Recent content by Tabiri

Yeah, I just did it that way and got the right answer. Thanks!

As in solve for ##y_s## in (1) and then plug that value for ##y_s## into (2)? Edit: Also, I know that the answer for ##\frac{\partial x}{\partial s}## is ##\frac{-19}{13}##

Homework Statement If ##xs^2 + yt^2 = 1## (1) and ##x^2s + y^2t = xy - 4,## (2) find ##\frac{\partial x}{\partial s}, \frac{\partial x}{\partial t}, \frac{\partial y}{\partial s}, \frac{\partial y}{\partial t}## at ##(x,y,s,t) = (1,-3,2,-1)##. Homework Equations Pretty much those just listed...

Nevermind, figured it out (integrate by parts a couple times).

Homework Statement Find vrms from the Maxwell speed distribution. Basically what I need to solve is the integral a\int_0^∞ \! v^4e^{xv^2} \, \mathrm{d}v Where ##a## and ##x## are constants. Homework Equations $$a\int_0^∞ \! v^4e^{xv^2} \, \mathrm{d}v $$ Where ##a## and ##x## are...

Thank you so much for the help!

Ah, so it would be ΔV=I(b-a). Then (b-a) cancels and everything works out, and it's E=I/Lr2\piσ.

Well, I=ΔV/R, so ΔV=RI. Also, E=ΔV/L, so ΔV = EL. Then RI=EL, and E=RI/L = I(b-a)/(L^2)r2\piσ Since electric field was what I was trying to find originally. The strange thing is what happens if I do it a different way. J = I/A = σE, so E=I/Aσ. Then it would be E=I/Lr2\piσ. What am I doing...

Well, the area of the shell would be the length * circumference, so A=2\piL. If it was a radial cross section, then the area of that would be dr*L. Since the current flows radially outward, then I think I would use the radial cross section. This would make the resistance R=L/Aσ = L/L(dr)σ =...

Would the net resistance just be R=(b-a)/Lr2\piσ? Since dθ is just a small segment, then if it were to be rotated all the way around the cylinder, then the total θ would be 2\pi.

I was putting what I got for A, r(dθ)L into R=L/Aσ, which goes to R=L/Lr(dθ)σ. Unless, for resistivity, would it be R=(r-b)/Aσ? Because since the current radiates outwards from the inner surface to the outer surface, then the current would only exist in the cylinder. Then it would be...

So since A=r(dθ)L, and R=L/Aσ, then by substitution L cancels and R=1/r(dθ)σ for the segment. If we were to extend that all the way around the cylinder, then the total θ would be 2\pi, so R=1/2\pirσ. Then substituting into the equation E=I/RL, E=I/2\pir(L)σ. Does that look right?

Would it have to do with arc length? Assuming θ is in radians, then the arc length would be s=r(dθ). Then the cross section area that the current passes through would be A=r(dθ)L.

Well, if it's an infinitesimally small angle, then wouldn't the cross section just be a rectangle, and the area would just be outer radius - inner radius * L, so (b-a)L. Edit with the resistance: R=L/(b-a)Lσ, and L would cancel, so it would just be R=1/(b-a)σ.

Well, the resistance would be R=L/Aσ, and since A=\pir^2, then R=L/σ\pir^2. To relate it to the whole cylinder, would you just multiply the whole thing by L? Because R=L/σ\pir^2 would just be the resistance for the cross section, so if you multiplied it by L then you'd get the resistance for...