My Math Forum  

Go Back   My Math Forum > College Math Forum > Calculus

Calculus Calculus Math Forum


Thanks Tree1Thanks
  • 1 Post By romsek
Reply
 
LinkBack Thread Tools Display Modes
September 15th, 2018, 07:31 PM   #1
Member
 
Joined: Apr 2017
From: New York

Posts: 77
Thanks: 6

parametrization

Has anybody got idea how parametrization happens? any kind will be helpful for me to initiate this topic.
thanks
Attached Images
File Type: jpg Screen Shot 2018-09-15 at 11.29.38 PM.jpg (21.9 KB, 11 views)
Leonardox is offline  
 
September 15th, 2018, 08:33 PM   #2
Senior Member
 
romsek's Avatar
 
Joined: Sep 2015
From: USA

Posts: 2,091
Thanks: 1087

I'm not sure what you mean by "how it happens". The surface dictates the parameterization. You have some number of free parameters and the rest of the elements of the parameter vector are functions of these free parameters.

a) $y = \dfrac{x+4}{5}$ so $\left(x,~ \dfrac{x+4}{5}\right)$ is a parameterization

d) $y^2 +z^2 = 1+x^2$

Look at the form of the equation. It's a circle in the yz plane whose radius squared is $1+x^2$

It's generally clever to parameterize circles in polar coordinates so we end up with

$\left(x,~\sqrt{1+x^2}\cos(\theta),~\sqrt{1+x^2} \sin(\theta)\right),~x\in \mathbb{R},~0 \leq \theta < 2\pi$

g) $x^2 - y^2+z^2-4x-2y-2z+4 = 0$

well let's complete all these squares first

$(x^2 - 4x ) - (y^2 +2y) + (z^2 -2z) = -4$

$(x-2)^2 - 4 - (y+1)^2 + 1 + (z-1)^2 -1 = -4$

$(x-2)^2 - (y+1)^2 +(z-1)^2 = 0$

$(x-2)^2 + (z-1)^2 = (y+1)^2$

This is a circle cross section in the xz plane with radius $y+1$ and again we can get

$\left((y+1) \cos(\theta),~y,~(y+1) \sin(\theta)\right),~y \in \mathbb{R},~0 \leq \theta < 2\pi$

See if you can get the rest of them. Do you have a way of graphing these things in 3D? It really helps to be able to see them.
Thanks from topsquark
romsek is offline  
September 17th, 2018, 06:46 AM   #3
Senior Member
 
Joined: Mar 2015
From: New Jersey

Posts: 1,431
Thanks: 105

Looked up definition:

Curve: parameter t determines a point in space:
x=x(t), y=y(t), z=z(t)

Surface: parameters u,v determine a point in space:
x=x(u,v), y=y(u,v), z=z(u,v)

Surface Examples:

1) x=x, y=y, z=z(x,y)

2) x=ucosv, y=u+v, z=uv. in principle you can solve this for u and v in terms of x and y to get z=z(x,y).

As for OP examples, the parametrization can be anything consistent with definition. I suspect the text had something specific in mind. Otherwise just solving for z as in Example 1) works.

EDIT
Curve on a surface:
x=x(u,v), y=y(u,v), z=z(u,v)
u=u(t), v=v(t)
Example:
z=z(x,y), x=x(t), y=y(t)

Last edited by zylo; September 17th, 2018 at 07:19 AM.
zylo is offline  
September 18th, 2018, 06:22 AM   #4
Senior Member
 
Joined: Mar 2015
From: New Jersey

Posts: 1,431
Thanks: 105

You could parametrize z = z(x,y) with a coordinate transformation:
$\displaystyle x=r\cos\theta$, $\displaystyle y=r\sin\theta$, $\displaystyle z=z(r\cos\theta,r\sin\theta)=z(r,\theta)$

Last edited by skipjack; September 18th, 2018 at 07:24 AM.
zylo is offline  
September 18th, 2018, 07:15 AM   #5
Math Team
 
Joined: Dec 2013
From: Colombia

Posts: 7,355
Thanks: 2469

Math Focus: Mainly analysis and algebra
Quote:
Originally Posted by zylo View Post
1) x=x, y=y, z=z(x,y)
$x=x$ is meaningless - it doesn't define $x$ at all. You should pick different names for your parameters.

$$x=s, \, y=t, \, z=z(s,t)$$

Note that $z=z(s,t)$ is less meaningless, because it tells us that the third coordinate $z$ of every point is the value of a function $z(x,y)$ evaluated at the point $(x,y)=(s,t)$. The difference in reusing the symbol $z$ is that the notation differentiates between the number $z$ and the function $z(x,y)$.
v8archie is online now  
Reply

  My Math Forum > College Math Forum > Calculus

Tags
parametrization


« UTV and UNV | - »

Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
Parametrization of curves sarajoveska Calculus 3 June 22nd, 2017 07:11 AM
Rational Parametrization Help numberguru1 Number Theory 5 February 4th, 2016 03:44 AM
Curve parametrization tmsgru Algebra 2 June 20th, 2013 10:28 AM
Parametrization of a manifold Robert Lownds Real Analysis 3 June 18th, 2013 09:21 AM
FLT parametrization mathbalarka Number Theory 4 April 29th, 2012 10:55 PM





Copyright © 2018 My Math Forum. All rights reserved.