Curve in parameter form

**lynch-mob** · June 11th 2008, 12:26 AM

A curve is given in this form

$x=t^3-3t$ , $y=t^2$ and $-2\leq t \leq 2$

Eliminate the t-variable so you get a funktion in the form $x=f(y)$ and draw the curve as accurately as you can!

Good luck!

**topsquark** · June 11th 2008, 03:54 AM

Originally Posted by lynch-mob

A curve is given in this form

$x=t^3-3t$ , $y=t^2$ and $-2\leq t \leq 2$

Eliminate the t-variable so you get a funktion in the form $x=f(y)$ and draw the curve as accurately as you can!

Good luck!

Well, $y = t^2$ so $t = \pm \sqrt{y}$

For the + branch:
$x = y^{3/2} - 3y^{1/2}$

and for the - branch:
$x = -(y^{3/2} - 3y^{1/2})$

Sketch both of these and you'll have your graph.

-Dan

**Mathstud28** · June 11th 2008, 07:11 AM

Originally Posted by lynch-mob

A curve is given in this form

$x=t^3-3t$ , $y=t^2$ and $-2\leq t \leq 2$

Eliminate the t-variable so you get a funktion in the form $x=f(y)$ and draw the curve as accurately as you can!

Good luck!

Just to be more general, given

$y=f(\xi)$ and $x=g(\xi)$ where $\xi$ is some parameter,

Then converting to rectangular coordinates we need to get y as a function of x so we see that

$x=g(\xi)\Rightarrow{\xi=g^{-1}(x)}$

So now making a substituion

$y=f(\xi)\Rightarrow{y=f(g^{-1}(x))}$

**lynch-mob** · June 16th 2008, 02:51 AM

How do I convert the interval values. I mean should I change it to an interval between to x-values or between two y-values. And how do I do it?

**kalagota** · June 16th 2008, 03:22 AM

yes, you have to.. just check what will the values of $x$ or $y$ when you substitute the values $a \leq t \leq b$ to $x(t)$ and/or $y(t)$

in your case, $-2 \leq x \leq 2$

**lynch-mob** · June 16th 2008, 04:04 AM

Thanks. A little problem. I´ve made a table with t,x and y. I have 9 different values for t and get nine different values for x and so on. But the problem is y values. My calculator tells me the graph will look like an S. But the table values for y tells me something else. When t=-2 then x=-2 and y=+4 but it should be y=-4. This goes on and on. I suppose it has something to do with $t=\pm\sqrt{y}$

How do I figure this out?

**kalagota** · June 16th 2008, 04:13 AM

look at what topsquark posted..

**lynch-mob** · June 16th 2008, 04:54 AM

Is this the correct graph? And by the way. Is there a way to draw a graph on a calculator when the function is given as x=f(y)?

Thanks man!

**topsquark** · June 16th 2008, 04:58 AM

Originally Posted by lynch-mob

Is this the correct graph? And by the way. Is there a way to draw a graph on a calculator when the function is given as x=f(y)?

Thanks man!

Edit: I just graphed the original parametric function (and the pair of implicit functions that I solved for) and got a different graph than you did. See if you can find out what went wrong with yours.

As to graphing on a calculator, no most calculators will not graph an implicit function. You might find this graphing program to be useful, however. It will graph implicit functions, as well as a host of other graphing methods.

-Dan

**kalagota** · June 16th 2008, 05:01 AM

i don't have a graphing software here, but i think it is correct..

yes there is.. note that $x = f(y) \Longleftrightarrow f^{-1}(x) = y$

Thread: Curve in parameter form

LinkBack

Thread Tools

Display

Curve in parameter form

Interval values

little problem

is this it?

Similar Math Help Forum Discussions

[SOLVED] Defining tangent line using new parameter to a curve

hypothesis testing: parameter 0 against parameter positive.

Inverse of a six-parameter asymmetric sigmoidal curve

sketch parameter curve

How can you convert from noraml to parameter form?

Search Tags