Lesson 1 of 12 · 10 min

Rate of change

NCERT §6.1–6.2

Kavya drops a washer into the rooftop water tank and a ring of ripples spreads out. The ring's radius grows steadily, yet the patch of disturbed water seems to spread faster and faster. Is that an illusion?

The story this chapter follows: Kavya's rooftop tin-box workshop

Kavya cuts tin sheets into open boxes on the roof of her building and sells them. Her ripple-filled water tank, the lamp she walks past, her costs and profits and the sheets she folds give every rate, rise and fall, peak and best choice in this chapter.

The lesson in notes

In short

If y = f(x), then dy/dx is the rate at which y changes per unit change in x, and its value at x = x₀ is the rate at that instant.

Worked example: the area of a circle is A = πr², so dA/dr = 2πr. When r = 5 cm, the area is growing at 10π cm² for each extra centimetre of radius.

When both x and y depend on time t, the chain rule links their rates: dy/dt = (dy/dx)·(dx/dt). Equally, dy/dx = (dy/dt)/(dx/dt) whenever dx/dt ≠ 0.

A positive rate means the quantity is increasing at that moment and a negative rate means it is decreasing. The units are the unit of y per unit of t.

Worked example: a stone makes circular ripples whose radius grows at 4 cm/s. With A = πr², dA/dt = 2πr·dr/dt, so at r = 10 cm the area grows at 2π × 10 × 4 = 80π cm²/s.

Worked example: a disc's radius grows at 0.05 cm/s. At r = 3.2 cm, dA/dt = 2π × 3.2 × 0.05 = 0.32π cm²/s.

Worked example: a particle moves with x = t²(2 − t/3). Its velocity is v = 4t − t², which is zero at t = 4; by then it has covered x(4) = 16 × 2/3 = 32/3 m.