Exercises in Environmental Physics (2006)(en)(330s) by Valerio Faraoni

By Valerio Faraoni

This can be the 1st publication in particular dedicated to routines at the program of physics to explain the surroundings together with human influence on it. it's a necessary software for college kids to improve abilities within the manipulation of actual suggestions and strategies whereas studying environmental technological know-how. The workouts are drawn from the author's instructing adventure and the necessity for exciting perform difficulties in quite a few environmental physics classes. A bankruptcy on mathematical tools utilized in the e-book vitamins the cloth.

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Vector calculus Vector calculus is covered in many excellent mathematics textbooks. , Refs. [59, 22]). 1 (C, B) Explain the difference between scalars and vectors and provide examples of scalar and vectorial physical quantities. What is a scalar field? What is a vector field? Solution 1 (level C) Physically, a scalar quantity is completely characterized by its magnitude while a vector quantity is characterized by its magnitude and direction. Examples of scalars are mass, temperature, time, while position with respect to a fixed origin, velocity, acceleration, force, electric and magnetic field are vector quantities.

A scalar field is a scalar function f (x) of the position x = (x, y, z), while a vector field is a vector quantity that depends on position, a = a (x). Solution 2 (level B) A mathematically more precise definition of scalars and vectors can be given by using the transformation properties of their components under a coordinate transformation i i xi −→ x = x (xj ). , s = s, while a covariant vector a with components ai transforms according to ∂x i j i ai −→ a = a . ∂xj j A 1-form or contravariant vector with components ωi instead transforms according to ωi −→ ω i = l ∂xl ωl .

B) To solve the Verhulst model, we write the logistic equation as dP 1 =a P (1 − αP ) dt and decompose the fraction on the left-hand side as follows: B 1 A + = . P 1 − αP P (1 − αP ) The appropriate values of A and B that make this decomposition possible are easily found to be A = 1 and B = α, hence 1 dP dP α + = a. P dt 1 − αP dt Integration gives P P0 dP − P (αP −1) (αP0 −1) and ln P P0 − ln d (αP − 1) = at, αP − 1 αP − 1 = at. αP0 − 1 By using the properties of logarithms and taking the exponential of both sides, we obtain P (αP0 − 1) = eαt P0 (αP − 1) and finally, after straightforward algebraic manipulations, P (t) = P0 P0 eat = , at 1 − αP0 + αP0 e αP0 + (1 − αP0 ) e−at where again, the integration constant has the meaning P0 = P (0).

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