1. Let's start with a classic related rates problem: "A balloon is rising vertically at a rate of 5 meters per second. A person is walking away from the balloon's launch point at 3 meters per second. How fast is the distance between the person and the balloon increasing when the person is 4 meters away from the launch point and the balloon is 10 meters high?" 2. To solve this, we use the Pythagorean theorem to relate the distances: $$s^2 = x^2 + y^2$$ where $s$ is the distance between the person and the balloon, $x$ is the horizontal distance of the person from the launch point, and $y$ is the height of the balloon. 3. Differentiate both sides with respect to time $t$ to find the rate of change of $s$: $$2s \frac{ds}{dt} = 2x \frac{dx}{dt} + 2y \frac{dy}{dt}$$ 4. Simplify by dividing both sides by 2: $$s \frac{ds}{dt} = x \frac{dx}{dt} + y \frac{dy}{dt}$$ 5. Calculate $s$ at the given moment using the Pythagorean theorem: $$s = \sqrt{x^2 + y^2} = \sqrt{4^2 + 10^2} = \sqrt{16 + 100} = \sqrt{116}$$ 6. Plug in the known values: $x=4$, $\frac{dx}{dt}=3$, $y=10$, $\frac{dy}{dt}=5$, and $s=\sqrt{116}$. 7. Substitute into the differentiated equation: $$\sqrt{116} \frac{ds}{dt} = 4 \times 3 + 10 \times 5 = 12 + 50 = 62$$ 8. Solve for $\frac{ds}{dt}$: $$\frac{ds}{dt} = \frac{62}{\sqrt{116}} = \frac{62}{2\sqrt{29}} = \frac{31}{\sqrt{29}}$$ meters per second. 9. Therefore, the distance between the person and the balloon is increasing at a rate of $$\frac{31}{\sqrt{29}}$$ meters per second at that instant.