Parametric speed

We have $\textbf{s}=\begin{pmatrix} { t }^{ 2 } \\ 6\ln { t } \end{pmatrix}$ , so $\textbf{v}=\cfrac { d \textbf{s} }{ dt } =\begin{pmatrix} 2t \\ { 6 }/{ t } \end{pmatrix}$ .

At $t=2$ , this gives us $\textbf{v}=\begin{pmatrix} 4 \\ 3 \end{pmatrix}$ , thus the speed is $\sqrt { { 4 }^{ 2 }+{ 3 }^{ 2 } } =\large \color{#20A900}{\boxed{5}}$ .

$\frac {dx}{dt}=2t\\ \frac {dy}{dt}=\frac {6}{t}\\ \frac {ds}{dt}=\sqrt {(\frac {dx}{dt})^2+(\frac {dy}{dt})^2}\\=\sqrt {(2t)^2+(\frac {6}{t})^2}\\=\sqrt {4^2+3^2}\\=\sqrt {25}\\=5$