Post by: Zebra on November 10, 2011, 03:35:20 pm
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....
Post by: dc302 on November 10, 2011, 04:00:18 pm
Post by: Zebra on November 10, 2011, 04:02:36 pm
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
Post by: dc302 on November 10, 2011, 04:12:52 pm
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?
Post by: Zebra on November 10, 2011, 04:16:14 pm
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
Post by: dc302 on November 10, 2011, 04:21:56 pm
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).
Post by: Zebra on November 10, 2011, 04:38:02 pm
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
Post by: Zebra on November 10, 2011, 04:42:35 pm
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.
Post by: Zebra on November 10, 2011, 04:56:44 pm
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
Post by: baws on November 10, 2011, 05:00:33 pm
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.'
Post by: Zebra on November 10, 2011, 05:06:51 pm
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?
Post by: dc302 on November 10, 2011, 05:10:31 pm
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.
Post by: abeybaby on November 10, 2011, 05:14:24 pm
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 :P
Post by: baws on November 10, 2011, 05:32:34 pm
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.
Post by: Zebra on November 10, 2011, 05:35:31 pm
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!
Post by: dc302 on November 10, 2011, 05:56:35 pm
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!
Well, for f''(x) to change signs at x=a, we need that f''(x) is positive on one side, negative on the other side, and of course 0 at x=a.
Ie.
y = x^3, y' = 3x^2, y'' = 6x
We want y'' to change signs at x = 0
So, y''(0) = 0, y''(1) = 6 > 0, y''(-1) = -6 < 0. So thus it changes sign and we have a POI. I may have been unclear but to 'change signs at x=a' automatically means it must be 0 at x=a anyway.
Post by: Zebra on November 10, 2011, 06:07:23 pm
sorry i'm really struggling
Post by: abeybaby on November 10, 2011, 06:12:51 pm
so we are only considering the sign changes in f''(x) not f'(x)?
sorry i'm really struggling
thats correct. what f'(x) does around x=a doesnt mean anything for points of inflexion
Post by: Mao on November 11, 2011, 01:33:35 am
so we are only considering the sign changes in f''(x) not f'(x)?
sorry i'm really struggling
thats correct. what f'(x) does around x=a doesnt mean anything for points of inflexion
Extension:
To see if f''(x) changes sign, you can simply evaluate f'''(x). If
Post by: TrueTears on November 11, 2011, 01:36:43 am
Post by: Mao on November 11, 2011, 01:45:18 am
also just to emphasize the point, plugging in values and checking whether f''(x) changes sign is a good idea in practice however to be more pedantic, evaluating f'''(x) would be a bulletproof method, this is because if you plug in values either side of 0, you might select "too big" of a number and thus lead to incorrect conclusions.
but if f'''(x) turns out to be inconclusive, we'll end up having to evaluate
Here, have the general n-order derivative test for shits and giggles. :P
http://en.wikipedia.org/wiki/Higher-order_derivative_test
Post by: TrueTears on November 11, 2011, 01:48:06 am
Post by: Mao on November 11, 2011, 01:50:57 am
LOL, i'd rage if they did that in a VCE paper haha
...
:P
We should probably bring it back to Specialist level now.
Post by: dc302 on November 11, 2011, 03:16:52 am
Post by: Zebra on November 11, 2011, 07:45:51 am