Question #278191

Find




the Curvature and




torsion of the Curves r = (u, (1 + u)/u, (1 - u ^ 2)/u)

1
Expert's answer
2021-12-13T17:58:06-0500

The curvature of the curve given by the vector function r\bm r is


k(u)=∣r′(u)×r′′(u)∣∣r′(u)∣3k(u)=\dfrac{|\bm r'(u)\times\bm r''(u)|}{|\bm r'(u)|^3}

r′(u)=⟨1,u−(1+u)u2,−2u2−(1−u2)u2⟩\bm r'(u)=\langle 1 , \dfrac{u-(1+u)}{u^2}, \dfrac{-2u^2-(1-u^2)}{u^2}\rangle

=⟨1,−1u2,−(1+1u2⟩=\langle 1 , -\dfrac{1}{u^2}, -(1+\dfrac{1}{u^2}\rangle

r′′(u)=⟨0,2u3,2u3⟩\bm r''(u)=\langle 0 , \dfrac{2}{u^3}, \dfrac{2}{u^3}\rangle

r′(u)×r′′(u)=∣ijk1−1/u2−1−1/u202/u32/u3∣\bm r'(u)\times\bm r''(u)=\begin{vmatrix} \bm i & \bm j & \bm k \\ 1 & -1/u^2 & -1-1/u^2 \\ 0 & 2/u^3 & 2/u^3 \\ \end{vmatrix}

=(−2/u5+2/u3+2/u5)i−(2/u3)j+()2/u3k=(-2/u^5+2/u^3+2/u^5)\bm i-(2/u^3)\bm j+()2/u^3\bm k

=(2/u3)i−(2/u3)j+(2/u3)k=(2/u^3)\bm i-(2/u^3)\bm j+(2/u^3)\bm k

∣r′(u)∣=1+1/u4+1+2/u2+1/u4|\bm r'(u)|=\sqrt{1+1/u^4+1+2/u^2+1/u^4}

=2u2u4+u2+1=\dfrac{\sqrt{2}}{u^2}\sqrt{u^4+u^2+1}


∣r′(u)×r′′(u)∣=6u2∣u∣|\bm r'(u)\times\bm r''(u)|=\dfrac{\sqrt{6}}{u^2|u|}

k(u)=6u2∣u∣⋅u622(u4+u2+1)3/2k(u)=\dfrac{\sqrt{6}}{u^2|u|}\cdot\dfrac{u^6}{2\sqrt{2}(u^4+u^2+1)^{3/2}}

k(u)=3u2∣u∣2(u4+u2+1)3/2k(u)=\dfrac{\sqrt{3}u^2|u|}{2(u^4+u^2+1)^{3/2}}

The torsion of the curve given by the vector function r\bm r is


τ(u)=(r′(u)×r′′(u))⋅r′′′(u)∣r′(u)×r′′(u)∣2\tau(u)=\dfrac{(\bm r'(u)\times\bm r''(u))\cdot\bm r'''(u)}{|\bm r'(u)\times\bm r''(u)|^2}

r′′′(u)=⟨0,−6u4,−6u4⟩\bm r'''(u)=\langle 0 , -\dfrac{6}{u^4}, -\dfrac{6}{u^4}\rangle

(r′(u)×r′′(u))⋅r′′′(u)=0+12/u7−12/u7=0(\bm r'(u)\times\bm r''(u))\cdot\bm r'''(u)=0+12/u^7-12/u^7=0

Ï„(u)=0\tau(u)=0


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS