Let \( P(t) \) be the population (in millions) of a certain city \( t \) years after 2015 , and suppose that \( P(t) \) satisfies the differential equation \( P^{\prime}(t)=0.02 P(t), P(0)=6 \). (a) Use the differential equation to determine how fast the population is growing when it reaches 8 million people. (b) Use the differential equation to determine the population size when it is growing at a rate of 400,000 people per year. (c) Find a formula for \( P(t) \).

To solve each part of the problem based on the differential equation \(P'(t) = 0.02 P(t)\) with the initial condition \(P(0) = 6\), we can follow the steps below. **(a)** To determine how fast the population is growing when it reaches 8 million people, we can use the given differential equation. We know: \[ P'(t) = 0.02 P(t) \] When \(P(t) = 8\): \[ P'(t) = 0.02 \times 8 = 0.16 \text{ million people per year} \] Therefore, the population is growing at a rate of 0.16 million people per year (or 160,000 people per year) when it reaches 8 million. **(b)** To find the population size when it is growing at a rate of 400,000 people per year, we note that: \[ P'(t) = 0.02 P(t) = 0.4 \text{ million} \] Setting this equal to our equation: \[ 0.02 P(t) = 0.4 \] Solving for \(P(t)\): \[ P(t) = \frac{0.4}{0.02} = 20 \text{ million people} \] Thus, the population size when it is growing at a rate of 400,000 people per year is 20 million. **(c)** To find a formula for \(P(t)\), we can solve the differential equation. This is a separable equation: \[ \frac{dP}{P} = 0.02 dt \] Integrating both sides: \[ \ln |P| = 0.02t + C \] Exponentiating to solve for \(P\): \[ P = e^{0.02t + C} = e^C e^{0.02t} \] Let \(e^C = K\), thus: \[ P(t) = K e^{0.02t} \] Using the initial condition \(P(0) = 6\): \[ 6 = K e^{0} = K \implies K = 6 \] The formula for \(P(t)\) is: \[ P(t) = 6 e^{0.02t} \]