Recent content by neilparker62

@robphy Thanks for your kind comment and desmos graph. As mentioned above, I'm working on developing kinematics lessons along the lines you suggest. In addition to the trapezoid area, I also split the area into two triangles leading to the equation s=(u+v)/2 x (u-v)/a. I also have done some...

Not at all. Will try to implement with emphasis on vector rather than scalar quantities. Thanks very much for the feedback. The 'presentation' hasn't been used yet. Am busy developing a set of lessons on kinematics. I thought I would pick that topic that as a 'testing ground' for rapid...

Thanks very much for your detailed comment - no apologies needed. Yes - for this particular lesson on interval kinematics we should indeed be using ##\Delta t## and probably ##\Delta v## and ##\Delta s## as well. I think the use of this equation is indeed conceptual since one can visualize...

For interest - comments welcome.

Some snapshots taken from the live stream. Times as recorded on my cellphone. GMT+2 23:49 01:54 02:07 02:09

ditto from Nairobi, Kenya!

May I respectfully submit that the article did not have any intention to compare human solutions with AI solutions nor indeed to compare solutions generally. AI came up with a couple of interesting additions to an already interesting thread and those were presented in the article alongside the...

The roots of the equation in part (i) are the same as the roots in part (ii). Since in both cases we have equations of the form ax^2 + bx + c = 0 with a=1 , 'c' = -c we must also have -p=(1+c)/c.

Thanks for the comment(s). Re functions: Yes - the AI engine wanted to go that way but I got a little confused by all the "defs" it came up with and couldn't keep track. As I mention in the article it's so easy for AI to make subtle programming errors which you won't notice unless you know...

"All in one" derivation of compound angle formulae - based on the video construction.

Neat geometric derivation of ##\tan(A+B)=\frac{\tan A + \tan B}{1-\tan A \tan B}## if we divide all terms in ##\frac{ay+bx}{by-ax}## by ##by##.

$$\tan x=\frac{1}{5} \implies \tan2x=\frac{5}{12} \implies \tan4x=\frac{120}{119}$$ $$\tan \left( \tan^{-1} \left( \frac{1}{239} \right)+\frac{\pi}{4}\right)=\frac{1+\frac{1}{239}}{1-\frac{1}{239}}=\frac{120}{119}$$