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March 28, 2024, 10:13:13 pm

Author Topic: VCE Specialist 3/4 Question Thread!  (Read 2164184 times)  Share 

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Opengangs

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9765 on: December 05, 2020, 10:17:23 pm »
+6
Hi guys,
Could someone explain this question to me please?

Let w = 2z. Describe the locus of w if z describes a circle with centre 1 + 2i and
radius 3.
soln is If w = 2z then w describes a circle with
centre (2, 4) and radius 6.

I don't get why u can just double everything and get the answer (esp the centre of the circle)
thanks!
Recall that if \(z\) describes the locus of a circle with radius \(r\), then we have the relation \[ \lvert z - z_0 \rvert = r,\] where \(z_0\) represents the centre of the circle. So we have \[ \lvert z - (1 + 2i) \rvert = 3. \tag{1}\] We want to see what the transformation \(w = 2z\) maps to.

Using \((1)\) and the fact that \(w = 2z \iff z = \frac{1}{2}w\), it's not hard to see the result unfolding.
\begin{align*}
\lvert z - (1 + 2i) \rvert = 3 &\implies \left\lvert \frac{1}{2}w - (1 + 2i) \right\rvert = 3 \\
&\implies \left\lvert \frac{1}{2}\left(w - 2(1 + 2i)\right)\right\rvert = 3 \\
&\implies \frac{1}{2}\left\lvert w - 2(1 + 2i)\right\rvert = 3 \\
&\implies \lvert w - 2(1 + 2i) \rvert = 6.
\end{align*}

This represents a circle with centre \(2(1 + 2i)\) and radius 6.

TheEagle

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9766 on: December 06, 2020, 01:35:00 am »
+3
Perfectly said by Opengangs.


It might also be worthy to note that the centre of a circle is always described by the real and imaginary component of z.
i.e. if Z = x + yi,  then the centre of the circle will be (x,y)

cutiepie30

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9767 on: December 19, 2020, 09:02:59 pm »
+1
Hey Guys,

I just had a question regarding the Cambridge Specialist Maths Worked Example.

Can anyone please explain this to me, as attached in the image below.

Thanks :)

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9768 on: December 20, 2020, 11:17:04 am »
+6
Hey Guys,

I just had a question regarding the Cambridge Specialist Maths Worked Example.

Can anyone please explain this to me, as attached in the image below.

Thanks :)

I believe that this question is focusing on the symmetry property in the unit circle which means that there are two quadrants where sin is positive (1st and 2nd) and where sin is negative (3rd and 4th).

So firstly, from inspection of the graph, it can be seen that a would be in the first quadrant as it is before the turning point and it has a positive y coordinate. So from here, the question asks for all points from between [0, 2π], so you want to find all the other points around the unit circle for when a would be positive which would be in the 2nd quadrant. So to get to the 2nd quadrant, you would need a reference angle which is a. From there, you would need to do π - a (based on the symmetry property), thus providing the other solution for between [0, 2π].

Secondly, upon inspection of the graph, it can be seen that b is in the 2nd quadrant as it's positive and after the turning point. So for the other solution, you would want to get an answer in the first quadrant. since b is in the 2nd quadrant, you can just use b as the reference angle and do π - b to get the angle symmetric to b in the first quadrant.

Thirdly, for c, it can be inspected that c is negative but before the minimum turning point, therefore, it would be in the 3rd quadrant. So you would be looking for another answer in the 4th quadrant as sin is also negative in the 4th quadrant. To get an answer in the 4th quadrant, you would need to get a reference angle (the base angle). The answer gets this by doing c - π. Then based on the symmetry property, to get an answer in the 4th quadrant, you would do 2π - x, for this solution it would be 2π - (c - π) and that simplifies to 3π - c.

Lastly, for x = d, this is in the 4th quadrant as it's negative and past the minimum turning point. Just like for c, you would need to find a solution in the 3rd quadrant. So firstly, you would find a reference angle, which would be 2π - d. Then to get to the 3rd quadrant, you would do π + (2π - d), based on the symmetry property. Thus, this would simplify to 3π - d.

I believe that's how they got all the answers for the values, so overall, it's just based on the symmetry property.

cutiepie30

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9769 on: December 20, 2020, 04:19:40 pm »
0
Thanks so much, @Danzorr, it makes so much sense now :)

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9770 on: January 02, 2021, 05:40:25 pm »
0
Hello,

Please see attached. The question is on left (Question 4C) and the solution on the right. I have a few questions:

1. Wouldn't the general solution be x = 76 - 5t since the original equation is 3x -5y =38?

2. Wouldn't the general solution be y = 38 - 3t since the general rule for a diophantine equation is y =y_0 -(a/d)t. In this case a = 3, and d = 1, so I do not understand how they got + 3t.

