Exciting... I'm seeing all of these ideas I know about and heaps I have no idea about and understanding how all this fits together is a boggle. Can't wait till the holidays :D
Did some thinking, and you are right; this is wrong. However, the rest of the proof is still correct with one additional observation: R*u_R = u_x Rcos(t) + u_y Rsin(t) One can take the constant R out of the integral and divide both sides (0 on the LHS and the integral on the RHS) by R. From there the proof can continue as in the video. That is why Dr. C gets the same result, he just missed a little calculational step.
Something isn't quite right. at 9:40 you claim that d/dR by the chain rule would give you the integrand but it doesn't. You have an extra R in the integrand. I think the reason is that you didn't divide by |r'(t)| = R which is necessary to give you the unit normal since you are not parameterizing by arc length. Or, I could be wrong.
It seems he includes a R to much when he takes the chain rule derivative of Rcos(t). When he does the subsequent inverse derivation on R, he does it as if the above R does not exist. Thus, the derivation is OK, it seems despite a "typgraphic" error. These Videos are extremly valueable, they make an almost forbidding area of mathematics approachable wirhout the fuss of complex incomprihensible formalities of greek letters etc.
On furher reflexion, he explicitly declares R as a constant. Thus, he can divide og multiply both sides by R without changing the equation. However, he treats R in one contest as a constant, in the other as a variable. I suppose this is OK
Because (Ux, Uy) is projected against the normal to the curve, and not against the tangent, as you usually would do to compute the circulation... He is interested in the outward flux in this case
![KSI - Dirty [Official Visualiser]](http://i.ytimg.com/vi/VoDh1h1dPUE/mqdefault.jpg)
Learned more than the class I paid $1200 to attend... Thanks
Exciting... I'm seeing all of these ideas I know about and heaps I have no idea about and understanding how all this fits together is a boggle. Can't wait till the holidays :D
Minute 10:28 ... an R should come out in front of the partial d/dr ... Right?
9:40 if you use "chain rule" R disapear, you will have ux cost+uy sint , doesn't it?
Same here? Anyone have an answer for this or is this just a little mistake?
Did some thinking, and you are right; this is wrong. However, the rest of the proof is still correct with one additional observation:
R*u_R = u_x Rcos(t) + u_y Rsin(t)
One can take the constant R out of the integral and divide both sides (0 on the LHS and the integral on the RHS) by R. From there the proof can continue as in the video. That is why Dr. C gets the same result, he just missed a little calculational step.
My question to you, Chris, is why did you change "r" (radius of the ball), for "R", in the proof.
thanks, can You please explain why the solutions of Laplace's equations are continuous?
Great Video, Thank you!
I think it would be easier to introduce the directional derivative of direction of radial vector instead of parameterising.
Great video, ty!
Something isn't quite right. at 9:40 you claim that d/dR by the chain rule would give you the integrand but it doesn't. You have an extra R in the integrand. I think the reason is that you didn't divide by |r'(t)| = R which is necessary to give you the unit normal since you are not parameterizing by arc length. Or, I could be wrong.
It seems he includes a R to much when he takes the chain rule derivative of Rcos(t).
When he does the subsequent inverse derivation on R, he does it as if the above R does not exist. Thus, the derivation is OK, it seems despite a "typgraphic" error. These Videos are extremly valueable, they make an almost forbidding area of mathematics approachable wirhout the fuss of complex incomprihensible formalities of greek letters etc.
On furher reflexion, he explicitly declares R as a constant. Thus, he can divide og multiply both sides by R without changing the equation. However, he treats R in one contest as a constant, in the other as a variable. I suppose this is OK
why is it Ux*dy -Uy*dx and not Ux*dx + Uy*dy ??
good video!
Because (Ux, Uy) is projected against the normal to the curve, and not against the tangent, as you usually would do to compute the circulation... He is interested in the outward flux in this case
@@gustavobagu7156 I think it is the opposite, (dx,dy) is normal to the curve and (dy,-dx) is tangential to the curve.
Thank you, but still I don’t understand it. ㅜㅜ;;;
OMEGA!!!