Hi, could I have some help??
question d: I thought the inclusive point of (0,0) by the circle overrides the exclusive point of the Arg(z) ray? Since in continuous probability the "elsewhere" x-endpoint gets overridden, so I applied a similar logic here.
Note in a previous part it said: "Arg(z) = 11pi/12 has the Cartesian equation y=ax, x<0, a E R". This doesn't really mean anything since by definition the argument ray is undefined at (0,0) and can't extend through the origin (correct me if I'm wrong).
Also, for argument rays in general is the end/start point always excluded, even when translated off the origin?
question e: does anyone know why my method does not get the correct area? The correct answer is 5pi/3 -1, my working shown does not get this.
Q3 c: why must we use velocity vectors here? The q says “the paths of the two bodies”, not the direction of the paths? So I thought position vectors?
https://imgur.com/a/cAqzf
I'll try to answer best I can, but can't say I'm completely confident in my responses.
Your reasoning for question d) where the inclusive part of the circle 'overrides' the open end point can really only be applied to composite functions where technically it's all one combined function. In this case, you have two separate functions, one of which includes (0,0) and one which does not. Imagine if you drew the graphs separately, and then overlapped them to find the points of intersection: (0,0) cannot be one because it doesn't exist on the ray.
You noted that the x<0, a E R doesn't mean anything but I'd be inclined to disagree. Without it, the implication is that the y=ax cartesian equation represents Arg(z) = 11pi/12 completely, which isn't accurate. You're right in saying that the argument ray is undefined at (0,0). However y=ax without any domain restrictions does not have this property.
Yes all argument rays have the start point excluded.
I can't quite see what you're writing in question e) but what I can see is the term Area of sector, and looking at the graph that section isn't a sector. A sector, from what I know originates from the centre. The bit you're taking away from the semicircle does is not a sector.
Honestly, my circle geometry is horrendous so don't take my word for it and I would have gone for a antidiff(f(x)-g(x)) from x= (-root(3)-2) to x=0, which is (probably?) what the solutions suggested.
Last thing, I disagree with your interpretation of path. Path implies direction, so velocity seems like the natural way to go. Also, it makes little sense for position vectors to be 'perpendicular' in this question. Like sure we represent them as vectors but the vector shows where a particle is relative to the origin, not its path e.g. particle a could be at point i + j at time = 1. And we'd represent it by a line pointing to (1,1), but it isn't necessarily the case that the particle directly moved from the origin to that point, it may have gone in a circle . If we had particle b at t = 1 at point -i-j, technically the vectors a and b are perpendicular but the vectors themselves aren't representatives of the paths a and b took to get to those points.
(a.b = 0 means that their positions relative to the origin are perpendicular).
Hopefully this makes sense... if it doesn't there are many much more intelligent people on this forum who could probably help you out better than I could.
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