1. **State the problem:** Two planes leave an airport at 1 pm. One travels north at 150 mph, the other west at 200 mph. We want to find the rate at which the distance between them is changing at 3 pm. 2. **Set variables:** Let $x(t)$ be the westward distance and $y(t)$ be the northward distance from the airport at time $t$ hours after 1 pm. 3. **Write expressions for distances:** Since the westbound plane travels at 200 mph, $x(t) = 200t$. The northbound plane travels at 150 mph, so $y(t) = 150t$. 4. **Distance between planes:** The distance $z(t)$ between the planes is given by the Pythagorean theorem: $$z(t) = \sqrt{x(t)^2 + y(t)^2} = \sqrt{(200t)^2 + (150t)^2} = \sqrt{40000t^2 + 22500t^2} = \sqrt{62500t^2} = 250t$$ 5. **Differentiate distance with respect to time:** To find the rate of change of distance, differentiate $z(t)$: $$\frac{dz}{dt} = \frac{d}{dt}(250t) = 250$$ 6. **Evaluate at 3 pm:** Since $t$ is hours after 1 pm, at 3 pm, $t=2$. The rate of change of distance is constant at 250 mph. **Final answer:** The distance between the two planes is increasing at a rate of $250$ mph at 3 pm.