Login

Welcome, Guest. Please login or register.

March 29, 2024, 10:06:42 pm

Author Topic: VCE Specialist 3/4 Question Thread!  (Read 2164633 times)  Share 

0 Members and 3 Guests are viewing this topic.

geminii

  • Victorian
  • Forum Leader
  • ****
  • Posts: 787
  • Do or do not, there is no try.
  • Respect: +42
Re: Specialist 3/4 Question Thread!
« Reply #8670 on: May 12, 2017, 12:02:28 am »
0
How do I sketch a locus?

The problem is z = -125i, and I've found that the roots are 5cis(-pi/6), 5cis(pi/2), and 5cis(7pi/6).

Thanks in advance!!
2016-17 (VCE): Biology, HHD, English, Methods, Specialist, Chemistry

2018-22: Bachelor of Biomedical Science @ Monash Uni

Quantum44

  • Victorian
  • Forum Leader
  • ****
  • Posts: 756
  • Respect: +313
Re: Specialist 3/4 Question Thread!
« Reply #8671 on: May 12, 2017, 07:22:58 am »
+1
How do I sketch a locus?

The problem is z = -125i, and I've found that the roots are 5cis(-pi/6), 5cis(pi/2), and 5cis(7pi/6).

Thanks in advance!!

Plot the points on an argand diagram and draw a line from the origin to each point. The angle between each line should be 2pi/3. Then draw a circle passing through all the points.
UAdel MBBS

Ahmad_A_1999

  • Trendsetter
  • **
  • Posts: 110
  • Respect: +1
Re: Specialist 3/4 Question Thread!
« Reply #8672 on: May 13, 2017, 06:48:34 am »
0
Can someone please help me with this, I don't understand how to find the domain of (12c) given the restriction is t is greater than or equal to zero, what I have been doing is getting x=f(t) rearranging that so that 't' is the subject, then solving that inequality to find when it is greater than or equal to zero, in order to find the domain for the Cartesian equation.

Is there a better method as that will take too long  :'(

http://imgur.com/a/26gnR

http://imgur.com/a/h2k86
Is it even possible to get the derivative in to what the question wants? I'm struggling to progress past what I've written. Thanks

Hey Student, with that, I am confused with the notation used, as I write (sin(x))^2 as sin^2(x), does this apply to negative powers as well? Or is your question saying inverse sine  :-\
2017:
Chem [41] Bio [44] Spesh / Methods / EngLang 
ATAR: 95.65
2018-2020: Bachelor of Biomedical Science @ Monash
2021-2024: Doctor of Medicine @ Monash

zsteve

  • ATAR Notes VIC MVP - 2016
  • Forum Leader
  • ****
  • Posts: 748
  • The LORD is my shepherd, I shall not want - Ps. 23
  • Respect: +218
Re: Specialist 3/4 Question Thread!
« Reply #8673 on: May 13, 2017, 04:56:17 pm »
0
Can someone please help me with this, I don't understand how to find the domain of (12c) given the restriction is t is greater than or equal to zero, what I have been doing is getting x=f(t) rearranging that so that 't' is the subject, then solving that inequality to find when it is greater than or equal to zero, in order to find the domain for the Cartesian equation.

Is there a better method as that will take too long  :'(

http://imgur.com/a/26gnR

Hey Student, with that, I am confused with the notation used, as I write (sin(x))^2 as sin^2(x), does this apply to negative powers as well? Or is your question saying inverse sine  :-\

Hey there! $$x = 4\sec(t) \Rightarrow x \in (-\infty, -1]\cup[1, \infty)$$.

Also, we have

$$y = 3\tan(t) \Rightarrow y \in \mathbb{R}$$.

Thus, the domain and range are respectively $$(-\infty, -1]\cup[1, \infty)$$ and $$\mathbb{R}$$
« Last Edit: May 13, 2017, 04:59:43 pm by zsteve »
~~ rarely checking these forums these days ~~

2015: Specialist [47] | Methods [48] | Chemistry [50] | Physics [48] | English Language [46] | UMEP Mathematics [5.0] | ATAR - 99.95
Premier's Award Recipient 2016: Top All-Round VCE High Achiever
2016-2019: University of Melbourne : Bachelor of Science (Biochemistry & Molecular Biology), Diploma in Mathematics (Applied)
2019-: University of British Columbia

Ahmad_A_1999

  • Trendsetter
  • **
  • Posts: 110
  • Respect: +1
Re: Specialist 3/4 Question Thread!
« Reply #8674 on: May 13, 2017, 07:28:12 pm »
0
Hey there! $$x = 4\sec(t) \Rightarrow x \in (-\infty, -1]\cup[1, \infty)$$.

Also, we have

$$y = 3\tan(t) \Rightarrow y \in \mathbb{R}$$.

Thus, the domain and range are respectively $$(-\infty, -1]\cup[1, \infty)$$ and $$\mathbb{R}$$

Hey zsteve!

Thanks for the response!

I still don't understand how you managed to do that, could you please explain with more detail  :P
2017:
Chem [41] Bio [44] Spesh / Methods / EngLang 
ATAR: 95.65
2018-2020: Bachelor of Biomedical Science @ Monash
2021-2024: Doctor of Medicine @ Monash

RuiAce

  • ATAR Notes Lecturer
  • Honorary Moderator
  • Great Wonder of ATAR Notes
  • *******
  • Posts: 8814
  • "All models are wrong, but some are useful."
  • Respect: +2575
Re: Specialist 3/4 Question Thread!
« Reply #8675 on: May 13, 2017, 07:32:07 pm »
0
Hey zsteve!

Thanks for the response!

