F fifthrapiers May 2006 369 2 Feb 2, 2007 #1 Consider the following surface: x = 4y^2 + 4z^2 What's this surface called? What coordinate plane does z=0 define? What's the trace of this surface in the xz plane look like? What's the trace of this surface in the planes x=k look like?

Consider the following surface: x = 4y^2 + 4z^2 What's this surface called? What coordinate plane does z=0 define? What's the trace of this surface in the xz plane look like? What's the trace of this surface in the planes x=k look like?

earboth Jan 2006 5,854 2,553 Germany Feb 2, 2007 #2 Ideasman said: Consider the following surface: x = 4y^2 + 4z^2 What's this surface called? What coordinate plane does z=0 define? What's the trace of this surface in the xz plane look like? What's the trace of this surface in the planes x=k look like? Click to expand... Hello, I've attached a diagram of this surface. a) It is a rotation paraboloid. The axis of rotation is the positive x-axis. b) It's a parabola with the vertex in the origin: \(\displaystyle x = 4y^2\) c) It's a circle: \(\displaystyle k = 4x^2+4y^2 \Longleftrightarrow x^2+y^2=\left( \frac{1}{2} \sqrt{k} \right)^2\) EB Attachments rotat_paraboloid1.GIF 13.4 KB Views: 23 Reactions: fifthrapiers

Ideasman said: Consider the following surface: x = 4y^2 + 4z^2 What's this surface called? What coordinate plane does z=0 define? What's the trace of this surface in the xz plane look like? What's the trace of this surface in the planes x=k look like? Click to expand... Hello, I've attached a diagram of this surface. a) It is a rotation paraboloid. The axis of rotation is the positive x-axis. b) It's a parabola with the vertex in the origin: \(\displaystyle x = 4y^2\) c) It's a circle: \(\displaystyle k = 4x^2+4y^2 \Longleftrightarrow x^2+y^2=\left( \frac{1}{2} \sqrt{k} \right)^2\) EB

S Soroban May 2006 12,028 6,341 Lexington, MA (USA) Feb 2, 2007 #3 Hello, Ideasman! I use the standard orientation of the three coordinate axes. Code: z | | | | * - - - - - - y / / / x Consider the following surface: .\(\displaystyle x \:= \:4y^2 + 4z^2\) What's this surface called? Click to expand... It is a paraboloid. What coordinate plane does \(\displaystyle z=0\) define? Click to expand... \(\displaystyle z = 0\) is the \(\displaystyle xy\)-plane (the "floor" of the graph). What's the trace of this surface in the \(\displaystyle xz\)-plane look like? Click to expand... The \(\displaystyle xz\)-plane is the "left wall" of the graph. Let \(\displaystyle y = 0\) and we have: .\(\displaystyle x = 4z^2\) This is a parabola on the "left wall", vertex at the origin, . . opening in the positive \(\displaystyle x\)-direction. What's the trace of this surface in the planes \(\displaystyle x=k\) look like? Click to expand... Let \(\displaystyle x = k\). .We have: .\(\displaystyle 4y^2 + 4z^2\:=\:k\quad\Rightarrow\quad y^2 + z^2\:=\:\frac{k}{4}\) These are circles: centered on the \(\displaystyle x\)-axis with radius \(\displaystyle \frac{\sqrt{k}}{2}\) Reactions: fifthrapiers

Hello, Ideasman! I use the standard orientation of the three coordinate axes. Code: z | | | | * - - - - - - y / / / x Consider the following surface: .\(\displaystyle x \:= \:4y^2 + 4z^2\) What's this surface called? Click to expand... It is a paraboloid. What coordinate plane does \(\displaystyle z=0\) define? Click to expand... \(\displaystyle z = 0\) is the \(\displaystyle xy\)-plane (the "floor" of the graph). What's the trace of this surface in the \(\displaystyle xz\)-plane look like? Click to expand... The \(\displaystyle xz\)-plane is the "left wall" of the graph. Let \(\displaystyle y = 0\) and we have: .\(\displaystyle x = 4z^2\) This is a parabola on the "left wall", vertex at the origin, . . opening in the positive \(\displaystyle x\)-direction. What's the trace of this surface in the planes \(\displaystyle x=k\) look like? Click to expand... Let \(\displaystyle x = k\). .We have: .\(\displaystyle 4y^2 + 4z^2\:=\:k\quad\Rightarrow\quad y^2 + z^2\:=\:\frac{k}{4}\) These are circles: centered on the \(\displaystyle x\)-axis with radius \(\displaystyle \frac{\sqrt{k}}{2}\)