To understand the rate of change of the population over time , we differentiate

with respect to$P$ :$t$

this is fine as is , but since we already know

, we can simplify to :$P$

The explicit version

gives the population size at any time$(\text{}P={P}_{0}\ast {e}^{r\ast t}\text{})$ , assuming exponential growth.$\left(\text{}t\text{}\right)$ The differential version

, describes how the population's rate of change at any moment is proportional to the current population size.$(\text{}\frac{dP}{dt}=r\ast P\text{})$ "the change in population size over an infinitesimally small time interval is directly proportional to the size of the population at that moment"

The explicit version gives a specific solution given some initial conditions

The differential version provides enables modeling at any moment in time , and is not tied to a specific initial condition.

= population size at time$P(t)$ $t$ = carrying capacity of the environment$K$ = initial population size${P}_{0}$ = intrinsic rate of increase$r$ This equation is used for "natural" growth models , where a population grows rapidly at first , but then slows down as it approaches the carrying capacity

$K$ To understand the dynamics of the logistic growth rate at any point in time , we differentiate

with respect to$P$ $t$ To make life easier , lets define a new variable

to replace most of the terms on the denominator$A$ Let

$A=\frac{K-{P}_{0}}{{P}_{0}}\ast {e}^{-r\ast t}$

Now differentiating both sides :

$$\frac{dP}{dt}=\frac{d}{dt}\left(\text{}\frac{K}{1+A}\text{}\right)$$ Pull out constant

:$K$

Applying Chain Rule :

differentiate

with respect to$\frac{1}{1+A}$ :$t$ $$\frac{d}{dt}\left(\text{}\frac{1}{1+A}\text{}\right)=\frac{d}{dA}\left(\text{}\frac{1}{1+A}\text{}\right)\ast \frac{d}{dt}$$ compute derivative of

with respect to$\frac{1}{1+A}$ :$A$ $$\frac{d}{dA}\left(\text{}\frac{1}{1+A}\text{}\right)=\frac{-1}{{(\text{}1+A\text{})}^{2}}$$ substitute

and the derivative of$\frac{dA}{dt}$ into the chain rule formula :$\frac{1}{1+A}$ $$\frac{dP}{dt}=K\ast \left(\text{}\frac{-1}{{(\text{}1+A\text{})}^{2}}\text{}\right)\ast (\text{}-r\ast \left(\text{}\frac{K-{P}_{0}}{{P}_{0}}\text{}\right)\ast {e}^{-r\ast t}\text{})$$ simplify :

$$\frac{dP}{dt}=\frac{K\ast r\ast \left(\text{}\frac{K-{P}_{0}}{{P}_{0}}\text{}\right)\ast {e}^{-r\ast t}}{{(\text{}1+A\text{})}^{2}}$$ now "un-substitute" stuff for

back in :$A$ $$\frac{dP}{dt}=r\ast P\ast (\text{}1-\frac{P}{K}\text{})$$ or written in another way :

$$\frac{dP}{dt}=r\ast P\ast \left(\text{}\frac{K-P}{K}\text{}\right)$$

**The rate of change of bears ,** , in a population is directly proportional to$N$ . Time ,$350-N$ , is measured in years. When$t$ , the population is$t=0$ and when$100$ , the population has increased to$t=2$ .$250$ **A.) Write and solve a differential equation that describes this scenario.**we are going to assume a logistical growth model because the question prompt had "rate of change ...... is directly proportional"

this indicates the growth rate is not constant , but rather changes with the size of the population

so we are going to use the logistical model instead of any other model ( like exponential )

however , we don't need the full original equation that includes a term that represents part of the carrying capacity that is not yet being used :

$(\text{}1-\frac{P}{K}\text{})$ they told us there is a linear/direct relationship in the question prompt between the growth rate and the difference from the maximum limit.

