May 24, 2017, 03:54:43 pm

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##### Re: Specialist 3/4 Question Thread!
« Reply #8700 on: May 13, 2017, 09:13:32 pm »
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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

#### peanut

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##### Re: Specialist 3/4 Question Thread!
« Reply #8701 on: May 14, 2017, 11:18:15 am »
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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
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#### RuiAce

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##### Re: Specialist 3/4 Question Thread!
« Reply #8702 on: May 14, 2017, 12:28:05 pm »
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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?
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#### j.wang

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##### Re: Specialist 3/4 Question Thread!
« Reply #8703 on: May 17, 2017, 04:48:51 pm »
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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

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##### Re: Specialist 3/4 Question Thread!
« Reply #8704 on: May 20, 2017, 01:13:05 am »
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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

$\text{impDif}(3x^3-y^2 +k ⋅ x+5y-2x ⋅ y=4,x,y)$
$\text{NOTE: the multiplication dot between two variables}$
$\text{this is essential}$
$\text{i fiddled around and i end up getting your answer if i don't put the dot in}$
$\text{which yields}$
$\frac{9x^2-2y+k}{2x+2y-5}$
$\text{i.e. equivalent to the "correct" answer given}$
« Last Edit: May 20, 2017, 01:20:35 am by Sine »
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#### peanut

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##### Re: Specialist 3/4 Question Thread!
« Reply #8705 on: May 20, 2017, 04:26:56 pm »
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Is there a way of directly sketching a graph like |z - 1| = |z - 2i| on CAS without converting to Cartesian form by hand?
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#### Sine

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##### Re: Specialist 3/4 Question Thread!
« Reply #8706 on: May 20, 2017, 04:39:49 pm »
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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
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#### j.wang

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##### Re: Specialist 3/4 Question Thread!
« Reply #8707 on: May 20, 2017, 08:44:19 pm »
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Sorry for the late response :/

I end up getting the latter answer

my syntax input is

$\text{impDif}(3x^3-y^2 +k ⋅ x+5y-2x ⋅ y=4,x,y)$
$\text{NOTE: the multiplication dot between two variables}$
$\text{this is essential}$
$\text{i fiddled around and i end up getting your answer if i don't put the dot in}$
$\text{which yields}$
$\frac{9x^2-2y+k}{2x+2y-5}$
$\text{i.e. equivalent to the "correct" answer given}$

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