Logistic Differential Equations Previously, we have studied exponential growth and decay. Recall that the “rate of change of a quantity y is proportional to the quantity y” can be translated to the differential equation: ! ! ! ! _____________________________ The solution to the above differential equation can be solved using the separation of variables kt technique and yields the equation: y = Ce . If the constant k is negative, then this equation models exponential decay. If k is positive, the equation models exponential growth. Note the graphs of each below. ! ! y exponential decay! ! t ! ! ! ! ! ! ! y exponential growth t Since the exponential growth equation describes unlimited growth (i.e., as t → ∞ , y → ∞ ), this model may be unrealistic for many situations. When there exists some upper limit L past which growth cannot occur, a logistic growth model may prove useful. For example, for a lake’s growing fish population, there might be a cap on the number of fish that the lake can sustain. Similarly, as the number of people arriving for a Packer game at Lambeau field increases, there is again a cap to this amount of possible growth -- the number of seats in the stadium. This upper limit of growth L is typically referred to as the carrying capacity. In the logistic differential equation, “the rate of change of a quantity y is proportional to the quantity y times the difference of L and y” which can be mathematically translated as: ! ! ! ! _____________________________ The solution to the above differential equation can again be found by separation of variables, but doing so requires the use partial fractions, a technique we have not yet studied (but will L y= 1 + ce− Lkt (with its eventually). Anyway, the solution to the logistic differential equation is graph shown on the next page). Both the solution and the logistic differential equation need to be memorized. ! ! ! ! ! logistic growth model, y= L 1 + ce− Lkt y ! ! ! ! t ! Note that y = L (recalling that L is the carrying capacity) is a horizontal asymptote of the graph (and so is y = 0 for that matter). Also, although this logistic growth model equation is always dy dy → _______ increasing (and thus, dt is always positive), as t → ∞ , dt . Finally, consider the point on the graph where y is increasing the fastest. In the below space, determine the y-value where y is increasing the fastest. PROBLEMS: 1. The number of bears in a national park is modeled by the function B that satisfies the dB B ⎞ ⎛ = 0.8B ⎜ 1 − ⎟ ⎝ 150 ⎠ where t is the time in years and B ( 0 ) = 23 . logistic differential equation dt (a) What is lim B ( t ) t →∞ ? ! ! ! ! (b) What is lim B′ ( t ) t→∞ ? (c) How many bears are in the park when their population in the park is increasing most rapidly? 2. MULTIPLE CHOICE. The rate of change, dP , of the number of people in a gym is dt modeled by a logistic differential equation. The maximum number of people allowed in the gym is 200. At 11AM, the number of people in the gym is 10 and is increasing at the rate of 70 people per hour. Which of the following differential equations describes the situation? WORK MUST BE SHOWN FOR FULL CREDIT. (A) ! dP 1 = P ( 200 − P ) ! dt 70 ! (D) ! (B) dP = 70P ( 200 − P ) ! dt dP 1 = P ( 200 − P ) !! dt 1900 ! (E) ! (C) dP 7 = P ( 200 − P ) dt 190 dP 1 = P ( 200 − P ) + 10 dt 1900 3. The big toy this coming Christmas season is the radian time clock. Research shows that the rate of sales is proportional to the number sold and the number in the target audience who have not purchased it. Assume that the target audience number is 4 million. If 950,000 have been sold at time, t = 0 , and 3.5 million have been sold 10 days later, after how many days did the company sell to half of its target audience?
© Copyright 2024 ExpyDoc
