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The next kind of problem seems as the population problem. Back in the first order modeling section we looked at several population problems. In such problems we noticed a single population and frequently involved some form of predation. The problem in this section was we supposed that the amount of predation would be constant. It though clearly won't be the case in most situations. The amount of predation will depend upon the population of the predators and the population of the predators will partially depend as least, upon the population of the prey.
Therefore, in order to more exactly (well at least more correct than what we originally did) we truly require to set up a model that will cover both populations, both the prey and the predator. These kinds of problems are usually termed as predator-prey problems. Now there are the assumptions as we'll make while we build up this model.
1. The prey will grow at a rate which is proportional to its recent population if there are no predators.
2. The population of predators will reduce at a rate proportional to its present population if there is no prey.
3. The number of encounters in between prey and predator will be proportional to the product of the populations.
4. Each encounter among the predator and prey will raise the population of the predator and reduce the population of the prey.
In a class, all pupils take Mathematics (M), 18 take Chemistry (C), 17 take Biology (B) and 24 take Physics (P) of those taking 3 subjects only, 5 take Physics and Chemistry, 7 ta
how to describe the locus of the equation x^2+6xy+y^2+z^2=1 in cylindrical polar coordinates?
A sphere and a cube have equal surface areas. Show that the ratio of the volume of the sphere to that of the cube is √6 : √π. Ans: S.A. of sphere = S.A of cube 4π r 2
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A bucket of height 8 cm and made up of copper sheet is in the form of frustum of right circular cone with radii of its lower and upper ends as 3 cm and 9 cm respectively. Calculate
tan45 degrees=tan(90degrees-45degrees)
Find the middle term of the AP 1, 8, 15....505. A ns: Middle terms a + (n-1)d = 505 a + (n-1)7 = 505 n - 1 = 504/7 n = 73 ∴ 37th term is middle term a 37
sequencing problem
While we first looked at mechanical vibrations we looked at a particular mass hanging on a spring with the possibility of both a damper or/and external force acting upon the mass.
how to evaluate the sums
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