Find the curvature and torsion of the curve x=acos2t, y=asin2t,z=2asint.

Question

I have tried to use the formula of curvature=|r'×r''|/|r'|^3 but the calculation is too long. Same for torsion too. Then I tried another approach as calculating unit tangent vector by using dr/ds. Then curvature×n=dt/ds but it is also very long and calculative. Provide the calculations too please

Answer 1

Given the curve $\gamma(t)=\left(\cos (2 t),\sin (2 t),2 \sin (t)\right)$, we have the derivatives.

$ \gamma'(t)=\left(-2 a \sin (2 t),2 a \cos (2 t),2 a \cos (t)\right)\\ \gamma''(t)=\left(-4 a \cos (2 t),-4 a \sin (2 t),-2 a \sin (t)\right)\\ \gamma'''(t)=\left(8 a \sin (2 t),-8 a \cos (2 t),-2 a \cos (t)\right) $

$$\kappa(t) =\frac {\|{\boldsymbol {\gamma }}'\times {\boldsymbol {\gamma }}''\|}{\|{\boldsymbol {\gamma }}'\|^{3}}$$

$$\gamma '\times \gamma ''=\left(2 a^2 (3 \sin (t)+\sin (3 t)),-8 a^2 \cos ^3(t),8 a^2\right)$$

$$\|{\boldsymbol {\gamma }}'\times {\boldsymbol {\gamma }}''\|=2 \sqrt{2} a^2 \sqrt{3 \cos (2 t)+13}$$

$$\|{\boldsymbol {\gamma }}'\|^{3}=2 \sqrt{2} \left(a^2 (\cos (2 t)+3)\right)^{3/2}$$ $$\kappa=\frac{\sqrt{3 \cos (2 t)+13}}{a (\cos (2 t)+3)^{3/2}}$$

$$\tau ={\frac {\det \left(\gamma',\gamma'',\gamma'''\right)}{\left\|\gamma'\times\gamma''\right\|^{2}}}$$

$$\det\left(\gamma',\gamma'',\gamma'''\right)=\det\left| \begin{array}{lll} -2 a \sin (2 t) & 2 a \cos (2 t) & 2 a \cos (t) \\ -4 a \cos (2 t) & -4 a \sin (2 t) & -2 a \sin (t) \\ 8 a \sin (2 t) & -8 a \cos (2 t) & -2 a \cos (t) \\ \end{array} \right| =48 a^3 \cos (t)$$

$$\left\|\gamma'\times\gamma''\right\|^{2}=8 a^4 (3 \cos (2 t)+13)$$

$$\tau=\frac{6 \cos (t)}{3 a \cos (2 t)+13 a}$$