# Thread: Parameterization of a curve.

1. ## Parameterization of a curve.

Hi all,

I have the following problem for homework:

A curve is represented by the two equations$y=x^2, z=x^3$. Parameterize this curve such that your parameter increases from (0, 0, 0) to (1, 1, 1). What is your parameter value for the point (−1, 1, −1)?

How do I start this question?

I tried to make x the subject of both and solved for y.

$x=\sqrt{y}, x=\sqrt[3]{z}$

$y=(\sqrt[3]{z})^{2}$

then made z=t and substituted in getting:
$x=\sqrt[3]{t}, y=(\sqrt[3]{t})^{2}, z=t$

Is this at correct? if not how do I solve such a question?

Thanks.

2. ## Re: Parameterization of a curve.

Hey mcleja.

If you parameterize using t, then the function is (t,t^2,t^3) for x,y,z respectively. If you are using some extra constant to give a specific parameterization to get a particular vector when t = 1, then you need to account for this. (In your case, you don't need to change anything).

What you have is spot on.

3. ## Re: Parameterization of a curve.

Isn't it already parameterized with parameter x?
When x=-1, y=1 and z=-1

4. ## Re: Parameterization of a curve.

Originally Posted by mcleja
Hi all,

I have the following problem for homework:

A curve is represented by the two equations$y=x^2, z=x^3$. Parameterize this curve such that your parameter increases from (0, 0, 0) to (1, 1, 1).
Go back and reread the problem. What you have written makes no sense. The parameter is a number and so cannot "increase from (0, 0, 0) to (1, 1, 1). I suspect they are asking you to find a parameterisation so that when the parameter is 0, the point is (0, 0, 0) and when the parameter is 1, the point is (1, 1, 1).

What is your parameter value for the point (−1, 1, −1)?

How do I start this question?

I tried to make x the subject of both and solved for y.

$x=\sqrt{y}, x=\sqrt[3]{z}$
Yes, what you give is a good parameterisation. Personally, seeing that y and z are given as functions of x, I would have used x itself as parameter: x= t, y= t^2, z= t^3.
$y=(\sqrt[3]{z})^{2}$

then made z=t and substituted in getting:
$x=\sqrt[3]{t}, y=(\sqrt[3]{t})^{2}, z=t$

Is this at correct? if not how do I solve such a question?

Thanks.

5. ## Re: Parameterization of a curve.

Originally Posted by HallsofIvy
Go back and reread the problem. What you have written makes no sense. The parameter is a number and so cannot "increase from (0, 0, 0) to (1, 1, 1). I suspect they are asking you to find a parameterisation so that when the parameter is 0, the point is (0, 0, 0) and when the parameter is 1, the point is (1, 1, 1).
That's what post #3 says.

6. ## Re: Parameterization of a curve.

The question states "Come up with your own parametrisation for this curve, such that your parameter increases as you proceed from the point (0, 0, 0) to (1, 1, 1). What is your parameter value for the point (−1, 1, −1)?"

Have I done the question correctly?

Thanks.

7. ## Re: Parameterization of a curve.

Originally Posted by mcleja
The question states "Come up with your own parametrisation for this curve, such that your parameter increases as you proceed from the point (0, 0, 0) to (1, 1, 1). What is your parameter value for the point (−1, 1, −1)?"

Have I done the question correctly?

Thanks.
Yes. You can make either x, y or z the parameter. x is easiest because it is already done. Setting the one you choose equal to t is unnecessary- it's just renaming the parameter. It changes the variables x,y,z to x,y,t.

8. ## Re: Parameterization of a curve.

So if i parameterise in x i get (x,x^2,x^3). Is this correct? I dont understand what the question means by "What is your parameter value for the point (−1, 1, −1)?"? Does it mean substitute -1 for x?

Thanks

9. ## Re: Parameterization of a curve.

Basically it means you need to solve for the particular value of your parameter.

In my post I used t and in your post above you used x. Basically its asking you that given a point, you give the corresponding value of t (or x).

10. ## Re: Parameterization of a curve.

Would the value of the parameter x be -1? giving (-1,-1^2,-1^3)=(-1,1,-1)?

Thanks.

11. ## Re: Parameterization of a curve.

Yes that is correct.

12. ## Re: Parameterization of a curve.

OK.

Thanks everyone!