VCE Specialist 3/4 Question Thread!

[ahat]

Hey,
If something is an element of Q, what does this mean?
Cheers

Rational numbers I believe where a rational number is an integer (whole number) or fraction.

[zvezda]

Rational numbers I believe where a rational number is an integer (whole number) or fraction.

Ahh i see. Thanks for that

[ahat]

Ahh i see. Thanks for that

No worries!

Hey, I'm wondering what these types of questions would be classified as. Would they be vector calculus/vectors etc.? I'm asking so I can do some internet research for similar questions/examples. Much appreciated.

[lzxnl]

I detest the name vector calculus...partly because what I associate with vector calculus is forbidding, messy and still not well understood by me

but yeah, it's calculus in vector form. It's not like the vector form is really necessary though.

[sin0001]

No worries!

Hey, I'm wondering what these types of questions would be classified as. Would they be vector calculus/vectors etc.? I'm asking so I can do some internet research for similar questions/examples. Much appreciated.

Vector Calculus/Vector Equations/Parametric Equations (or even Kinematics)
Edit: Beaten

[ahat]

^ Thanks guys!
Has anyone learnt about cross product for vectors?

[lzxnl]

If you're asking someone else, I presume lots of them have.

I have.
If you ever get stuck, all you have to remember are these few rules: ixj=k, axa=0, bx(c+d)=bxc+bxd and most importantly, axb=-bxa

You'll find that you can work out any cartesian cross product from that. I think.
Other than that, remember that |axb|=absin theta and you're done...

[ahat]

If you're asking someone else, I presume lots of them have.

I have.
If you ever get stuck, all you have to remember are these few rules: ixj=k, axa=0, bx(c+d)=bxc+bxd and most importantly, axb=-bxa

You'll find that you can work out any cartesian cross product from that. I think.
Other than that, remember that |axb|=absin theta and you're done...

I never learnt it, but thanks for the info! I just had a depressing epiphany! I don't know the projectile motion equations D: Gosh I hope they aren't on the exam.

Btw, does 'x' in your post denote multiplication or something else?

[b^3]

If you're asking someone else, I presume lots of them have.

I have.
If you ever get stuck, all you have to remember are these few rules: ixj=k, axa=0, bx(c+d)=bxc+bxd and most importantly, axb=-bxa

You'll find that you can work out any cartesian cross product from that. I think.
Other than that, remember that |axb|=absin theta and you're done...

From memory, it's not on the course/won't be examined, which is a shame really since it's not that much harder than the dot product and is quite useful.

[lzxnl]

I never learnt it, but thanks for the info! I just had a depressing epiphany! I don't know the projectile motion equations D: Gosh I hope they aren't on the exam.

Btw, does 'x' in your post denote multiplication or something else?

x denotes the cross product
cross products aren't on the exam. Projectile motion = 2 D constant acceleration

[ahat]

x denotes the cross product
cross products aren't on the exam. Projectile motion = 2 D constant acceleration

Woah, I knew something was up with post numbers! I remember the day well when you hit the 1000 posts mark. I swear I had more posts too, but just assumed they were a figment of my imagination. Does anyone know why they went down?
And thanks again.

[ahat]

If where asked to graph a relation in the complex plane and it has a restriction Arg(z) <  π/2, do we graph it between - π < Arg(z) <  π/2?

Considering this graph: y = x/2 + 1 where Arg(z) <  π/2. This line in graphed in the first and third quadrant.
When it's restricted to Arg(z) <  π/4, it's only graphed in quadrant one. Could someone please explain why?

[belsebob]

how did you guys go on the exam

[Stevensmay]

Woah, I knew something was up with post numbers! I remember the day well when you hit the 1000 posts mark. I swear I had more posts too, but just assumed they were a figment of my imagination. Does anyone know why they went down?

Year 12 2013 discussion thread got moved to Casual Games. Posts made in that subforum don't count towards your post count.
So if you made any posts in that thread, they are no longer counted thus a drop in your post count.

[papertowns]

Could someone tell me what the equation of the asymptotes are? I know for the normal hyperbolas where it's x^2 - y^2 it's y = +-b/a but what happens when it's the other way around? I've read somewhere that the b becomes the one underneath the x and the a becomes the one with the y but then answers says otherwise. I'm so confused does it matter if it's switched around?