How many days until the population doubles

Question

I have a function called $P(t)$ that is the number of the population at time $t$. $t$ being in days.

We know the growth rate is $P'(t) = 2t + 6$

We also know that $P(0) = 100$. How many days till the population doubles?

edit: $P(t) = t^2 + 6t$ edit: $P(t) = t^2 + 6t = 200$ edit: $t^2 + 6t - 200 = 0$

Answer 1

You integrated (almost) correctly; however \begin{eqnarray} P(t) &=& \int 2t + 6 dt \\ &=& t^{2}+6t+C \end{eqnarray} $C$ is a constant of integration; now your intital condition is $P(0)=100$ meaning at $t=0$ the population was $100$. Plug it into the formula for $P(t)$ \begin{eqnarray} P(0) &=& 100 \\ &=& 0^{2} + 6(0) + C \\ &=& C \end{eqnarray} Meaning $C=100$; therefore \begin{equation} P(t) = t^{2}+6t+100 \end{equation}

Answer 2

Hint: You are given $P'(t)=\frac {dP}{dt}=2t+6$ Integrate both sides with respect to $t$ to get $P(t)$ At this stage you will have a constant of integration which you can assess using your data on $P(0)$ What is the population when the doubling is over? Solve $P(t)=$this