14.5 curvature and normal vectors not started created less than a minute ago · last updated less than a…

Explanation:

Step1: Recall curvature formula

The curvature formula for a vector - valued function $\mathbf{r}(t)=\langle x(t),y(t),z(t)\rangle$ is $\kappa(t)=\frac{\left|\mathbf{r}'(t)\times\mathbf{r}''(t)\right|}{\left|\mathbf{r}'(t)\right|^{3}}$. First, we have $\mathbf{r}'(t)=\langle - 2,4,12t\rangle$ and $\mathbf{r}''(t)=\langle0,0,12\rangle$.

Step2: Calculate the cross - product $\mathbf{r}'(t)\times\mathbf{r}''(t)$

The cross - product of two vectors $\mathbf{a}=\langle a_1,a_2,a_3\rangle$ and $\mathbf{b}=\langle b_1,b_2,b_3\rangle$ is given by $\mathbf{a}\times\mathbf{b}=\langle a_2b_3 - a_3b_2,a_3b_1 - a_1b_3,a_1b_2 - a_2b_1\rangle$. For $\mathbf{r}'(t)=\langle - 2,4,12t\rangle$ and $\mathbf{r}''(t)=\langle0,0,12\rangle$, we have: [ \begin{align*} \mathbf{r}'(t)\times\mathbf{r}''(t)&=\langle4\times12-12t\times0,12t\times0 - (-2)\times12,(-2)\times0 - 4\times0\rangle\ &=\langle48,24,0\rangle \end{align*} ]

Step3: Calculate the magnitudes

The magnitude of a vector $\mathbf{v}=\langle v_1,v_2,v_3\rangle$ is $\left|\mathbf{v}\right|=\sqrt{v_1^{2}+v_2^{2}+v_3^{2}}$. $\left|\mathbf{r}'(t)\right|=\sqrt{(-2)^{2}+4^{2}+(12t)^{2}}=\sqrt{4 + 16+144t^{2}}=\sqrt{20 + 144t^{2}}$. $\left|\mathbf{r}'(t)\times\mathbf{r}''(t)\right|=\sqrt{48^{2}+24^{2}+0^{2}}=\sqrt{2304 + 576}=\sqrt{2880}=24\sqrt{5}$.

Step4: Evaluate at $t = 1$

When $t = 1$, $\left|\mathbf{r}'(1)\right|=\sqrt{20+144}=\sqrt{164}=2\sqrt{41}$. Then $\kappa(1)=\frac{\left|\mathbf{r}'(1)\times\mathbf{r}''(1)\right|}{\left|\mathbf{r}'(1)\right|^{3}}=\frac{24\sqrt{5}}{(2\sqrt{41})^{3}}=\frac{24\sqrt{5}}{8\times41\sqrt{41}}=\frac{3\sqrt{5}}{41\sqrt{41}}=\frac{3\sqrt{205}}{1681}$.

Answer:

$\frac{3\sqrt{205}}{1681}$