Finding the signed curvature of $(t, t^3-3t)$.

Question

I'm looking to find the signed curvature of $(t, t^3-3t)$. Right now, I have evaluated the following:

The unit tangent vector is given by $\boldsymbol{T} = \left(\dfrac{1}{\sqrt{9t^4 - 18t^2 + 10}}, \dfrac{3t^2 - 3}{\sqrt{9t^4 - 18t^2 + 10}}\right)$

The unit normal vector is given by $\boldsymbol{N} = \left(-\dfrac{3t^2 - 3}{\sqrt{9t^4 - 18t^2 + 10}}, \dfrac{1}{\sqrt{9t^4 - 18t^2 + 10}}\right)$

But from here, it seems incredibly painful to continue working with these to find the signed curvature. Have I made a mistake? Is there an easier way to do this? What do I do from here?

Answer 1

Can you use this formula:

$k=\frac{x'y''-x''y'}{(x'^2+y'^2)^{\frac{3}{2}}}$

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