Answer to Question #331413 in Differential Geometry | Topology for rimsha

Question #331413

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

1
Expert's answer
2022-04-22T02:38:05-0400

r(u)=a(3uu3),  3au2,  a(3u+u3)r(u)=\langle a(3u-u^3),\ \ 3au^2,\ \ a(3u+u^3)\rangle

r(u)=3a(1u2),  6au,  3a(1+u2)r^\prime (u)=\langle 3a(1-u^2),\ \ 6au,\ \ 3a(1+u^2)\rangle

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

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




r×r=ijk3a(1u2)6au3a(1+u2)6au6a6au=18a2(u21)i36a2uj+18a2(u2+1)kr^\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×r=18a2(u21)2+(2u)2+(u2+1)2=182a2(u2+1)|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=3a(1u2)2+(2u)2+(1+u2)2=32a(u2+1)|r^\prime|=3a\sqrt{(1-u^2)^2+(2u)^2+(1+u^2)^2}=3\sqrt{2}a(u^2+1)


(r×r)r=3a(1u2)6au3a(1+u2)6au6a6au6a06a=216a3(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




κ=r×rr3=182a2(u2+1)542a3(u2+1)3=13a(u2+1)2\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}


τ=(r×r)rr×r2=216a3648a4(u2+1)2=13a(u2+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, κ=τ\kappa =\tau .


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