Welcome, Guest. Please login or register.

May 24, 2017, 03:54:43 pm

Author Topic: Specialist 3/4 Question Thread!  (Read 569555 times)  Share 

0 Members and 1 Guest are viewing this topic.

Ahmad_A_1999

  • Forum Regular
  • **
  • Posts: 53
  • Respect: 0
Re: Specialist 3/4 Question Thread!
« Reply #8700 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

peanut

  • Forum Regular
  • **
  • Posts: 56
  • Respect: 0
Re: Specialist 3/4 Question Thread!
« Reply #8701 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
2016: Biology [48] | Maths Methods [43]
2017: Chemistry | English Language | Specialist Maths | Physics

RuiAce

  • HSC Moderator
  • ATAR Notes Legend
  • *****
  • Posts: 3939
  • It will be ok
  • Respect: +102
Re: Specialist 3/4 Question Thread!
« Reply #8702 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?
Interested in Mathematics Tutoring? Message for details!

ATAR: 98.60
Currently studying: Bachelor of Actuarial Studies/Bachelor of Science (Advanced Maths) @ UNSW

j.wang

  • Forum Regular
  • **
  • Posts: 60
  • Respect: +5
Re: Specialist 3/4 Question Thread!
« Reply #8703 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

  • Moderator
  • Forum Leader
  • *****
  • Posts: 931
  • Respect: +68
Re: Specialist 3/4 Question Thread!
« Reply #8704 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 »
2017-2019: Biomedicine (scholars program) [Monash University]
2020- ???

BMS1011, BMS1021, BMS1031 & MTH1030 Subject Reviews Pending

peanut

  • Forum Regular
  • **
  • Posts: 56
  • Respect: 0
Re: Specialist 3/4 Question Thread!
« Reply #8705 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?
2016: Biology [48] | Maths Methods [43]
2017: Chemistry | English Language | Specialist Maths | Physics

Sine

  • Moderator
  • Forum Leader
  • *****
  • Posts: 931
  • Respect: +68
Re: Specialist 3/4 Question Thread!
« Reply #8706 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 :)
2017-2019: Biomedicine (scholars program) [Monash University]
2020- ???

BMS1011, BMS1021, BMS1031 & MTH1030 Subject Reviews Pending

j.wang

  • Forum Regular
  • **
  • Posts: 60
  • Respect: +5
Re: Specialist 3/4 Question Thread!
« Reply #8707 on: May 20, 2017, 08:44:19 pm »
0
Sorry for the late response :/

I end up getting the latter answer

my syntax input is









Thanks so much Sine, I was staring at the cas for what felt like hours :P