VCE Specialist 3/4 Question Thread!

[scribble]

i'm not b^3, but you just need to find the y value for when x = r.
from the previous parts, you found the equation of the line to be y=H/b-a) * (x-a) right?
so when x=r, y = H/(b-a) * (r-a)
then if you put (H/(b-a) * (r-a)) as your upper terminal, and 0 as your lower terminal for the same volume integral as you used in parts a and b, you should get it

[shogiking]

Hi!

I just have one question which I'm having difficulty with from some revision I've been doing.

I had a graph and had to find the equation of the hyperbola and got (x+2)^2/1 + (y-1)^2/4 = 1

The next part said list a sequence of transformations that when applied to this hyperbola would give x^2 - y^2 = 1

As well as doing this, could someone also just tell me the transformations on that hyperbola? Ie how you would get from x^2 - y^2 = 1 to that hyperbola I think is how you could also word it..maybe, which would be the exact opposite I guess haha but yeah if someone could help me out that'd be great!

[Homer]

hey im not quite sure if im right but

translation of 2 units to the negative x-axis, and 1 in the positive y-axis
dilation of 4 from the x-axis?

[sin0001]

i'm not b^3, but you just need to find the y value for when x = r.
from the previous parts, you found the equation of the line to be y=H/b-a) * (x-a) right?
so when x=r, y = H/(b-a) * (r-a)
then if you put (H/(b-a) * (r-a)) as your upper terminal, and 0 as your lower terminal for the same volume integral as you used in parts a and b, you should get it

Thanks!

[lzxnl]

(x+2)^2/1 + (y-1)^2/4 = 1 is NOT a hyperbola. I am assuming you mean
(x+2)^2/1 - (y-1)^2/4 = 1
to
x'^2 - y'^2 = 1
dashes are mine

so we have x' = x+2
y' = (y-1)/2

so just reading off, translation in positive x direction of two units
translation in one unit in negative y direction
dilation factor 1/2 from x axis

Homer, be careful. Translating by one unit in the positive y direction means replace y with y - 1. In the original equation, that would yield a y-2 term. Also, it's ((y-1)/2)^2; dilation factor 1/2, not four. It's not from x^2-y^2=1 to the hyperbola, it's the other way around.

[dpagan]

Hi could someone take a look at this question.  I don't think I'm doing it right:

A large decorative concrete pond is to be sunk into the courtyard of a new building complex.  This curve is to be rotated about the y-axis to form the shape of the pond which is of the form: f(x)= (m/x) + n where m and n are constants.

The bottom and top is circular with diameters 6 m and 8 m respectively.  The middle of the bottom of the pond corresponds to the origin of the Cartesian axes and the pond is to have a depth of 6 m.

Find values of m and n.

[SocialRhubarb]

Okay, let's see what we have:
Top has a diameter of 8m, and radius of 4m.
Bottom has a diameter of 6m, and radius of 3m.
The bottom of the lake sits on the x-axis, since the middle of the bottom is on the origin and the bottom is circular.
The top of the pond must sit on the line y=6, since the depth of the pond is 6 metres.
The edges of the cross-section then correspond to and

Now if we sub these in to the form given:





Subtracting the second equation from the first:





Substituting back into the original equations:

[dpagan]

yeh that's right.  I just forgot the -ve 72.

Thanks very much

[Will T]

Help pls:

Sketch {z:z=x^2, for all x in R}






Also
Therefore, the sketch is the two points (0,0) and (1,0), right...?
I am so confused....

[zvezda]

Help pls:

Sketch {z:z=x^2, for all x in R}






Also
Therefore, the sketch is the two points (0,0) and (1,0), right...?
I am so confused....

z is only real so the solution is on the real axis. the values that x^2 can take is [0, infinity). so on the argand diagram its a straight line on the real axis from 0 inclusive to infinity

[brightsky]

z is only real so the solution is on the real axis. the values that x^2 can take is [0, infinity). so on the argand diagram its a straight line on the real axis from 0 inclusive to infinity

i beg to differ. i think the graph is literally just two points, as willt said in his post. counterexample: take the point (2,0). sub it into the equation: 2=4.

[zvezda]

i beg to differ. i think the graph is literally just two points, as willt said in his post. counterexample: take the point (2,0). sub it into the equation: 2=4.

z and x are different though, didn't you just have them equalling each other in that counterexample?

edit: dw i see it now.

edit again: ive come across this q before and it has what ive done.

[Will T]

z and x are different though, didn't you just have them equalling each other in that counterexample?

edit: dw i see it now.

edit again: ive come across this q before and it has what ive done.

Yeah the solutions for that hand-out have it as how you've done it, but you know how poor the quality of Haileybury homework is. *cough Methods revision cough*.
I think what it is saying is:
z is generated by x^2, I.e. put in values for x and you get a value for z
For example, x=2, z = 4, i.e. z= 4+0i so (4,0).... x=3, z=9, z=9+0i so (9,0).....

Yeah I don't know though, it kind of goes against reasoning..... because z literally is defined as being x +iy.........

[BubbleWrapMan]

You shouldn't assume unless it's given in the question. The more correct definition is ; we only use and as an abbreviation. On that note, if you use this abbreviation in a question, you should really write "Let ".

So for the set of complex numbers such that , it literally just means , so it's half the real axis.

[Will T]

You shouldn't assume unless it's given in the question. The more correct definition is ; we only use and as an abbreviation. On that note, if you use this abbreviation in a question, you should really write "Let ".

So for the set of complex numbers such that , it literally just means , so it's half the real axis.

That clarifies matters. For reference, the question did say 'Given z=x+iy, show the following on the complex plane:' so in this case, would it be the two points?