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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: keltingmeith on February 11, 2016, 10:16:41 pm

Title: Specialist 1/2 Question Thread!
Post by: keltingmeith on February 11, 2016, 10:16:41 pm
Given just how different specialist 1/2 and specialist 3/4 are, we've now made a second question thread for all your 1/2 needs!! This will help guarantee that your question doesn't get lost amongst all the other questions being asked, just in case anyone answering knows a lot about 3/4, but not 1/2.
Title: Re: Specialist 1/2 Question Thread!
Post by: qazser on February 11, 2016, 10:50:04 pm
Ty Dickens for making this  ;D
Title: Re: Specialist 1/2 Question Thread!
Post by: Gogo14 on February 16, 2016, 10:17:17 pm
My teacher goes thru the content quite fast in class, and I feel I need more practice as I am struggling. I did the qs in the textbook, but I found it hard and would like more practice.
What are some free resources that I can utilise?
Thnx
Title: Re: Specialist 1/2 Question Thread!
Post by: MightyBeh on February 17, 2016, 05:41:59 am
My teacher goes thru the content quite fast in class, and I feel I need more practice as I am struggling. I did the qs in the textbook, but I found it hard and would like more practice.
What are some free resources that I can utilise?
Thnx

For learning:
Khan Academy (and practice)
Paul's Online Math Notes
Better Explained

If you're running out of fresh questions to do, maybe try a methods textbook? Obviously not all the content will be the same but there's plenty of relevant stuff in there. You should ask your teacher too, they might have more resources. Personally, I learned the most during practice (and actual ::) ) SACs. This might also be helpful :)
Title: Re: Specialist 1/2 Question Thread!
Post by: Adequace on February 23, 2016, 06:45:40 pm
I need some help on a couple questions http://m.imgur.com/a/rEeRc

For Q5b) what's wrong with my equation/method to working this out? I'm lost on how to solve this problem even though it's almost exactly the same to the others.

For 8b) the answer says 72 minutes, but I got 90. x=50 is correct as well.

For 8a and 10, I did these with a scientific calculator because the square root looked pretty messy. Are these questions intended to be solved by hand?

Thanks
Title: Re: Specialist 1/2 Question Thread!
Post by: tysh on February 23, 2016, 08:24:50 pm
I need some help on a couple questions http://m.imgur.com/a/rEeRc

For Q5b) what's wrong with my equation/method to working this out? I'm lost on how to solve this problem even though it's almost exactly the same to the others.

For 8b) the answer says 72 minutes, but I got 90. x=50 is correct as well.

For 8a and 10, I did these with a scientific calculator because the square root looked pretty messy. Are these questions intended to be solved by hand?

Thanks

Hey Adequace, they were just minor errors:
For 5b, it was 600/x = 600/(x-5.5) - 220, instead of 600/x = 600/(x-5.5) + 220: remember that the plane's speed is faster, thus we substract the 220 from the plane.
For 8b, instead of 75/x (which is the usual time), they are looking for 75/(x+12.5), as this is the actual time taken for that particular journey. The question was probably a bit vague anyway though.
Questions like 10 and 8a are definitely doable by hand (although they ARE a bit tedious!). Sometimes, to speed up the process (such as in question 10) you can use km/min or another unit to avoid fractions during working out, then convert it back to km/h at the end.
Title: Re: Specialist 1/2 Question Thread!
Post by: Adequace on February 23, 2016, 09:03:17 pm
Hey Adequace, they were just minor errors:
For 5b, it was 600/x = 600/(x-5.5) - 220, instead of 600/x = 600/(x-5.5) + 220: remember that the plane's speed is faster, thus we substract the 220 from the plane.
For 8b, instead of 75/x (which is the usual time), they are looking for 75/(x+12.5), as this is the actual time taken for that particular journey. The question was probably a bit vague anyway though.
Questions like 10 and 8a are definitely doable by hand (although they ARE a bit tedious!). Sometimes, to speed up the process (such as in question 10) you can use km/min or another unit to avoid fractions during working out, then convert it back to km/h at the end.
Thanks man, appreciate it a lot!
Title: Re: Specialist 1/2 Question Thread!
Post by: Adequace on February 26, 2016, 08:01:33 pm
Another question.

http://imgur.com/a/vZeP5

For Q16) the answer just has p=10 but I can't get to answer. I've attached my working as well.

For Q12) I struggle greatly with worded problems. I don't think the denominators in my pipe equations are correct. Is there anything I did wrong? Haven't really tried solving this question due to my lack of confidence..

Thanks
Title: Re: Specialist 1/2 Question Thread!
Post by: StupidProdigy on February 26, 2016, 09:20:50 pm
Another question.

http://imgur.com/a/vZeP5

For Q16) the answer just has p=10 but I can't get to answer. I've attached my working as well.

For Q12) I struggle greatly with worded problems. I don't think the denominators in my pipe equations are correct. Is there anything I did wrong? Haven't really tried solving this question due to my lack of confidence..

Thanks
16. You're working is good, you're reading of the question not so much haha. p has to be a positive integer. 2sqrt(21) is roughly 9.16 therefore p must =10 since that is the closest integer that is both positive and larger than 9.16
12..can't be bothered sorry
Title: Re: Specialist 1/2 Question Thread!
Post by: Adequace on February 26, 2016, 10:06:41 pm
16. You're working is good, you're reading of the question not so much haha. p has to be a positive integer. 2sqrt(21) is roughly 9.16 therefore p must =10 since that is the closest integer that is both positive and larger than 9.16
12..can't be bothered sorry
Thanks regardless, but yeah the worded one is a pain.
Title: Re: Specialist 1/2 Question Thread!
Post by: nerdgasm on February 27, 2016, 12:05:35 am
I have a solution for Q12, but it is by no means elegant, and I think there's probably a better way to do it that I've missed:

Spoiler
Let C = the capacity of the tank. Let r1 and r2 be the initial flow rates through the first and second pipes respectively. Let r1_new and r2_new be the new flow rates. Let t1 and t2 be the initial time it takes to fill the tank using the initial flow rates of the first and second pipes respectively.

Using the formula Capacity = Rate * Time (and its rearrangements), we can form the following six equations:
Equation 1: r1 + r2 = 3C/20   
Equations 2 and 3: r1 = C/t1 ; r2 = C/t2
Equations 4 and 5: r1_new = C/(t1 - 1) ; r2_new = C/(t2 + 2)
Equation 6: r1_new + r2_new = C/7.

I now substitute Eqs. 2 and 3 into 1, and Eqs. 4 and 5 into 6, to get the following:

Now I factor out C from both equations (C > 0, or else the question makes no sense). We arrive at:
Our goal is therefore to work out the value of t1 and t2, so we can work out (t1-1) and (t2+2), which is the ultimate aim.

Algebraically manipulating the fractions, we get the following:;. I now treat this as a system of (non-standard) simultaneous equations, and try to solve. We get:
. But notice that we previously had Therefore, we may subtract these two equations to get We can substitute this back into 5t1-4t2=27 to get This quadratic can be factorised to give . If you substitute these values back into 5t1-4t2=27, you will get the corresponding solutions Of course, we can't have a negative time value, so it follows that . Those are the initial times to fill the tank for each pipe. To get the new times, we simply evaluate t1-1 and t2+2, to get new times of 14 minutes for both pipes! Therefore, under the new flow rates, each pipe will take 14 minutes to fill the tank!

In summary: Initially, the first pipe took 15 minutes to fill the tank, and the second pipe took 12 minutes. Hence the first pipe fills 1/15 of the tank per minute, and the second pipe fills 1/12 of the tank per minute. If you combine the two pipes, they fill 1/15 + 1/12 = 3/20 of the tank per minute, and hence it takes 20/3 minutes to fill the tank. After the flow rates were adjusted, the two pipes both take 14 minutes to fill the tank individually, and hence both fill 1/14 of the tank per minute. So when you combine the two pipes, they fill 1/14 + 1/14 = 1/7 of the tank per minute, and hence it takes 7 minutes to fill the tank.

There has seriously got to be a better way to do this :-\.

Title: Re: Specialist 1/2 Question Thread!
Post by: Adequace on February 27, 2016, 11:40:35 am
I have a solution for Q12, but it is by no means elegant, and I think there's probably a better way to do it that I've missed:

Spoiler
Let C = the capacity of the tank. Let r1 and r2 be the initial flow rates through the first and second pipes respectively. Let r1_new and r2_new be the new flow rates. Let t1 and t2 be the initial time it takes to fill the tank using the initial flow rates of the first and second pipes respectively.

Using the formula Capacity = Rate * Time (and its rearrangements), we can form the following six equations:
Equation 1: r1 + r2 = 3C/20   
Equations 2 and 3: r1 = C/t1 ; r2 = C/t2
Equations 4 and 5: r1_new = C/(t1 - 1) ; r2_new = C/(t2 + 2)
Equation 6: r1_new + r2_new = C/7.

I now substitute Eqs. 2 and 3 into 1, and Eqs. 4 and 5 into 6, to get the following:

Now I factor out C from both equations (C > 0, or else the question makes no sense). We arrive at:
Our goal is therefore to work out the value of t1 and t2, so we can work out (t1-1) and (t2+2), which is the ultimate aim.

Algebraically manipulating the fractions, we get the following:;. I now treat this as a system of (non-standard) simultaneous equations, and try to solve. We get:
. But notice that we previously had Therefore, we may subtract these two equations to get We can substitute this back into 5t1-4t2=27 to get This quadratic can be factorised to give . If you substitute these values back into 5t1-4t2=27, you will get the corresponding solutions Of course, we can't have a negative time value, so it follows that . Those are the initial times to fill the tank for each pipe. To get the new times, we simply evaluate t1-1 and t2+2, to get new times of 14 minutes for both pipes! Therefore, under the new flow rates, each pipe will take 14 minutes to fill the tank!

In summary: Initially, the first pipe took 15 minutes to fill the tank, and the second pipe took 12 minutes. Hence the first pipe fills 1/15 of the tank per minute, and the second pipe fills 1/12 of the tank per minute. If you combine the two pipes, they fill 1/15 + 1/12 = 3/20 of the tank per minute, and hence it takes 20/3 minutes to fill the tank. After the flow rates were adjusted, the two pipes both take 14 minutes to fill the tank individually, and hence both fill 1/14 of the tank per minute. So when you combine the two pipes, they fill 1/14 + 1/14 = 1/7 of the tank per minute, and hence it takes 7 minutes to fill the tank.

There has seriously got to be a better way to do this :-\.
Thanks a lot man! Also, when you factorised 5t^2 -87t +180 =0. Is this expected to be done by hand?
Title: Re: Specialist 1/2 Question Thread!
Post by: keltingmeith on February 27, 2016, 12:59:25 pm
Thanks a lot man! Also, when you factorised 5t^2 -87t +180 =0. Is this expected to be done by hand?

