chezmarylou1i

2021-12-14

I am struggling with the concept of parameterizing curves. I am not even sure if I know what it means so I tried to look some things up.

Since I didnt

Since I didnt

Jeffery Autrey

Beginner2021-12-15Added 35 answers

The idea of parameterization is that you have some equation for a subset X of a space (often $R}^{n$ ), e.g., the usual equation

${x}^{2}+{y}^{2}=1$

for the unit circle C in$R}^{2$ , and you want to describe a function $\gamma \left(t\right)=(x\left(t\right),y\left(t\right))$ that traces out that subset (or sometimes, just part of it) as t varies.

With a parameterization in hand, you can then specify a point on X just by giving a single value of t, which corresponds to the point$\gamma \left(t\right)$ on X. One can still give points on X, say, (x,y), directly, of course, but this has the disadvantage that often one needs to check that a given point (x,y) is on X, that is, that it satisfies the equation defining X, whereas by construction a point $\gamma \left(t\right)$ is always on X. Provided that the function $\gamma \left(t\right)$ traces out all of X, we say that X is the image of $\gamma$ .

In your example, we can parameterize the unit circle C by the parametric function

$\gamma \left(t\right)=(x\left(t\right),y\left(t\right)):=(\mathrm{cos}t,\mathrm{sin}t)$

We can check that the points specified by$\gamma \left(t\right)$ really do lie on C just by substituting cost for x and sint for y; indeed:

${\left(\mathrm{cos}t\right)}^{2}+{\left(\mathrm{sin}t\right)}^{2}=1$ .

It's not too hard to show that$\gamma$ actually traces out the full circle t (in fact, this is an immediate consequence of the usual geometric definitions of cos and sin). Note too that this parameterization traces over the circle infinitely many times, and in particular, there is more than one t value corresponding to any point on the circle. In fact, since the components cost and sint of $\gamma \left(t\right)$ have period $2\pi$ , we have $\gamma (t+2\pi )=\gamma \left(t\right)$ for all t.

There are many other parameterizations for all or part of the circle, too, and which is best depends on the context. Substituting the components in${x}^{2}+{y}^{2}=1$ (try this!) shows that for all t,

$\alpha \left(t\right):=(\frac{2t}{{t}^{2}+1},\frac{{t}^{2}-1}{{t}^{2}+1})$

is on the unit circle, and with some more work we can show that (1)$\alpha$ traces out the full circle with the single exception of the point (0,1) (because we have $\frac{{t}^{2}-1}{{t}^{2}+1}<1$ for all t), and (2) it is injective, that is, it only traces over the (punctured) circle once. This parameterization looks qualitatively different from the trigonometric parameterization $\gamma \left(t\right)$ above, but they are related by a clever and important change of variable related to Pythagorean triples and which proves to be extremely helpful in evaluating certain integrals.

One can, by the way, also parameterize surfaces (and even higher-dimensional objects); the most important difference is that (at least sensible) parameterizations of surfaces require two parameters, as a consequence of the fact that on surfaces one can move in two independent directions. A simple example is the parameterization$r(\varphi ,\theta )$ of the unit sphere

${x}^{2}+{y}^{2}+{z}^{2}=1$

by latitude$\varphi$ and longitude $\theta$ :

$r(\varphi ,\theta ){\textstyle \phantom{\rule{0.222em}{0ex}}}=(\mathrm{cos}\varphi \mathrm{cos}\theta ,\mathrm{cos}\varphi \mathrm{sin}\theta ,\mathrm{sin}\varphi )$ .

When giving latitude an longitude of a point on Earth, we typically specify points with latitude$-{90}^{\circ}\le \varphi \le {90}^{\circ}$ and longitude $-{180}^{\circ}\le \theta \le {180}^{\circ}$ . (Here, points with $\varphi ={0}^{\circ}$ comprise the equator, and points with $\theta ={0}^{\circ}$ the "prime meridian").

Many common shapes (lines, circles, other conic sections, planes, spheres, etc.) have well-known parameterizations, and graphs of functions$R}^{m}\Rightarrow {R}^{n$ have canonical parameterizations that are easy to write down, but like you say, for sufficiently complicated shapes parameterization can be a very difficult problem.

for the unit circle C in

With a parameterization in hand, you can then specify a point on X just by giving a single value of t, which corresponds to the point

In your example, we can parameterize the unit circle C by the parametric function

We can check that the points specified by

It's not too hard to show that

There are many other parameterizations for all or part of the circle, too, and which is best depends on the context. Substituting the components in

is on the unit circle, and with some more work we can show that (1)

One can, by the way, also parameterize surfaces (and even higher-dimensional objects); the most important difference is that (at least sensible) parameterizations of surfaces require two parameters, as a consequence of the fact that on surfaces one can move in two independent directions. A simple example is the parameterization

by latitude

When giving latitude an longitude of a point on Earth, we typically specify points with latitude

Many common shapes (lines, circles, other conic sections, planes, spheres, etc.) have well-known parameterizations, and graphs of functions

Tiefdruckot

Beginner2021-12-16Added 46 answers

Often a curve $\gamma$ in the plane is defined as the set of points (x,y) satisfying a certain geometric or algebraic condition. An example is

$\gamma :=\{(x,y)\mid \frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1\}$ , (1)

whereby the values of$a>0\text{}{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}\text{}b0$ are given. Such a description is implicit; it just provides a quick test whether a trial point (x,y) is lying on $\gamma$ or not.

When we really want to geometrically analyze the curve$\gamma$ , which means calculating its length or the enclosed area, etc., then we need a parametric representation. This is a production scheme that produces all points of $\gamma$ in a systematic and analytically manageable way. In this way the points of the curve $\gamma$ in (1) are produced by the vector-valued function

$$t\Rightarrow \{\begin{array}{ll}x(y)& :=a\mathrm{cos}t\\ y(t)& :=b\mathrm{sin}t\end{array}(0\le t\le 2\pi )(2)$$

in a 1:1 way, whereby$t=0\text{}{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}\text{}t=2\pi$ produce the same point.

Since an implicit representation of a curve$\gamma :F(x,y)=0$ does not determine a parametric representation (a "timetable") $t\Rightarrow (x\left(t\right),y\left(t\right))$ in an unique way there is no automatic procedure (like multiplication of two polynomials, or calculating a derivative) that generates a parametric representation from an equation $F(x,y)=0$ ; whence experience and intuition is asked for. Sometimes you can solve such an equation for y and obtain a representation as a graph in the form $x\Rightarrow (x,y\left(x\right))$ .

That deep mathematics is involved here can be seen from the following simple example: The innocent looking equation${x}^{2}+{y}^{2}=1$ determines a curve γ that encloses an area of $\pi$ and whose length is $2\pi$ .

whereby the values of

When we really want to geometrically analyze the curve

in a 1:1 way, whereby

Since an implicit representation of a curve

That deep mathematics is involved here can be seen from the following simple example: The innocent looking equation

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