Simulation · Maths · Class 12

How fast a ripple's area grows

From the lesson Rate of change in Application of Derivatives. Change the values and watch what happens.

The idea behind it

NCERT §6.1–6.2

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.

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