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# Lecture 16: February 28, 2023 ## Example 1 Consider the curve $r = \langle \cos t, \sin t, t \rangle$, $0 \leq t \leq 4\pi$ 1. Find the unit tangent vector $\mathbf{T}(t)$. 2. Find the normal vector $\mathbf{N}(t)$. 3. Find the binormal vector $\mathbf{B}(t)$. 4. Compute the curvature $\kap...

# Lecture 16: February 28, 2023 ## Example 1 Consider the curve $r = \langle \cos t, \sin t, t \rangle$, $0 \leq t \leq 4\pi$ 1. Find the unit tangent vector $\mathbf{T}(t)$. 2. Find the normal vector $\mathbf{N}(t)$. 3. Find the binormal vector $\mathbf{B}(t)$. 4. Compute the curvature $\kappa(t)$. ### Solution 1. $\mathbf{r}'(t) = \langle -\sin t, \cos t, 1 \rangle$ $\| \mathbf{r}'(t) \| = \sqrt{(-\sin t)^2 + (\cos t)^2 + 1^2} = \sqrt{\sin^2 t + \cos^2 t + 1} = \sqrt{1 + 1} = \sqrt{2}$ $\mathbf{T}(t) = \frac{\mathbf{r}'(t)}{\| \mathbf{r}'(t) \|} = \frac{1}{\sqrt{2}} \langle -\sin t, \cos t, 1 \rangle = \langle -\frac{\sin t}{\sqrt{2}}, \frac{\cos t}{\sqrt{2}}, \frac{1}{\sqrt{2}} \rangle$ 2. $\mathbf{T}'(t) = \langle -\frac{\cos t}{\sqrt{2}}, -\frac{\sin t}{\sqrt{2}}, 0 \rangle$ $\| \mathbf{T}'(t) \| = \sqrt{\frac{\cos^2 t}{2} + \frac{\sin^2 t}{2}} = \sqrt{\frac{1}{2} (\cos^2 t + \sin^2 t)} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$ $\mathbf{N}(t) = \frac{\mathbf{T}'(t)}{\| \mathbf{T}'(t) \|} = \frac{\langle -\frac{\cos t}{\sqrt{2}}, -\frac{\sin t}{\sqrt{2}}, 0 \rangle}{\frac{1}{\sqrt{2}}} = \langle -\cos t, -\sin t, 0 \rangle$ 3. $\mathbf{B}(t) = \mathbf{T}(t) \times \mathbf{N}(t) = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -\frac{\sin t}{\sqrt{2}} & \frac{\cos t}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ -\cos t & -\sin t & 0 \end{vmatrix} = \mathbf{i} \begin{vmatrix} \frac{\cos t}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ -\sin t & 0 \end{vmatrix} - \mathbf{j} \begin{vmatrix} -\frac{\sin t}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ -\cos t & 0 \end{vmatrix} + \mathbf{k} \begin{vmatrix} -\frac{\sin t}{\sqrt{2}} & \frac{\cos t}{\sqrt{2}} \\ -\cos t & -\sin t \end{vmatrix}$ $= \mathbf{i} \left( 0 - (-\frac{\sin t}{\sqrt{2}}) \right) - \mathbf{j} \left( 0 - (-\frac{\cos t}{\sqrt{2}}) \right) + \mathbf{k} \left( \frac{\sin^2 t}{\sqrt{2}} - (-\frac{\cos^2 t}{\sqrt{2}}) \right)$ $= \langle \frac{\sin t}{\sqrt{2}}, -\frac{\cos t}{\sqrt{2}}, \frac{\sin^2 t + \cos^2 t}{\sqrt{2}} \rangle = \langle \frac{\sin t}{\sqrt{2}}, -\frac{\cos t}{\sqrt{2}}, \frac{1}{\sqrt{2}} \rangle$ 4. $\kappa(t) = \frac{\| \mathbf{T}'(t) \|}{\| \mathbf{r}'(t) \|} = \frac{\frac{1}{\sqrt{2}}}{\sqrt{2}} = \frac{1}{2}$ ## Curvature Formula $\kappa = \frac{\| \mathbf{r}'(t) \times \mathbf{r}''(t) \|}{\| \mathbf{r}'(t) \|^3}$ ### Example 2 $\mathbf{r}(t) = \langle t, t^2, t^3 \rangle$ at $(0, 0, 0)$ ### Solution $\mathbf{r}'(t) = \langle 1, 2t, 3t^2 \rangle \implies \mathbf{r}'(0) = \langle 1, 0, 0 \rangle$ $\mathbf{r}''(t) = \langle 0, 2, 6t \rangle \implies \mathbf{r}''(0) = \langle 0, 2, 0 \rangle$ $\mathbf{r}'(0) \times \mathbf{r}''(0) = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 0 & 0 \\ 0 & 2 & 0 \end{vmatrix} = \langle 0, 0, 2 \rangle$ $\| \mathbf{r}'(0) \times \mathbf{r}''(0) \| = \sqrt{0^2 + 0^2 + 2^2} = 2$ $\| \mathbf{r}'(0) \| = \sqrt{1^2 + 0^2 + 0^2} = 1$ $\kappa(0) = \frac{\| \mathbf{r}'(0) \times \mathbf{r}''(0) \|}{\| \mathbf{r}'(0) \|^3} = \frac{2}{1^3} = 2$