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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**Example text**

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).