Answer: (a) The exact length of the curve is \(4\sqrt{3}\).
(b) \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} \gt 0\) for \(0\lt t\lt 1\).
(a) For a parametric curve, its length is
\(\displaystyle \int_0^3 \sqrt{\left(\frac{\mathrm{d}x}{\mathrm{d}t}\right)^2+\left(\frac{\mathrm{d}y}{\mathrm{d}t}\right)^2}\,\mathrm{d}t.\)
Differentiate:
\(\displaystyle \frac{\mathrm{d}x}{\mathrm{d}t}=\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{2}{3}t^{3/2}-2t^{1/2}\right)=t^{1/2}-t^{-1/2},\qquad \frac{\mathrm{d}y}{\mathrm{d}t}=2.\)
So
\(\displaystyle \left(\frac{\mathrm{d}x}{\mathrm{d}t}\right)^2+\left(\frac{\mathrm{d}y}{\mathrm{d}t}\right)^2=(t^{1/2}-t^{-1/2})^2+4=t+2+t^{-1}=(t^{1/2}+t^{-1/2})^2.\)
Since \(t\gt 0\),
\(\displaystyle \sqrt{\left(\frac{\mathrm{d}x}{\mathrm{d}t}\right)^2+\left(\frac{\mathrm{d}y}{\mathrm{d}t}\right)^2}=t^{1/2}+t^{-1/2}.\)
Hence the length is
\(\displaystyle \int_0^3 (t^{1/2}+t^{-1/2})\,\mathrm{d}t=\left[\frac{2}{3}t^{3/2}+2t^{1/2}\right]_0^3=4\sqrt{3}.\)
Therefore the exact length is \(\boxed{4\sqrt{3}}\).
(b) First,
\(\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\frac{\mathrm{d}y}{\mathrm{d}t}}{\frac{\mathrm{d}x}{\mathrm{d}t}}=\frac{2}{t^{1/2}-t^{-1/2}}.\)
Differentiate with respect to \(t\):
\(\displaystyle \frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{2}{t^{1/2}-t^{-1/2}}\right)=\frac{-2\left(\frac12 t^{-1/2}+\frac12 t^{-3/2}\right)}{(t^{1/2}-t^{-1/2})^2}.\)
Now use
\(\displaystyle \frac{\mathrm{d}^2y}{\mathrm{d}x^2}=\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)\cdot \frac{\mathrm{d}t}{\mathrm{d}x},\)
so
\(\displaystyle \frac{\mathrm{d}^2y}{\mathrm{d}x^2}=\frac{-2\left(\frac12 t^{-1/2}+\frac12 t^{-3/2}\right)}{(t^{1/2}-t^{-1/2})^3}.\)
Simplifying,
\(\displaystyle \frac{\mathrm{d}^2y}{\mathrm{d}x^2}=-\frac{t+1}{(t-1)^3}.\)
We want
\(\displaystyle -\frac{t+1}{(t-1)^3}\gt 0.\)
For \(0\lt t\le 3\), we have \(t+1\gt 0\), so this is positive exactly when \((t-1)^3\lt 0\), i.e. when \(t\lt 1\).
Given the domain, the required set is \(\boxed{0\lt t\lt 1}\).
