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March 29, 2024, 02:48:31 am

Author Topic: VCE Specialist 3/4 Question Thread!  (Read 2164271 times)  Share 

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Harrycc3000

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9780 on: January 24, 2021, 08:19:08 pm »
0
Hi Guys,
Just asking about this q (16e) from the 3/4 Cambridge textbook (chapt 2F has diagram)
16 The points A, B, C, D and E shown in the diagram have
position vectors
a = i + 11 j b = 2i + 8 j c = −i + 7 j
d = −2i + 8 j e = −4i + 6 j
respectively. The lines AB and DC intersect at F as shown.
a Show that E lies on the lines DA and BC.
b Find −−→AB and −−→DC.
c Find the position vector of the point F.
d Show that FD is perpendicular to EA and that EB is perpendicular to AF.
e Find the position vector of the centre of the circle through E, D, B and F

The solution said that because angle EDF and angle EBF were 90 degree angles it implied EF was the diameter (One of the circle theorems) and that you just needed to find the midpoint to get the vector for the centre of the circle.
I was wondering that if I just didn't realise this (didn't catch it) in the exam would subbing in x and y values for (x-h)^2 + (y-k)^2 = r^2 for each of the position vectors E, D, B and F and then simultaneously solving each of these equations work? This was how I tried to solve the equation but it didn't work but I'm wondering whether this is an arithmetic error because this method has worked for a past non-vector q that asked for the eqn of a circle from 3 known points so I want to see if it could be applied to a q like this. Also idk if I want to go through an hour of pain looking for whether I did an arithmetic error so I'm just gonna ask the q here lol.

Any responses are much appreciated!!!!!
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Harrycc3000

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9781 on: January 25, 2021, 12:38:49 am »
0
Hi again guys lol
I've spent an extremely long time on this question (19 b) and i still can't quite get it. 
Would appreciate any help!
Thanks
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kinslayer

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9782 on: January 25, 2021, 03:30:52 am »
+2
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 :)


If Y is the sum of the lifetimes of the batteries, then saying that , and by extension is saying that the actual (ie. deterministic) sum of the lifetimes is equal to 20 times the lifetime of the first battery (or any single battery), which is obviously wrong and I believe the reason for your confusion. In other words,

Given the information that the Xi are independent and identically distributed (so they all have the same mean and variance and are uncorrelated), you'll get the correct variance by putting:

Var(Y) = Var(X1 + X2 + ... + X20)
    = Var(X1) + Var(X2) + ... + Var(X20)
    = 20*Var(X1) = 5

This gives the correct standard deviation and you can then use the normal approximation to solve the problem. The key is that X1 is not equal to X2, but Var(X1) = Var(X2). Same for expectation, but it doesn't matter in this case. See below.

PS. For the expectation, both "ways" give 140 since expectation is linear whereas variance is bilinear. The fact that E[Y] = E[20*X1] is simply coincidence in this case.





« Last Edit: January 25, 2021, 04:35:28 am by kinslayer »

fun_jirachi

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9783 on: January 25, 2021, 02:10:39 pm »
+7
--snip--

Hey there :)

If possible, please try and include the diagram - really helps people skimming the forum and especially people that want to answer your questions, not everyone has the textbook in question! Just something to note for next time.

The method you suggest should work - we can look at your arithmetic for you if you still can't spot an error. However, this method is often redundant when there's a quicker method like the solution has.

Given distinct points \((x_1, y_1), (x_2, y_2), (x_3, y_3)\) all lie on the same circle, you can set up two equations

then solve for h and k, which basically gives you the position vector for the centre.

Hi again guys lol
I've spent an extremely long time on this question (19 b) and i still can't quite get it. 
Would appreciate any help!
Thanks


Only going to give the full working if you really really can't do it. That being said, here are a few hints that you might want to consider (they do go progressively, so look at them in order).

What do you notice about \(\triangle EFX\) and \(\triangle CDX\)?
Note that you are given that BE = AF = BC - is there a convenient way to link this to the fact that \(|\vec{AB}| = k|\vec{BC}|\)?
Can you set up an equation that uses one of EC and EX or FX and FD in terms of k?

« Last Edit: January 25, 2021, 02:23:09 pm by fun_jirachi »
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Harrycc3000

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9784 on: January 26, 2021, 06:25:38 pm »
0
Hey there :)

If possible, please try and include the diagram - really helps people skimming the forum and especially people that want to answer your questions, not everyone has the textbook in question! Just something to note for next time.

The method you suggest should work - we can look at your arithmetic for you if you still can't spot an error. However, this method is often redundant when there's a quicker method like the solution has.

Given distinct points \((x_1, y_1), (x_2, y_2), (x_3, y_3)\) all lie on the same circle, you can set up two equations

then solve for h and k, which basically gives you the position vector for the centre.

Only going to give the full working if you really really can't do it. That being said, here are a few hints that you might want to consider (they do go progressively, so look at them in order).

