How to find the curvature K of the curve?

$r \left(t\right) = t i + {t}^{2} j + {t}^{2} k$

Apr 14, 2018

See below

Explanation:

Using a vector approach to curvature, $\kappa$:

$\kappa \left(t\right) = \frac{| m a t h b f r ' \times m a t h b f r ' ' |}{| m a t h b f r ' {|}^{3}}$

$m a t h b f r \left(t\right) = {\left[t , {t}^{2} , {t}^{2}\right]}^{T}$

$m a t h b f r ' \left(t\right) = {\left[1 , 2 t , 2 t\right]}^{T}$

$m a t h b f r ' ' \left(t\right) = {\left[0 , 2 , 2\right]}^{T}$

$| m a t h b f r ' \times m a t h b f r ' ' | = | \det \left(\begin{matrix}\hat{x} & \hat{y} & \hat{z} \\ 1 & 2 t & 2 t \\ 0 & 2 & 2\end{matrix}\right) | = | {\left[0 , - 2 , 2\right]}^{T} | = 2 \sqrt{2}$

$| m a t h b f r ' {|}^{3} = {\left(m a t h b f r ' \cdot m a t h b f r '\right)}^{\frac{3}{2}} = {\left(1 + 8 {t}^{2}\right)}^{\frac{3}{2}}$

$\implies \kappa = \frac{2 \sqrt{2}}{{\left(1 + 8 {t}^{2}\right)}^{\frac{3}{2}}}$

Now, if you want to param in terms of the arc length, $s$, because $\kappa \left(s\right) = \frac{d m a t h b f \left(\hat{T} \left(s\right)\right)}{\mathrm{ds}}$, then:

$m a t h b f T \left(t\right) = m a t h b f r ' \left(t\right)$ and so $\hat{m a t h b f T} = \frac{m a t h b f r '}{| m a t h b f r ' |}$

Plus:

$s = {\int}_{\Delta t} \setminus \dot{s} \setminus \mathrm{dt} = {\int}_{\Delta t} \sqrt{m a t h b f r ' \cdot m a t h b f r '} \setminus \mathrm{dt}$

You can work from there.