Yes - you are expected to be able to solve all quadratics by hand. Don't forget about techniques like completing the square and the quadratic equation!
Title: Re: Specialist 1/2 Question Thread!
Post by: Adequace on February 27, 2016, 09:44:54 pm
Yes - you are expected to be able to solve all quadratics by hand. Don't forget about techniques like completing the square and the quadratic equation!
Ah, thanks!

I have another question http://m.imgur.com/a/iVNip

For parts f and h
For f: why can't you solve it using (Ax+B)/x^2 -x)
For h, I can't tell what I'm doing wrong but the answer has x/(x^2 +3) for the 2nd part of the partial fraction.
Don't worry about j, I just realised what I did wrong.

Sorry about all the questions  :'( hopefully in 2 years time I can look back on this and be happy with what I've achieved  :)
Title: Re: Specialist 1/2 Question Thread!
Post by: Callum@1373 on February 27, 2016, 09:52:32 pm
Ah, thanks!

I have another question http://m.imgur.com/a/iVNip

For parts f and h
For f: why can't you solve it using (Ax+B)/x^2 -x)
For h, I can't tell what I'm doing wrong but the answer has x/(x^2 +3) for the 2nd part of the partial fraction.
Don't worry about j, I just realised what I did wrong.

Sorry about all the questions  :'( hopefully in 2 years time I can look back on this and be happy with what I've achieved  :)
For part f, x^2 -x isn't an irreducible quadratic factor because it can be written as x(x-1) so you would write A/x + B/x-1

I can't help with h at the moment as I don't have pen and paper with me sorry
Title: Re: Specialist 1/2 Question Thread!
Post by: keltingmeith on February 28, 2016, 12:43:49 pm
Ah, thanks!

I have another question http://m.imgur.com/a/iVNip

For parts f and h
For f: why can't you solve it using (Ax+B)/x^2 -x)
For h, I can't tell what I'm doing wrong but the answer has x/(x^2 +3) for the 2nd part of the partial fraction.
Don't worry about j, I just realised what I did wrong.

Sorry about all the questions  :'( hopefully in 2 years time I can look back on this and be happy with what I've achieved  :)


For h, they've probably mucked up, because neither of the denominators should have x^2+3 there.
Title: Re: Specialist 1/2 Question Thread!
Post by: Adequace on February 28, 2016, 07:30:39 pm
For h, they've probably mucked up, because neither of the denominators should have x^2+3 there.
Whoops I meant x^2 +4 for the denominator
Title: Re: Specialist 1/2 Question Thread!
Post by: One Step at a Time on February 29, 2016, 12:43:09 pm
Hi everyone!

Doing an exercice on Arithmetic Series and I'm struggling a lot so I would appreciate any help :)

1) For the sequence 4,8,12,…4,8,12,…, find {n:Sn=180}. (My answer: 10, textbook's answer: 9)

2) There are 110 logs to be put in a pile, with 15 logs in the bottom layer, 14 in the next, 13 in the next, and so on. How many layers will there be? (Textbook's answer: 11)

3) Dora’s walking club plans 15 walks for the summer. The first walk is a distance of 6 km, the last walk is a distance of 27 km, and the distances of the walks form an arithmetic sequence.
How far is the 8th walk? (Textbook's answer: 16.5km)
Title: Re: Specialist 1/2 Question Thread!
Post by: Syndicate on February 29, 2016, 04:57:14 pm
Hi everyone!

Doing an exercice on Arithmetic Series and I'm struggling a lot so I would appreciate any help :)

1) For the sequence 4,8,12,…4,8,12,…, find {n:Sn=180}. (My answer: 10, textbook's answer: 9)

2) There are 110 logs to be put in a pile, with 15 logs in the bottom layer, 14 in the next, 13 in the next, and so on. How many layers will there be? (Textbook's answer: 11)

3) Dora’s walking club plans 15 walks for the summer. The first walk is a distance of 6 km, the last walk is a distance of 27 km, and the distances of the walks form an arithmetic sequence.
How far is the 8th walk? (Textbook's answer: 16.5km)

1) well, the difference between each sequence is 4.

If you add all the numbers upto 36 you will get 180

(4+8+12+16+20+24+28+32+36 = 180)

Your textbook is correct, there are only 9 numbers required.

2) well as they have told you the sequence the logs are piled up in. you would keep subtracting the squence number from the previous value

Working out: 110-15=95, 95-14=81, 81-13 = 68, 68-12=56, 56-11=45, 45-10=35, 35-9=26, 26-8= 18, 18-7=11, 11-6=5, and 5-5 = 0

Therefore there are 11 layers of logs in the pile

3) well I found that there is a difference of 1.5km between each walk

(27-6)/14 = 1.5

As we know that the first walk is 6km, we would need to workout the rest of the 7 walks ahead.

7 x 1.5 = 10.5km

and if you add 10.5 and 6 to get the distance of the 8th walk, you will get 16.5km
Title: Re: Specialist 1/2 Question Thread!
Post by: tysh on February 29, 2016, 10:49:28 pm
Hi everyone!

