How fast a ripple's area grows
"How fast a ripple's area grows" is a free, interactive Class 12 Maths simulation for JEE, built on the lesson "Rate of change" in Application of Derivatives. The circle's radius is r and it grows at dr/dt. The orange ring is the area it gains in the next tenth of a second. A bigger circle has a longer rim, so the same growth in r adds more area: dA/dt = 2πr · dr/dt. Rates are linked through the chain rule: dA/dt = (dA/dr)(dr/dt). Write the relation between the quantities first, differentiate it with respect to t, and substitute the instant's values last. JEE Main asked 47 questions from this chapter across the 42 shifts Lumi has analysed (2017–2026).
From the lesson Rate of change in Application of Derivatives. Change the values and watch what happens.
In JEE Main
42 shifts analysed
47 questions from Application of Derivatives across the 42 JEE Main shifts Lumi has analysed (2017–2026).
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.
Questions
3 answers
- What does the "How fast a ripple's area grows" simulation show?
- The circle's radius is r and it grows at dr/dt. The orange ring is the area it gains in the next tenth of a second. A bigger circle has a longer rim, so the same growth in r adds more area: dA/dt = 2πr · dr/dt.
- Does JEE Main ask questions from Application of Derivatives?
- Yes. JEE Main asked 47 questions from this chapter across the 42 shifts Lumi has analysed (2017–2026).
- Is this simulation free?
- Yes. It is free and runs in the browser on a phone or a laptop. It belongs to the lesson "Rate of change" in Application of Derivatives (Class 12 Maths).
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