How twisty is it?

Calculus Level 3

Evaluate the curvature of the point $(0,0,0)$ on a twisted cubic curve $\mathbf{r}(t) = t\ \mathbf{i} + t^2\ \mathbf{j} + t^3\ \mathbf{k}$ .

Details and assumptions

The curvature of a curve given by a vector function r can be found as follows: $\kappa(t) = \frac{|\mathbf{r}^{(1)}(t)\times\mathbf{r}^{(2)}(t)|}{|\mathbf{r}^{(1)}(t)|^3}$

where $\mathbf{r}^{(1)}(t)$ denotes the first derivative of $\mathbf{r}(t)$ , $\mathbf{r}^{(2)}(t)$ denotes the second derivative of $\mathbf{r}(t)$ and $\mathbf{a} \times \mathbf{b}$ denotes the cross product of vectors $\mathbf{a}$ and $\mathbf{b}$ .

The answer is 2.

2 solutions

Ricky Escobar
Dec 17, 2013

First, we note that the point $(0,0,0)$ corresponds to $t=0$ . Now, we find the first derivative of $\mathbf{r}(t)$ : $$\mathbf{r}'(t)=\langle 1,2t,3t^2 \rangle.$$ So, $$\mathbf{r}'(0)=\langle 1,0,0 \rangle.$$ The second derivative of $\mathbf{r}(t)$ is $$\mathbf{r}''(t)=\langle 0,2,6t \rangle.$$ So, $$\mathbf{r}''(0)=\langle 0,2,0 \rangle.$$ Because $\mathbf{r}'(0) \cdot \mathbf{r}''(0) =0$ (i.e., they are orthogonal), the magnitude of their cross product is simply the product of their magnitudes $$|\mathbf{r}'(0) \times \mathbf{r}''(0)| = |\mathbf{r}'(0)| |\mathbf{r}''(0)| = (1)(2) = 2.$$ So, the curvature $$\kappa(0)=\frac{2}{(1)^3}=\fbox{2}.$$

Vipul Panwar
Dec 18, 2013

here r(t)= ti+t^2j+t^3 r'(t)=i+2tj+3t^2k r''(t)=2j+6tk
so(r'(t)xr''(t))= (12t^2-6t^2)i-6tj+2k so |r'(t)xr''(t)|=2 at (0,0,0) similarly |r'(t)|^3=1 so ( |r'(t)xr''(t)| / |r'(t)|^3 ) = 2

How twisty is it?

The answer is 2.

2 solutions

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