Answer to Question #351032 in Differential Geometry | Topology for Anna

Question

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

Answer 1

$$\vec{x}(t)=\langle t,t^2,t^4\rangle$$

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

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

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

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

a.

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

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

b.

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

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

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

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

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

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

c.

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

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

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

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

d.

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

$t=1$

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

e.

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

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

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

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

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

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

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

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

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

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

$t=1$

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

$$=\dfrac{48}{404}=\dfrac{12}{101}$$