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So I have very little background in physics since I'm a mathematical sciences major, but upon being exposed to some physics I've had some difficulties in understanding how to infer the derivatives and integrals (and specifically, why are they used and when), because I assume that in physics they're occasionally chosen because of some changes in units or quantities. E.g. that the first derivative of displacement is velocity.

For an example I draw an equation I found from an answer to one of our homeworks,

$$\frac{\partial}{\partial t} \int_{\Omega} E_{T}(r,t)dr = \int_{\Omega} [E_C(r,t) + \Phi(r,t)]dr$$

where $E_C(r,t)$ denotes thermal energy production in a composting process as a function of unit Jordan measure and unit of time and $\Phi(r,t)$ is a negative quantity that describes thermal energy leak to the environment. $E_T(r,t)$ is the total thermal energy in the system.

Now why does the L.H.S. have a derivative over time of an integral over some set $\Omega$? Or why does the R.H.S. have to be integrated over $\Omega$?

I have difficulties in "visualizing" these equations or knowing how to make the correct unit conversions (using derivatives/integrals), because to me $E_C(r,t)$ is enough to give place/coordinate dependent function values, but why does one want to integrate it to get some sort of area? Is the intuition about summing all those individual $E_C(r,t)$s over all $r \in \Omega$?

Are there any good resources for studying physics or reading about some well known "physics derivatives and integrals heuristics" from a mathematics point of view or do you think that the intuitions in mathematics work in physics reasonably well and I'm just inexperienced in reading and forming physics equations?

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closed as too broad by Qmechanic Nov 1 '16 at 12:51

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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All physics "integrals and derivatives" arise from physical laws which are expressed as differential equations. For example $F=ma$, Newton's law, is a differential equation $$F=m{d^2 x\over dt^2}.$$ If you have a constant force such as gravity (at the earth's surface) you get the familiar relations like $x=ut+1/2 at^2$ from integrating this twice with respect to time.

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  • $\begingroup$ So are you suggesting that to learn how integrals and derivatives are used in physics one must always consider the domain of application and the specific physical theories and their formulation? $\endgroup$ – mavavilj Oct 14 '16 at 8:12
  • $\begingroup$ Lots of physical laws look like each other for reasons of symmetry, for example the inverse square law is the same for electrostatics and for gravity. $\endgroup$ – Suzu Hirose Oct 14 '16 at 8:39
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In general, derivatives and integrals are used in physics to compute the rate of change of one quantity with respect to another quantity- a good example of this is velocity, which is the rate of change of an object's position over time. $$ v=(dx/dt) $$As far as integration goes, consider this example: Gauss' Law of electricity states that the electric flux out of any closed surface is equal to the charge enclosed by that surface divided by the permittivity of free space:$$ \iint_S \vec E \cdot d\vec S =Q/\epsilon $$ So the surface that must be integrated over is the surface through which the field lines are flowing, and both the electric field and the normal vector to the surface vary. You are no doubt familiar with the concept of flux through a surface from way back in multivariable calculus. As far as your specific example goes, the left hand side represents the rate of change of the total thermal energy and the right hand side is the integral of two rates of change. Basically, there is a derivative on the left side and not the right because both the quantities inside the integrand are already change-related in nature, at least if I'm understanding the expression correctly.

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