Doing an exercice on Arithmetic Series and I'm struggling a lot so I would appreciate any help :)

1) For the sequence 4,8,12,…4,8,12,…, find {n:Sn=180}. (My answer: 10, textbook's answer: 9)

2) There are 110 logs to be put in a pile, with 15 logs in the bottom layer, 14 in the next, 13 in the next, and so on. How many layers will there be? (Textbook's answer: 11)

3) Dora’s walking club plans 15 walks for the summer. The first walk is a distance of 6 km, the last walk is a distance of 27 km, and the distances of the walks form an arithmetic sequence.
How far is the 8th walk? (Textbook's answer: 16.5km)

It's a good idea to use arithmetic series formula for the questions: Sn=(n/2)(2a+(n-1)d), or sometimes (n/2)(a+l) since l=a+(n-1)d.
1) a=4, d=4, so 180=(n/2)(8+4n-4), so you get a quadratic and factorise to (n+10)(n-9), obviously n=9 since it's positive.
2) Same thing, a=15, d=-1, so 110=(n/2)(30-n+1) and you will get a quadratic (n-11)(n-20). Slightly trickier here, you take n=11 because n=20 actually means that the piles goes ABOVE 110, then arrives back at 110 when the series starts decreasing.
3) a=6, d=27, so 27=6 + (14)d, n=(27-6)/14=1.5 (you can use a more direct approach as Syndicate has outlined above). 8th=6+7*1.5=16.5km
Title: Re: Specialist 1/2 Question Thread!
Post by: qwertyqwerty on March 17, 2016, 08:55:17 pm
assistance needed please, got an outcome tomorrow
Title: Re: Specialist 1/2 Question Thread!
Post by: TheProphetPancake on March 17, 2016, 10:03:42 pm
assistance needed please, got an outcome tomorrow
Hi there,
I believe the angle ACD is 47 degrees.
I got this by first recognising that since AB=AD, then triangle BAD is isosceles, therefore angle ABD and angle ADB must be same and must equal, when added together, 94 degrees (I got this by considering triangle BAD and simply taking the 86 degree angle from 180). Therefore, angle ABD is equal to 47 degrees. Then by using the 'same-angle-in-same-segment' theorem, you can ascertain the angle ACD to be 47 degrees as well.
Title: Re: Specialist 1/2 Question Thread!
Post by: Sil on March 19, 2016, 09:49:30 pm
Have no idea how to start this. I will probably work this out eventually, but I'll just leave this here in case I don't :P Thanks!
Title: Re: Specialist 1/2 Question Thread!
Post by: IndefatigableLover on March 19, 2016, 09:56:55 pm
Have no idea how to start this. I will probably work this out eventually, but I'll just leave this here in case I don't :P Thanks!
If you have real coefficients then you can utilise the conjugate-root theorem to solve this problem :)
Title: Re: Specialist 1/2 Question Thread!
Post by: qwertyqwerty on March 21, 2016, 02:32:21 pm
Solve for x:
|x-4| - |x+2| = 6
Title: Re: Specialist 1/2 Question Thread!
Post by: tysh on April 13, 2016, 10:03:28 pm
Solve for x:
|x-4| - |x+2| = 6

One possible approach is this:

Form magnitude values:
|x-4|=x-4 if x>4, |x-4|=4-x if x<4.
|x+2|=x+2 if x>-2, |x+2|=-x-2 if x<-2.

If x>-2 and x>4, so x-4-(x+2)=6, but -6=6 is false.
If x<-2 and x<4, so 4-x-(-x-2)=6, we end up with an equation with infinite solutions (x=x), thus x<-2 for these solutions.
If x>-2 and x<4, so 4-x-(x+2)=6, 2x=-4, x=-2 is a solution.
We have x=-2 and x<-2, so x≤-2.

Title: Re: Specialist 1/2 Question Thread!
Post by: bk.fuse on April 15, 2016, 12:02:31 am
Hey guys
I just had my second spesh test and I had done all the school text book questions and additional questions from other textbooks and i can do all the questions off by heart, even with the practise tests my teacher gave us, i could do really easily. Even though i could do all of that, the test questions were a lot harder and even though i knew how to do all the questions, i ran out of time. Do any of you guys know any resources that i could use to practise with? Such as websites or any text books that could help so i can just practise a lot more.
Thanks!  :)
Title: Re: Specialist 1/2 Question Thread!
Post by: Adequace on April 16, 2016, 09:42:37 pm
http://imgur.com/a/hN7pz

I need some help for part a. My working was a random method but I'm not sure why this doesn't work?
Another way which seems more likely is finding AB then using the cosine rule to find the magnitude of the angle, my problem is how do I find the length AB?

Thanks
Title: Re: Specialist 1/2 Question Thread!
Post by: Syndicate on April 17, 2016, 12:18:48 pm
http://imgur.com/a/hN7pz

I need some help for part a. My working was a random method but I'm not sure why this doesn't work?
Another way which seems more likely is finding AB then using the cosine rule to find the magnitude of the angle, my problem is how do I find the length AB?

Thanks

I believe you meant to ask: how would you find the length of O'B (because only then you could use the cosine formula to workout the angle between AO'B).

As the triangles share the chord lengh equally, it must mean that they are isosceles triangles. Hence, O'B = O'A.