3. The entire simplifying part makes no sense. If you replaced t with t - 15, are not you changing the value of the equation without balancing it? I don't get this part at all. Why not t - 10, t - 200, or t - 0.37919?  I don't understand.

Thanks.

keltingmeith

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9771 on: January 04, 2021, 01:33:30 am »
+2
Hello,

Please see attached. The question is on left (Question 4C) and the solution on the right. I have a few questions:

1. Wouldn't the general solution be x = 76 - 5t since the original equation is 3x -5y =38?

2. Wouldn't the general solution be y = 38 - 3t since the general rule for a diophantine equation is y =y_0 -(a/d)t. In this case a = 3, and d = 1, so I do not understand how they got + 3t.

3. The entire simplifying part makes no sense. If you replaced t with t - 15, are not you changing the value of the equation without balancing it? I don't get this part at all. Why not t - 10, t - 200, or t - 0.37919?  I don't understand.

Thanks.

Preface: I'm not that great with my integer algebra, so sorry if this explanation feels a bit funny. Lmk if you need clarifications

For 1 and 2 - you're right, but it looks to me like there's some symmetry. For example:

If t=1:
x=76+5
y=38+3

If t=-1:
x=76+(-5)=76-5
y=38+(-3)=38-3

You can verify that these both satisfy the original equation of you'd like. Now, you're proposing the REAL solutions should be these:

x=76-5t
y=38-3t

Well, for t=1:
x=76-5
y=38-3

And t=-1:
x=76-(-5)=76+5
y=38-(-3)=38+3

Notice something? The values x and y take are exactly the same - just for reversed values of t! There's a symmetry that means everything still works, as long as you make the change for both values of t. This actually leads to your question number 3:

You could make any of those substitutions that you described. t is what we call a dummy variable, and all it does is control which answers correspond to t=0,1,2,3, etc. For example, look at what happens if I plug in t=16 with the new equation they provided:

x=1+5(16)=81
y=-7+3(16)=41

Which match the same values we got for t=1 in the original equation. Remember - these equations define a line of solutions for x and y. If you change the value of t by the same amount in both equations, it won't change the actual line that x and y make, it'll just change where the first value of t puts you on that line. I can write up a desmos graph later if this idea confuses you, still.

You might find it easier to think of this as a substitution - instead of changing t with t-15, think of it as making the substitution t=u-15. Then your dummy variable becomes u, and when u=16, t=16-15=1, and so u=16 will give the same (x,y) values as t=1 (which is what we saw)

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9772 on: January 04, 2021, 07:25:22 pm »
0


Thank you for your response. I understand it now  :D

miyukiaura

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9773 on: January 08, 2021, 01:22:09 pm »
0
Does anyone know how to factorise this:


it's one of the examples from the study design. I tried expanding (z+a)(z+b)(z_c) to get a general expression for the trinomial and then equated coefficients to get 3 simultaneous equations, but that didn't work :P
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Re: VCE Specialist 3/4 Question Thread!
« Reply #9774 on: January 08, 2021, 01:58:33 pm »
+1
Does anyone know how to factorise this:


it's one of the examples from the study design. I tried expanding (z+a)(z+b)(z_c) to get a general expression for the trinomial and then equated coefficients to get 3 simultaneous equations, but that didn't work :P

In order to solve this, I just let b=2-i in order to turn the equation into a simpler form. Then it becomes clear that z-b is a factor. Hope this helps :)
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miyukiaura

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9775 on: January 08, 2021, 03:44:13 pm »
0
Thank you! Idk how I didn't think of that hahah
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Re: VCE Specialist 3/4 Question Thread!
« Reply #9776 on: January 08, 2021, 03:55:02 pm »
0
Thank you! Idk how I didn't think of that hahah

TBF, your approach should have worked, it's just difficult since you haven't effectively covered solving simultaneous equations of 3 unknowns in VCE. The other approach requires you to recognise that z=2-i is a solution to that equation, which I wouldn't expect people to get instantly (I certainly didn't notice it!)

