$\frac{dP}{dt}=-145e^{\frac{-t}{30}}$ Where $t$ is the time in days and when $t=0$, we have that the water squirrel population is $4,350$

Question

5.3

Because of an insufficient oxygen supply, the population of water squirrels in a lake is going down. The population's rate of change can be modeled by:

$\frac{dP}{dt}=-145e^{\frac{-t}{30}}$

Where $t$ is the time in days and when $t=0$, we have that the water squirrel population is $4,350$

a) Find an equation to model the population

b) What is the water squirrel population after 11 days?

c) How long will it take until there are no water squirrels left? (assume the population dies off when the population is less than 1)

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My Solution:

$a)$ They gave us an equation to solve for $\frac{dP}{dt}$... this means that we can find the rate of change in which the population is changing with respect to time. But Part (a) is asking for an equation describing the population, so we need to solve for $P$, so we need to integrate:

$\frac{dP}{dt}=-145e^{\frac{-t}{30}}$

$\rightarrow dP =-145e^{\frac{-t}{30}}dt$

$\rightarrow P = \int dP = -145 \int e^{\frac{-t}{30}}dt$

Ugh. So in general, we have the identity that for any number, say $7$, we have $\int e^{7x}dx = \frac{e^{7x}}{7} + C$... So in our case we have $\frac{-1}{30}$ instead of $7$... SO in the future you can just use that identity, but if you can't remember then you can always just figure it out yourself by using $u$ substitution as follows:

$\rightarrow P = -145 \int e^{\frac{-t}{30}}dt$

Let $u=\frac{-t}{30}$. then $\frac{du}{dt}=\frac{-1}{30}$ and so $du = \frac{-1}{30}dt$

Thus we have $-30du = dt$ and thus:

$=-145 \int e^{\frac{-t}{30}}dt$

$=-145 \int e^{u} (-30)du$

$= 4350 \int e^u du$

$= 4350 e^u + C$

$=4350 e^{\frac{-t}{30}} + C$

So $P = 4350 e^{\frac{-t}{30}} + C$

Okay, but they told us that when $t=0$ that we have the total population is $4,350$. Thus we have:

$4,350 = 4,350 e^0 + C = 4,350 (1) +C$

And it follows that $C=0$

And so we have the answer to part (a):

$P = 4350 e^{\frac{-t}{30}}$

PART b):

So we know $P = 4350 e^{\frac{-t}{30}}$, so lets just plug $t=11$ and solve for $P$. you end up getting 3015

PART c):

Note that this is sort of an interesting question because if we model population by:

$P = 4350 e^{\frac{-t}{30}}$

Then $P$ will NEVER equal zero for any $t$. That's why they added the bit about we consider the population to have died off once $P$ is less than one. So we just need to solve for the time when $P=1$ as so:

$1 = 4350 e^{\frac{-t}{30}}$

$\frac{1}{4350} = e^{\frac{-t}{30}}$

$ln(\frac{1}{4350}) = ln(e^{\frac{-t}{30}})$

$ln(\frac{1}{4350}) = \frac{-t}{30}$ because $ln(x)$ and $e^x$ are inverse operations

Thus we have:

$\rightarrow -30 ln(\frac{1}{4350}) = t$

Answer 1

On (a) I would say

• $\frac{d}{dx} e^x = e^x$
• so $\frac{d}{dx} e^{kx} = ke^{kx}$ by the chain rule
• so $\int e^{kx} \,dx = \frac{1}{k} e^{kx}+C$ by the fundamental rule of calculus
• and here $k=-\frac1{30}$
• leading to $P=4350 e^{\frac{-t}{30}} + C$ with the starting condition to $C=0$ etc.

(b) looks fine

For (c) I would say you want to solve $4350 e^{\frac{-t}{30}} \lt 1$ which leads to

• $e^{\frac{-t}{30}} \lt \frac1{4350}$
• ${\frac{-t}{30}} \lt \log_e\left(\frac1{4350}\right)$
• $t \gt -30\log_e\left(\frac1{4350}\right)$
• $t \gt 30\log_e\left({4350}\right)$

Answer 2

As said Paul for (c) :

Solve P=1 using the logarithm.

You have $$ \ln(4350e^{\frac{-t}{30}})=\ln(4350)-\frac{t}{30} $$

Then solve for $P=1$ and you get the result, because your population is supposed diying when it remains 1 individual.