1. **State the problem:** Two planes leave the 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. **Define variables:** Let $t$ be the time in hours after 1 pm. - Distance north plane has traveled: $y = 150t$ - Distance west plane has traveled: $x = 200t$ - Distance between planes: $z$ 3. **Relationship between distances:** By the Pythagorean theorem, $$z^2 = x^2 + y^2$$ 4. **Differentiate with respect to time $t$:** $$2z \frac{dz}{dt} = 2x \frac{dx}{dt} + 2y \frac{dy}{dt}$$ Simplify by dividing both sides by 2: $$z \frac{dz}{dt} = x \frac{dx}{dt} + y \frac{dy}{dt}$$ 5. **Substitute known values at $t=2$ hours (3 pm):** - $x = 200 \times 2 = 400$ - $y = 150 \times 2 = 300$ - $\frac{dx}{dt} = 200$ - $\frac{dy}{dt} = 150$ Calculate $z$: $$z = \sqrt{x^2 + y^2} = \sqrt{400^2 + 300^2} = \sqrt{160000 + 90000} = \sqrt{250000} = 500$$ 6. **Plug values into differentiated equation:** $$500 \frac{dz}{dt} = 400 \times 200 + 300 \times 150 = 80000 + 45000 = 125000$$ 7. **Solve for $\frac{dz}{dt}$:** $$\frac{dz}{dt} = \frac{125000}{500} = 250$$ **Answer:** The distance between the planes is increasing at 250 mph at 3 pm.