So, O'A = O'B = 6cm, and AB = 8 cm


cosO' = (6^2 + 6^2 - 8^2)/ (2 x 6 x 6)

plug it in your calculator, and you will end up with 83.62 degrees.
Title: Re: Specialist 1/2 Question Thread!
Post by: Adequace on April 17, 2016, 12:51:02 pm
Thanks, I just realised as well!
Title: Re: Specialist 1/2 Question Thread!
Post by: mtDNA on April 18, 2016, 08:06:59 pm
Not necessarily content related, however I was just wondering, from the Cambridge textbook which chapters are similar to the 3/4 specialist course? Which chapters would be have to pay particular attention to?
Thanks!
Title: Re: Specialist 1/2 Question Thread!
Post by: Gogo14 on April 21, 2016, 06:28:33 pm
How do you simplify the cosine rule in terms of sine? Teacher said it a trig identity , but someone explain to me what is and how you do it. Thnx
Title: Re: Specialist 1/2 Question Thread!
Post by: Syndicate on April 21, 2016, 07:04:23 pm
How do you simplify the cosine rule in terms of sine? Teacher said it a trig identity , but someone explain to me what is and how you do it. Thnx

Cosine rule:


Sine rule:


Trigonometric Identity:



So now you can write the cosine rule in terms of sine. Your final formula will look like:
Title: Re: Specialist 1/2 Question Thread!
Post by: Syndicate on April 24, 2016, 04:01:21 pm
Two questions, see the attached pictures.

1) I believe e is not the right answer in this case. As you are trying to trying to find x^2 +1/x (were x = 2+/3), you should be getting 9 - 3 * sqrt{3}



and.....


Therefore....

and even if you try to somehow expand it, it is nowhere close to the correct answer.

2) Assuming your set contains 5 elements (A, B, C, D, E), you should end up with 32 subsets (inc the empty subset). The formula you should be using in thise case is:


where n is the number of elements in the set (A).

More direct way of approaching the situation is:

empty subset (containing 0 elements) - 1
{}
subsets containing 1 element - 5
{A}
{B}
{C}
{D}
{E}

subsets containing 2 elements - 10
{AB}
{AC}
{AD}
{AE}
{BC}
{BD}
{BE}
{CD}
{CE}
{DE}

subsets containing 3 elements - 10
{ABC}
{ABD}
{ABE}
{ACD}
{ACE}
{ADE}
{BCD}
{BCE}
{BDE}
{CDE}

subsets containing 4 elements - 5
{ABCD}
{ABCE}
{ABDE}
{ACDE}
{BCDE}

subsets containing 5 elements - 1
{ABCDE}

Add the amount of subsets and you will get 32, which is d

Title: Re: Specialist 1/2 Question Thread!
Post by: anotherworld2b on May 21, 2016, 10:34:26 am
Hello
I was wondering if I could have help in how to solve Q9.  :)
Title: Re: Specialist 1/2 Question Thread!
Post by: anotherworld2b on June 04, 2016, 09:44:17 am
Hello
I was wondering if I could get advice in what to put on the notes that we are given for
the specialist exam. I know that the calc free section you can't use notes or calc and that
for calc assumed you can. But I am confused on what I should know how to do without notes and
what I should put on the notes. We've covered vector proof, component form ,dot product, scalar projection, permutation, combination, Pascal's triangle/ proofs, converse, contrapositive and inverse as well as circle proofs. Help is greatly appreciated for my study  ;D
Title: Re: Specialist 1/2 Question Thread!
Post by: StupidProdigy on June 05, 2016, 10:36:53 am
http://imgur.com/a/eYMxC quick question. I had to copy the 2nd line from the book to solve this but on the RHS, why don't they have a y^2 in the square root as well since (y-0)^2.

Ty
Only had a quick look but for the line x=-1, it has the coordinates (-1,y), and what you've said is that it is a single point (-1,0) instead of a vertical line that can have infinitely many y values. So effectively you get (y-y)^2 which is why the term is absent.  :)
Title: Re: Specialist 1/2 Question Thread!
Post by: keltingmeith on June 09, 2016, 11:37:34 pm
Hello
I was wondering if I could get advice in what to put on the notes that we are given for
the specialist exam. I know that the calc free section you can't use notes or calc and that
for calc assumed you can. But I am confused on what I should know how to do without notes and
what I should put on the notes. We've covered vector proof, component form ,dot product, scalar projection, permutation, combination, Pascal's triangle/ proofs, converse, contrapositive and inverse as well as circle proofs. Help is greatly appreciated for my study  ;D

Put everything in there. But also have everything memorised. There is, technically, no limit on what they can give you on calc free or calc able, so best to be prepared for anything. You'll probably find that in the process of writing your reference book for your calc able exam, things will start to stick in your mind, even without you trying to memorise them. ;D
Title: Re: Specialist 1/2 Question Thread!
Post by: Adequace on June 15, 2016, 09:08:21 pm
http://imgur.com/HaMD8la

For part b, I'm a bit stuck. Before calculating p, the answer states that the number of male swimmers can be 0,1,2,3,4. Can't there also be 5?

They then calculate P=0.8 and 1 (just listed 2 of the answers), how did they get these 2 answers. Isnt the sample size 20?

Thanks
Title: Re: Specialist 1/2 Question Thread!
Post by: Clockwork on June 28, 2016, 06:19:59 pm
Hi everyone,

I have the following question and I'm just wondering if my answer and working is all correct.
"Find the locus of the point P(x ,y) such that AP=6, given point A(3 ,2)"

This is circle with a centre of (3, 2) and a radius of 6.
Title: Re: Specialist 1/2 Question Thread!
Post by: zsteve on June 28, 2016, 06:32:43 pm
Hi everyone,

I have the following question and I'm just wondering if my answer and working is all correct.
"Find the locus of the point P(x ,y) such that AP=6, given point A(3 ,2)"

This is circle with a centre of (3, 2) and a radius of 6.

