VCE Specialist 3/4 Question Thread!

[polar]

Does anyone know what mutually perpendicular vectors are? Say you had 3 vectors: a=4i+3j-k, b=2i-j+5k and c=yi+zj-2k
Find y and z, such that a,b,c are mutually perpendicular.
All i know is that if you scalar product 2 vectors, it should equal 0 for it to be perpendicular.

yeah so a.b=0, a.c=0 and b.c=0 (each vector is perpendicular to the other two) from that you can set up a system of equations

[jono88]

Cool thanks i got it

[abcdqd]

question about asymptotes:
the equation of the graph is (x^2-5x+6) / (x^2-x-2), and after factorising everything the denominator (x-2)(x+1) suggests that vertical asymptotes are x=2 and x=-1. but if you cancel (x-2) with the (x-2) in the numerator, you get a hyperbola which doesn't have asymptote at x=2. how would i sketch the graph? do i just sketch the hyperbola normally, showing one asymptote at x=-1?

[Daenerys Targaryen]

No. Must have asymptote at x=2 as well. Because if you think about it, if you put x=2 into the equation you're going to end up with somthing over 0. CANNOT DIVIDE BY '0'

[abcdqd]

No. Must have asymptote at x=2 as well. Because if you think about it, if you put x=2 into the equation you're going to end up with somthing over 0. CANNOT DIVIDE BY '0'

so just a vertical dotted line? or do I have to draw an open dot as well

[lzxnl]

I'll explain it this way. If you sketch the graph of y=x/x, you get the same line as y=1 except that when x=0, you have an open dot at where (0,1) should be.
(x^2-5x+6) / (x^2-x-2) = (x-3)/(x+1) for all x not equal to -1 or 2
The two equations are equivalent at all other x values. x=-1 really is an asymptote. x=2 is what we call a "removable singularity"; you would have an open dot at where x=2 would be on (x-3)/(x+1), but not an asymptote.

VCE doesn't teach asymptotes formally enough. Just because the denominator is zero does NOT mean that the function has an asymptote there. y=x/x is a simple example. Strictly speaking, y=1/x^2 only has an asymptote at x=0 because when y=>infinity, x=>0 and vice versa. You let each of x,y approach infinity to find asymptotes.

[TrueTears]

lol vce needs a formalisation of limits -> resolves all queries regarding asymptotes and what not

[b^3]

Lost my post due to net dc...

Anyways, was going to point out along the lines of what nliu1995 has.

The function doesn't actually go off to infinity or - infinity at x=2, rather it is not defined at x=2.
i.e.


Had the example y=x^2/x typed up, but nliu1995 has kinda covered that.

i.e. for y=x, x=/=0, the graph doesn't become really large in either the positive or negative direction near x=0, rather it is not defined at x=0.

[abcdqd]

I'll explain it this way. If you sketch the graph of y=x/x, you get the same line as y=1 except that when x=0, you have an open dot at where (0,1) should be.
(x^2-5x+6) / (x^2-x-2) = (x-3)/(x+1) for all x not equal to -1 or 2
The two equations are equivalent at all other x values. x=-1 really is an asymptote. x=2 is what we call a "removable singularity"; you would have an open dot at where x=2 would be on (x-3)/(x+1), but not an asymptote.

VCE doesn't teach asymptotes formally enough. Just because the denominator is zero does NOT mean that the function has an asymptote there. y=x/x is a simple example. Strictly speaking, y=1/x^2 only has an asymptote at x=0 because when y=>infinity, x=>0 and vice versa. You let each of x,y approach infinity to find asymptotes.

cheers mate, cleared it up for me!

[Daenerys Targaryen]

An ellipse has a horizontal semi-axis length of 3 and a vertical semi-axis length of 2.
Given that the centre of the ellipse has coordinates (1,3), a possible parametric form for the ellipse is:

a) and

b) and

c) and

d) and

e) and

[e^1]

The parametric equations of an ellipse are:

and

Spoiler

As the centre is , then and .

Furthermore, and . Now we have and . Therefore the answer is E.

[Daenerys Targaryen]

Are the parametric equations for different graphs different?

I thought that they were interchangable; just depended what was provided

[NoticeMeSenpai]

Are the parametric equations for different graphs different?

I thought that they were interchangable; just depended what was provided

Circle: x=acos(t)   y=asin(t)
Ellipse: x=acos(t)    y=bsin(t)
Hyperbola: x=asec(t)   y=btan(t)

[Holmes]

I'm really confused with sketching rational functions. Is there a good way to sketch them other then using addition of ordinates. I know how to do reciprocal function sketching, but sometimes it's really hard to use that.

For example, I'm having trouble especially with this one:



My teacher said that we can look at what happens at x approaches +/- infinity, but I'm still really confused.

[Homer]

dont know how to