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# Suppose a population $P(t)$ satisfies $\frac {dP}{dt} = 0.4 P - 0.001P^2 P(0) = 50$where $t$ us measured in years.(a) What is the carrying capacity?(b) What is $P'(0)?$(c) When will the population reach $50%$ of the carrying capacity?

## a) 400b) 17.5c 4.86 years

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Differential Equations

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hands. Claire is the one you married here. So for part A, we have dp over G t is equal to four over T P minus one over 1000 piece square. We're gonna factor out of for over 10 p. And it's one minus one over 400 p. And we see that the carrying capacity is equal to 400 for part B. We have the equation for over 10 p. Times one minus one over 400 p hands. Claire is the one you married here. So for part A, we have dp over G t is equal to four over t p minus one over 1000 piece square. We're gonna factor out before we get P of tax zero and we get four times 50 over 10 times one minus 50 over 400 over 10 p. And it's one minus one over 400 p. And we see that the carrying capacity is equal to 400. For part B. We have the equation for over 10 p. Times one minus one over 400 p. Then we get to be equal to 17.5 for part c. We know the carrying capacity equation when we get pfft is equal to we get P of tax zero and we get four times 50 over 10 times one minus 50 over 400 400 all over one plus 400 minus 50 over 50 e to the negative 0.4 key. Then we just simplify to get 400 over one plus seven e to the then we get to be equal to 17.5 for part C. We know the carrying capacity equation when we get pfft is equal to negative 0.40. We know this is equal to 200 so we have to have the carrying capacity and we have to solve for tea. Then we get one plus seven e to the negative. Zero point for P is equal to 400 over 200. When we saw for this, we get tea is about 400 all over one plus 400 minus 50 over 50 e to the negative 0.40. Then we just simplify to get 400 over one plus seven e to the 4.86 years

#### Topics

Differential Equations

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