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Explanation of common calculus examples 
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#1
Feb613, 10:29 AM

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[itex]Work (J) =\int Force(N) .dx(m)[/itex]
I understand this one, work done = force * distance moved in direction due force, but the force can change in magnitude or direction so less/more component of the force is in the direction of movement, so in this case the area under the graph of force vs distance is the work done. [itex] Force(N) = \int GPE(J) *\frac{1}{r(m)} .dr'(m)[/itex] I do not understand this one at all. I;ve never even seen dr' before. source of equation: http://hyperphysics.phyastr.gsu.edu.../gpot.html#gpi And then there are maxwells equations, which make absolutely no sense to me what so ever [itex]Charge (C) = \epsilon_{0} \oint \stackrel{\rightarrow}{Electric Field Strength}(NC^{1} or Vm^{1}) .d \stackrel{\rightarrow}{A}(m^{2})[/itex] I study electrical engineering, not physics, so when this was taught in our electromagnetism lectures it was the first time I had seen a "closed loop integral", and I still don't understand how this is different from a normal integral. Also what do the arrows mean? Is this basically saying that the charge that an object has is proportional to the electric field strength at the surface of the object, multiplied by the surface area of the object? And the reason its an integral and not just a multiplication is that the electric field strength can be different at different places on the surface area of the object? I have the same problem with the second of maxwell's equations. http://hyperphysics.phyastr.gsu.edu...ric/maxeq.html (it takes a while to write them out so i will just reference this list) The 3rd equation looks similar to what I learnt in A level Physics, that the emf induced in a conductor is proportional to the rate of change of flux, where flux is the magnetic field inside a conductor. What is wrong with this? I can see why a differential is needed to describe the rate of change of flux, but why is an integral needed? The 4th looks completely new to me, I do not know what this is trying to explain. I don't think we have ever covered this in our lectures, maybe this one isn't as important to electrical engineers. Also, quick extra question, can any of these be used to derive the electric field strength for a radial field. [itex]E = \frac{1}{4\pi\epsilon} * \frac{1}{r^{2}} * Q [/itex] Thanks! 


#2
Feb613, 02:04 PM

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1. The r' variable is a dummy variable, used to distinguish a general radial coordinate from a particular value of a radius. In the Hyperphysics link, the Gravitational Potential Energy (GPE) = U(r) and is the same at any radius r from the Earth. In order to calculate U(r), the integral must be evaluated everywhere from r above the earth out to infinity.
2. The arrows signify that the variables are vectors, that is they have magnitude and direction. Because of this, the the context of the calculus, vector variables are treated a little differently from regular variables. Certain theorems and identities, for example, can be employed to simplify the evaluation of the integrals. The loop integral, as you call it, is a path integral. It is given a special sign to indicate that it is to be evaluated on a particular closed curve, rather than a simple interval for a 'regular' integral. 3. Your questions about the Maxwell equations and electric fields I cannot answer in a simple way. Without having studied vector calculus, there may not be a simple verbal description which would make sense. 


#3
Feb613, 03:01 PM

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#4
Feb613, 03:36 PM

P: 887

Explanation of common calculus examples
The notation used in the hyperphysics page is mostly useful as a reminder to someone who already knew Maxwell's laws and just needed a refresher. For additional clarity, the integrals in equations I, II, and the rhs of IV should be written as double integrals, because they are integrations over area, while the integrals for III and lhs of IV should remain single integrals because they are integrals along a curve. The limits of the integrals are left out since you are supposed to remember what they are.
The limits for the integrals in I and II are any choice of closed surface (without edge), and the loop around the integral is just a reminder that this is a closed surface. The charge q is the total net charge contained inside the chosen closed surface. The limits for the integrals in III and IV are any choice of closed loop. For the RHS of IV, the integral is over a surface patch whose boundary is the closed loop. Since this isn't a closed surface in 2D, there is no loop in the integral symbol. The LHS integral contains a loop because the loop is closed in 1D. The current i is the total through the chosen loop. It should work for any choice of loop, as long as you use the same choice on both sides of the equation. Note that [itex]\frac{d\Phi_B}{dt}[/itex] just means [itex]\frac{\partial}{\partial t}∫\vec{B}\cdot d\vec{A}[/itex] so actually, equations III and IV are pretty similar. 


#5
Feb713, 12:38 PM

P: 202

I still don't really understand this. I think my problem is more of a maths problem than physics problem, If fully understood what all the notation physically means I could start to understand the physics behind these equations. Thanks for your time though, its much appreciated.
Also, I am currently studding vector calculus, so far we have covered dot, cross, grad, div, and curl functions, but I do not fully understand the last three 


#6
Feb713, 01:52 PM

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Hi CraigH!
Very briefly:f(r) = ∫ g(r') dr' [or f(r) = ∫ g(s) ds] is the correct way of writing what is usually sloppily written, ##\int \vec{E}\cdot\vec{dA}## is basically the same as ∫∫ E(x,y).n dxdy An ordinary ∫ or ∫∫ is between limits, a looped ∫ has no limits … it's a loop! To understand grad div and curl, you need to wait until you've studied electromagnetic fields in detail … then you'll get an intuitive idea of all of them (loosely speaking, grad is the gradient as on a hill, div is the amount that is being created per unit area, which is usually zero, and curl measures how a vector field loops around a line) 


#7
Feb713, 02:11 PM

P: 887

tinytim, that's not quite right. Indefinite integrals have an arbitrary constant because you don't know the starting point, but definite integrals don't need one.
##\int_0^x## x' dx' = x^{2}/2 


#8
Feb713, 02:16 PM

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oops!
Thanks, Khashishi! … I'll edit that! 


#9
Feb713, 03:13 PM

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Most upperdivision E&M textbooks for physics students start with a chapter about vector calculus that is tailored to the needs of physics students in such a course. The general approach and emphasis is intended to fit in with the physics that follows, and tends to be different from what you find in vector calculus taught as a math course. Examples are Griffiths, "Introduction to Electrodynamics", or Pollack & Stump, "Electromagnetism". If you don't want to spend money for one of these, you can probably find a copy in your university library. 


#10
Feb713, 04:59 PM

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Don't future (electrical, mechanical, ...) engineers in your school/country take a serious course in (multivariate) calculus up to PDE's before or at the same time with theire general physics / electromagnetism lectures ? In my country/hometown they do. 


#11
Feb713, 05:28 PM

P: 202

Thanks tinytim and Khashishi, thats the best explanation I've seen so far.
And jtbell I've just bought "Introduction to Electrodynamics" Griffiths, reading text books isn't my preferred style of learning, but I guess I'm gunna have to give it a go. dextercioby I studied implicit differentiation and PDE's of the form dz/dx and dz/dy in A level maths in college before I came to university, but all we did was differentiate multi variable functions and use the PDE to find gradients, we never put together complex PDE's and solved them. I am currently studying vector calculus in my maths lectures alongside my EM lectures. Problem is the maths lectures aren't going very well, and therefore I'm also finding the EM lectures difficult. 


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