Hutchoo's spesh thread.

[Hutchoo]

Ahhhhhhhhhhhhhhhhhhhhh, so you can only REMOVE CONSTANTS out of an integral, but not equations like 2x +6?
If that's the case, then I 18239123183213% understand.

[pi]

Yep, ONLY constant

[Hutchoo]

http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2007specmaths1.pdf

I don't know why, but I can't seem to get question 2b. Whenever I have a complex equation with a complex coefficients, I turn into a carrot and just sit there.

Thanks for the help <3

[Lasercookie]

Conjugate root theorem won't apply. There'll be three solutions. We know that one solution from 2a is and therefore is a factor.

We could do long division, but yucky. Factorising is a bit more elegant, this one is really easy to factorise as well. But wait, didn't we do this in 2a? Well you might not have if you used a different method for that question, so I'll go through the whole thing.

From the equation given, we can factorise like this (edit: if it's unclear, we're looking at the first half z^3 - (\sqrt{5} - i)z^2  and the second half of the equation 4z - 4\sqrt{5} + 4i separately here)

and then factorise that a bit more
The step in 2a that I'd be continuing from is this



Solve that:

[Hutchoo]

Can you show me the working for a)? I don't fully understand this "factorising method" /my method seems a bit shit.
Thanks

[Lasercookie]

The working for a starts from this

From the equation given, we can factorise like this (edit: if it's unclear, we're looking at the first half z^3 - (\sqrt{5} - i)z^2  and the second half of the equation 4z - 4\sqrt{5} + 4i separately here)

and then factorise that a bit more

And the null factor law and solve for z. I don't think that's what you were confused on though. I'll go through what I did at the start a bit more slowly.



Factorise the z^2 from the first bit:

Factorise the 4 from the second bit (also expanding out the brackets from the first bit)

(z - \sqrt{5} + i) is a common factor of both of those two bits there, so factorise that out

[Jenny_2108]

If you can't recognise the common factor like Laseredd did, you may do this way

From 2a) is a solution so



Now we find a

[Hutchoo]

You guys are awesome! Thanks.

[Hutchoo]

Hey guys
Can someone please help me with these?

[Phantom-II]

Horizontally along plane:

Fnet = 12gsin30 - friction
=6g-((sqrt3/10)*12gcos30)
=6g-(18g/10)
=4.2g
Fnet/total mass = accel.
 so /22 = (4.2/22)g as req.

b) either whole system about to move up or down. If down:
Fnet = 0
12gsin30 = P+friction
p= 6g-((sqrt3/10)12gcos30)
=6g - 1.8g
=4.2g

if up:
Fnet = 0
12gsin30 + friction = P
6g+ ((sqrt3/10)12gcos30) = P
=7.8g

so P can be between those values

alternatively, you know the constant down-the-slope force component is 12gsin30, so you can + or - the max. friction to get the limits

[Hutchoo]

http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2009specmaths2-w.pdf

Question 15 MC. Menounderstand D=

[paulsterio]

There you go - you can just use geometry to figure out the resultant (use the cosine rule - you have two sides and the angle between)



Solve for F

Edit - Made a mistake with the Angle before, thanks Laseredd!

[Phy124]

I'm not sure what is the most efficient method having not done spesh and not knowing what formulae you guys given, but one way of doing it would be splitting the two forces into horizontal and vertical components and the finding the hypotenuse (/) of a triangle with bottom edge being that of the sum of the horizontal forces and the right edge being that of the sum of the vertical forces.

That is;







edit: Ah yeah, do it Pauls way if you're given the cosine rule.

[Hutchoo]

AHHHHHHHH, THE COSINE RULE!
Thanks Paul

Also, Ranga, that method is boss.

[paulsterio]

Btw, just a little tip. Back when I used to do a crapload of practice exams and then my real VCAA exams, I was always sick of MC questions (they're annoying as hell) - so I always tried to answer the question without actually answering the question (i.e. use a little bit of intuition). If you use intuition, you really can answer this question in around 5 seconds.

Here's how I would do it in a real exam.

So we have 4N NE and 3N E. We want the magnitude of the resultant, we know that this magnitude has to be between 4N and 7N. Look at the answers.



Now, obviously the answer can't be 7, it HAS TO BE smaller than 7 (vectors are in different directions). Not A.
It can't be B, that's more than 25, which is well above 7.
C is possible - keep C present
D is not  possible, that's close to 2.7-ish, which is below 4 (how did I know that? the square root of 2 is around 1.4, so close to 1.5 - so 25 - 12 x 1.5 = 7, so sqrt has to be between 2 and 3)
E is not possible because that's GREATER THAN 7 (square root of 2 is 1.4 and 12 x 1.4 < 18)

So thus, the answer can only really be C. If you have quick eyes, you could have really done this question in a matter of seconds, or even better, during reading time

You answered the question, without actually answering the question asked