1. **Stating the problem:** We are given the function describing the depth of fluid in a tank as a function of time $t$ hours: $$y = 5 \left(2 - \frac{t}{15}\right)^2$$ We need to find the rate of change of the depth with respect to time, i.e., $\frac{dy}{dt}$. 2. **Formula used:** To find the rate of change, we differentiate $y$ with respect to $t$ using the chain rule. 3. **Differentiation steps:** Let $u = 2 - \frac{t}{15}$, so $y = 5u^2$. Then, $$\frac{dy}{dt} = 5 \cdot 2u \cdot \frac{du}{dt} = 10u \cdot \frac{du}{dt}$$ Since $u = 2 - \frac{t}{15}$, $$\frac{du}{dt} = -\frac{1}{15}$$ Therefore, $$\frac{dy}{dt} = 10 \left(2 - \frac{t}{15}\right) \left(-\frac{1}{15}\right) = -\frac{10}{15} \left(2 - \frac{t}{15}\right) = -\frac{2}{3} \left(2 - \frac{t}{15}\right)$$ 4. **Simplify the expression:** $$\frac{dy}{dt} = -\frac{2}{3} \left(2 - \frac{t}{15}\right) = -\frac{2}{3} \cdot 2 + \frac{2}{3} \cdot \frac{t}{15} = -\frac{4}{3} + \frac{2t}{45}$$ 5. **Interpretation:** The rate of change of the fluid depth decreases initially and then changes linearly with time. **Final answer:** $$\boxed{\frac{dy}{dt} = -\frac{4}{3} + \frac{2t}{45}}$$