How to Draw the Hyperbolic Parametric Curve on the Left side and Confirm the Animation

General Usage
1

Hi all,

I am plotting a parametric hyperbolic curve, but I cannot draw the curve on the left side with my codes. I see from this :

I want to know whether the animation is true, for each radians/degree the location of the ball is correct? isn’t cos hyperbolic and sine hyperbolic goes to infinity as t goes to infinity?

using Plots, LaTeXStrings, Plots.PlotMeasures
gr()

θ1 = 0:0.01:2π
x1 = 1 .*cos.(θ1)
y1 = 1 .*sin.(θ1)

θ2 = -2π:0.01:2π
x2 = 1 .*cosh.(θ2)
y2 = 1 .*sinh.(θ2)

θ3 = 0:0.01:π/6
x3 = 1 .*cosh.(θ3)
y3 = 1 .*sinh.(θ3)

x4 = 1 .*cos.(θ3)
y4 = 1 .*sin.(θ3)

plot(x1, y1, xlims=(-4,4), ylims=(-1.2,2),
	size=(900, 600), label=L"(x,y)=(\cos \ t, \sin \ t)", 
	framestyle=:zerolines, legend=:bottomright,
	bottom_margin=3mm)

plot!(x2, y2, 
	size=(900, 600), label=L"(x,y)=(\cosh \ t, \sinh \ t)", 
	framestyle=:zerolines,
	bottom_margin=3mm)
plot!(x3, y3, 
	size=(900, 600), label=L"(\cosh \ t, \sinh \ t), \ 0 \ ≤ \ t \ ≤ \ \pi/6", 
	linecolor=:green, arrow=true, linewidth=3,
	bottom_margin=3mm)
plot!(x4, y4, 
	size=(900, 600), label=L"(\cos \ t, \sin \ t), \ 0 \ ≤ \ t \ ≤ \ \pi/6", 
	linecolor=:green3, arrow=true, linewidth=3,
	bottom_margin=3mm)

scatter!([1], [0], color = "red1", label="", markersize = 3)
scatter!([cosh(pi/6)], [sinh(pi/6)], color = "red1", label="", markersize = 3)
scatter!([cos(pi/6)], [sin(pi/6)], color = "red1", label="", markersize = 3)

annotate!([(-0.9,0.7, (L"x^{2} + y^{2} = 1", 10, :black))])
annotate!([(1.4,-0.6, (L"x^{2} - y^{2} = 1", 10, :black))])

Thanks.

Capture d’écran_2023-06-22_05-13-31
Capture d’écran_2023-06-22_05-13-31889×575 26.2 KB

2

This should help

1 Like
3

I think the animation is correct, and your statement also. There are (more than) two ways of parametrizing the curves

  • one with the normal (non-hyperbolic) \sec{\theta} and \tan{\theta} for an angle \theta between 0 and 2\pi
  • one with the hyperbolic functions \cosh t and \sinh t, but then the curve parameter t is not the same as \theta, instead it has to be in all of \mathbb{R}.

There should be also be some way of expressing t in terms of \theta, basically solving \sec \theta = \cosh t etc.

This might be useful:

1 Like