💬 👋 We’re always here. Join our Discord to connect with other students 24/7, any time, night or day.Join Here!

# Biologist stocked a lake with 400 fish and estimated the carrying capacity (the maximal population for the fish of that species in that lake) to be 10,000. The number of fish tripled on the first year.(a) Assuming that the size of the fish population satisfies the logistic equation, find an expression for the size of the population after $t$ years.(b) How long will it take for the population to increase to 5000?

## a) $P(t)=\frac{10000}{1+24 \cdot\left(\frac{11}{36}\right)^{t}}$b) $t \simeq 2.68$ years

#### Topics

Differential Equations

### Discussion

You must be signed in to discuss.

Lectures

Join Bootcamp

### Video Transcript

Hey, it's clear. So when you married here so far apart, they we make my m B equal to 10,000 in peace of zero equal to 400. No. So when we find for a it's 10,000 minus 400 all over 400 is equal to 24. You only get pft is equal to 10,000 all over one plus 24 eats the negative. Katie only see that the population has tripled after the first year. So using piece of one is equal to 2012 1200 which is equal to the equation we found above negative K times one, we end up getting e to the negative. K is equal to 11 over 36. So we get P of tea to be equal to 10,000. Pull over one plus 24 e to the negative K to the T power, which is equal to 10,000. We just plug in need to the negative k to the T power for part B. We're going to to the used occlusion PR team that we got from per e and we're gonna make this be equal to 5000 is equal to 10,000 all over one plus 24 terms. 11 over 36 to the T power. And we end up getting e to the to the negative 1.18 56 be equal to 10,000. Pull over 5000 minus one times one over 24. Now we're gonna multiply both sides by one plus 24 e to the negative 1.186 over 5000 to get the T value for about 2.68 years.

#### Topics

Differential Equations

Lectures

Join Bootcamp