Do different parametrizations of the same curve in R³ give the same tangent vector?

[ayalam]

Do different parametrizations of the same curve in R^3 result in identical tangent vectors at a given point on the same curve? Example may be helpful.

 

[benorin]

Here are two different parameterizations of the twisted cubic (a space curve):

$$r_1(t)=\left< t,t^2,t^3\right> ,\mbox{ for } 0\leq t\leq 2$$
and $$r_2(u)=\left< e^u,e^{2u},e^{3u}\right> ,\mbox{ for } 0\leq u\leq \ln 2$$

their parameters are related by $t=e^u$.

Try it.

 

[ayalam]

I got for r(t) r'(t)=<1,2t,3t^2> and for r'(0)=<1,0,0>

But for r'(u)=<e^u,2e^2u,3e^3u> and r'(0)=<1,2,3>


So i would assume the answer would be no?

 

[benorin]

Sorry, I made a typo above: it should read

$$r_1(t)=\left< t,t^2,t^3\right> ,\mbox{ for } 1\leq t\leq 2$$

Since t=0 and u=0 generate different points on the curve in R^3, which are, $r_1(0)=\left< 0,0,0\right>$, and $r_2(0)=\left< 1,1,1\right>$, they shouldn't have the same tangent vectors.

But when evaluated at the same point on the curve, they should be the same, that is

$$r_1^{\prime}(t)=\left< 1,2t,3t^2\right> ,\mbox{ for } 1\leq t\leq 2$$
and $$r_2^{\prime}(u)=\left< e^u,2e^{2u},3e^{3u}\right> ,\mbox{ for } 0\leq u\leq \ln 2$$

Notice that the points generated by the parameterizations are related such that:
$$r_1\left( e^u\right) =r_2(u) /mbox{ and } r_2\left( \ln(t) \right) =r_1(t)$$

and hence the relations:$$r_1^{\prime}\left( e^u\right)e^u =r_2^{\prime}(u) /mbox{ and } r_2^{\prime}\left( \ln(t) \right) \frac{1}{t} =r_1^{\prime}(t)$$ hold for the ranges $1\leq t\leq 2, 0\leq u\leq \ln 2$

So, I suppose the answer to your question is: no, they're not the same, but there does exist some relation between them (these two, that is). I think... no, I can't figure it out. Different parameterizations of a curve, still describe the same curve, right? Hmmm. Geometry: try parameterizing w.r.t. arc length along the curve, see calculus book: try Stewart.

 

[benorin]

Sorry, I made a typo above: it should read

$$r_1(t)=\left< t,t^2,t^3\right>$$ for 1 $$\leq t\leq 2$$

Since t=0 and u=0 generate different points on the curve in R^3, which are, $r_1(0)=\left< 0,0,0\right>$, and $r_2(0)=\left< 1,1,1\right>$, they shouldn't have the same tangent vectors.

But when evaluated at the same point on the curve, they should be the same, that is

$$r_1^{\prime}(t)=\left< 1,2t,3t^2\right> ,\mbox{ for } 1\leq t\leq 2$$
and $$r_2^{\prime}(u)=\left< e^u,2e^{2u},3e^{3u}\right> ,\mbox{ for } 0\leq u\leq \ln 2$$

Notice that the points generated by the parameterizations are related such that:

$$r_1\left( e^u\right) =r_2(u) /mbox{ and } r_2\left( \ln(t) \right) =r_1(t)$$

and hence the relations:

$$r_1^{\prime}\left( e^u\right)e^u =r_2^{\prime}(u) \mbox{ and } r_2^{\prime}\left( \ln(t) \right) \frac{1}{t} =r_1^{\prime}(t)$$ hold for the ranges $1\leq t\leq 2, 0\leq u\leq \ln 2$

So, I suppose the answer to your question is: no, they're not the same, but there does exist some relation between them (these two, that is). I think... no, I can't figure it out. Different parameterizations of a curve, still describe the same curve, right? Hmmm. Geometry: try parameterizing w.r.t. arc length along the curve, see calculus book: try Stewart.

 

[HallsofIvy]

One way of thinking about this is "dynamically". Think of an object moving along the curve with t equal to the time it is at each point. The "tangent" vector would then be the velocity vector of the object. Different parametrizations can be interpreted as objects moving along the same curve at different speeds. The tangent vector, for different parametrizations, will, of course, always point in the same direction but its length would be the speed. The length of the tangent vector may change for different parametrizations. If you want to be completely "geometric" you should use arclength as parametrization. That way the tangent vector always has length 1.