14.5 curvature and normal vectors\nnot started\ncreated less than a minute ago · last updated less than a…

Explanation:

Step1: Recall curvature formula

The formula for the curvature $\kappa(t)$ of 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 $\mathbf{r}'(t)=\langle1,5e^{5t}, - 5e^{-5t}\rangle$ and $\mathbf{r}''(t)=\langle0,25e^{5t},25e^{-5t}\rangle$.

Step2: Calculate the cross - product

$\mathbf{r}'(t)\times\mathbf{r}''(t)=\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\1&5e^{5t}&- 5e^{-5t}\0&25e^{5t}&25e^{-5t}\end{vmatrix}=\mathbf{i}(125e^{0}+125e^{0})-\mathbf{j}(25e^{-5t}-0)+\mathbf{k}(25e^{5t}-0)=\langle250,-25e^{-5t},25e^{5t}\rangle$.

Step3: Calculate the magnitudes

When $t = 0$, $\mathbf{r}'(0)=\langle1,5, - 5\rangle$, so $\left|\mathbf{r}'(0)\right|=\sqrt{1 + 25+25}=\sqrt{51}$. $\mathbf{r}'(0)\times\mathbf{r}''(0)=\langle250,-25,25\rangle$, so $\left|\mathbf{r}'(0)\times\mathbf{r}''(0)\right|=\sqrt{250^{2}+(-25)^{2}+25^{2}}=\sqrt{62500 + 625+625}=\sqrt{63750}=25\sqrt{102}$.

Step4: Calculate the curvature at $t = 0$

$\kappa(0)=\frac{\left|\mathbf{r}'(0)\times\mathbf{r}''(0)\right|}{\left|\mathbf{r}'(0)\right|^{3}}=\frac{25\sqrt{102}}{(\sqrt{51})^{3}}=\frac{25\sqrt{102}}{51\sqrt{51}}=\frac{25\sqrt{2\times51}}{51\sqrt{51}}=\frac{25\sqrt{2}}{51}$.

Answer:

$\frac{25\sqrt{2}}{51}$