Rate of Change Problems
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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.