EXAM 2 QUESTIONS

[Zebra]

1. 2006 VCAA exam 2 question 3 )a ) i)

asks to show that MAGNITUDE of acc right?

l a l < do i need to actually square and square root what i get?

:S hope that makes sense..


just in case it didnt.... so..

105600=48000 x a is the motion equation i used but obtaining acc thru this method is actually not showing the magnitude of acc....

[dc302]

Actually, yes it is.

[Zebra]

I don't get what you mean? could you elaborate... and also..

when im given a +bi (a,b positive real constants) and asked to plot -i (a +bi)
why can't i just rotate 90 degrees anticlockwise and reflect it in the x axis?

for some reason, it works if i expand the point -i(a+bi) to -a i + b

but, it doesn't work when i do transformation :S

[dc302]

I mean, 105600=48000 x a will give you the magnitude of the acceleration. They say magnitude just to clarify so you don't end up trying to write it in terms of a vector or something.

Also, you are right about rotating 90 deg anticlockwise, but the negative sign does not mean you reflect in the x-axis. Since the neg sign applies to both the real and imaginary part of the complex number, you reflect it in both the x and y axes.

And what do you mean by, 'when you do transformation' it doesn't work?

[Zebra]

thanks. I get it, so you don't always have to find the 'magnitude' sometimes it will be there for clarifying... hmm i see.

since i don't have much time till tmrw exam, i will just expand everything. I realised it's all about getting the right answer using any method.... so yeah i won't worry about it.

in 2006 vcaa mc 20, why is it wrong to say

F1 / sin (60) = F2 / sin (150) = F3/ sin(150)

lami's theorem right?

AND

lami's theorem only involves THREE forces ?? thanks man

[dc302]

Sorry, I worded that badly. What I'm saying is that, the value 2.2 IS the magnitude. But you were looking too deeply in thinking that you had to do square roots or whatever, so I was trying to say that none of that is necessary.

And, which answer to 20 are you referring to?

Yes it only works if there are three forces--it's like the sine rule (for triangles).

[Zebra]

sorry it was a calculation error. silly me.

2008 mc question 15 ! could you please check this?
i used cos (60 ) because force q is in the direction i+ sq root of 3 j. and i keep getting A

[Zebra]

okay i get the correct answer using scalar resolute.

we used scalar resolute because we are not interested in the direction of force right? we want the MAGNITUDE of the force. hmm i still don't understand why cos (60) doesn't get me the correct answer.

[Zebra]

proving a particular point as POI.

f''(x)=0
f'(x) doesn't equal zero.

is this sufficient?

OR

f''(x)=0
f'(x) doesn't equal zero.
either side of f'(x) changes sign

[baws]

proving a particular point as POI.

f''(x)=0
f'(x) doesn't equal zero.

is this sufficient?

OR

f''(x)=0
f'(x) doesn't equal zero.
either side of f'(x) changes sign

If either side of f'(x) changes sign, then f'(x)=0 which is a turning point. Your first option is correct.'

[Zebra]

wait, so rry my bad i meant,

f''(x)=0
f'(x) doesn't equal zero.
either side of f'(x) changes sign     <-----------------NO, I meant either side f'(x) sign DOES NOT CHANGE.

now, which option is better?

[dc302]

okay i get the correct answer using scalar resolute.

we used scalar resolute because we are not interested in the direction of force right? we want the MAGNITUDE of the force. hmm i still don't understand why cos (60) doesn't get me the correct answer.

If you want to resolve the forces, you can.

We have 4N in one direction and 1N in the other.

So split the 4N force into i, j components:

Q = 4cos60i + 4sin60j
P = 1i + 0j

Add the two up, we get total force F = 3i + 2sqrt3 j

Then the magnitude  = sqrt21



As for POI, neither is what you want.

You only need: f"(x) changes sign at x = a, where a is the point of inflection. You don't need f'(x) at all, unless they want you to prove its a SPOI, in which case you need f'(a) = 0.

[abeybaby]

I don't get what you mean? could you elaborate... and also..

when im given a +bi (a,b positive real constants) and asked to plot -i (a +bi)
why can't i just rotate 90 degrees anticlockwise and reflect it in the x axis?

for some reason, it works if i expand the point -i(a+bi) to -a i + b

but, it doesn't work when i do transformation :S

i=cis(pi/2). so you rotate pi/2 radians anticlockwise about the origin. then shoving a negative sign in front of everything doesnt reflect in the x-axis (there IS no x-axis!), it rotates by pi radians. -1=cis(pi), so its a rotation pi radians anticlockwise about the origin.

and in showing P.O.I, the two things you need to show are this:

1) f''(x)=0
2) f''(x) changes sign at x.

if you want a stationary P.O.I, everythings the same but f'(x)=0 as well.



EDIT: dc beat me

[baws]

wait, so rry my bad i meant,

f''(x)=0
f'(x) doesn't equal zero.
either side of f'(x) changes sign     <-----------------NO, I meant either side f'(x) sign DOES NOT CHANGE.

now, which option is better?

Showing f'(x) does not equal zero subsequently shows that the sign does not change either side of it. Showing f''(x)=0 and f'(x) =/= is sufficient.

[Zebra]

oh boy... what do you guys mean by 'f''(x) changes sign at x'?

f''(x) should be zero!


so...... oh god i've been looking for how to verify POI / SPOI for like the last few days but I still don't get it.
ARGH!