Mathematics: Consider a population p of field mice that grows at a rate proportional to the current population, so that dp dt = rp. (note: remember that, as in the text, t is measured in months, not days. one month is 30 days.) (a) find the rate constant r if the population doubles in 210 days. (round your answer to four decimal places.) r = (b) find r if the population doubles in n days. r = ?

7% = 7/100 so,86,000 / 100 = 860Then,860 x 7 = 6020.Hope I ve helped....

Consider a population p of field mice that grows

nssjdjsididiodododod

15.01.2022

a) $r = \frac{ln(2)}{210}=0.00330070086$

b) $r = \frac{ln(2)}{N}$

Step-by-step explanation:

For this case we assume the followin differential equation:

$\frac{dp}{dt}= rp$

Where is is the consttant growth/decay rate , p represent the population and the the time.

For this case we can rewrite this expression like this:

$\frac{dp}{p}= rdt$

And now we can apply integrals on both sides like this:

$\int \frac{dp}{p} r\int dt$

ln |p|= rt+C

If we apply exponential on both sides we got:

$p(t) = e^{rt} *e^c = p_o e^{rt}$

And from the previous equation p_o represent the initial population.

Part a

For this case we are assuming that the population doubles in t=210 so then we can set the following equation:

$2p_o = p_o e^{210r}$

We can cancel p_o in both sides and we got:

$2 = e^{210r}$

We can apply natural log on both sides and we got:

ln (2) = 210 r

$r = \frac{ln(2)}{210}=0.00330070086$

Part b

For this case we are assuming that the population doubles in t=N so then we can set the following equation:

$2p_o = p_o e^{Nr}$

We can cancel p_o in both sides and we got:

$2 = e^{Nr}$

We can apply natural log on both sides and we got:

ln (2) = N r

$r = \frac{ln(2)}{N}$

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