M Mail Fizer, Imani outlo... X Math 132 Section 11.8 X G how to screenshot on a
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M Mail Fizer, Imani outlo... X Math 132 Section 11.8 X G how to screenshot on a su... X C Accounting question Ch X D www.webassign.net/web/Student/Assignment-Responses/submit?dep C a Search submit?dep-13223523 5. 013 points l Previous Answers HHCalcCustom1 11.8.020 My Notes Ask Your Teacher Let w be the number of worms (in millions) and r the number of robins (in thousands living on an island. Suppose w and r satisfy the following differential equations, which correspond to the slope field in the figure below dw dr E w wr, r Wr dt dt r (predator) W (prey Assume w 1.5 and r 1.5 when t 0. Estimate the maximum and the minimum values of the robin population X thousand robins maximum X thousand robins minimum How many worms are there at the time when the robin population reaches its maximum? On worms Submit Answer Save Progress Practice Another Version 6. 11 points l Previous Answers HHCalcCustom1 11.8.022. My Notes Ask Your Teacher Let w be the number of worms (in millions) and r the number of robins (in thousands living on an island. Suppose w and r satisfy the following differential equations, which 4:35 PM HE Search the web and Windows 4/29/2016Explanation / Answer
w(0) = initial number of worms = 1.5 million ; r(0) = 0
dw/dt = w since r(t) = 0.
Solution: w(t) = w(0) e^t (the worms have food, but are
eaten by the robins; no robins means that worm population grows exponentially
(at least for a while -- eventually the equation is unrealistic.)
r(t) = 0 then satisfies the second equation with the initial condition r(0) = 0.
No worms: w(t) = 0. Then the robins have nothing to eat and starve
to death: dr/dt = -r , r(t) = r(0) e^(-t) ,this implies it is < 1, which means no more
robins.
So, r(0) e^(-t) < 1 means ln(r(0)) - t < 0 or t > ln r(0) = ln 3000 = 8.01 units
The unit of time is not specified. If you just look at r(0) e^(-t) , this is positive
but decreases rapidly as t --> oo. The robin population is halved whenever t increases
by T, with e^(-T) = 1/2 or T = ln(2) = 0.69 time units.
So after time T there are 1500 robins; after 2T, 750 robins; after 3T, 375 robins = 1/8
the initial robin population; after 9T, (1/8)^3 = 1/2^9 = 1/512 the original number,
which is < 6 robins – so, 5 of the robins left.
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