# Are the two curve being equivalent....

• Nov 4th 2011, 04:01 PM
Are the two curve being equivalent....
dear all, are the following two curves
$\displaystyle r(t)=(t+\sqrt{3}t,2\cos t,\sqrt{3}t-\sin t)$
and
$\displaystyle \tilde r(t)=(2\cos\frac{t}{2},2\sin\frac{t}{2},-t)$
being equivalent?

Here, equivalence means that the two curves coincide after a rigid motion.

I find the second curve has curvature $\displaystyle 1/4$, and torsion $\displaystyle -1/4$. However, it is difficult to solve the first one, and seems the result is not the same as the second one.

3x.
• Nov 6th 2011, 04:01 PM
Drexel28
Re: Are the two curve being equivalent....
Quote:

Originally Posted by xinglongdada
dear all, are the following two curves
$\displaystyle r(t)=(t+\sqrt{3}t,2\cos t,\sqrt{3}t-\sin t)$
and
$\displaystyle \tilde r(t)=(2\cos\frac{t}{2},2\sin\frac{t}{2},-t)$
being equivalent?

Here, equivalence means that the two curves coincide after a rigid motion.

I find the second curve has curvature $\displaystyle 1/4$, and torsion $\displaystyle -1/4$. However, it is difficult to solve the first one, and seems the result is not the same as the second one.

3x.

This is the correct approach, so how did you try to calculate the torsion of the first one? Don't tell me you just didn't feel like doing the formula.
• Nov 8th 2011, 12:02 AM