VCE Specialist 3/4 Question Thread!

[BLACKCATT]

If it covered the portion in the 4th quadrant then the complex number would have coordinates (a,-b) would it not?, But the complex number u is (-a,-b) which is in the 3rd quadrant only.

[Bestie]

can someone please help me with these questions?
19/20

Thank you

[Valyria]

If it covered the portion in the 4th quadrant then the complex number would have coordinates (a,-b) would it not?, But the complex number u is (-a,-b) which is in the 3rd quadrant only.

Never-mind, I'm pretty sure it's related to the implied restriction; -pi<Arg(z)</=pi 

[keltingmeith]

can someone please help me with these questions?
19/20

Thank you

This can be shown through the product/quotient law of complex numbers.

If z1=rcis(t) and z2=pcis(s), then:

z1z2=rpcis(t+s) and z1/z2=(r/p)cis(t-s)

[allstar]

vcaa 2012 q7
please?

also from vcaa 2012 q11?
How would I find the points of inflection?

[casbanjo]

vcaa 2012 q7
please?

also from vcaa 2012 q11?
How would I find the points of inflection?

q7 B, since Im(z) doesn't include the letter i

[casbanjo]

vcaa 2012 q7
please?

also from vcaa 2012 q11?
How would I find the points of inflection?

q11

Since dy/dx and d^2y/dx^2 both equal 0 at x=0, this is a stationary p.o.i.

Since d^2y/dx^2=0 at x=1, there is another p.o.i at x=1, but it is not stationary

Since dy/dx=0 at x=3/2, there is a stationary point at x=3/2, question states this is a minimum.

Answer is E

[Edward Elric]

For the 2013 vcaa paper 1, i need to clarify some things that I did slightly different

1. Instead if labelling the forces from the surface if the object, I did it from the cebtre of mass, is this wrong?
2. And for part b of the same question, I left tension as '5g' and did not evaluate it to '49' would I lose marks?

Thanks

[Conic]

Since dy/dx and d^2y/dx^2 both equal 0 at x=0, this is a stationary p.o.i.\

This is a common misconception. Both of these conditions will be true at a stationary point of inflection, but they don't imply that the stationary point is a stationary point of inflection. If we define , we obtain and , both of which are zero at , but the stationary point is a local minimum rather than a stationary point of inflection.

[lzxnl]

This is a common misconception. Both of these conditions will be true at a stationary point of inflection, but they don't imply that the stationary point is a stationary point of inflection. If we define , we obtain and , both of which are zero at , but the stationary point is a local minimum rather than a stationary point of inflection.

Indeed; more accurately, f'(a) = 0 and f''(x) changes sign at x=a is better

[speedy]

Could someone explain what they're doing in the second line? Thanks

[keltingmeith]

Could someone explain what they're doing in the second line? Thanks
(Image removed from quote.)

Making it a difference of two squares, by multiplying by (which, of course, is one)

[speedy]

Making it a difference of two squares, by multiplying by (which, of course, is one)

What do you think the best way to go about these questions in the exam is?

[casbanjo]

This is a common misconception. Both of these conditions will be true at a stationary point of inflection, but they don't imply that the stationary point is a stationary point of inflection. If we define , we obtain and , both of which are zero at , but the stationary point is a local minimum rather than a stationary point of inflection.

Oops :/ Sorry, I'll leave this thread to the professionals 

[lzxnl]

Oops :/ Sorry, I'll leave this thread to the professionals 

It's ok; you're welcome to make a contribution, but other people will add their part if they feel like it. That's how this forum works. Besides, you learn by contributing too.