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9777 on: January 22, 2021, 05:12:24 pm »
+1
Hi, I have some questions, any help would be appreciated :)

It's from the Exercise 4.3 of Maths Quest 12 Specialist, if that helps, but the questions are here below.

11 From a hot air balloon rising vertically upward with a speed of 8 m/s, a sandbag is
dropped which hits the ground in 4 seconds. Determine the height of the balloon
when the sandbag was dropped.

12 A missile is projected vertically upward with a speed of 73.5 m/s, and 3 seconds
later a second missile is projected vertically upward from the same point with the
same speed. Find when and where the two missiles collide.

13 A flare, A, is fired vertically upwards with a velocity of 35 m/s from a boat. Four
seconds later, another flare, B, is fired vertically upwards from the same point with
a velocity of 75 m/s. Find when and where the flares collide.

Thanks guys!!! :D
Nani?
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Re: VCE Specialist 3/4 Question Thread!
« Reply #9778 on: January 22, 2021, 11:13:16 pm »
+6
Hi, I have some questions, any help would be appreciated :)

It's from the Exercise 4.3 of Maths Quest 12 Specialist, if that helps, but the questions are here below.

11 From a hot air balloon rising vertically upward with a speed of 8 m/s, a sandbag is
dropped which hits the ground in 4 seconds. Determine the height of the balloon
when the sandbag was dropped.

12 A missile is projected vertically upward with a speed of 73.5 m/s, and 3 seconds
later a second missile is projected vertically upward from the same point with the
same speed. Find when and where the two missiles collide.

13 A flare, A, is fired vertically upwards with a velocity of 35 m/s from a boat. Four
seconds later, another flare, B, is fired vertically upwards from the same point with
a velocity of 75 m/s. Find when and where the flares collide.

Thanks guys!!! :D

I haven't done any maths in over two months now so if my solutions are incorrect someone please correct me.
For all of these questions I would use the suvat equations with a=9.8(m/s). In fact, for all of these questions I believe only is needed.

11. In this question you are simply solving for height, or s
The question tells you the initial velocity (u) is 8m/s and the time taken to traverse the distance between the position of the hot air balloon when the sandbag is dropped and the ground is 4 seconds, so you now have your u and t values respectively.
Remember that the hot air balloon is rising vertically, and thus you must assign opposite values to u and a (I personally would assign the positive value to the upwards direction).
So your equation would go from:

to

Hence s=-46.4m, and thus the height of the hot air balloon when the sandbag was dropped was 46.4m (in this case the negative sign is indicating that the bag fell in the direction of gravity, ie. downwards).

12. This time you have two objects and are trying to find the point at which they collide, thus I would recommend equating their s values.
If we let t be the time passed (in seconds) since the first missile was launched, then the time passed since the second missile was launched would be t-3.
Both missiles have the same initial velocity of 73.5m/s, and both have gravity as their constant acceleration.
Therefore we can equate the values of s for each missile to create the equation

where u=73.5, a=-9.8 and we are solving for t.
Therefore t=9 (seconds) and by substituting that into the original s=ut+1/2at^2 equation we can find s=264.6m
Hence the missiles collide 264.6m above the ground 9 seconds after the launch of the first missile.

13. Though the flares don't start with the same initial velocity, we can apply the same methodology in this question as the last. The only difference here would be that the values of u on each side of the equation would differ (we'll use b and c to denote these different initial velocities).
Again, by letting t be the time since the launch of the first flare, the time since the launch of the second flare would be t-4
So the equation would go from

to

Thus t=43/9 secs, and by substituting that in s=44849/810m or 55.37m to 2 d.p.
Therefore the flares collide 43/9 (or 4.78 to 2 d.p.) seconds after the launch of the first flare 55.37m above the ground.

Again it's late rn and I haven't done anything remotely mathematical for two months now so if these are incorrect please correct me and I apologise in advance.
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Re: VCE Specialist 3/4 Question Thread!
« Reply #9779 on: January 23, 2021, 12:06:33 pm »
0
Hi! Just hoping to get some help for this question here :)
I think I have an idea as to what I did wrong.
The answers say that the mean of the 20 batteries is 140 and the standard deviation of the 20 batteries is 2.236.
If you use the method of...
E(20X) = 20E(X) and Var(20X) = 400Var(X0, then you'll get the wrong answer.
If you use this method of adding X 20 times instead, you'll get
E(X + X + X...+X) = 20E(X) and Var(X + X + X+....+X) = 20Var(X).
When do I know which method to use?
Any help or advice would be appreciated! Thanks :)