How do you plot x = 2sinh(u), y = v, z = 2cosh(u) in Wolfram Language or Mathematica? Everything I try fails and I’m ready to shoot my computer. Do I need a different CAS? Why is this so hard?
what is `v` in `y = v` ? assuming this is a typo and it is meant to be `u` then
ClearAll[u]
ParametricPlot3D[{2 Sinh[u], u, 2 Cosh[u]}, {u, -Pi/4, Pi/4}, PlotStyle -> {Thick, Red}, AxesLabel -> {x, y, z}]
For y=v then
ClearAll[u, v]
ParametricPlot3D[{2 Sinh[u], v, 2 Cosh[u]}, {u, -Pi/4, Pi/4}, {v, -Pi/4, Pi/4}, PlotStyle -> {Thick, Red}, AxesLabel -> {x, y, z}]
No, y is independent of x and z. I’m trying to plot a parametric surface. I can see the example in Mathematica but it doesn’t accept my input. It recognizes v as an independent variable in the example but doesn’t like my input. it should be a hyperbolic parabloid of some form.
Thank you, this program is confusing to me. I just wanted a simple 3D graphing program.
You wouldn’t happen to know how to graph a TNB frame?
it should be a hyperbolic parabloid of some form
see it there http://mathworld.wolfram.com/HyperbolicParaboloid.html , please and display it
ParametricPlot3D[{u, v, 1/3 u v}, {u, -2, 2}, {v, -3, 4}]
your parametrization was wrong. It can’t be simpler, right? In any case you should follow the syntax (cosh(u) from school is
Cosh[u]
in Mathematica) if you want to keep your computer healthy and yourself productive.
You wouldn’t happen to know how to graph a TNB frame?
a tangent-normal-binormal frame on a curve in 3D you will hardly do with a simple 3D graphing program. With Mma you can do it. Again, read a bit, the mathematical name is Frenet-Formulae
http://mathworld.wolfram.com/FrenetFormulas.html
and here is a simulation
[http://demonstrations.wolfram.com/FrenetFrame/
](http://demonstrations.wolfram.com/FrenetFrame/)you can download the notebook, see all formulae and experiment with it. You need to compute the tangential vector, the normal vector and the cross product of the two former and you are done, it’s necessary that the curve has 2 continous derivatives at least.