Q.A ladder $5\ \text{m}$ ($500\ \text{cm}$) long is leaning against a vertical wall. The bottom of the ladder is being pulled away from the wall along the ground at the rate of $2\ \text{cm/s}$. Find the rate at which the height of the ladder on the wall is decreasing when the foot of the ladder is $4\ \text{m}$ ($400\ \text{cm}$) from the wall.

Many quantities in geometry and physics are functions of time (or of some other varying quantity), and the derivative measures how fast one quantity changes as another changes. If $y=f(x)$ and both $x$ and $y$ vary with time $t$, then $dy/dt=(dy/dx)\cdot(dx/dt)$ by the chain rule — this lets us relate the rate of change of $y$ to the rate of change of $x$ once we know how $y$ depends on $x$. Typical situations combine a geometric formula (area of a circle, volume of a sphere or cone, area of a triangle, etc.) with a given rate for one variable to find the rate of another variable at a specific instant. For problems involving similar triangles (a lamp-post and a walking person's shadow, a ladder sliding down a wall), first set up an algebraic relation between the two changing lengths using similar triangles or the Pythagorean theorem, then differentiate that relation with respect to time before substituting the known numerical rate and the instant in question. Velocity is the rate of change of displacement with time, acceleration is the rate of change of velocity (the second derivative of displacement), and jerk is the rate of change of acceleration (the third derivative) — each one differentiates the previous quantity with respect to time.