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Pie
3 years ago
14

Consider a population p of field mice that grows at a rate proportional to the current population, so that dp dt = rp. (Note: Re

member 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 = ?
Mathematics
1 answer:
Vadim26 [7]3 years ago
8 0

Answer:

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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