The rate of change (dP)/(dt) of the number of bears at a national park is modeled by the following differential equation: (dP)/(dt)=(7280)/(71839)P(1-(P)/( 560)) At t=0, the number of bears at the national park is 199 and is increasing at a rate of 13 bears per day. Find lim_(t rarr oo)P^(')(t). Answer:

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The rate of change  (dP)/(dt) of the number of bears at a national park is modeled by the following differential equation:  (dP)/(dt)=(7280)/(71839)P(1-(P)/( 560)) At  t=0, the number of bears at the national park is 199 and is increasing at a rate of 13 bears per day. Find  lim_(t rarr oo)P^(')(t). Answer:

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Equilibrium Points Analysis: The given differential equation is:  (dP)/(dt) = (7280)/(71839)P(1 - (P)/(560))  This equation models the rate of change of the population of bears in a national park. To find the limit of P'(t) as t approaches infinity, we need to analyze the behavior of the differential equation as P grows large.  First, let's look at the equilibrium points of the differential equation, which occur when (dP)/(dt) = 0. Setting the right-hand side of the equation to zero gives us:  0 = (7280)/(71839)P(1 - (P)/(560))  This equation has two solutions for P: P = 0 and P = 560. These are the points where the population will not change over time.  Since we are interested in the limit as t approaches infinity, we need to consider the stability of these equilibrium points. If P = 560 is a stable equilibrium, then as t approaches infinity, the population P will approach 560, and the rate of change P'(t) will approach 0.Stability Analysis: To determine the stability of the equilibrium point P = 560, we can analyze the sign of the derivative (dP)/(dt) around this point. If the derivative is negative for P just above 560 and positive for P just below 560, then P = 560 is a stable equilibrium.  For P slightly less than 560, the term (1 - (P)/(560)) is positive, and since P is also positive, the whole expression for (dP)/(dt) is positive. This means the population is increasing when it is just below 560.  For P slightly more than 560, the term (1 - (P)/(560)) is negative, and since P is positive, the whole expression for (dP)/(dt) is negative. This means the population is decreasing when it is just above 560.  This behavior indicates that P = 560 is indeed a stable equilibrium point.Approaching Infinity: Since P = 560 is a stable equilibrium, as t approaches infinity, the population P will approach 560, and the rate of change P'(t) will approach 0.  Therefore, the limit of P'(t) as t approaches infinity is 0.

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