Looks good to me
Title: Re: Specialist 1/2 Question Thread!
Post by: Clockwork on June 28, 2016, 11:53:48 pm
Looks good to me

Thank you!
Title: Re: Specialist 1/2 Question Thread!
Post by: anotherworld2b on July 05, 2016, 11:47:38 pm
Hi i was wondering if i fould recieve help on solving these proofs. I am still relatively new to proving trig identities so i didnt know what i can and should do
Title: Re: Specialist 1/2 Question Thread!
Post by: Syndicate on July 06, 2016, 03:09:56 pm
Hi i was wondering if i fould recieve help on solving these proofs. I am still relatively new to proving trig identities so i didnt know what i can and should do

10)











11)













Title: Re: Specialist 1/2 Question Thread!
Post by: anotherworld2b on July 06, 2016, 10:56:14 pm
10)











11)














Thank you so much for your help :D
I wanted to ask some questions if that's okay
For Q10
How did It go from

to






Further explanation would be greatly appreciated :D
I am still greatly lacking in understanding how to apply the trig identities and manipulating them
I also wanted to ask about Q11. How did you prove it from this step onwards?
I am still relatively new to trig identities proofs











Title: Re: Specialist 1/2 Question Thread!
Post by: Syndicate on July 07, 2016, 10:52:02 am
Thank you so much for your help :D
I wanted to ask some questions if that's okay
For Q10
How did It go from

to


For these kind off questions, I would recommend you to remember the exact values of trigonmetric functions and its identities. So basically:


I basically coverted cos 45 and sin 45 into its exact value (which was 1/sqrt(2)), and as they had the same denominator, I was able to put sinx/sqrt(2) and cosx/sqrt(2) as one fraction.

 
(http://www.mathwords.com/t/t_assets/t80.gif)
I would really recommend you to remember these^





Subsequent to multiplying the two equations together, you should end up sin^2x-cos^2x. Then, take out the negative factor, in order for you to get -cos(2x).

cos(2x) has three different variations.
= cos^2x - sin^2x
= 2cos^2x-1
= 1- 2sin^2x

Further explanation would be greatly appreciated :D
I am still greatly lacking in understanding how to apply the trig identities and manipulating them
I also wanted to ask about Q11. How did you prove it from this step onwards?
I am still relatively new to trig identities proofs













Well, when you have fractions situated within fractions, you would multiply the numerator with the reciprocal of the denominator. Hence, when you are multiplying them together, you would begin by canceling out the cosx, in both fractions.

To simplify the fraction, you would multiply the numerator and the denomirator, with the invert of the denominator (which in this case was cosx + sinx).

Then using your knowledge of trig identities, you would instantly convert the sin^2x + cos^2x to 1 (as sin^2x + cos^2x = 1).

Likewise, you would also convert the cos^2x - sin^2x to cos2x, and then to 1 - 2sin^2x, in order to prove what your are given.

Title: Re: Specialist 1/2 Question Thread!
Post by: anotherworld2b on July 10, 2016, 12:52:44 am
I am still a bit confused

-1 was taken out

what happened here? why is it 2x?

Title: Re: Specialist 1/2 Question Thread!
Post by: RuiAce on July 10, 2016, 10:31:09 am
I am still a bit confused

-1 was taken out

what happened here? why is it 2x?




Title: Re: Specialist 1/2 Question Thread!
Post by: Syndicate on July 10, 2016, 10:37:25 am
These are all the double angle formulas you are required to remember.

(http://schoolnet.org.za/PILAfrica/en/webs/17119/media/3_201.gif)
Title: Re: Specialist 1/2 Question Thread!
Post by: Clockwork on July 20, 2016, 05:50:21 pm
I have a question on changing Cartesian equations to their polar form and I'm a bit stuck on how to do this one.
The Cartesian equation is , so far I've converted it to .

I have a feeling I'm way off with this question, could someone please provide some advice with it? If I have somehow done it correctly so far, can I simpilify the equation any further than this?
Title: Re: Specialist 1/2 Question Thread!
Post by: RuiAce on July 20, 2016, 05:55:31 pm
I have a question on changing Cartesian equations to their polar form and I'm a bit stuck on how to do this one.
The Cartesian equation is , so far I've converted it to .

I have a feeling I'm way off with this question, could someone please provide some advice with it? If I have somehow done it correctly so far, can I simpilify the equation any further than this?
Are you sure that this was the question

Because that's about as simplified as it can get.
Title: Re: Specialist 1/2 Question Thread!
Post by: Clockwork on July 20, 2016, 06:02:46 pm
Are you sure that this was the question

Because that's about as simplified as it can get.

Thanks for that! Yeah, that was the question I was given. I just wasn't 100% sure on the answer because it seemed a bit 'messy' compared to the other questions we've done so far.
Title: Re: Specialist 1/2 Question Thread!
Post by: RuiAce on July 20, 2016, 06:06:54 pm
Thanks for that! Yeah, that was the question I was given. I just wasn't 100% sure on the answer because it seemed a bit 'messy' compared to the other questions we've done so far.


Title: Re: Specialist 1/2 Question Thread!
Post by: Adequace on July 20, 2016, 09:30:31 pm
Hi, I need help with this question, http://m.imgur.com/a/WlmIG

For part c) I'm not sure how to do it but the answer went AB - [ans of part c]. However, I don't get how that gets us the shortest distance?

