Answer
$$\sqrt{\frac{76}{14^3}}$$
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 $t=1$.
$$
[\kappa]_{t=1}=\frac{\sqrt{76}}{\sqrt{14^3}}=\sqrt{\frac{76}{14^3}}
$$
