Explain basic calculus in physics to me...

Factual Questions
1

I’m a second-year college student majoring in mechanical engineering and I’m currently in the midst of some intro-level classes and am seemingly struggling with physics equations and their perplexing use of calculus to me. Now I’m no stranger to calculus (currently in Calc 2 and doin’ fine), and I understand it in its own entity.

What I don’t get is the use of various calculus symbols and whatnot in physics formula. For example, in physics we’re currently doing electricity and magnetism. Take the Biot-Savart law for example:

Now how the heck can I solve that integral? I don’t see an I anywhere in that integral to solve… Or are integrals just used in physics formulas to symbolize the addition of several elements? Whenever a physics formulas use a d(x) (x being whatever variable) is that just representing an infintestimal amount of something?

I’ve also seen my share of formulas involving closed-line integrals and double integrals but I have no idea what I’m supposed to do with them either. It seems there’s a distinct line drawn between my calc classes and physics classes and neither seems to want to explain the other. Help?

2

That’s a vector integral, and not something you’d usually see until third-semester calculus or later. I’m surprised they don’t list that as a prerequisite for whatever class you’re in.

3

Hmm maybe that would explain it… Our calculus book doesn’t start involving vectors until much later than where we are at (just finished trig substitution and improper integrals) so perhaps that would be why I look at these physics formulas and am completely lost. I’m just grateful when I come across an equation with no integral signs or differentials.

4

Often times basic E&M is given in physics courses before Calculus III(where vector calc and multi-dimensional calc is introduced). They simpley use instances of Maxwell’s equations where you can use symmetry arguments and the like to figure out the necessary integrals without having to actually know anymore then the basic concepts.

Brandon, I’d go find a copy of “Div, Grad, Curl and all That” by Schey which gives a good, informal introduction to Calc III topics applied to E&M that outta tide you over till its introduced more formally in your math classes. It’s short (there’s like four chapters, and you could probably do a single chapter in one sitting of a few hours), and if memory serves, I don’t think it assumes any knowledge beyond Calc I.

5

I agree with ultrafilter. Vector integrals are definitely calculus III material.

If your physics class requires working that equation, then look ahead in your calculus book (use the index) for the terms “cross product” and “line vector” or “line vector calculus.” Or, find the resource suggested by Malodorous.

This identity should help you get started on integrating an equation liked the one linked in the OP:

A x B = n|A||B|sinΘ where n is the unit vector ┴ to the plane containing vectors A and B and Θ is the angle between A and B.

Speaking from the POV of someone who just earned a bachelor’s in engineering a few months ago, I recommend switching majors or getting accustomed to working physics equations with differentials and integrals. For fuck’s sake, you’ll be taking fluid dynamics and thermodynamics later on if you continue with mechanical engineering!

6

The Biot-Savart law is a pain in the ass to deal with, so let’s drop it for something easier. Let’s say that you have a simple acceleration equation, and you want the velocity:

acceleration=t^2+3t+3

Now remember that acceleration is nothing more than the instantaneous change in velocity. In other words, dV/dt. So we have:

dV/dt=t^2+3t+3

Move dt over to the right:

dV=(t^2+3t+3)dt

From there you ought to be able to do that integral, but let’s look closer at that equation. In words what does it say? Well, on the right side we have a part that tells us the acceleration at any given time. In other words, for any given t I know that I am accelerating at some rate. We know that acceleration*time yields us a velocity, but we can’t just do that becuase our acceleration is constantly changing. So we need to take our acceleration at every single t, and multiply it by the infitesimally small amount of time that the object is accelerating at that rate.

The tool we use for that is calculus, and more specifically the integral. In essence the relationship between calculus and physics is that calculus is a tool for us to solve a certain type of problem.

I am going to go out on a limb here and say that the problem you have with the Biot-Savant law is not calculus, it’s geometry. Or maybe you just have the equation written down wrong…

Let’s assume that we have an infinitely long wire that begins directly below the point we want to find B at. Let’s call the distance, the 90 degree one, between the wire and the point Z. Basically we have:

P
:
:
:________________________

Where P is the point we are trying to find dB, the dotted line is Z, and the line is the wire. The wire starts at 0 on the left, and goes on to infinity to the right.

The Biot-Savant law is:

dB=Uo/(4pi)[(IdL x R)/r^2]

Where Uo is the magnetic constant, I is the current, and R is the unit vector to the point, and r is the distance between dL and the point
You should know already that the cross product is:

A x B=|A|*|B|*sin(theta)

in our case it’s:

dL x R=|dL|*|R|*sin(theta)

Now, R is a unit vector so its magnitude is 1. So we have:

dL x R=dL*sin(theta)

Now, draw a line from P to anywhere on point on the wire, with theta forming the inside angle. The length of that line, i.e. the distance from P to the point on the wire, which we called r is:

sqrt(Z^2+L^2), by the pythagorium theorem. Z being the vertical component of the triangle, and L being the horizontal component of the triangle.

The definition of the sin is opposite over hypotenuse. In this case Z is the opposite side of the triangle, and we just found the hypotenuse to be sqrt(Z^2+L^2).

Great, now we have

dL x R=dL*Z/sqrt(Z^2+L^2)

Recall that our equation is:

dB=Uo/(4pi)[(IdL x R)/r^2]

Let’s plug in what we have determined:

dB=Uo/(4pi)I[(dLZ/sqrt(Z^2+L^2))/(Z^2+L^2)]

Like the acceleration example I went through we have a magnitude of B for every L. Now we need to add them all up using calculus. In other words we have to integrate the left side from 0 to B, and the right side from 0 to infinity. You should be able to do the integral on the right side. I can’t becuase I haven’t done an integral by hand in 2 years, but my trusty calculator tells me it comes out to:

B=Uo/(4pi)(1/Z)*I

Basically every single equation you see in ME will look scary like the Biot-Savant equation. The battle is using appropiate geometry to get your integral to something that you know how to do. The trick is to get every part of your equation into one single variable (or two, or three, depending on if you are doing a single, double, or triple integral).

7

Missed this, but yes that’s exactly what it means. The dx you see in Calc class is no different from the dx you see in Physics.

8

Thanks a ton everyone! Something must be bass ackwards with the system since both physics courses (at two different colleges) have done it this way and the physics courses usually just list Calc I or II as a prereq and no more.

9

I think the problem here is not just the vectors… The reason the equation doesn’t tell you what I is, is the equation doesn’t know yet. You could, in principle, arrange wires in any crazy way, and have an I corresponding to that arrangement of wires and the power sources they’re attached to. Once you have the currents, then you can use the Biot-Savart Law to solve for all of the magnetic fields (in principle, at least: The calculation might get rather hairy). But the equation doesn’t tell you what arrangement of currents to use, since the same equation applies to all arrangements of currents.