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Numerade Educator



Problem 2 Easy Difficulty

A population grows according to the given logistic equation, where $ t $ is measured in weeks.
(a) What is the carrying capacity? What is the value of $ k? $
(b) Write the solution of the equation.
(c) What is the population after 10 weeks?
$ \frac {dP}{dt} = 0.02P - 0.0004P^2, P(0) = 40 $


a) $k=0.02$ and carrying capacity is 50
b) $P(t)=\frac{50}{1+0.25 e^{-0.02 t}}$
c) 41


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Video Transcript

hey is clear. So when you right here. So we're given the DP over. DT was equal to zero point zero to pee minus zero point 00 Thorough for peace where we got factor out A 0.2 p, you know, one minus 0.2 and this is equal to 0.2 p times one minus p over 50 for part B. We know that the carrying capacity is gonna be part is gonna be 50 part, eh? And we know K is equal to point 02 So we know that M is equal to 50 and p of zero is equal to 40. So for our A value, which is equal to 50 minus 40 over 40 which is equal to point 25 So our solution is P of tea is equal to 50 over one plus 0.25 eats the negative 0.2 tea for part C. We have the equation from part B. So we're looking at the population 10 weeks after, so we just plug in p of tea and this is around 41