Answer: The arc length is \(\displaystyle \int_0^3 \sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}\,dt\).

Here \(\frac{dx}{dt}=\mathrm{e}^t-1\) and \(\frac{dy}{dt}=2\mathrm{e}^{t/2}\), so

\(\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2=(\mathrm{e}^t-1)^2+4\mathrm{e}^t=(\mathrm{e}^t+1)^2\).

For \(0\le t\le 3\), this gives \(\sqrt{(\mathrm{e}^t+1)^2}=\mathrm{e}^t+1\), so the length is

\(\displaystyle \int_0^3 (\mathrm{e}^t+1)\,dt = \big[\mathrm{e}^t+t\big]_0^3 = (\mathrm{e}^3+3)-(1+0)=\mathrm{e}^3+2\).

Answer: \(\mathrm{e}^3+2\)

For a curve given parametrically by \(x=x(t)\) and \(y=y(t)\), the arc length from \(t=a\) to \(t=b\) is

\(\displaystyle L=\int_a^b \sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}\,dt\).

Here

\(x=\mathrm{e}^t-t\), so \(\displaystyle \frac{dx}{dt}=\mathrm{e}^t-1\),

and

\(y=4\mathrm{e}^{t/2}\), so \(\displaystyle \frac{dy}{dt}=4\cdot \frac12 \mathrm{e}^{t/2}=2\mathrm{e}^{t/2}\).

Hence

\(\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2=(\mathrm{e}^t-1)^2+(2\mathrm{e}^{t/2})^2\).

Expand and simplify:

\(\begin{aligned} (\mathrm{e}^t-1)^2+(2\mathrm{e}^{t/2})^2 &= \mathrm{e}^{2t}-2\mathrm{e}^t+1+4\mathrm{e}^t \\ &= \mathrm{e}^{2t}+2\mathrm{e}^t+1 \\ &=(\mathrm{e}^t+1)^2. \end{aligned}\)

Therefore the speed is

\(\sqrt{(\mathrm{e}^t+1)^2}=\mathrm{e}^t+1\)

since \(\mathrm{e}^t+1>0\) for all \(t\).

So the required arc length is

\(\begin{aligned} L&=\int_0^3 (\mathrm{e}^t+1)\,dt \\ &= \left[\mathrm{e}^t+t\right]_0^3 \\ &=(\mathrm{e}^3+3)-(1+0) \\ &=\mathrm{e}^3+2. \end{aligned}\)

So the length of the arc is \(\boxed{\mathrm{e}^3+2}\).