r ⃗ ( t ) = { 2 sin t , 5 t , 2 cos t } . \vec{r}(t) = \lbrace 2\sin t, 5t, 2\cos t \rbrace. r ( t ) = { 2 sin t , 5 t , 2 cos t } .
Let us take derivatives with respect to t:
r ⃗ ′ ( t ) = { 2 cos t , 5 , − 2 sin t } , r ⃗ ′ ′ ( t ) = { − 2 sin t , 0 , − 2 cos t } . \vec{r}'(t) = \lbrace 2\cos t, 5, -2\sin t \rbrace, \;\; \vec{r}''(t) = \lbrace -2\sin t, 0, -2\cos t \rbrace. r ′ ( t ) = { 2 cos t , 5 , − 2 sin t } , r ′′ ( t ) = { − 2 sin t , 0 , − 2 cos t } .
The tangent is τ ⃗ ( t ) = r ⃗ ′ ( t ) = { 2 cos t , 5 , − 2 sin t } . \vec{\tau}(t) = \vec{r}'(t) = \lbrace 2\cos t, 5, -2\sin t \rbrace. τ ( t ) = r ′ ( t ) = { 2 cos t , 5 , − 2 sin t } . We should normalize this vector (norm is 4 cos 2 t + 25 + 4 sin 2 t = 29 \sqrt{4\cos^2t + 25 + 4\sin^2t} = \sqrt{29} 4 cos 2 t + 25 + 4 sin 2 t = 29 ), so the unit tangent is T ⃗ = { 2 cos t 29 , 5 29 , − 2 sin t 29 } \vec{T} =\left \lbrace \dfrac{2\cos t}{\sqrt{29}}, \dfrac{5}{\sqrt{29}}, \dfrac{-2\sin t}{\sqrt{29}} \right\rbrace T = { 29 2 cos t , 29 5 , 29 − 2 sin t } .
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) ) . β ( t ) = r ′ ( t ) × ( r ′ ( t ) × r ′′ ( t )) .
r ⃗ ′ ( t ) × r ⃗ ′ ′ ( t ) = ∣ i ⃗ j ⃗ k ⃗ 2 cos t 5 − 2 sin t − 2 sin t 0 − 2 cos t ∣ = { − 10 cos t , 4 , 10 sin t } . \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 ) = ∣ ∣ i 2 cos t − 2 sin t j 5 0 k − 2 sin t − 2 cos t ∣ ∣ = { − 10 cos t , 4 , 10 sin t } .
r ⃗ ′ ( t ) × ( r ⃗ ′ ( t ) × r ⃗ ′ ′ ( t ) ) = ∣ i ⃗ j ⃗ k ⃗ 2 cos t 5 − 2 sin t − 10 cos t 4 10 sin t ∣ = { 58 cos t , 0 , 58 sin t } . \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. r ′ ( t ) × ( r ′ ( t ) × r ′′ ( t )) = ∣ ∣ i 2 cos t − 10 cos t j 5 4 k − 2 sin t 10 sin t ∣ ∣ = { 58 cos t , 0 , 58 sin t } .
The unit normal is N ⃗ = { cos t , 0 , sin t } . \vec{N} = \lbrace \cos t, 0, \sin t \rbrace. N = { cos t , 0 , sin t } .
The curvature is k = ∣ ∣ r ⃗ ′ ( t ) × r ⃗ ′ ′ ( t ) ∣ ∣ ∣ ∣ r ⃗ ′ ( t ) ∣ ∣ 3 = 116 29 3 = 2 29 . k = \dfrac{||\vec{r}'(t)\times\vec{r}''(t)||}{||\vec{r}'(t)||^3} = \dfrac{\sqrt{116}}{\sqrt{29}^3} = \dfrac{2}{29}. k = ∣∣ r ′ ( t ) ∣ ∣ 3 ∣∣ r ′ ( t ) × r ′′ ( t ) ∣∣ = 29 3 116 = 29 2 .
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