ELI5: two quantities are tied together by an equation; if one changes over time, the other is forced to change too. The trick: differentiate the relationship with respect to time, plug in the instant you care about, and solve for the rate you don't know.
A 10-ft ladder leans on a wall. Its base is yanked outward at 1 ft/s. When the base is 6 ft out (so the top is 8 ft up), how fast is the top sliding down? The legs obey x² + y² = 10²; differentiating in time gives x·x′ + y·y′ = 0, so y′ = −x·x′ / y.
xlad=6 base distance from the wall (ft) ylad=8 height of the top up the wall (ft) xrate=1 base slides out at 1 ft/s yrate=ylad−xlad⋅xrate = −0.7500 top's rate: −x·x′/y = −0.75 ft/s (downward) ✓ pass abs(xlad⋅xrate+ylad⋅yrate)<10−9 satisfies the differentiated relation x·x′ + y·y′ = 0 Real-world hook: related rates are everywhere motion couples to size — a lengthening shadow as you walk from a lamppost, the rising level in a filling tank, the closing distance between two cars at an intersection.
Try it yourself: a spherical balloon inflates at dV/dt = 36π cm³/s. Its volume is V = (4/3)π r³, so dV/dt = 4π r²·(dr/dt). How fast is the radius growing when r = 3 cm?
rbal=3 radius at this instant (cm) dVdt=36⋅π = 113.0973 inflation rate (cm³/s) drdt=1 ✏️ Your turn: replace 0 with dr/dt (cm/s) at r = 3. The check verifies 4π r²·(dr/dt) = dV/dt — so you must invert it yourself. ✓ pass abs(4⋅π⋅rbal2⋅drdt−dVdt)<0.000001 green when your radius-rate reproduces the inflation rate