Calculate the curvature $k(t)$, for the curve $r(t)=\langle 1t^{-1},-5,3t \rangle$

Question

I have that $k(t)=\frac{\mid r'(t)\times r''(t) \mid}{\mid r'(t)\mid^3}$.

So first, $r'(t)=\langle -\frac{1}{t^2},0,3 \rangle$.

$r''(t)=\langle \frac{2}{t^3},0,0 \rangle$.

$\mid r'(t)\mid = \sqrt{t^{-4}+9}$.

Then I did $r'(t) \times r''(t)$ to get $\langle 0,\frac{6}{t^{3}}, 0 \rangle$ and took the magnitude of this to get $\sqrt{\frac{36}{t^{6}}}$.

I then put that all over $\left(\frac{1}{t^4}+9\right)^{\frac{3}{2}}$

Answer 1

What you have thus far looks correct. I'm not sure what your question is, but putting it all together yields: $$k(t)=\frac{\big \lvert\frac{6}{t^3}\big \rvert}{{\left(t^{-4}+9\right)}^{\frac{3}{2}}}=\frac{6}{\left(1+9t^4\right)^{\frac{3}{2}}}$$