VCE Specialist 3/4 Question Thread!

[lzxnl]

If in doubt, don't do that, because we're talking VCE where anything outside the course will be harshly penalized if one step is wrong.

[Jaswinder]

{z:|z-2i|=1} find the exact maximum and minimum values of Arg(z)

[Phy124]

{z:|z-2i|=1} find the exact maximum and minimum values of Arg(z)

Start at this post and read the few that come after it 

[Jaswinder]

hmm i tried, but i cant understand, is there a different way to approach this question?

[lzxnl]

So let's rephrase the question.We have a circle x^2+(y-2)^2=1
Arg(z) is really asking for arctan(m) where m is the line from the origin to any point (Re(z),Im(z)) on the circle.
What are the extreme points? If you use some geometric intuition, these points occur where a line from the origin is a tangent to the circle. Find these points. Then find the gradients of these lines. You should then be able to work out Arg z from these gradients.

[Jaswinder]

Find these points.

could someone help me do that

[lzxnl]

x^2 + (y-2)^2 = 1
Intersects with y = kx
so x^2 + (kx-2)^2 = 1
x^2(k^2 + 1) - 4kx + 4 = 1
x^2(k^2+1) - 4kx + 3 = 0
We want a tangent, so only one intersection point  for a given k. Discriminant = 0
16k^2 - 4*(k^2 + 1) * 3 = 0
16k^2 - 12k^2 - 12 = 0
k^2 = 3
k=+-sqrt(3)
so from this, solve for x
and hence y
You'll have your xy coordinates and then you can convert that to Arg z values.

[Alwin]

{z:|z-2i|=1} find the exact maximum and minimum values of Arg(z)

could someone help me do that

Well, personally I can see two methods, algebraic and geometric, but there may be more.

Algebric (long way)
As nliu1995 pointed out before, we have a circle


Now, the ray Arg(z) can be represented by the Cartesian line:


Personally, I like diagrams to help illustrate the point:

Image source

http://www.wolframalpha.com/input/?i=plot+x%5E2%2B%28y-2%29%5E2%3D1+and+y%3Dmx+where+m%3D%7B-3%2C+-2%2C+-1%2C+-0.5%2C+0%2C+0.5%2C+1%2C+2%2C+3%7D

Note that wolfram alpha has not drawn the y-axis at x=0.

So clearly we want to find the values of m where there is only one point of intersection:

Working

Geometric (short, easy way I prefer)
Again, we begin with a diagram:

Image source

My amazing paint skills, years of hard work right there. Oh, and http://www.wolframalpha.com/input/?i=plot+x%5E2%2B%28y-2%29%5E2%3D1+and+y%3Dmx+where+m%3D%7B-sqrt%283%29%2C+sqrt%283%29%7D

This is a purely geometric way of solving


Clearly the two methods give the same result.



Note, we can also use the geometric method on harder questions, for example:

EXAMPLE

{z:|z+1-2i|=1} find the exact maximum and minimum values of Arg(z)

What would we do? Firstly we draw a diagram:


Then repeat the same process

Hope it helps!

EDIT 1: beaten! I think my paint skills need to be sharpened and quickened in order to out ninja nliu
EDIT 2: put working in spoilers so its less monstrous

[Homer]

Could someone please verify these answers? I dont have the solutions

where

1) Find the co-ordinated of the maximum turning point of the graph of g and use the second derivative to verify its nature.

What I got: (2,0.5)=maximum turning point since g'(2)=0 and g''(2) is negative

2) Find the co-ordinates of the point of inflection of the graph of g. Give your answers correct to two decimal places.

What I got: (-3.76,-0.15) (0.69,0.22) (3.06,0.42)

3) The region enclosed by the graph of g and the straight line with equation is rotated 360degrees about the x-axis to form a solid of revolution. Find, correct to two decimal places, the volume of the resulting solid.

What I got: 0.22 (Dont think thats right though :s)


 THANKS

[Jeggz]

Has anyone else done any of the Extended Response questions in Checkpoints, relating to Calculus?
I just want to know if I'm the only one finding them quite challenging 

[Alwin]

Has anyone else done any of the Extended Response questions in Checkpoints, relating to Calculus?
I just want to know if I'm the only one finding them quite challenging 

HI JEGGZ!

Depends what Checkpoints. I did the 2011 one (much, much more qs than the 2013 one).

It was okay, but that's just a personal opinion. remember you're asking an educational forum

[Jeggz]

HI JEGGZ!

Depends what Checkpoints. I did the 2011 one (much, much more qs than the 2013 one).

It was okay, but that's just a personal opinion. remember you're asking an educational forum

Alwin!!
I have the 2012 Checkpoints. It was okay? WHAAT? :\
I swear I've needed to look at the solutions for every second question

[Homer]

Could someone please verify these answers? I dont have the solutions

where

1) Find the co-ordinated of the maximum turning point of the graph of g and use the second derivative to verify its nature.

What I got: (2,0.5)=maximum turning point since g'(2)=0 and g''(2) is negative

2) Find the co-ordinates of the point of inflection of the graph of g. Give your answers correct to two decimal places.

What I got: (-3.76,-0.15) (0.69,0.22) (3.06,0.42)

3) The region enclosed by the graph of g and the straight line with equation is rotated 360degrees about the x-axis to form a solid of revolution. Find, correct to two decimal places, the volume of the resulting solid.

What I got: 0.22 (Dont think thats right though :s)


 THANKS

anyone?

[b^3]

You might want to check your domain restriction for part b

[Homer]

You might want to check your domain restriction for part b

oh yeahhh, thanks B^3. is 3) right?