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

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 $$\left(\frac{1}{t^4}+9\right)^{\frac{3}{2}}$$

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