Q.A man of height 1.5 meters walks toward a lamp post of height 4.5 meters, at the rate of $\dfrac34$ meter/sec. Find the rate at which

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 wi …