Answer
$$\frac{3}{\sqrt{10}}$$
$$y=\frac{3}{\sqrt{10}} t+\frac{1}{\sqrt{10}}$$
Work Step by Step
Let $f(t)=\sqrt{t^{2}+1} .$ Then
\begin{aligned}
f^{\prime}(3) &=\lim _{h \rightarrow 0} \frac{f(3+h)-f(3)}{h}\\
&=\lim _{h \rightarrow 0} \frac{\sqrt{10+6 h+h^{2}}-\sqrt{10}}{h} \\
&=\lim _{h \rightarrow 0} \frac{\sqrt{10+6 h+h^{2}}-\sqrt{10}}{h} \cdot \frac{\sqrt{10+6 h+h^{2}}+\sqrt{10}}{\sqrt{10+6 h+h^{2}}+\sqrt{10}} \\
&=\lim _{h \rightarrow 0} \frac{6 h+h^{2}}{h(\sqrt{10+6 h+h^{2}}+\sqrt{10})}\\
&=\lim _{h \rightarrow 0} \frac{6+h}{\sqrt{10+6 h+h^{2}}+\sqrt{10}}=\frac{3}{\sqrt{10}}
\end{aligned}
The tangent line at $a=3$ is
\begin{aligned}
y&=f^{\prime}(3)(t-3)+f(3)\\
&=\frac{3}{\sqrt{10}}(t-3)+\sqrt{10}\\
&=\frac{3}{\sqrt{10}} t+\frac{1}{\sqrt{10}}
\end{aligned}
