# Chapter 13 - Vector Functions - 13.3 Arc Length and Curvature - 13.3 Exercises: 25

$$\sqrt{\frac{76}{14^3}}$$

#### Work Step by Step

Given that $\mathbf{r}(t)=\langle{t, t^2, t^3}\rangle$, evaluate curvature $\kappa$ at $P(1, 1, 1)$ 1) Determine value of $t$ using given point $P(1, 1, 1)$. Since $t$ (the x-component of $\mathbf{r}(t)$) $=1$(the x-component of point $P$) at $t=1$, $1$ is the value of $t$. Check by verifying that $t^2=1$ at $t=1$ and that $t^3=1$ at $t=1$. 2) Find $\mathbf{r}'(t)$ by differentiating $\mathbf{r}(t)$ using power rule. $$\mathbf{r}'(t)=\frac{d(\langle{t, t^2, t^3}\rangle)}{dt}=\langle{1, 2t, 3t^2}\rangle$$ 3) Find $\mathbf{r}''(t)$ by differentiating $\mathbf{r}'(t)$ using power rule. $$\mathbf{r}''(t)=\frac{d(\langle{1, 2t, 3t^2}\rangle)}{dt}=\langle{0, 2, 6t}\rangle$$ 4) Substitute $t=1$ into $\mathbf{r}'(t)$ and $\mathbf{r}''(t)$ to find $\mathbf{r}'(1)$ and $\mathbf{r}''(1)$ $$\mathbf{r}'(1)=\langle{1, 2, 3}\rangle\\ \mathbf{r}''(1)=\langle{0, 2, 6}\rangle$$ 5) Find $\mathbf{r}'(1)\times\mathbf{r}''(1)$ using determinant notation. $$\mathbf{r}'(1)\times\mathbf{r}''(1)=\left|\begin{matrix} \mathbf{\hat{i}} & \mathbf{\hat{j}} & \mathbf{\hat{k}}\\ 1 & 2 & 3 \\ 0 & 2 & 6 \end{matrix}\right|\\ =\left|\begin{matrix} 2 & 3 \\ 2 & 6 \end{matrix}\right|\mathbf{\hat{i}}- \left|\begin{matrix} 1 & 3 \\ 0 & 6 \end{matrix}\right|\mathbf{\hat{j}}+ \left|\begin{matrix} 1 & 2 \\ 0 & 2 \end{matrix}\right|\mathbf{\hat{k}}\\ =[(2)(6)-(3)(2)]\mathbf{\hat{i}}-[(1)(6)-(3)(0)]\mathbf{\hat{j}}+[(1)(2)-(2)(0)]\mathbf{\hat{k}}\\ =6\mathbf{\hat{i}}-6\mathbf{\hat{j}}+2\mathbf{\hat{k}}=\langle{6, -6, 2}\rangle$$ 6) Find magnitude $|\mathbf{r}'(1)\times\mathbf{r}''(1)|$ of $\mathbf{r}'(1)\times\mathbf{r}''(1)$ $$|\mathbf{r}'(1)\times\mathbf{r}''(1)|=|\langle{6, -6, 2}\rangle|\\ =\sqrt{6^2+(-6)^2+2^2}=\sqrt{76}\\ \therefore |\mathbf{r}'(1)\times\mathbf{r}''(1)|=\sqrt{76}$$ 7) Find and cube magnitude $|\mathbf{r}'(1)|$ of $\mathbf{r}'(1)$. $$|\mathbf{r}'(1)|=|\langle{1, 2, 3}\rangle|\\ |\mathbf{r}'(1)|=\sqrt{1^2+2^2+3^2}=\sqrt{14}\\ |\mathbf{r}'(1)|^3=\sqrt{14^3}$$ 8) Recall that curvature $\kappa$ is defined as $$[\kappa]_t=\frac{|\mathbf{r}'(t)\times\mathbf{r}''(t)|}{|\mathbf{r}'(t)|^3}$$ 9) Substitute values of $|\mathbf{r}'(1)\times\mathbf{r}''(1)|$ and $|\mathbf{r}(1)|^3$ into curvature formula to find $[\kappa]_{t=1}$, the curvature of $\mathbf{r}(t)$ at $t=1$. $$[\kappa]_{t=1}=\frac{\sqrt{76}}{\sqrt{14^3}}=\sqrt{\frac{76}{14^3}}$$

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