Trouble finding tangential component for $r=(3t-t^3)i+(3t^2)j$

Question

I have $$r=(3t-t^3)i+(3t^2)j$$ and need to find the tangential component. I got: $$r'=(3-3t^2)i+6tj$$ $$r'*r"= 18t^3+18t$$ $$|r'|=\sqrt{9t^2(2+t^2)}=3t\sqrt{2_t^2}$$ $$\frac{r'*r"}{|r'|}=\frac{6(t^2+1)}{\sqrt{2+2t}}$$

The answer says it should be just 6t.

Answer 1

$$|r'| = \sqrt{(3-3t^2)^2+36t^2} = \sqrt{9t^4+18t^2+9} = 3\sqrt{(t^2+1)^2} = 3(t^2+1)$$