Question

for the curve x=a(3u-u³), y=3au², z=a(3u+u³) prove that k=t

Answer 1

$r(u)=\langle a(3u-u^3),\ \ 3au^2,\ \ a(3u+u^3)\rangle$

$r^\prime (u)=\langle 3a(1-u^2),\ \ 6au,\ \ 3a(1+u^2)\rangle$

$r^{\prime \prime}(u)=\langle -6au,\ \ 6a,\ \ 6au\rangle$

$r^{\prime\prime\prime}(u)=\langle -6a,\ \ 0,\ \ 6a\rangle$

$r^\prime \times r^{\prime \prime }=\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k} \\ 3a(1-u^2)&6au&3a(1+u^2) \\ -6au&6a&6au \end{vmatrix}=18a^2(u^2-1)\mathbf{i}-36a^2u\mathbf{j}+18a^2(u^2+1)\mathbf{k}$

$|r^\prime\times r^{\prime \prime}|=18a^2\sqrt{(u^2-1)^2+(-2u)^2+(u^2+1)^2}=18\sqrt{2}a^2(u^2+1)$

$|r^\prime|=3a\sqrt{(1-u^2)^2+(2u)^2+(1+u^2)^2}=3\sqrt{2}a(u^2+1)$

$(r^\prime \times r^{\prime \prime})\cdot r^{\prime\prime\prime}=\begin{vmatrix} 3a(1-u^2)&6au&3a(1+u^2) \\ -6au&6a&6au\\ -6a&0&6a \end{vmatrix}=216a^3$

$\kappa =\frac{|r^\prime \times r^{\prime \prime }|}{|r^\prime|^3}=\frac{18\sqrt{2}a^2(u^2+1)}{54\sqrt{2}a^3(u^2+1)^3}=\frac{1}{3a(u^2+1)^2}$

$\tau =\frac{(r^\prime \times r^{\prime \prime })\cdot r^{\prime \prime\prime}}{|r^\prime \times r^{\prime \prime}|^2}=\frac{216a^3}{648a^4(u^2+1)^2}=\frac{1}{3a(u^2+1)^2}$

So, $\kappa =\tau$ .