# Rate of Change Problems > Rate of change problems for JEE Mathematics: related rates for sphere, cone, cube and ladder setups, the differentiate-then-substitute rule and sign traps. - Canonical URL: https://prepelephant.com/topics/jee/mathematics/rate-of-change-problems - Exam / course: JEE · Subject: Mathematics - Publisher: PrepElephant (https://prepelephant.com) — Prepared and reviewed by the PrepElephant Academic Review Team - First published: 2026-10-02 - Last updated: 2026-10-02 - How to cite: "Rate of Change Problems", PrepElephant, https://prepelephant.com/topics/jee/mathematics/rate-of-change-problems ## Direct answer Rates arrive chained: air pumped into a balloon feeds the radius, the radius feeds the surface area, and the chain rule connects the visible rate to the wanted one as dV/dt = (dV/dr)(dr/dt). The method is uniform — write the geometric relation between the quantities, differentiate it with respect to time t, and only then substitute the instantaneous values. The workhorses: for a sphere, dV/dt = 4πr^2 dr/dt and dA/dt = 8πr dr/dt; for a cube, dV/dt = 3a^2 da/dt; for a sliding ladder, x(dx/dt) + y(dy/dt) = 0 from x^2 + y^2 = L^2. Differentiate first, substitute last — in that order, always. ## What you must remember - **Sphere:** V = (4/3)πr^3 gives dV/dt = 4πr^2 dr/dt; A = 4πr^2 gives dA/dt = 8πr dr/dt. - **Cube:** V = a^3 gives dV/dt = 3a^2 da/dt — at edge 10 cm growing at 3 cm/s, the volume grows at 900 cubic cm/s. - **Circle:** A = πr^2 gives dA/dt = 2πr dr/dt; the area's rate depends on the current radius, not the radius's rate alone. - **Cone:** V = (1/3)πr^2h; when the cone's semi-vertical angle is fixed, r and h stay proportional by similar triangles, converting one rate into the other. - **Ladder:** x^2 + y^2 = L^2 differentiated gives x(dx/dt) + y(dy/dt) = 0; as the foot slides out at a steady rate, the top accelerates downward. - **Sign discipline:** shrinking quantities carry negative rates; answers frequently hinge on the sign, not the magnitude. - **Order of operations:** differentiate with symbols and substitute numbers at the instant — substituting early freezes the chain and produces wrong answers. ## A balloon and its skin, differentiated Air enters a spherical balloon at 100 cubic centimetres per second; find how fast the radius and the surface area grow at the instant r = 5 cm. Differentiate V = (4/3)πr^3 with respect to time: dV/dt = 4πr^2 dr/dt. Substitute the instant: 100 = 4π(25) dr/dt, so dr/dt = 1/π centimetres per second. Now the chain the question really wants: A = 4πr^2, so dA/dt = 8πr dr/dt = 8π × 5 × (1/π) = 40 square centimetres per second. Two observations worth carrying forward: the π cancelled only because the substitution was made after differentiation (the chain remained intact), and the surface rate needed the radius rate as an intermediate — no formula connects dA/dt to dV/dt without passing through r. Any related-rates problem in JEE Main is this skeleton wearing different geometry: a cone filling with water (similar triangles link r to h), a stone dropped (link via s = ut + at^2/2), a shadow lengthening (similar triangles again). ## Where marks leak JEE Main owns this chapter numerically, and the losses are procedural rather than conceptual. First, substitution before differentiation: putting r = 5 into V = (4/3)πr^3 before differentiating yields d/dt of a constant — zero — and a confidently wrong answer. Second, sign blindness: a melting snowball has dV/dt negative, and the question asking "how fast is the radius decreasing" wants the positive magnitude of a negative dr/dt; options carry both signs. Third, the cone's hidden step: forgetting that r and h are proportional in a fixed-angle cone, and differentiating as if they were independent. Units deserve a glance too — centimetres cubed per second for volumes, centimetres per second for lengths — because a mis-scaled answer often survives a candidate's own checking. ## Frequently asked questions ### How fast does the radius grow when air enters at 100 cm^3/s and r = 5 cm? dr/dt = 100/(4π × 25) = 1/π cm/s, from dV/dt = 4πr^2 dr/dt. ### Why must numbers be substituted after differentiating? Early substitution freezes the geometric relation into a constant, killing the chain that connects the rates. ### How is the sliding ladder problem set up? From x^2 + y^2 = L^2, differentiate to get x(dx/dt) + y(dy/dt) = 0 and solve with the current x and y. ### What is the area rate of an expanding circle? dA/dt = 2πr dr/dt — it depends on both the instantaneous radius and its rate. ### Why do cone problems need similar triangles? In a cone of fixed angle, r and h remain proportional, so one rate determines the other before the volume formula is differentiated.