I still don't understand how you managed to do that, could you please explain with more detail  :P


Ahmad_A_1999

  • Trendsetter
  • **
  • Posts: 110
  • Respect: +1
Re: Specialist 3/4 Question Thread!
« Reply #8676 on: May 13, 2017, 07:44:56 pm »
0



Thanks man!

I understand all that, so would this likely come up on a non-calculator task, as it would take too long for me to draw the sec and tan functions just to figure out the domain and range  :'(
2017:
Chem [41] Bio [44] Spesh / Methods / EngLang 
ATAR: 95.65
2018-2020: Bachelor of Biomedical Science @ Monash
2021-2024: Doctor of Medicine @ Monash

RuiAce

  • ATAR Notes Lecturer
  • Honorary Moderator
  • Great Wonder of ATAR Notes
  • *******
  • Posts: 8814
  • "All models are wrong, but some are useful."
  • Respect: +2575
Re: Specialist 3/4 Question Thread!
« Reply #8677 on: May 13, 2017, 08:03:46 pm »
+1
Thanks man!

I understand all that, so would this likely come up on a non-calculator task, as it would take too long for me to draw the sec and tan functions just to figure out the domain and range  :'(
From my understanding, these are graphs that you are expected to know off by heart. A calculator should not be necessary.

Ahmad_A_1999

  • Trendsetter
  • **
  • Posts: 110
  • Respect: +1
Re: Specialist 3/4 Question Thread!
« Reply #8678 on: May 13, 2017, 09:13:32 pm »
0
From my understanding, these are graphs that you are expected to know off by heart. A calculator should not be necessary.

I've forgotten what they look like lol, I'll get to memorising them  ;D
2017:
Chem [41] Bio [44] Spesh / Methods / EngLang 
ATAR: 95.65
2018-2020: Bachelor of Biomedical Science @ Monash
2021-2024: Doctor of Medicine @ Monash

peanut

  • Forum Regular
  • **
  • Posts: 89
  • Respect: +1
Re: Specialist 3/4 Question Thread!
« Reply #8679 on: May 14, 2017, 11:18:15 am »
0
Shouldn't the domain of the first part of d'(t) be:
t E [0,2) U (6, 10]

In other words, I don't see why the derivative doesn't exist at t = 0 and t = 10

RuiAce

  • ATAR Notes Lecturer
  • Honorary Moderator
  • Great Wonder of ATAR Notes
  • *******
  • Posts: 8814
  • "All models are wrong, but some are useful."
  • Respect: +2575
Re: Specialist 3/4 Question Thread!
« Reply #8680 on: May 14, 2017, 12:28:05 pm »
0

Shouldn't the domain of the first part of d'(t) be:
t E [0,2) U (6, 10]

In other words, I don't see why the derivative doesn't exist at t = 0 and t = 10
Derivatives don't exist at the end points (or boundaries) of intervals altogether. Are you taught this in VCE?

j.wang

  • Forum Regular
  • **
  • Posts: 76
  • Respect: +7
Re: Specialist 3/4 Question Thread!
« Reply #8681 on: May 17, 2017, 04:48:51 pm »
0
can someone check if you get the same thing on your cas?
When you implicit diff 3x^3-y^2+kx+5y-2xy=4, i get 9x^2/ 2y-5 on the cas but the correct answer is (9x^2 + k-2y)/ (2y-5+2x)
what am I doing wrong? thanks :)

Sine

  • Werewolf
  • National Moderator
  • Great Wonder of ATAR Notes
  • *****
  • Posts: 5135
  • Respect: +2103
Re: Specialist 3/4 Question Thread!
« Reply #8682 on: May 20, 2017, 01:13:05 am »
+3
can someone check if you get the same thing on your cas?
When you implicit diff 3x^3-y^2+kx+5y-2xy=4, i get 9x^2/ 2y-5 on the cas but the correct answer is (9x^2 + k-2y)/ (2y-5+2x)
what am I doing wrong? thanks :)
Sorry for the late response :/

I end up getting the latter answer

my syntax input is







« Last Edit: May 20, 2017, 01:20:35 am by Sine »

peanut

  • Forum Regular
  • **
  • Posts: 89
  • Respect: +1
Re: Specialist 3/4 Question Thread!
« Reply #8683 on: May 20, 2017, 04:26:56 pm »
0
Is there a way of directly sketching a graph like |z - 1| = |z - 2i| on CAS without converting to Cartesian form by hand?

Sine

  • Werewolf
  • National Moderator
  • Great Wonder of ATAR Notes
  • *****
  • Posts: 5135
  • Respect: +2103
Re: Specialist 3/4 Question Thread!
« Reply #8684 on: May 20, 2017, 04:39:49 pm »
0
Is there a way of directly sketching a graph like |z - 1| = |z - 2i| on CAS without converting to Cartesian form by hand?
You should be able to recognise that graph :) or you will be able to with time.
The graph |z - 1| = |z - 2i|
means the distance of some point from 1 is the same distance from the point 2i on the complex plane.
Therefore the graph required is the line that is the perpendicular bisector of the points 1 and 2i.
Now let's consider the cartesian plane where the points are (1,0) and (0,2)
the mid point is M (1/2,1)
The gradient between the points is rise/run = -2/1 = -2
We need the perpendicular bisector
y-y = m(x-x)
It is perpedicular so m = -1/-2 = 1/2
and passes through (1/2,1)
so
y - 1 = 1/2 (x -1/2)
y = 1/2x -1/4 + 1
y = 1/2x +3/4

Note you normally wouldn't actually need to find the equation of the line just graph it. make sure you do so indicating that the line  is perpendicular and the same distance away as the image suggests. In this case A is 1 and B is 2i




hope this helps :)