calculus - Where did I go wrong? Calculating $k(t)$ for $mathbf r(t) = langlecos(2t), -sin(2t), 4trangle$.

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!