4/9/2004, LINE INTEGRALS II Math21a,O. KnillSection 13.2, Problems 20,24,40REVIEW LINE INTEGRALS.RCF · dr =RbaF (~r(t)) · ~r0(t) dtis called the line integral of F along the curve C. If F is a force, then the line integral is work.A RIDDLE ABOUT WORK.1) If you accelerate a car from 0m/s to10m/s = 2.24miles/hour the kinetic energyneeded is mass times 102/2 = 50.2) If you accelerate a car from 10m/s to20m/s = 4.48miles/hour the kinetic energy hasincreased by 202/2 − 102/2 = 200 − 50 = 150.3) Now watch situation 2) from a moving coordinatesystem which moves with constant 10m/s. In thatsystem, the car accerates from 0 to 10 meter persecond.Because all physical laws are the same in differentcoordinate systems (going from one to the other isa Galilei transformation), the energy should be thesame when accelerating from 0 to 10 or from 10 to 20.This is in contradiction to the fact that acceleratingthe car from 0 to 10 needs three times less energythan accelerating the car from 10 to 20. Why?The picture above shows the Swatch car, anextremly energy friendly car seen often inEurope, where gas prizes are much higherthan here (even now)EXAMPLE. Compute the line integral for the vector field F (x, y) = (x2, y2) along the boundary of a square inthe counter clockwise direction.We split up the path into four paths:γ1: r(t) = (t, 0) for t ∈ [0, 1], γ2: r(t) = (1, t) for t ∈ [0, 1],γ3: r(t) = (1 − t, 1) for t ∈ [0, 1], γ4: r(t) = (0, 1 − t) for t ∈ [0, 1],then γ = γ1+ γ2+ γ3+ γ4.The integral is thenZγF · dr =Zγ1F · dr +Zγ2F · dr +Zγ3F · dr +Zγ4F · drxyPlugging in F · dr = F (r(t)) · r0(t) givesZ10(t2, 0) · (1, 0) dt +Z10(1, t2) · (0, 1) dt +Z10((1 − t)2, 1) · (−1, 0) dt +Z10(0, (1 − t)2) · (0, −1) dtand end up withR102t2− 2(1 − t)2dt = 0.AMPERES LAW. A wire along the z-axes carries a currentI. What is the strength |B| of the magnetic field~B(x, y, z) =(−By, Bx, 0) in distance r from the z-axes if we know that Am`erelawZC~B ·~dr = µ0Iholds for a closed curve C : ~r(t) = (r cos(t), r sin(t), 0) winding indistance r once around the x axes and where µ0is a constant.Remark. The Ampere law actually follows from a basic identitywhich is one of the Maxwell equations and which we will see laterin this course.Solution: compute B(r(t)) = (−B sin(t), B cos(t), 0) and r0(t) = (−r sin(t), r cos(t)) The line integral is 2πrBand because we know this is µI, we have B = µI/2πr.We see that the magnetic field decays like 1/r. One can actuallyderive this result also from special relativity: A striking applica-tion of the Lorentz transformation is that if you take two wiresand let an electric current flow in the same direction, then the dis-tance between the electrons shrinks: the positively charged ionsin the wire see a larger electron density than the ion density. Theother wires appears negatively ”charged” for the ions and attracteach other. If the currents flow in different directions and we gointo a coordinate system, where the electrons are at rest in thefirst wire, then the ion density of the ions in the same wire appearsdenser and also the electron density is denser in the other wire.The two wires then repell each other. The force is proportional to1/r because two charged wires in distance r attract or repel eachother with the force 1/r.A PERPETUUM MOTION MACHINE. Before entering high school, one of my passions was to construct”perpetuum motion machines”. Physics classes in school quickly killed that dream. But still, these machinesstill fascinate me, especially if they work! Will will see more about these machines next week, but here is amodel which allows us to practice line integrals:Problem: by arranging cleverly charged plates (see lecture), we realize a static electric field~E = (0, 0, x). Weconstruct a wire along the elliptical path C : r(t) = (cos(t), 0, 3 sin(t)). What is the VoltageV =ZC~E(r(t)) · r0(t) dtwe measure at the wire?Answer:R2π03 cos2(t) dt = 3π.30 second background info in electricity: when an electron is moved around in an electric field alonga path C, it gains the potentialRC~E ·~dr. This potential is also called ”voltage” and measured in ”Volts”.When moving a charge through a voltage differenence, it gains some energy. This energy is proportional tothe amount of charge going through the wire as well as the voltage V . If the charge is It which is ”current”times ”time”, then the energy is V It, which can also be seen as the power V I (”volt times ampere = Watts)multiplied with time. For example, through a light bulb of 100W , a current of about one ampere flows if thevoltage is 110 Volts. If we let the lamp burn for 10 hours, we use the energy 10 · 100W h which is one Watthour. In Massachusetts, a Kilowatt hour costs about 10 − 15 cents.Running a 100 Watt light bulb for 24 hours costs a
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