1
$\begingroup$

A cone is the union of a set of half-lines that start at a common apex point and go through a base which can be any parametric curve. Show that the graph of z = ( x ^2 + 4y ^2)^(1/2) is a cone. What can be chosen as its base? Sketch the base.

I can see how I show this is cone by hand, just squaring and dividing across by 4 to get (z^2)/4=(x^2)/4 + y^2 but not sure how I can sketch it mathematica, including the base. Do I have to change it from this form? Is it Plot 3-D or is it parametric plot as I'll need to use something as a base. Anyone got any advice?

$\endgroup$
5
  • $\begingroup$ Plot3D[(z^2)/4= (x^2)/4 + y^2), {x, -5, 5}, {y, -5, 5}, {z, -5, 5}] is returning error for the brackets? anyone know why this code wont plot this cone? $\endgroup$
    – user24907
    Feb 5, 2015 at 19:51
  • $\begingroup$ you mean something like ContourPlot3D[ x^2/4 + y^2 == z^2/4, {x, -5, 5}, {y, -5, 5}, {z, -5, 0}]? $\endgroup$
    – kglr
    Feb 5, 2015 at 20:00
  • $\begingroup$ Perfect! just still unsure about when I should use ContourPlot3D or Plot3D or Parametric plot. So if I want to now add a parametric curve as base what function must I use? $\endgroup$
    – user24907
    Feb 5, 2015 at 20:05
  • $\begingroup$ mathopenref.com/coordparamellipse.html $\endgroup$ Feb 5, 2015 at 20:08
  • $\begingroup$ See e.g. Cone image refinement $\endgroup$
    – Artes
    Feb 5, 2015 at 20:29

3 Answers 3

0
$\begingroup$

An example of using MMA to "talk" about your problem

cone1 = (x^2 + 4 y^2)^(1/2)

Plot3D[cone1, {x, -3, 3}, {y, -3, 3}]

ContourPlot[cone1, {x, -3, 3}, {y, -3, 3}, Contours -> {1, 2, 3}]

base1 = #^2 & /@ (cone1 == 1)

ContourPlot[Evaluate[base1], {x, -1, 1}, {y, -1, 1}]

parametricBase1 = {x -> Cos[t], y -> (1/2) Sin[t]}

base1 /. parametricBase1

coneXPlane = cone1 /. {x -> 0}

Plot[coneXPlane, {y, -3, 3}]

coneYPlane = cone1 /. {y -> 0}

Plot[coneYPlane, {x, -3, 3}]

cone MMA example

$\endgroup$
3
  • $\begingroup$ Thank you, do you mind explaining what this base function is what the symbols you've listed do? And why you haven't used z^2 in your cone expression? $\endgroup$
    – user24907
    Feb 5, 2015 at 20:12
  • $\begingroup$ less about doing your homework for you, more about examples of "natural" ways to refer to things in MMA, also added parametric form of base $\endgroup$ Feb 5, 2015 at 20:18
  • $\begingroup$ Yes I'm trying to understand of these "natural" ways. mainly confused as to #^2 &/ @ in the base function is. Is there any particular reason for theses variables? $\endgroup$
    – user24907
    Feb 5, 2015 at 20:24
2
$\begingroup$
options = {PlotRange -> {{-5, 5}, {-5, 5}, {-5, 0}}, Mesh -> None, 
 ColorFunction -> "Rainbow", BaseStyle -> Opacity[.8], BoxRatios -> 1, Lighting -> "Neutral"};

ParametricPlot3D[{Cos[t] z, Sin[t] z/2, z}, {t, 0, 2 \[Pi]}, {z, -5,  0}, Evaluate@options]

enter image description here

ParametricPlot3D[z {Cos[t], Sin[2 t]/2, 1}, {t, 0, 2 \[Pi]}, {z, -5, 0}, Evaluate@options]

enter image description here

ParametricPlot3D[ z {Cos[t], Cos[3 t] Sin[t]/2, 1}, {t, 0, 2 \[Pi]}, {z, -5, 0}, 
    Evaluate@options]

enter image description here

ParametricPlot3D[z { Cos[t] Cos[2 t], Sin[ t] Cos[ 2 t], 1}, {t, 0, 2 \[Pi]}, {z, -5, 0}, 
  Evaluate@options]

enter image description here

$\endgroup$
2
  • $\begingroup$ Thank you, so i would put this after ContourPlot3D[ x^2/4 + y^2 == z^2/4, {x, -5, 5}, {y, -5, 5}, {z, -5, 0}] ? all in a pair of [] brackets? $\endgroup$
    – user24907
    Feb 5, 2015 at 20:26
  • $\begingroup$ @user24907, the posted examples show how to use ParametricPlot3D to get cones with various parametric bases. $\endgroup$
    – kglr
    Feb 5, 2015 at 20:56
0
$\begingroup$

A simpler/improved parametrization (t,h) for an elliptic cone. It is more convenient from apex as it is at the apex that all generators converge.

 ParametricPlot3D[h {2 Sin[t],Cos[t],2 },{t,0,2 Pi},{h,0,5},PlotStyle-> {Yellow}]

EllipticCone

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.