October 24, 2017, 10:08:11 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 »
"Sine"

#### 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
"Sine"

#### 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

#### Gogo14

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##### Re: Specialist 3/4 Question Thread!
« Reply #8708 on: May 24, 2017, 06:20:10 pm »
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Help with question 4.Answer is 8pi. When you rotate x^2=4-y, arent you rotating both branches? So the volume will overlap, so shouldnt you half the volume?

#### Syndicate

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##### Re: Specialist 3/4 Question Thread!
« Reply #8709 on: May 25, 2017, 12:02:44 am »
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Help with question 4.Answer is 8pi. When you rotate x^2=4-y, arent you rotating both branches? So the volume will overlap, so shouldnt you half the volume?

Yes, there is an overlap, but does this really affect the amount of volume you will get? (No, as you are basically finding the same volume over the same region)
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#### Joseph41

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##### Re: Specialist 3/4 Question Thread!
« Reply #8710 on: June 01, 2017, 03:20:06 pm »
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Hey there! Welcome to the VCE Specialist Maths Question Thread.

If you have a Spesh question you'd like answered, you'll first need to make an ATAR Notes account - it should take about four seconds. Then, simply scroll down and ask your questions in the "Quick Reply" box. I've included a picture below to show you what it looks like.

Alternatively, you have in this thread almost 600 pages of Specialist questions and answers! All the best!

#### vasuk

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##### Re: Specialist 3/4 Question Thread!
« Reply #8711 on: June 03, 2017, 01:00:48 pm »
+3
Could someone please explain why it would be wrong to see this question as a hemisphere, minus the volume of the solid of revolution formed between the curve y = x^2 and the x axis between zero to one?

#### RuiAce

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##### Re: Specialist 3/4 Question Thread!
« Reply #8712 on: June 03, 2017, 06:22:54 pm »
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Could someone please explain why it would be wrong to see this question as a hemisphere, minus the volume of the solid of revolution formed between the curve y = x^2 and the x axis between zero to one?
$\text{Where is the hemisphere?}$
$\text{The equation of a circle with radius 1 and centre (1,0) is}\\ (x-1)^2 + y^2 = 1\\ \text{Note the presence of }\textit{both }x^2\text{ and }y^2$
$\text{What you have are a pair of parabolas concaving perpendicularly to each other}\\ y=x^2\text{ and }x = y^2\\ \text{Make sure you didn't confuse }x=y^2\text{ with }(x-1)^2+y^2=1$
A parabola does not look like a circle. When a circle is rotated, the solid of revolution will be a sphere. When a parabola is rotated, the solid of revolution will be a paraboloid
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#### vasuk

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##### Re: Specialist 3/4 Question Thread!
« Reply #8713 on: June 03, 2017, 07:19:21 pm »
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$\text{Where is the hemisphere?}$
$\text{The equation of a circle with radius 1 and centre (1,0) is}\\ (x-1)^2 + y^2 = 1\\ \text{Note the presence of }\textit{both }x^2\text{ and }y^2$
$\text{What you have are a pair of parabolas concaving perpendicularly to each other}\\ y=x^2\text{ and }x = y^2\\ \text{Make sure you didn't confuse }x=y^2\text{ with }(x-1)^2+y^2=1$
A parabola does not look like a circle. When a circle is rotated, the solid of revolution will be a sphere. When a parabola is rotated, the solid of revolution will be a paraboloid

thanks a lot!!

#### KLRah

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##### Re: Specialist 3/4 Question Thread!
« Reply #8714 on: June 03, 2017, 07:51:52 pm »
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For the function f(x)= 2cos(ax)+1, where x is between [0,4]. Find the largest value of a to three decimal places for which their is only one point on the graph at which the gradient is equal to the average gradient of the function over the specified domain. Any help would be appreciated