Answer to Question #265375 in Differential Equations for joanne

Question

1. A certain substance was placed inside a room where the temperature is 17°C. it is observed that after 30 seconds, the temperature of the substance drops to 27°C and after 1 minute, the temperature drops to 20°C. what is the initial temperature of the body? Ans. 50.33°C
2. The bureau of census record in 1975 shows that the population in the country doubles compared to that of 1955. In what year will the population quadruple? Ans. 1995

Answer 1

$\frac{}{}$ $\text{The change in temperature is given by }\\ \frac{d\theta}{dt}= k(17-\theta)\\ \implies \frac{d\theta}{17-\theta}=kdt\\ \implies \theta = 17-ce^{-kt}-(1)\\ \text{at t = 60 seconds and $\theta = 20$, we have that}\\ -3=ce^{-60k}-(3)\\ \text{at t = 30seconds and $\theta = 27$, we have that}\\ \text{Dividing (3) by (2), we have}\\ 0.3 = e^{-30k}\\ \implies k = 0.0401\\ \text{Putting k =0.0401 in 3 we have that}\\ c = -33.33\\ \text{Next, we substitute k and c in (1), at time 0 to get our solution}\\ \therefore \theta = 17+33.33e^0=50.33^0C\\ \frac{dp}{dt}=kt\\ \implies \frac{dp}{p}=kt\\ \implies \ln p =kt +A\\ \implies p = Ce^{kt}\\ \text{At t = 20, p=2C, as given in the question}\\ \implies 2 = e^{20k}\\ \implies \ln 2 = 20k\\ \implies k = 0.0347\\ \text{When p=4C, we have that}\\ 4 = e^{0.0347t}\\ \implies t = 39.95 \text{ approximately 40 years. Hence p quadruples at}\\ \text{1955 + 40years=1995}$