1. **State the problem:** A puppy's weight $W(t)$ grows proportionally to its current weight over time $t$ (in months). Given $W(0)=2.0$ pounds and $W(2)=3.5$ pounds, find $W(3)$. 2. **Model the problem:** The growth rate proportional to weight means $$\frac{dW}{dt} = kW,$$ where $k$ is a constant. 3. **Solve the differential equation:** Separate variables: $$\frac{dW}{W} = k dt.$$ Integrate both sides: $$\int \frac{1}{W} dW = \int k dt \Rightarrow \ln|W| = kt + C,$$ where $C$ is the integration constant. 4. **Rewrite the solution:** Exponentiate both sides: $$W = e^{kt + C} = e^C e^{kt} = Ce^{kt},$$ where $C = e^C$ is a positive constant. 5. **Use initial condition $W(0)=2.0$:** $$2.0 = Ce^{k \cdot 0} = C \Rightarrow C = 2.0.$$ So, $$W(t) = 2.0 e^{kt}.$$ 6. **Use $W(2) = 3.5$ to find $k$:** $$3.5 = 2.0 e^{2k} \Rightarrow e^{2k} = \frac{3.5}{2.0} = 1.75.$$ Take natural log: $$2k = \ln(1.75) \Rightarrow k = \frac{\ln(1.75)}{2}.$$ 7. **Find $W(3)$:** $$W(3) = 2.0 e^{3k} = 2.0 e^{3 \cdot \frac{\ln(1.75)}{2}} = 2.0 e^{\frac{3}{2} \ln(1.75)} = 2.0 (e^{\ln(1.75)})^{\frac{3}{2}} = 2.0 (1.75)^{1.5}.$$ Calculate: $$(1.75)^{1.5} = 1.75^{1} \times 1.75^{0.5} = 1.75 \times \sqrt{1.75} \approx 1.75 \times 1.3229 = 2.315.$$ So, $$W(3) \approx 2.0 \times 2.315 = 4.63 \text{ pounds}.$$ 8. **Answer:** The puppy weighs approximately 4.6 pounds at 3 months old, which corresponds to choice B. **Summary:** We used the exponential growth model $W(t) = Ce^{kt}$, found constants from given data, and evaluated at $t=3$ months.