Calculate the curvature and torsion of
$$x= e^t\sin(t),\quad y= e^t\cos(t),\quad z= e^t$$
I'm not sure if I am doing this correctly since I am getting quite complicated results.
But I understand that curvature $$\kappa= \frac{\left|X' \times X"\right|}{v^3}$$ and $$\tau=\frac{(X' \times X")\cdot X'''}{v^3\kappa^2}.$$
Please confirm that these formulas are correct and how to use them properly.
Thank you!
Best Answer
In these formulas, we hav that $X = (x(t), y(t), z(t))$. Therefore, we begin by calculating the derivatives of the vectors component-wise with respect to $t$ and have
$$X' = (x'(t), y'(t),z'(t)) = \left( e^t \sin(t) + e^t \cos(t) , e^t \cos(t) - e^t \sin(t), e^t\right)$$ $$X^{\prime\prime} = \left(2e^t \cos(t) , - 2e^t \sin(t), e^t\right)$$ $$X^{\prime\prime\prime} = \left( 2e^t \cos(t) - 2e^t \sin(t), -2e^t \sin(t) - 2e^t \cos(t) , e^t\right)$$
To compute the curvature, we start by computing
\begin{align} \left| X' \times X'' \right| & = \left| (e^{2 t}\cos(t)+e^{2 t} \sin(t), e^{2 t} \cos(t)-e^{2 t} \sin(t), -2 e^{2 t} \cos^2(t)-2 e^{2 t} \sin^2(t)) \right| \\ &= \sqrt{6}e^{2t}\end{align}
Also, $$v^3 = \left| X' \right|^3 = \sqrt{(e^t \sin(t) + e^t \cos(t))^2 + (e^t \cos(t) - e^t \sin(t))^2 + (e^t)^2}^3 = 3\sqrt{3}e^{3t}$$
Therefore we have that the curvature is given by $$\kappa = \frac{\left| X' \times X'' \right| }{v^3 } = \frac{\sqrt{6}e^{2t}}{3\sqrt{3}e^{3t}} = \frac{\sqrt{2}}{3}e^{-t}$$
For the torsion, we have a few more calculations. Now
$$ (X' \times X^{\prime\prime} ) \cdot X^{\prime\prime\prime} = (e^{2 t}\cos(t)+e^{2 t} \sin(t), e^{2 t} \cos(t)-e^{2 t} \sin(t), -2 e^{2 t} \cos^2(t)-2 e^{2 t} \sin^2(t)) \cdot ( 2e^t \cos(t) - 2e^t \sin(t), -2e^t \sin(t) - 2e^t \cos(t) , e^t ) = -2e^{3t}$$
Also
$$v^3 \kappa^2 = 3\sqrt{3}e^{3t} \left( \frac{\sqrt{2}}{3}e^{-t} \right)^2 = \frac{2\sqrt{3}}{3} e^t$$
Therefore the torsion is $$\tau = \frac{-2e^{3t}}{\frac{2\sqrt{3}}{3} e^t} = -\sqrt{3}e^{2t}$$