Question

Evaluate the Curvature at the given point: r(t) = <1/t,1/t^2,t^2> t=-1

Answer 1

\[\kappa(t)=\frac{\left| \dot{\vec{r}}(t) \times \ddot{\vec{r}}(t)\right|}{\left| \dot{\vec{r}}(t)\right|^3}\]
\[\text{ with } \dot{\vec{r}}(t) = \frac{d \vec{r}(t)}{dt}\]

Answer 2

You take those derivatives and find that ratio for t = -1 and you have your curvature.

Answer 3

$$\kappa(t)=\frac{\|\mathbf{r}'(t)\times\mathbf{r}''(t)\|}{\|r'(t)\|^3}$$At our point we have:$$\mathbf{r}'(t)=\left(-\frac1{t^2},-\frac2{t^3},2t\right)\quad\implies\mathbf{r}'(-1)=(-1,2,-2)\\\mathbf{r}''(t)=\left(\frac2{t^3},\frac6{t^4},2\right)\quad\quad\quad\implies \mathbf{r}''(-1)=(-2,6,2)\\\mathbf{r}'(t)\times\mathbf{r}''(t)=(4+12,4+2,-6+4)=(16,6,-2)$$... so we compute curvature:$$\kappa(-1)=\frac{\sqrt{16^2+6^2+2^2}}{\sqrt{1^2+2^2+2^2}^3}=\frac{2\sqrt{8^2+3^2+1^2}}{3^3}=\frac{2\sqrt{74}}{27}$$

Answer 4

thank you @oldrin.bataku ! Clears it up so well!