Section 2.5 Population Dynamics: Problem 1 Problem ListNext (1 point) The siope

Question

Section 2.5 Population Dynamics: Problem 1 Problem ListNext (1 point) The siope field tor a population P modeled by d.P/dt-3P-3P Is shown in the fgure below (a) On a print-out of the slope neld, sketch three non-zero solution curves showing difterent types of behavior for the popuslation P Give an intiail condition that will produce eadh Pto)- P(O) Pto)- (b) is there a stable value of the population? it so. give the value if not, enter none evalue e) Considering the shape of soutions for the population, give any intervals for which the tolowing are true. it no such intervat exists, solutions are ncreasing wnen Pis between 1 and 3, enter (,3 for that answer, t they are decreasing wnen act that Ps a population j enter none, and if there are mutiple intervais, give them as a tist f P is between f and 2 or between 3 and 4 enter (1,2.(34) Note that your answers may re Pis increasing ahen Ps in Pis decreasing when Psn Think about what these conditions mean for the population, and be sure that you are abie to explain that

Explanation / Answer

To solve this we will solve like that ,

First solve for P:
3p-3p^2=0
so....
P=0 and P=1

For part (a) simply choose three arbitrary points between 0 and 1

(b) the point where the population reaches zero, so it's 1

(c)when we sketch the graph of the equation... it's a downward parabola, where P is the x-axis with intercepts 0 and 1.
---------it increases at an interval (0,1)
---------it decreases at an interval (1, infinity)

the outcome is 1 because that's where the population is approaching

inflection points are the points where the graph switches its direction.... so since 1/4 is on the peak of the curve, that's the answer

(d) the same interval as above... increasing is positive, decreasing is negative

when dP/dt is zero, P=0,P=1 **since they are the P-intercepts
when dP/dt is at max, P=2 **since that's the vertex

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