What do you notice about \(\triangle EFX\) and \(\triangle CDX\)?
Note that you are given that BE = AF = BC - is there a convenient way to link this to the fact that \(|\vec{AB}| = k|\vec{BC}|\)?
Can you set up an equation that uses one of EC and EX or FX and FD in terms of k?
Hi!
I managed to get a solution and will attach here, just wondering if it was the same method as yours?
Thanks for the tips! Really helped
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fun_jirachi

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9785 on: January 26, 2021, 08:53:53 pm »
+5
It's similar but not the same - you don't have to worry about solutions being 'bang on' as long as they make logical sense and result in the correct answer. I'm glad you worked it out for yourself! :D
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jasmine24

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9786 on: January 29, 2021, 08:37:50 am »
0
Hi, would anyone be able to explain where the sec = 5/4 came from?
Thank you! :)

fun_jirachi

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9787 on: January 29, 2021, 08:48:35 am »
+4
Since we are calculating the rate of change at h = 30, \(\tan (\alpha) = \frac{h}{40} = \frac{3}{4}\). Drawing up a right angled triangle with sides of length 3, 4, and 5 with corresponding angle \(\alpha\) shows that \(\sec (\alpha) = \frac{5}{4}\) :D
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Re: VCE Specialist 3/4 Question Thread!
« Reply #9788 on: February 01, 2021, 08:04:20 pm »
0
Hello

Please see attached, Question C.

The answer says 8454.02, but I got 8454.01.

What did I do wrong?

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9789 on: February 14, 2021, 09:58:59 am »
+2
Hello

Please see attached, Question C.

The answer says 8454.02, but I got 8454.01.

What did I do wrong?

I think you meant "8854.01", but anyway:

I got 8854.02 by rounding the balance of the investment to 2 decimal places after each $400 deposit is made.

(I wouldn't say you've done anything wrong, but that accounts for the difference between the two answers).

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9790 on: February 15, 2021, 11:51:56 am »
0
Heyo!

I was stuck on this particular question (attached below)…
I am mainly confused about how the gradient of the asymptotes should be calculated… I calculated them to be plus or minus 7/6, but the answer section of my textbook says it should be plus or minus 6/7…

Thanks!  :)

fun_jirachi

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9791 on: February 15, 2021, 12:46:04 pm »
+1
Hey! :D

When drawing hyperbolae, you have to be a bit more careful with asymptotes if they're vertical hyperbolae like in part c), as opposed to horizontal like in part a). It's not just as simple as 'the asymptotes are \(y = \pm \frac{b}{a}x\)', though this is true for horizontal hyperbolae - for vertical hyperbolae the asymptotes are \(y = \pm \frac{a}{b}x\) - see if you can figure out why!

Hope this helps :)
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Re: VCE Specialist 3/4 Question Thread!
« Reply #9792 on: February 18, 2021, 05:41:20 pm »
0
hi, need help with the following question my teacher set on an assignment... i can't seem to get it

given that sin(x) + cos(x) = sqrt(2)/3, where x is in Q4, find:
i) the exact value of sin(2x)
ii) the exact value of sin(x) - cos(x)
iii) the exact value of tan(x)
thanks for the help in advance!
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fun_jirachi

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9793 on: February 18, 2021, 06:04:02 pm »
+5
hi, need help with the following question my teacher set on an assignment... i can't seem to get it

given that sin(x) + cos(x) = sqrt(2)/3, where x is in Q4, find:
i) the exact value of sin(2x)
ii) the exact value of sin(x) - cos(x)
iii) the exact value of tan(x)
thanks for the help in advance!

Hey :)

Assuming Q4 means the fourth quadrant? (it's not abundantly clear but makes sense from context clues) we know that \(\sin (x) < 0\) while \(\cos (x) > 0\). Since we also know that \(\cos^2 (x) = 1 - \sin^2 (x)\) and we're only considering the fourth quadrant, we can just take the positive root and see that \(\cos (x) = \sqrt{1 - \sin^2 (x)}\). Hence, we now have that \(\sin(x) + \sqrt{1 - \sin^2 (x)} = \frac{\sqrt{2}}{3}\). Doing a bit of work now:



We select the negative root because of what we noted up front.
Since we now have the value for \(\sin (x)\) we also have the value for \(\cos (x)\) since we know that \(\cos^2 (x) = 1 - \sin^2 (x)\).

The rest of the questions should follow reasonably simply now you know the values of \(\sin (x)\) and \(\cos (x)\).
Hints in the spoiler if you need them.

Hints
i) \(\sin(2x) = 2\sin (x)\cos (x)\)
ii) You have the values for sin and cos
iii) \(\tan (x) = \frac{\sin (x)}{\cos (x)}\)

hope this helps :)
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Re: VCE Specialist 3/4 Question Thread!
« Reply #9794 on: February 18, 2021, 06:53:44 pm »
+1
Hey :)

Assuming Q4 means the fourth quadrant? (it's not abundantly clear but makes sense from context clues) we know that \(\sin (x) < 0\) while \(\cos (x) > 0\). Since we also know that \(\cos^2 (x) = 1 - \sin^2 (x)\) and we're only considering the fourth quadrant, we can just take the positive root and see that \(\cos (x) = \sqrt{1 - \sin^2 (x)}\). Hence, we now have that \(\sin(x) + \sqrt{1 - \sin^2 (x)} = \frac{\sqrt{2}}{3}\). Doing a bit of work now:


thanks so much!!!! yes Q4 did mean quadrant 4, I would have never thought to use square roots on that basic sin squared(x) + cos squared(x) =1 formula  ;D I managed to get all the parts
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