I have $r(t)= (3t + 4\sin(t) + 4\cos(t))$ and I am tying to use the formula $$\frac{ |r' \times r''|}{|r'|^3}$$
I got $$r'= 3 + 4\cos(t) - 4\sin(t) \ \text{and} \ \ r'' = 0 - 4\sin(t) - 4\cos(t)$$
I did the cross product of those and came up with $$-32\cos(t) + 12\cos(t) - 12\sin(t) $$ Then when get the magnitude of the that cross product I came up with $\sqrt{(-32\cos(t))^2 + 144}$ and that's as far as I got.
The answer is suposed to be $4/25$ but I certainly don't see anyway to get from what I got to that so I don't know if I'm missing something at this point or messes up something earlier, I double checked my derivatives with an online calculator and didn't find a problem there.
The problem is that $r(t)$ is a vector in $\mathbb R^3$. So it should be written $r(t) = (3t, 4 \sin t, 4 \cos t)$.
Now if $r(t) = (3t, 4 \sin t, 4 \cos t)$ then $r'(t) = (3, 4 \cos t, -4\sin t)$ and $r''(t) = (0, -4\sin t, -4 \cos t)$. Thus
$$r' \times r'' = \begin{vmatrix}i & j & k \\ 3 & 4 \cos t & -4 \sin t \\ 0 & -4 \sin t & -4 \cos t \end{vmatrix} = -16i - (-12 \cos t) j + (- 12 \sin t) k$$
then $|r' \times r''| = \sqrt{ 256 + 144 (\cos^2 t + \sin^2 t) } = 20$. On the other hand $|r'| = \sqrt {9 + 16} = 5$. Conclusion
$$\frac{|r'\times r''|}{|r'|^3} = \frac{20}{125} =\color{red}{ \frac{4}{25}}$$