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September 15th, 2018, 11:43 AM   #1
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how to parametrize

I need to know how I can parametrize please.
If I can learn this I can do the others.
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September 15th, 2018, 12:31 PM   #2
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I am able to parametrize a plane now. so disregard the plane part please. and now still watching video classes
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September 15th, 2018, 09:58 PM   #3
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you can rewrite the quadric surface as

$4x^2 + 3z^2 = 1+2y^2$

and parameterize it as

$(2\sqrt{1+2y^2} \cos(\theta),~y,~\sqrt{3(1+2y^2)} \sin(\theta)),~y\in \mathbb{R},~0 \leq \theta < 2\pi$
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September 16th, 2018, 05:05 AM   #4
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Unfortunately, the coefficients aren't quite right in romsek's answer and the equation used was incorrect (it should have been $4x^2 + 3z^2 = 2y^2 - 1$).

Corrected (and using ±(1/√2)cosh(u) instead of y), one gets
((1/2)sinh(u)cos(θ), ±(1/√2)cosh(u), (1/√3)sinh(u)sin(θ)), u $\small\geqslant$ 0, 0 $\small\leqslant$ θ < 2$\pi$.
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September 16th, 2018, 04:34 PM   #5
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Since I don't know anything about how to parametrize a surface I did not understand anything from the answer
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September 16th, 2018, 05:37 PM   #6
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Math Focus: Wibbly wobbly timey-wimey stuff.
Quote:
Originally Posted by Leonardox View Post
Since I don't know anything about how to parametrize a surface I did not understand anything from the answer
Let's try this. Do you know what it means to parametrize? For example, a circle can be parametrized by putting it into polar coordinates: $\displaystyle x^2 + y^2 = 4 \equiv x = 2 \cos( \theta ),~y = 2 \sin( \theta )$. The parameter is $\displaystyle \theta$ because we can vary $\displaystyle \theta$, a single variable, to get the two x and y values from the original equation.

So all we are doing is finding a way to "cut down" on the number of variables we need to trace out a solution. (Or alternately, put the equation into a form that is simpler to work with.)

-Dan

Last edited by skipjack; September 16th, 2018 at 11:06 PM.
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September 16th, 2018, 06:31 PM   #7
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for circle yes
x= rcos(Theta)
y=rsin(Theta)

what about 3x-z=6y+3

or
2y^2-3z^2=1+4x^2



need to learn before test
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September 16th, 2018, 11:19 PM   #8
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Obtaining parametric equations for a plane (such as 3x - z = 6y + 3) is explained in detail in this article.
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September 17th, 2018, 06:50 AM   #9
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Quote:
Originally Posted by zylo View Post
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.
Solve for z.
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