My Math Forum parametrization

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September 15th, 2018, 08:31 PM   #1
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parametrization

Has anybody got idea how parametrization happens? any kind will be helpful for me to initiate this topic.
thanks
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 September 15th, 2018, 09:33 PM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,172 Thanks: 1142 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
 September 17th, 2018, 07:46 AM #3 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,603 Thanks: 115 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 08:19 AM.
 September 18th, 2018, 07:22 AM #4 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,603 Thanks: 115 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 08:24 AM.
September 18th, 2018, 08:15 AM   #5
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Quote:
 Originally Posted by zylo 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 $z(x,y)$.

 September 19th, 2018, 11:12 AM #6 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,603 Thanks: 115 x=t $\displaystyle \rightarrow$ t=x $\displaystyle \rightarrow$ x=x y=f(x). z=g(x), x=x, is a curve in space in terms of parameter (independent variable) x. If you prefer, you can write this as y=f(t), z=g(t), t=x. Whatever floats your boat, as long as you get the idea right. Last edited by zylo; September 19th, 2018 at 11:15 AM.

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