Answer to Question #134210 in Differential Geometry | Topology for Promise Omiponle

Question #134210
Find the unit tangent T, the unit normal N and the curvature K for the curve r(t) =<2 sint, 5t, 2 cost>.
1
Expert's answer
2020-09-24T17:04:39-0400

r(t)={2sint,5t,2cost}.\vec{r}(t) = \lbrace 2\sin t, 5t, 2\cos t \rbrace.

Let us take derivatives with respect to t:

r(t)={2cost,5,2sint},    r(t)={2sint,0,2cost}.\vec{r}'(t) = \lbrace 2\cos t, 5, -2\sin t \rbrace, \;\; \vec{r}''(t) = \lbrace -2\sin t, 0, -2\cos t \rbrace.


The tangent is τ(t)=r(t)={2cost,5,2sint}.\vec{\tau}(t) = \vec{r}'(t) = \lbrace 2\cos t, 5, -2\sin t \rbrace. We should normalize this vector (norm is 4cos2t+25+4sin2t=29\sqrt{4\cos^2t + 25 + 4\sin^2t} = \sqrt{29} ), so the unit tangent is T={2cost29,529,2sint29}\vec{T} =\left \lbrace \dfrac{2\cos t}{\sqrt{29}}, \dfrac{5}{\sqrt{29}}, \dfrac{-2\sin t}{\sqrt{29}} \right\rbrace .


The normal is β(t)=r(t)×(r(t)×r(t)).\vec{\beta}(t) = \vec{r}'(t)\times (\vec{r}'(t)\times\vec{r}''(t) ) .

r(t)×r(t)=ijk2cost52sint2sint02cost={10cost,4,10sint}.\vec{r}'(t)\times\vec{r}''(t) = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 2\cos t & 5 & -2\sin t \\ -2\sin t & 0 & -2\cos t \end{vmatrix} = \lbrace -10\cos t, 4 , 10\sin t \rbrace .

r(t)×(r(t)×r(t))=ijk2cost52sint10cost410sint={58cost,0,58sint}.\vec{r}'(t)\times (\vec{r}'(t)\times\vec{r}''(t) ) = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 2\cos t & 5 & -2\sin t \\ -10\cos t & 4 & 10\sin t \end{vmatrix} = \lbrace 58\cos t, 0, 58\sin t \rbrace.

The unit normal is N={cost,0,sint}.\vec{N} = \lbrace \cos t, 0, \sin t \rbrace.


The curvature is k=r(t)×r(t)r(t)3=116293=229.k = \dfrac{||\vec{r}'(t)\times\vec{r}''(t)||}{||\vec{r}'(t)||^3} = \dfrac{\sqrt{116}}{\sqrt{29}^3} = \dfrac{2}{29}.


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