1. **State the problem:** We are given parametric equations $x = 5t$ and $y = \sqrt{t}$, and we want to find the second derivative $\frac{d^2y}{dx^2}$ at $t = \frac{1}{9}$. 2. **Recall the formulas:** For parametric curves, the first derivative $\frac{dy}{dx}$ is given by $$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}.$$ The second derivative $\frac{d^2y}{dx^2}$ is given by $$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}.$$ 3. **Calculate derivatives with respect to $t$: ** - $\frac{dx}{dt} = 5$ - $y = t^{1/2}$ so $\frac{dy}{dt} = \frac{1}{2}t^{-1/2} = \frac{1}{2\sqrt{t}}$ 4. **Find $\frac{dy}{dx}$:** $$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{\frac{1}{2\sqrt{t}}}{5} = \frac{1}{10\sqrt{t}}.$$ 5. **Differentiate $\frac{dy}{dx}$ with respect to $t$: ** $$\frac{d}{dt}\left(\frac{dy}{dx}\right) = \frac{d}{dt}\left(\frac{1}{10}t^{-1/2}\right) = \frac{1}{10} \cdot \left(-\frac{1}{2}\right) t^{-3/2} = -\frac{1}{20t^{3/2}}.$$ 6. **Calculate $\frac{d^2y}{dx^2}$:** $$\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}} = \frac{-\frac{1}{20t^{3/2}}}{5} = -\frac{1}{100t^{3/2}}.$$ 7. **Evaluate at $t = \frac{1}{9}$:** $$t^{3/2} = \left(\frac{1}{9}\right)^{3/2} = \left(\frac{1}{9}\right)^{1} \cdot \left(\frac{1}{9}\right)^{1/2} = \frac{1}{9} \cdot \frac{1}{3} = \frac{1}{27}.$$ So, $$\frac{d^2y}{dx^2} = -\frac{1}{100} \cdot \frac{1}{\frac{1}{27}} = -\frac{1}{100} \cdot 27 = -\frac{27}{100} = -0.27.$$ **Final answer:** $$\boxed{-0.27}.$$