Recent content by Reflected

So after factoring it you get x=-x + 1 ± \sqrt{-3x^2+2x+7} / 2 and your saying to set it so.. x=-t + 1 ± \sqrt{-3t^2+2t+7} / 2 y = t z=1-(-t + 1 ± \sqrt{-3t^2+2x+7} / 2) - t Also, what do u do with the ± sign? edit: nevermind, i figured it out, thanks for your help.

Hmm... I'm still a bit confused. By y = f(x) do you mean y = -2x + 2 +- (\sqrt{(-12x^2+8x+28)})/4 and then just substitute all of those X's for T's?

Hmm okay.. so after you plug everything into the quad formula you get something like (-2x + 2 +- (\sqrt{(-12x^2+8x+28)}))/4 What would you do at this point? and yes, I am in your class apparently

yes, i tried doing that as well. How did you set that expression into the quadratic formula? You have 2x^2 + 2xy - 2x + 2y^2 -2y = 3 so how does that go into the quad form?

I don't understand how you parametrizted the final equation after all of the substitution and completing the square.

Ok, thanks for your help!

I'm really not sure how to attempt doing this: I tried messing around in Mathematica by trying to somehow equate the two parameterizations together such as... x = 2 Sin[u] Cos[v], y = 2 Sin[u] Sin[v] z = 1 - 2 Sin[u]Cos[v] - 2 Sin[u] Sin[v] Graphing this in Mathematica gives a slanted...

Since its a sphere of radius 2 it should be: x = 2sinΦcosΘ y = 2sinΦsinΘ z = 2cosΦ

By parametrizing the plane, are you referring to the second equation? If so, can't you just do something like: x = u y = v z = 1 - u - v

I'm actually not too sure; I want to graph them in Mathematica, but I don't know the command. edit: Ok, I graphed them in Mathematica as follows first = ContourPlot3D[{x^2 + y^2 + z^2 - 4}, {x, -10, 10}, {y, -10, 10}, {z, -10, 10}]; second = ContourPlot3D[{(x + y + z - 1)}, {x, -10, 10}...

Determine the curve of intersection of the surfaces x^2 + y^2 + z^2 = 4 and x + y + z = 1. The curve should be in parametric form. With this problem, I'm really not sure what direction to go in. I had thought about using the quadratic formula in some manner, but really unclear. Any advice...

It's kind of both, I have gotten somewhere though. secondcylinder = ParametricPlot3D[{(u*Sin[Pi] + Cos[Pi]*Cos[t]), Sin[t], (u*Cos[Pi] - Sin[Pi]*Cos[t])}, {t, 0, 2 Pi}, {u, -2, 2} , PlotStyle -> {Blue, Opacity -> .3}, Mesh -> None]; Show[secondcylinder] I don't know how to...

Below is some code that creates a semi-transparent cylinder. firstcylinder = ParametricPlot3D[{Cos[t], u, Sin[t]}, {t, 0, 2 Pi}, {u, -2, 2} , PlotStyle -> {Red, Opacity -> .3}, Mesh -> None]; Show[firstcylinder] Draw a second cylinder that is perpendicular to the given cylinder. Draw a...