Thanks
Title: Re: Specialist 1/2 Question Thread!
Post by: RuiAce on July 20, 2016, 09:39:22 pm
Hi, I need help with this question, http://m.imgur.com/a/WlmIG

For part c) I'm not sure how to do it but the answer went AB - [ans of part c]. However, I don't get how that gets us the shortest distance?

Thanks



Edit: Yes, this is a formula. If you want me to prove it that will be a slight bit harder.
Title: Re: Specialist 1/2 Question Thread!
Post by: Syndicate on July 20, 2016, 10:19:41 pm
Hi, I need help with this question, http://m.imgur.com/a/WlmIG

For part c) I'm not sure how to do it but the answer went AB - [ans of part c]. However, I don't get how that gets us the shortest distance?

Thanks

The shortest distance from point B to line AC is when the line between point B to line AC is perpendicular.
(https://bhs-specialist.wikispaces.com/file/view/9Eresolutes.gif/138703013/316x223/9Eresolutes.gif)

In this case, it's the red line^

In order to do your question, you would first find the vector resolute of AB in the direction of AC, (because that gives you the vector AD(assuming D is the point where the line red line meets on the vector AC).

Now in order for you to get the Vector BD (which is the shortest distance from point B to line AC), you would subtract AB by AD. Then find the modulus of BD in order to get the distance of that vector.
Title: Re: Specialist 1/2 Question Thread!
Post by: Adequace on July 21, 2016, 07:36:58 pm
Thanks Rui and Syndicate
Title: Re: Specialist 1/2 Question Thread!
Post by: Adequace on July 24, 2016, 06:56:47 pm
Could I have some help with these 2 questions for vector proofs http://m.imgur.com/a/KPQDE

For Q9) How does the linked solution prove that they bisect with their working? They proved that BM = MD, are we supposed to then assume that the other diagonal is just going to bisect it with logic..?

For Q10) I'm completely lost with their working. How did they form those 2 simultaneous equations and what are they actually representing?

Thanks
Title: Re: Specialist 1/2 Question Thread!
Post by: Syndicate on July 24, 2016, 07:26:29 pm
Could I have some help with these 2 questions for vector proofs http://m.imgur.com/a/KPQDE

For Q9) How does the linked solution prove that they bisect with their working? They proved that BM = MD, are we supposed to then assume that the other diagonal is just going to bisect it with logic..?

For Q10) I'm completely lost with their working. How did they form those 2 simultaneous equations and what are they actually representing?
Thanks
Q9) Since, AC and BD have the same midpoint, the diagonals of the parallelogram bisect (divide into two equal pieces) each other

(http://amsi.org.au/teacher_modules/C3/C3g4.png)
Q10) An altitude is a line segment, which is perpendicular to the base.
(https://encrypted-tbn3.gstatic.com/images?q=tbn:ANd9GcQyNHY3Hvxel2Ai1sWG4WDgmZk9mhumbda3EaPI7K-S10JnlfhnVA)
Since, they all are perpendicular and meet at the point O (assume it's the point where all the lines intersect in the image), you can assume that the dot product of A to BC and B to AC are equal. So here, you basically have to prove that OC (c) is perpendicular to AB (-a+b), by equating the two dot products, and showing that they equal c . (b-a)
Title: Re: Specialist 1/2 Question Thread!
Post by: RuiAce on July 24, 2016, 07:31:03 pm
Just to add on.

Note that M was defined as the midpoint of AC. Because it is a definition, BM, MD and BD automatically bisects AC.
Title: Re: Specialist 1/2 Question Thread!
Post by: Adequace on July 24, 2016, 08:16:01 pm
Thanks guys again!
Title: Re: Specialist 1/2 Question Thread!
Post by: Adequace on July 24, 2016, 08:41:55 pm
Q9) Since, AC and BD have the same midpoint, the diagonals of the parallelogram bisect (divide into two equal pieces) each other

(http://amsi.org.au/teacher_modules/C3/C3g4.png)
Q10) An altitude is a line segment, which is perpendicular to the base.
(https://encrypted-tbn3.gstatic.com/images?q=tbn:ANd9GcQyNHY3Hvxel2Ai1sWG4WDgmZk9mhumbda3EaPI7K-S10JnlfhnVA)
Since, they all are perpendicular and meet at the point O (assume it's the point where all the lines intersect in the image), you can assume that the dot product of A to BC and B to AC are equal. So here, you basically have to prove that OC (c) is perpendicular to AB (-a+b), by equating the two dot products, and showing that they equal c . (b-a)
Sorry for double post, but I'm still a bit unsure on why you can assume the dot product of A to BC and B to AC are equal even though they are perpendicular?

Wouldn't vector a and b be different lengths?

Thanks
Title: Re: Specialist 1/2 Question Thread!
Post by: Syndicate on July 24, 2016, 10:03:01 pm
Sorry for double post, but I'm still a bit unsure on why you can assume the dot product of A to BC and B to AC are equal even though they are perpendicular?

Wouldn't vector a and b be different lengths?

Thanks

They can equate since they equal the same dot product when you expand them out (the book hasn't shown that)

Personally, I would have done something like:










QED
Title: Re: Specialist 1/2 Question Thread!
Post by: anotherworld2b on July 24, 2016, 11:35:01 pm
Could i get help with these questions?
I've been trying to solve them the past 2 hours TTT ^ TTT
Title: Re: Specialist 1/2 Question Thread!
Post by: RuiAce on July 24, 2016, 11:46:51 pm
Could i get help with these questions?
I've been trying to solve them the past 2 hours TTT ^ TTT
Are these proofs?


