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| This extends Henrik's nice answer to include the polys and their normals. ind = Position[RegionIntersection[#, line] & /@ polygs, _Point]; faces = polygs[[#]] & /@ ind; mr = MeshRegion[RegionUnion[faces // Flatten], ... |
| You could also try the anonymous function (i.e., #&) approach where you replace x[t] with # like so (D[#, t, t] + #^3) &[A Sin[Omega t]] (* -(A*Omega^2*Sin[Omega*t]) + A^3*Sin[Omega*t]^3 *) |
| Hi Jake, I took another look at your problem and I think it is fundamentally ill-posed. Your reaction equations need to be linearly independent. You have 5 reactions, but only 3 are independent. You can check my reasoning below. I am... |
| I was having trouble with your definition of gravity, so I just replaced it with the internal definition of the physical constant. The manipulate appears to work as intended. gamma = ChemicalData["Water", "SurfaceTension"]; mu =... |
| Hi Sascha, You may want to take a quick look at my post to [this discussion](https://community.wolfram.com/groups/-/m/t/1560271?p_p_auth=7FWscD87) that shows how unit discovery can be used to create shorthand notation fairly painlessly . For... |
| Hi John, Here is an attempt to answer your questions. Generally, one has some variable of interest (e.g., temperature) that DEPENDS on INDEPENDENT variables of space (x, y, z) and time (t). The equations to describe this dependence usually... |
| Hi Natasha, Your diffusion coefficient is huge and that implies that the system responds very quickly. To see this response, you will need to look at times much closer to 0 and not begin at $t=1$. You probably do not need many terms to capture... |
| _Mathematica_ understands that the solution is complex and is probably trying to write it in the most compact form. To expand, you could use ExpToTrig like so roots = (s /. Solve[s^2 + s + 1 == 0, s]) // ExpToTrig ... |
| Thank you Rutton! Since $Z$ is a combined variable of z and t, I think you need to multiply the $\phi$ expression by $\sqrt{t}$ to make it equivalent to dimensionless $\mathbf{v}^*$. If we choose the $t=\frac{1}{256}$ again, then we can plot the... |
| Does this help point you in the right direction? legend = SwatchLegend[ Automatic, {"0\[Degree]C", "1\[Degree]C", "2\[Degree]C", "3\[Degree]C"}, LegendMarkers -> Graphics[{EdgeForm[Black], Opacity[1],... |