parametrization

**akhayoon** · April 21st 2008, 08:23 PM

how would you go about parameterizing

$x^{2/3}+y^{2/3}=1$ ??

**Soroban** · April 21st 2008, 09:01 PM

Hello, akhayoon!

Parametize: . $x^{\frac{2}{3}}+y^{\frac{2}{3}}\:=\:1$

. . . . $\begin{array}{ccc}x &=&\cos^3\!\theta \\ y &=&\sin^3\!\theta \end{array}$

**akhayoon** · April 21st 2008, 09:18 PM

so how do I know when to you cos and sin when parameterizing? is there a general rule of some sort?

**Soroban** · April 23rd 2008, 01:19 PM

Hello, akhayoon!

so how do I know when to use cos and sin when parameterizing?
is there a general rule of some sort?

Good questions . . . I wish knew the answers.

From experience, when I see $x^n + y^n$ (both to the same power),
. . I suspect that I can hammer it into the form: . $\sin^2\!\theta + \cos^2\!\theta \:=\:1$

Example: . $x^3 + y^3 \:=\:8$

Here is my rather primitive approach . . .

Divide by 8: . $\left(\frac{x}{2}\right)^3 + \left(\frac{y}{2}\right)^3 \:=\:1$

We want this to equal: . $\sin^2\!\theta + \cos^2\!\theta \:=\:1$

So we want: . $\begin{array}{ccc}\left(\dfrac{x}{2}\right)^3 &=&\cos^2\theta \\ \\ [-4mm] \left(\dfrac{y}{2}\right)^3 &=&\sin\theta \end{array}$

Solve for $x\text{ and }y$

. . $\left(\frac{x}{2}\right)^3 \:=\:\cos^2\theta \quad\Rightarrow\quad \frac{x}{2} \:=\:(\cos\theta)^{\frac{2}{3}}\quad\Rightarrow\qu ad \boxed{x \:=\:2(\cos\theta)^{\frac{2}{3}}}$

. . $\left(\frac{y}{2}\right)^3 \:=\:\sin^2\theta \quad\Rightarrow\quad \frac{y}{2} \:=\:(\sin\theta)^{\frac{2}{3}} \quad\Rightarrow\quad \boxed{y \:=\:2(\sin\theta)^{\frac{2}{3}}}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

If we have the form: . $x^n - y^n \:=\:C$

. . force it into the form: . $\sec^2\!\theta - \tan^2\!\theta \:=\:1$

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