Example 9: ₹1000 growing continuously at 5% a year doubles at t = 20 logₑ2उदाहरण 9: 5% वार्षिक की सतत दर से बढ़ते ₹1000, t = 20 logₑ2 पर दुगनेExample 9: 5% saalana continuous rate se badhte ₹1000, t = 20 logₑ2 par double

NCERT Example 9 · Class 12 Maths · Chapter 9 · interactive diagram

Parts and what they mean

• P = 1000\,e^{t/20} (P = 1000\,e^{t/20}) — Solution of \frac{dP}{dt} = \frac{5}{100}P with P = 1000 at t = 0. The more money there is, the faster it grows.
• Start: P = 1000 when t = 0 (आरंभ: t = 0 पर P = 1000) — This initial condition gives C = 1000 in P = Ce^{t/20}.
• Doubled: P = 2000 (दुगना: P = 2000) — 2000 = 1000\,e^{t/20} \Rightarrow e^{t/20} = 2.
• Level P = 2000 (स्तर P = 2000) — Double the starting ₹1000.
• Doubling time t = 20\log_e 2 (दुगना होने का समय t = 20\log_e 2) — NCERT answer: t = 20\log_e 2 years, which is about 13.86 years (using \log_e 2 \approx 0.6931).
• x-axis (x-अक्ष) — The horizontal number line; inputs x are read here.
• y-axis (y-अक्ष) — The vertical number line; outputs y = f(x) are read here.
• Origin O(0, 0) (मूल बिंदु O(0, 0)) — Where the two axes cross.

Original diagram drawn by Simple Learning Hub, based on NCERT Example 9. Report a mistake