2
$\begingroup$

I was thinking, it's possible to plot inverse an animation??

Clear["Global`*"]
k = 0.002;a = 0.68;
1/Sqrt[k] x Hypergeometric2F1[1/2, 1/(2 - 3 b), 
1 + 1/(2 - 3 b), -((x^(2 - 3 b) a)/k)] /. b -> 1
animation = Animate[Plot[1/Sqrt[k] x Hypergeometric2F1[1/2, 1/(2 - 3 b), 
 1 + 1/(2 - 3 b), -((x^(2 - 3 b) a)/k)] Sqrt[1 + (x^(2 - 3 b) a)/k], {x, 1, 100}], {b, 1, 2}]
Improve this question
$\endgroup$
2
  • $\begingroup$ AnimationDirection -> Backward $\endgroup$
    – cvgmt
    Mar 28 at 13:48
  • $\begingroup$ no, the idea is I would like to change x,y direction like to plot y^-1[x] $\endgroup$
    – Felipe Dura
    Mar 28 at 14:56

2 Answers 2

Reset to default
2
$\begingroup$

For Animate, it is generally best to fix the PlotRange to keep the axes fixed. For a ParametricPlot, specify an AspectRatio.

Clear["Global`*"]

k = 0.002; a = 0.68;

animation = Animate[
  ParametricPlot[
   {1/Sqrt[k] x Hypergeometric2F1[1/2, 1/(2 - 3 b), 
      1 + 1/(2 - 3 b), -((x^(2 - 3 b) a)/k)] Sqrt[1 + (x^(2 - 3 b) a)/k], x},
   {x, 1, 100},
   PlotRange -> {{0, 4000}, {0, 100}},
   AspectRatio -> 1],
  {b, 1, 2}]
Improve this answer
$\endgroup$
0
$\begingroup$

I guess this will work :

ParametricPlot[x, y[x], {x, 1, 100}]

y[x] = 1/Sqrt[k] x Hypergeometric2F1[1/2, 1/(2 - 3 b), 
   1 + 1/(2 - 3 b), -((x^(2 - 3 b) a)/k)]
```
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.