Question

Assume that COVID-19 growth rate of number of infections per day (c) in a certain population is represented by the following differential equation. \[ 100 \frac{dc}{dt} - 7c = 0 \] Where, t stands for time in days. Time taken for the number of infections per day to double, in days, is \(\underline{\hspace{2cm}}\) (round off to the nearest integer).

Hint:

For exponential growth, the time to double can be found using the formula \( t = \frac{\ln 2}{r} \), where \( r \) is the growth rate.

Answer 1

Correct Answer: 9

Given the differential equation: \[ 100 \frac{dc}{dt} - 7c = 0 \] Rearranging the terms: \[ \frac{dc}{dt} = \frac{7}{100} c \] This is a first-order linear differential equation, which we can solve by separation of variables: \[ \frac{dc}{c} = \frac{7}{100} dt \] Integrating both sides: \[ \ln c = \frac{7}{100} t + C \] Taking the exponential of both sides: \[ c = C e^{\frac{7}{100} t} \] To find the time \( t \) taken for the infections to double, we use the condition that \( c(t) = 2c(0) \), where \( c(0) \) is the initial infection rate. Substituting into the equation: \[ 2c(0) = c(0) e^{\frac{7}{100} t} \] Simplifying: \[ 2 = e^{\frac{7}{100} t} \] Taking the natural logarithm of both sides: \[ \ln 2 = \frac{7}{100} t \] Solving for \( t \): \[ t = \frac{100 \ln 2}{7} \approx 9.90 \, \text{days} \] Thus, the time taken for the number of infections per day to double is approximately: \[ \boxed{10 \, \text{days}} \]