VCE Specialist 3/4 Question Thread!

[speedy]

I was thinking about the shaded red section:

They give you the forth line Arg(z) = pi/3 (y=root(3)x) which means that the region has to be the one they shaded.

[magneto]

quick question:
A particle held in equilibrium by three concurrent coplanar forces P, Q and R
P is 5N east
Q is 5N north
therefore R is 5sqrt2N SW

but how does that show that they are concurrent?

thanks

[lzxnl]

I was thinking about the shaded red section:

Think outer sector - inner sector

quick question:
A particle held in equilibrium by three concurrent coplanar forces P, Q and R
P is 5N east
Q is 5N north
therefore R is 5sqrt2N SW

but how does that show that they are concurrent?

thanks

Three lines are concurrent if they intersect the same point. I think they mean three forces acting on a point.

[eagles]

Sorry I'm lost once again   

Just like you said "outer - inner" why is the red shaded region not possible?

I understand up to the annulus part but I think I am stuck with the angle and the line part.

Thanks for your help!

[lzxnl]

Sorry I'm lost once again   

Just like you said "outer - inner" why is the red shaded region not possible?

I understand up to the annulus part but I think I am stuck with the angle and the line part.

Thanks for your help!

I did not read the question that time LOL I thought you meant find the area. OK.

The reason why your graph isn't good enough is because 1 < |z| < 2 means |z| is between 1 and 2. But |z| = 1 is a FULL circle. So is |z| = 2. Hence, you need to shade in the region between two FULL circles. You haven't done that. You've shaded in the region between two parts of two full circles.

[speedy]

I did not read the question that time LOL I thought you meant find the area. OK.

The reason why your graph isn't good enough is because 1 < |z| < 2 means |z| is between 1 and 2. But |z| = 1 is a FULL circle. So is |z| = 2. Hence, you need to shade in the region between two FULL circles. You haven't done that. You've shaded in the region between two parts of two full circles.

It also has the bounds y=(1/root(3))x and Arg(z) = pi/3-> which means it can't be the red section and must be the section given in the assesors report

[eagles]

Hmm. Pls stay with me, I'll explain my train of thought:

The angle pi/3 extends from the line y=0 in the first quadrant to the second dotted line counting from the first quadrant.

The line y=(1/root3)x is the bolded line.

So to shade between the 2 circles (1 with radius 1 and the other with radius 2) and between the above given regions, I believe we can shade above the line and below the angle (so the assessors' answer) or else we can shade below the line but above the angle (so the red shaded part).

I am still wondering why the red one isn't accepted? Thanks again.

[speedy]

Hmm. Pls stay with me, I'll explain my train of thought:

The angle pi/3 extends from the line y=0 in the first quadrant to the second dotted line counting from the first quadrant.

The line y=(1/root3)x is the bolded line.

So to shade between the 2 circles (1 with radius 1 and the other with radius 2) and between the above given regions, I believe we can shade above the line and below the angle (so the assessors' answer) or else we can shade below the line but above the angle (so the red shaded part).

I am still wondering why the red one isn't accepted? Thanks again.

How is the red enclosed by all four lines? If it were Arg(z) < pi/3 I think the red region would work. 

[eagles]

How is the red enclosed by all four lines?

Thanks so much!! I get it now!!!

I just realised that I've been assuming Arg(z) is a region and not a line!!!!

[speedy]

Thanks so much!! I get it now!!!

I just realised that I've been assuming Arg(z) is a region and not a line!!!!

Yeah aha! I edited my post as I realised this

[eagles]

How is the red enclosed by all four lines? If it were Arg(z) < pi/3 I think the red region would work.

Just be careful about this because if that restriction was there then I think we would have to shade in the fourth and part of the third quadrant too.
Don't forget -pi < Arg(z) </= pi 

[speedy]

Just be careful about this because if that restriction was there then I think we would have to shade in the fourth and part of the third quadrant too.
Don't forget -pi < Arg(z) </= pi 

Yeah i know aha I didn't mean to imply that the red region would be the only region :~S

[devilsadvocate]

For general solutions of cos(x) = a, would I say x = 2npi +- arccos(a) or just 2npi + arccos(a)?

Also for the Argand diagram, if I have one inclusive line/curve and one exclusive (say |z|=<2 and |z-1|<|z|) would the intersection of these two lines have an open circle or do I not need to worry about it?

Thanks in advance

[magneto]

thank you! so much!

i have another one:
please refer to the attachment

[veryJasonn]

For general solutions of cos(x) = a, would I say x = 2npi +- arccos(a) or just 2npi + arccos(a)?

Also for the Argand diagram, if I have one inclusive line/curve and one exclusive (say |z|=<2 and |z-1|<|z|) would the intersection of these two lines have an open circle or do I not need to worry about it?

Thanks in advance

Definitely the +/-, since 2npi + arccos(a) would be removing half the solutions. For example, solutions to cos(x) = 1/2 would involve pi/3 and -pi/3, but 2npi + arccos(a) has no way of equalling -pi/3 since arccos(a) is pi/3.

Also for the second question, I don't see in what scenario you would require to mark the intersection with a dot. Intersections (or intercepts) are typically marked with a cross, as that is standard, so no dots