Is this a valid parametrisation of a curve?

[rickonstark999]

Hi, so I am learning how to parametrise curves.

For the curve y^2=16x, I have said let x=t^2. Then, we can say y^2=16t^2, so that we can take the root of this and get y=4t. What I wanted to ask was do we have to say "plus or minus" in front of the 4t, or do we just leave it as positive to get a valid parametrisation?

Thanks.

 

[PeroK]

Hi, so I am learning how to parametrise curves.

For the curve y^2=16x, I have said let x=t^2. Then, we can say y^2=16t^2, so that we can take the root of this and get y=4t. What I wanted to ask was do we have to say "plus or minus" in front of the 4t, or do we just leave it as positive to get a valid parametrisation?

Thanks.

The question to ask is what values can $t$ take?

 

[Ray Vickson]

Hi, so I am learning how to parametrise curves.

For the curve y^2=16x, I have said let x=t^2. Then, we can say y^2=16t^2, so that we can take the root of this and get y=4t. What I wanted to ask was do we have to say "plus or minus" in front of the 4t, or do we just leave it as positive to get a valid parametrisation?

Thanks.

Is $y$ allowed to be $<0$ along the curve? Is $t < 0$ allowed?

 

[rickonstark999]

Is $y$ allowed to be $<0$ along the curve? Is $t < 0$ allowed?

I suppose so? The question doesn't specify so I am not sure.

 

[Ray Vickson]

I suppose so? The question doesn't specify so I am not sure.

How would you draw a graph of the relation $x = \frac{1}{16} y^2$?

 

[rickonstark999]

How would you draw a graph of the relation $x = \frac{1}{16} y^2$?

You would have a sideways parabola where y can be any real number, but x cannot be below 0 because you are squaring.

 

[Ray Vickson]

You would have a sideways parabola where y can be any real number, but x cannot be below 0 because you are squaring.

OK, so now what does that tell you about $t$ in your parametrization?

 

[rickonstark999]

OK, so now what does that tell you about $t$ in your parametrization?

From my parametrisations, t can be both positive and negative. Would you say that is correct?

 

[Mark44]

Then, we can say y^2=16t^2, so that we can take the root of this and get y=4t.

No, that isn't right. If $y^2 = 16t^2$, then $y = \pm 4t$.

 

[rickonstark999]

No, that isn't right. If $y^2 = 16t^2$, then $y = \pm 4t$.

Thank you. That is what I was asking in the sentence after that.

 

[LCKurtz]

From my parametrisations, t can be both positive and negative. Would you say that is correct?

No, that isn't right. If $y^2 = 16t^2$, then $y = \pm 4t$.

Thank you. That is what I was asking in the sentence after that.

But @rickonstark999, you do understand that your first post above is correct and your parameterization $x=t^2,~y=4t$ with $t$ both positive and negative, ie $-\infty < t < \infty$ is a correct parameterization of your curve, right? You don't need any $\pm$ sign.