VCE Specialist 3/4 Question Thread!

[Alwin]

the question asks to label all the forces acting on the two masses. How do we know which direction the friction is?
The answers suggest that it should be down the plane.

thanks guys

Hmm, I don't like this question. I've done enough qs to know that in later parts it will give you values of m and perhaps even ask for the values of m such that the system is in equilibrium.

Friction always opposes the net force, but since we don't have a value of m it's not possible to deduce which way friction acts.

Eg if m is small, the friction can up act the plane and STILL be in equilibrium
Eg if m is large, the friction can act down the plane and STILL be in equilibrium

Can I ask where you got this question from ?

[ahat]

Hey guys, what method would you use for this question?

I tried using this formula:

Didn't work out too well

[Alwin]

Hey guys, what method would you use for this question?

I tried using this formula:

Didn't work out too well

Hint #1:


Hint #2:
Find the gradients of the asymptotes and then find which option matches it

Hope it makes sense

[Jaswinder]

Hey guys, what method would you use for this question?

I tried using this formula:

Didn't work out too well

Hint #1:
(Image removed from quote.)

Hint #2:
Find the gradients of the asymptotes and then find which option matches it

Hope it makes sense

when it says the 'angle between them' couldn't it be the other angle aswell? How do you know its that angle, they want?   

[ahat]

when it says the 'angle between them' couldn't it be the other angle aswell? How do you know its that angle, they want?   

:p This is a good point, because whilst I did what Alwin did the first time, I used the other angle and got stuck

[Alwin]

:p This is a good point, because whilst I did what Alwin did the first time, I used the other angle and got stuck

The angle between two lines / planes / etc is always defined as the smaller / smallest one even in R^3



Source

[LOLs99]

http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2008specmath1-w.pdf
Q9. a) why can't I find the area from 0 to pi/2 based on the y axis(dy) with cos(y)?
and can anyone please explain Q10.c.?

Many thanks

[ahat]

The angle between two lines / planes / etc is always defined as the smaller / smallest one even in R^3

(Image removed from quote.)

Source

Thank you Is this "assumed knowledge"?

[Jaswinder]

i think theres a bit more area they want you to calculate

[BasicAcid]

Do we need to know tan(theta) = modulus of m1m1 etc..?

[Stevensmay]

http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2008specmath1-w.pdf
Q9. a) why can't I find the area from 0 to pi/2 based on the y axis(dy) with cos(y)?
and can anyone please explain Q10.c.?

You would only be finding the area in the first quadrant of the graph, leaving out the area in the second.
Keep in mind this is a one mark question, therefore it's probably going to have a simple solution (which it does).

[sin0001]

I tried using this formula:

May I ask where you got this formula from? Is this assumed knowledge of some sort?
Never seen it o.O

[ahat]

 

http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2008specmath1-w.pdf
Q9. a) why can't I find the area from 0 to pi/2 based on the y axis(dy) with cos(y)?
and can anyone please explain Q10.c.?

Many thanks

I shall try my best to answer
If you took the value from the y axis, and multiplied it by 2, you would be finding this area (red - attached), which does not match the blue area. If you've read the solutions, you will notice that, like they say, it is half of a rectangle.

10c)
Hint 1: "Real constants" Conjuagate root theorem aplies. As the question says "z = w" then we already know (from the previous question) that two solution will be z = 1 + √3i and z = 1 - √3i

Hint 2: The solutions satisy |z³| = 8, therefore, solving this, z = 2 or z = -2.  Not both though. The complex number is to degree three, so can have a maximum of three solutions.

Therefore, z's that satisfy |z³| = 8 are z = 1 ± √3i and one of z = 2 or -2

Hint 3: P(z) = (z - (1 ± √3i)) (z - a) where a is going to be -2 or 2.
Expand, (z² - 2z + 4) (z - a)
Expand again.
z³ -z²(a + 2) + z(2a - 4) - 4a

Looking at the info we're given in the question, a + 2  ≠ 0 (because of non-zero real constants)
therefore, a  ≠ -2 (this satisfies the requirement "show these are the only values")

Subbing a = 2 back into (z² - 2z + 4) (z - a) we get:
z³ - 4z² + 8z - 8

May I ask where you got this formula from? Is this assumed knowledge of some sort?
Never seen it o.O

Googled "angle between two lines" :p

Edit: I have a question, if two particles are going to collide, do we equate their position vectors or velocity vectors? I always thought it was velocity, but this question says position. In what scenarios (not really sure how to describe this question) do we use velocity vector opposed to position vector for something? Thanks.

^ Don't worry, I remember why I thought this.

[Alwin]

I tried using this formula:

May I ask where you got this formula from? Is this assumed knowledge of some sort?
Never seen it o.O

Hope it makes sense

[LOLs99]

I shall try my best to answer
If you took the value from the y axis, and multiplied it by 2, you would be finding this area (red - attached), which does not match the blue area. If you've read the solutions, you will notice that, like they say, it is half of a rectangle.

10c)
Hint 1: "Real constants" Conjuagate root theorem aplies. As the question says "z = w" then we already know (from the previous question) that two solution will be z = 1 + √3i and z = 1 - √3i

Hint 2: The solutions satisy |z³| = 8, therefore, solving this, z = 2 or z = -2.  Not both though. The complex number is to degree three, so can have a maximum of three solutions.

Therefore, z's that satisfy |z³| = 8 are z = 1 ± √3i and one of z = 2 or -2

Hint 3: P(z) = (z - (1 ± √3i)) (z - a) where a is going to be -2 or 2.
Expand, (z² - 2z + 4) (z - a)
Expand again.
z³ -z²(a + 2) + z(2a - 4) - 4a

Looking at the info we're given in the question, a + 2  ≠ 0 (because of non-zero real constants)
therefore, a  ≠ -2 (this satisfies the requirement "show these are the only values")

Subbing a = 2 back into (z² - 2z + 4) (z - a) we get:
z³ - 4z² + 8z - 8

Googled "angle between two lines" :p

I get Q10.c. now, thanks to your detailed solutions . But can I use De moivre theorem to solve this?
As for Q9a. is there a way to solve the area using calculus?

i think theres a bit more area they want you to calculate

What about 0 to pi. Is it possible in that case? Im a bit confused