$$\frac{dN}{dt}=k\ast (\text{}350-N\text{})$$ this represents a form of the logistic growth rate ,

where the growth rate decreases as

approaches the carrying capcity$N$ = constant of proportionality = determines the rate at which the population's growth rate decreases as it approaches the threshold of 350.$k$ reflects how quickly the population approaches the limit

so now lets solve

:$\frac{dN}{dt}=k\ast (\text{}350-N\text{})$ separate variables / rearrange :

$$\frac{1}{350-N}\ast dN=k\ast dt$$ integrate both sides :

$$\int (\text{}\frac{1}{350-N}\ast dN\text{})=\int (\text{}k\ast dt\text{})$$ the standard form of the integral

$\int (\text{}\frac{1}{x}\ast dx\text{})=ln\left|x\right|+C$

$$-ln|350-N|=(\text{}k\ast t\text{})+C$$ solve for

, ( exponentiate both sides to get rid of natural log function ) :$N$

$${e}^{(\text{}-ln|350-N|\text{})}={e}^{(\text{}(\text{}k\ast t\text{})+C\text{})}$$ $$350-N={e}^{(\text{}(\text{}-k\ast t\text{})-C\text{})}$$ $$N=350-{e}^{(\text{}(\text{}-k\ast t\text{})-C\text{})}$$ is just a constant , so we can pull this out , and lets rename it to just like${e}^{-C}$ ${C}^{1}$

$$N=350-(\text{}{C}^{1}\ast {e}^{(\text{}-k\ast t\text{})}\text{})$$ now we can use the 2 time points we were given to find

and$k$ :${C}^{1}$ using

and$t=0$ to find$N=100$ :${C}^{1}$ $$100=350-(\text{}{C}^{1}\ast {e}^{(\text{}-k\ast 0\text{})})$$ $$100=350-(\text{}{C}^{1}\ast {e}^{\left(\text{}0\text{}\right)})$$ $$100=350-(\text{}{C}^{1}\ast 1.0\text{})$$ $${C}^{1}=350-100=250$$ now that we found

, substitute it back in :${C}^{1}$

$$N=350-(\text{}250\ast {e}^{(\text{}-k\ast t\text{})}\text{})$$ now lets use

and$t=2$ and our newly found$N=250$ to find${C}^{1}$ :$k$ $$250=350-(\text{}250\ast {e}^{(\text{}-k\ast 2\text{})})$$ solving for

:$k$

$$100=250\ast {e}^{(\text{}-k\ast 2\text{})}$$ $${e}^{(\text{}-k\ast 2\text{})}=\frac{2}{5}$$ take natural log of both sides :

$$-k\ast 2=ln\left(\text{}\frac{2}{5}\text{}\right)$$ simplify ;

$$k=\frac{-ln\left(\text{}\frac{2}{5}\text{}\right)}{2}$$

Final Solution :

$$N=350-(\text{}250\ast {e}^{(\text{}-\left(\text{}\frac{-ln\left(\text{}\frac{2}{5}\text{}\right)}{2}\text{}\right)\ast t\text{})}\text{})$$ **B.) Find the bear population in 3 years**$$N=350-(\text{}250\ast {e}^{(\text{}\left(\text{}\frac{ln\left(\text{}\frac{2}{5}\text{}\right)}{2}\text{}\right)\ast 3.0\text{})}\text{})=286.7544467966324$$ **C.) Find**$\underset{t\to \mathrm{\infty}}{lim}N(t)$ as

approaches infinity , the exponential component$t$ approaches zero${e}^{\left(\text{}(\text{}\frac{ln(\frac{2}{5})}{2}\ast t\text{})\text{}\right)}$ is negative , because$ln\left(\text{}\frac{2}{5}\text{}\right)$ is less than$2/5$ $1$ multiplying it by any positive number and dividing by 2 keeps it negative

to the power of any negative large number tends towards zero$e$

therefore :

$$\underset{t\to \mathrm{\infty}}{lim}N(t)=350-0=350$$