How long to double?
"How long to double?" is a free, interactive Class 12 Maths simulation for JEE, built on the lesson "Growth, decay and cooling" in Differential Equations. Rs 1000 grows continuously, so dP/dt = kP and P = 1000e^(kt). The green marks show when it reaches 2000 and then 4000. Notice that the second doubling takes exactly as long as the first. DP/dt = kP gives P = P₀e^(kt), and the doubling time (log 2)/k does not depend on how much you start with. JEE Main asked 57 questions from this chapter across the 42 shifts Lumi has analysed (2017–2026).
From the lesson Growth, decay and cooling in Differential Equations. Change the values and watch what happens.
In JEE Main
42 shifts analysed
57 questions from Differential Equations across the 42 JEE Main shifts Lumi has analysed (2017–2026).
The idea behind it
NCERT §9.4.1 (Example 9, applications)
- If a quantity grows at a rate proportional to its size, dP/dt = kP with k > 0. Separating gives P = P₀e^(kt), where P₀ is the value at t = 0.
- The time for P to double is found from 2 = e^(kT): T = (log 2)/k, the same from any starting value.
- Worked example: money grows continuously at 10% a year, so k = 0.1. It doubles in 10 log 2 ≈ 6.93 years, and Rs 1000 becomes 1000e ≈ Rs 2718 after 10 years. At 7% a year it doubles in about 9.9 years.
- Decay is the same equation with a negative constant: dN/dt = −λN gives N = N₀e^(−λt) and a half-life of (log 2)/λ.
- Worked example: a culture that triples in 5 hours has e^(5k) = 3. After 10 hours it is e^(10k) = 3² = 9 times its starting size.
- Worked example (cooling): a drink at 85 °C in a 25 °C room follows dT/dt = −k(T − 25), so T − 25 = 60e^(−kt). If it reaches 55 °C in 10 minutes, the gap 60 halves in 10 minutes, so after 20 minutes the gap is 15 and the drink is at 40 °C.
Questions
3 answers
- What does the "How long to double?" simulation show?
- Rs 1000 grows continuously, so dP/dt = kP and P = 1000e^(kt). The green marks show when it reaches 2000 and then 4000. Notice that the second doubling takes exactly as long as the first.
- Does JEE Main ask questions from Differential Equations?
- Yes. JEE Main asked 57 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 "Growth, decay and cooling" in Differential Equations (Class 12 Maths).
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