Variable Separable
A population grows so that its rate of change is proportional to its current size, modeled by \frac{dP}{dt} = kP; find the time for the population to double from its initial value.
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About This Question
- Subject
- mathematics
- Chapter
- differential equations
- Topic
- variable separable
- Difficulty
- Easy
- Year
- 2025
Solution
Correct Answer:
t=kln2
This is the classic exponential growth model, a separable first-order equation where the rate of change is proportional to the quantity itself. Separating variables gives \frac{dP}{P} = k,dt, and integrating both sides yields \ln P = kt + C. Exponentiating produces P = P_0 e^{kt}, where P_0 is the initial population determined by setting t = 0. The conceptual heart of the model is that constant proportional growth always produces an exponential curve, never a linear or polynomial one. To find the doubling time, set P = 2P_0, giving 2P_0 = P_0 e^{kt}, so 2 = e^{kt}. Taking natural logarithms gives \ln 2 = kt, hence t = \frac{\ln 2}{k}. Option t = \frac{2}{k} confuses the factor 2 with its logarithm. Option t = k\ln 2 inverts the role of k. Option t = \frac{\ln 2}{2k} introduces a spurious factor of 2 in the denominator. This is the standard JEE Advanced growth-decay application. As a final plausibility check, a larger growth rate k makes the doubling time smaller, which the formula t = \frac{\ln 2}{k} correctly reflects since t is inversely proportional to k.
This easy difficulty mathematics question is from the chapter differential equations, covering the topic of variable separable. It appeared in the 2025 exam.
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