Question #351032

Calculate T ; N; B; κ\kappa ; τ\tau of the curve x(t) = (t; t 2 ; t 4) at the point (1; 1; 1).


1
Expert's answer
2022-06-26T17:19:42-0400
x(t)=t,t2,t4\vec{x}(t)=\langle t,t^2,t^4\rangle

x(t)=1,2t,4t3\vec{x}'(t)=\langle 1,2t,4t^3\rangle

x(t)=(1)2+(2t)2+(4t3)2|\vec{x}'(t)|=\sqrt{(1)^2+(2t)^2+(4t^3)^2}

x(1)=1,2,4\vec{x}'(1)=\langle 1,2,4\rangle

x(1)=12+22+42=21|\vec{x}'(1)|=\sqrt{1^2+2^2+4^2}=\sqrt{21}

a.


T(t)=x(t)x(t)\vec{T}(t)=\dfrac{\vec{x}'(t)}{|\vec{x}'(t)|}

T(1)=x(1)x(1)=121,221,421\vec{T}(1)=\dfrac{\vec{x}'(1)}{|\vec{x}'(1)|}=\langle \dfrac{1}{\sqrt{21}},\dfrac{2}{\sqrt{21}},\dfrac{4}{\sqrt{21}}\rangle



b.


N(t)=T(t)T(t)\vec{N}(t)=\dfrac{\vec{T}'(t)}{|\vec{T}'(t)|}

T(t)=0,2,12t2\vec{T}'(t)=\langle 0,2,12t^2\rangle

T(t)=(0)2+(2)2+(12t2)2|\vec{T}'(t)|=\sqrt{(0)^2+(2)^2+(12t^2)^2}

T(1)=0,2,12\vec{T}'(1)=\langle 0,2,12\rangle

T(1)=(0)2+(2)2+(12)2=237|\vec{T}'(1)|=\sqrt{(0)^2+(2)^2+(12)^2}=2\sqrt{37}

N(1)=T(1)T(1)=0,137,637\vec{N}(1)=\dfrac{\vec{T}'(1)}{|\vec{T}'(1)|}=\langle 0,\dfrac{1}{\sqrt{37}},\dfrac{6}{\sqrt{37}}\rangle

c.


B(t)=T(t)×N(t)\vec{B}(t)=\vec{T}(t)\times \vec{N}(t)

B(1)=T(1)×N(1)=ijk1212214210137637\vec{B}(1)=\vec{T}(1)\times \vec{N}(1)=\begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ \dfrac{1}{\sqrt{21}} & \dfrac{2}{\sqrt{21}} & \dfrac{4}{\sqrt{21}} \\ 0 & \dfrac{1}{\sqrt{37}} & \dfrac{6}{\sqrt{37}} \end{vmatrix}

=8777i6777j+1777k= \dfrac{8}{\sqrt{777}}\vec{i}- \dfrac{6}{\sqrt{777}}\vec{j} + \dfrac{1}{\sqrt{777}}\vec{k}

B(1)=8777,6777,1777\vec{B}(1)=\langle \dfrac{8}{\sqrt{777}},- \dfrac{6}{\sqrt{777}}, \dfrac{1}{\sqrt{777}}\rangle

d.


κ=T(t)x(t)\kappa=\dfrac{|\vec{T}'(t)|}{|\vec{x}'(t)|}

t=1t=1

κ=T(1)x(1)=23721\kappa=\dfrac{|\vec{T}'(1)|}{|\vec{x}'(1)|}=\dfrac{2\sqrt{37}}{\sqrt{21}}


e.


x(t)=0,2,12t2\vec{x}''(t)=\langle 0,2,12t^2\rangle

x(1)=0,2,12\vec{x}''(1)=\langle 0,2,12\rangle

x(1)×x(1)=ijk1240212\vec{x'}(1)\times \vec{x''}(1)=\begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 1 & 2& 4\\ 0 & 2 & 12 \end{vmatrix}

=16i12j+2k=16\vec{i}- 12\vec{j} +2\vec{k}

x(1)×x(1)=16,12,2\vec{x'}(1)\times \vec{x''}(1)=\langle 16,-12,2\rangle

x(1)×x(1)=(16)+(12)2+(2)2=2101|\vec{x'}(1)\times \vec{x''}(1)|=\sqrt{(16)^+(-12)^2+(2)^2}=2\sqrt{101}



x(t)=0,0,24t\vec{x}'''(t)=\langle 0,0,24t\rangle

x(1)=0,0,24\vec{x}'''(1)=\langle 0,0,24\rangle

(x(1)×x(1))x(1)=00+48=48(\vec{x'}(1)\times \vec{x''}(1))\cdot\vec{x}'''(1)=0-0+48=48

τ=(x(t)×x(t))x(t)x(t)×x(t)2\tau=\dfrac{(\vec{x'}(t)\times \vec{x''}(t))\cdot\vec{x}'''(t)}{|\vec{x'}(t)\times \vec{x''}(t)|^2}

t=1t=1


τ=(x(1)×x(1))x(1)x(1)×x(1)2\tau=\dfrac{(\vec{x'}(1)\times \vec{x''}(1))\cdot\vec{x}'''(1)}{|\vec{x'}(1)\times \vec{x''}(1)|^2}

=48404=12101=\dfrac{48}{404}=\dfrac{12}{101}

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