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I'm calculating the curvature k(t) of the function

$$\mathbf r(t) = \langle \cos(2t), -\sin(2t), 4t\rangle.$$

The formula I'm using is

$$k(t) = \frac{||\mathbf r'(t) \times\mathbf r''(t)|| }{ ||\mathbf r'(t)||^3}.$$

Here's my work:

\begin{align*} \mathbf r'(t) &= \langle -2\sin(2t), -2 \cos(2t), 4\rangle;\\ \mathbf r''(t) &= \langle-4\cos(2t), 4\sin(2t), 0\rangle \end{align*}

Therefore:

$$\mathbf r'(t)\times\mathbf r''(t) = (0 - 16\sin(2t))\mathbf i - (0 + 16\cos(2t))\mathbf j + (-8)\mathbf k$$

(Note that I got $-8\mathbf k$ using the Pythagorean identity.)

So \begin{align*} ||\mathbf r'(t)\times\mathbf r''(t)|| &= \sqrt{16^2 + 8^2}\qquad \text{(Pythagorean again)}\\ &= 64\sqrt 5. \end{align*}

Then, I got \begin{align*} ||\mathbf r'(t)|| &= \sqrt{4 + 4}\qquad \text{(also Pythagorean identity)}\\ &= 2\sqrt 2, \end{align*} so $||\mathbf r'(t)||^3 = 2(\sqrt 2)^3 = 16\sqrt 2.$

So finally, I have \begin{align*} k(t)&= \frac{64\sqrt 5}{16\sqrt 2}\\ \implies \;\;k(t)&= 2\sqrt{10}. \end{align*}

Thoughts? I probably just misdifferentiated somewhere or took the cross-product incorrectly. Thanks!

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$\textbf{r}'(t)\times\textbf{r}''(t)=-16\sin(2t)\textbf{i}-16\cos(2t)\textbf{j}-8\textbf{k}$

$\left|\textbf{r}'(t)\times\textbf{r}''(t)\right|=\sqrt{16^2+8^2}=8\sqrt{5}$

$\left|\textbf{r}'(t)\right|=\sqrt{2^2+4^2}=2\sqrt{5}$

$\displaystyle\therefore \kappa=\frac{8\sqrt{5}}{40\sqrt{5}}=\frac{1}{5}$

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$$ \sqrt{16^2 + 8^2} = 8 \sqrt{5} \text{.} $$

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In fact, $r'(t)=\bigl(-2\sin(2t),2\cos(2t),4\bigr)$ and $r''(t)=\bigl(-4\cos(2t),-4\sin(2t),0\bigr)$. Therefore, $r'(t)\times r''(t)=\bigl(16\sin(2t),-16\cos(2t),8\bigr)$ and $\bigl\|r'(t)\times r''(t)\bigr\|=\sqrt{320}=8\sqrt5$. On the other hand, $\bigl\|r'(t)\bigr\|=2\sqrt5$. So, $k(t)=\frac15$.

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