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 
September 15th, 2018, 08:33 PM  #2 
Senior Member 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) $x^2  y^2+z^24x2y2z+4 = 0$ well let's complete all these squares first $(x^2  4x )  (y^2 +2y) + (z^2 2z) = 4$ $(x2)^2  4  (y+1)^2 + 1 + (z1)^2 1 = 4$ $(x2)^2  (y+1)^2 +(z1)^2 = 0$ $(x2)^2 + (z1)^2 = (y+1)^2$ 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. 
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. 
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. 
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  $x=x$ 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 $z(x,y)$. 

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