x, TraditionalForm[y == E^-x]],
   Tooltip[1, TraditionalForm[y == 1]]},
  {x, -1, 3}],
 Epilog -> Tooltip[Line[{{2, 0}, {2, 1.5}}],
   TraditionalForm[x == 2]]]

enter image description here

area = Integrate[1 - E^-x, {x, 0, 2}]

1 + 1/E^2

$\endgroup$
2
$\begingroup$

You might find the answers to an old question on StackOverflow useful

My suggested hack in that case involved Rotate:

ticks = {{None, ({#, Rotate[#, 90 Degree], {0.02, 0}} & /@ 
      Range[0, 4])}, {({#, Rotate[#, 90 Degree], {0.02, 0}} & /@ 
      Range[0, 1, 0.25]), None}};

Rotate[Plot[2, {x, 0, 1}, AspectRatio -> GoldenRatio, 
  AxesOrigin -> {1, 0}, Frame -> True, 
  FrameTicks -> ticks], -90 Degree]

I don't know why the previous version didn't rotate properly.

Of course, since you want to plot x=something and y=something simultaneously, this might not work for you, in which case I'd recommend Jens' answer, or hacking the setting for AxesOrigin to create a horizontal line as well as a vertical one.

$\endgroup$
1
$\begingroup$

ParametricPlot[{10, y}, {x, -10, 10}, {y, -10, 10}] works for me.

$\endgroup$
2
  • 2
    $\begingroup$ Is there any way to actually plot the equation x = 2, with the Plot function? ParametricPlot is still not what I'm looking for. I'm trying to enclose an area. $\endgroup$
    – Cory
    Sep 12, 2014 at 1:30
  • $\begingroup$ No, there is not. $\endgroup$
    – Igor Rivin
    Sep 12, 2014 at 1:48
1
$\begingroup$

First we construct some helpers:

f[x_] := E^(-x)

yval = f[2]

1/E^2

h[x_] := 1

v[t_] := 2

Vertical lines can be constructed with ParametricPlot:

ParametricPlot[{v[t], t}
 , {t, yval, 1.}
 , PlotStyle -> {Darker[Red], Thick}
 , PlotRange -> {{-.5, 2.5}, {0, 1.5}}]

enter image description here

Putting it all Together:

Show[Plot[{f[x], h[x]}
  , {x, 0., 2.}
  , Filling -> {2 -> {1}}]
 , ParametricPlot[{v[t], t}
  , {t, yval, 1.}
  , PlotStyle -> {Darker[Red], Thick}]
 , PlotRange -> {{-.5, 2.5}, {0, 1.5}}]

enter image description here

Edit

You can also work with Epilog, Line or Arrow

Plot[{f[x], h[x]}
 , {x, 0, 2}
 , PlotRange -> {{-.5, 2.5}, {0, 1.5}}
 , Epilog -> {Thick, Darker[Red], Line[{{2, yval}, {2, 1}}]}
 , Filling -> {2 -> {1}}
 , Frame -> True
 , Axes -> False
 ]

enter image description here

Plot[{f[x], h[x]}
 , {x, 0, 2}
 , PlotRange -> {{-.5, 2.5}, {0, 1.5}}
 , Epilog -> {Thick, Darker[Red], Arrow[{{2, yval}, {2, 1}}]}
 , Filling -> {2 -> {1}}
 , Frame -> True
 , Axes -> False
 ]

enter image description here

enter image description here

enter image description here

enter image description here

Thx @Jens and @Bob Hanlon for inspiration.

$\endgroup$
0
$\begingroup$

Use the GridLines option of the Plot command. You can place a single vertical line anywhere along the x-axis, color it, make it thick or thin, dash it, and maybe even annotate it. Why mess around with sticky equations? Use the built in Plot features. See the GridLines examples.

$\endgroup$
1
  • $\begingroup$ Answers (esp. on stack sites such as this one) are supposed to include code and present results. Please add such an example, otherwise this answer will be deemed of low quality and will eventually get deleted after due process. I encourage you to look at the other answers for a representative sample. $\endgroup$
    – Syed
    Apr 13 at 6:37

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.