Solve the problem.
-Assume that the rate of change of the population of a certain city is given by $\frac{d y}{d t}=6000 e^{.06 t}$, where $y$ is the population at time $t$, in years. The population was 100,000 in 1980. Assume $t=0$ is 1980. Find the population function.
A) $y=100,000 e^{.06 t}$
B) $y=100,000 e \cdot 06$
C) $y=10,000 e^{.06 t}+90,000$
D) $y=1,000,000 e^{.06}+900,000$

Question

Quizplus · Accepted Answer

The population function is found by integrating the rate of change of the population with respect to time.$$\int \frac{d y}{d t}= \int 6000 e^{.06 t} dt$$$$y=100,000 e^{.06 t}+C$$Using the initial condition, $y=100,000$ when $t=0$, we can solve for the constant $C$.$$100,000=100,000 e^{.06 (0)}+C$$$$C=0$$Therefore, the population function is $y=100,000 e^{.06 t}$.

Solve the Problem dydt=6000e.06t\frac{d y}{d t}=6000 e^{.06 t}dtdy​=6000e.06t , Where yyy Is the Population at Time

Solve the Problem $\frac{d y}{d t}=6000 e^{.06 t}$ , Where $y$ Is the Population at Time