Title: Re: Specialist 1/2 Question Thread!
Post by: anotherworld2b on July 25, 2016, 01:36:00 am
Are these proofs?




Thank you so much for your help i was stuck on them for hours  :-[ and yes they are proofs.
Title: Re: Specialist 1/2 Question Thread!
Post by: anotherworld2b on July 31, 2016, 07:22:48 pm
I was really confused on how to solve this question
Title: Re: Specialist 1/2 Question Thread!
Post by: zsteve on July 31, 2016, 07:31:14 pm
I was really confused on how to solve this question
This is a fairly routine method, outlined in the Cambridge textbook (SM 1/2)



Then solve on CAS. You'll find this method works as long as sin and cos have the same argument.
Title: Re: Specialist 1/2 Question Thread!
Post by: RuiAce on July 31, 2016, 07:51:41 pm
Do they teach the tangent half-angle substitution or the auxiliary angle method in this course?
Title: Re: Specialist 1/2 Question Thread!
Post by: Gogo14 on July 31, 2016, 07:55:25 pm
This is a fairly routine method, outlined in the Cambridge textbook (SM 1/2)



Then solve on CAS. You'll find this method works as long as sin and cos have the same argument.
Why do you set the domain for u between [0,pi]?
Title: Re: Specialist 1/2 Question Thread!
Post by: zsteve on July 31, 2016, 08:57:52 pm
I set \(u\in[0, \pi/2]\) so that both sin and cos are positive, so I can use \(\cos(x) = +\sqrt{1-\sin^2(x)}\) to avoid ambiguity regarding the sign of cos(x)
Title: Re: Specialist 1/2 Question Thread!
Post by: StupidProdigy on July 31, 2016, 09:08:18 pm
Do they teach the tangent half-angle substitution or the auxiliary angle method in this course?
Nope! (at least not in units 3/4 specialist)
Title: Re: Specialist 1/2 Question Thread!
Post by: excelsiorxlcr on August 04, 2016, 05:40:39 pm
Please help me ASAP! I have a test very soon on this. A tram decelerates uniformly from a speed of 60 km/h to rest in 60s. Find the time taken for it to travel half the total distance. Thank you so much in advance :)
Title: Re: Specialist 1/2 Question Thread!
Post by: Sine on August 05, 2016, 01:22:54 am
Please help me ASAP! I have a test very soon on this. A tram decelerates uniformly from a speed of 60 km/h to rest in 60s. Find the time taken for it to travel half the total distance. Thank you so much in advance :)
Use constant acceleration formulae

So to find the total displacement : s







so half the distance is 250m
and








now











Title: Re: Specialist 1/2 Question Thread!
Post by: anotherworld2b on August 06, 2016, 12:38:32 am
Are these proofs?


I am still a bit confused did you use the double angle formula for the 3rd line? Also where did the cos^2theta - sin^theta come from in line 4?

How would you prove q19?
Title: Re: Specialist 1/2 Question Thread!
Post by: Syndicate on August 06, 2016, 10:29:33 am
I am still a bit confused did you use the double angle formula for the 3rd line? Also where did the cos^2theta - sin^theta come from in line 4?


How would you prove q19?

sin4x can be simplified down to 2sin2xcos2x.


Here we use the sin addition formula.

since sin2xcos2x is the same as cos2xsin2x, you are able to add them together.



In the fourth line, Rui has used the trigonometric double angle formulas.

(http://www.sosmath.com/trig/Trig5/trig5/img7.gif)


Q19)




- Expand this using the double angle formula.



- Now we apply the double angle formula once again.


- Expand (2cos^2(x) - 1)^2 (Note (2cos^2(x)-1)^2 = (2cos^2(x)-1)(2cos^2(x) -1) and can be expanded using the rule, (a+b)^2 = a^2 = 2ab + b^2)






............... QED


Title: Re: Specialist 1/2 Question Thread!
Post by: anotherworld2b on August 06, 2016, 06:04:27 pm
thank you so much  ;D syndicate
Title: Re: Specialist 1/2 Question Thread!
Post by: anotherworld2b on August 07, 2016, 12:12:41 am
Could i get further explanation please? I am still quite the novice in how to approach this question. Would you use the formula for sin (a+x)?
This is a fairly routine method, outlined in the Cambridge textbook (SM 1/2)



Then solve on CAS. You'll find this method works as long as sin and cos have the same argument.
Title: Re: Specialist 1/2 Question Thread!
Post by: Syndicate on August 07, 2016, 12:10:14 pm
Could i get further explanation please? I am still quite the novice in how to approach this question. Would you use the formula for sin (a+x)?

Simlifying 2sin(x) + 5cos(x) = 3 to rcos(x-A) or rsin(x+B) will yield the same result. However, since sin(x)sin(u) + cos(x)cos(u) can easily be simplified down to cos(x-u), it would be more relevant to convert it to cos(x-u) than to sin(x+u) (as then, it would require more working out).

Title: Re: Specialist 1/2 Question Thread!
Post by: Gogo14 on September 06, 2016, 08:53:18 pm
In a lift that is accelerating upwards at 2 m/s2, a spring balance shows the apparent
weight of an object to be 2.5 kg wt. What would be the reading if the lift were at rest?
Title: Re: Specialist 1/2 Question Thread!
Post by: Syndicate on September 07, 2016, 05:40:56 pm
In a lift that is accelerating upwards at 2 m/s2, a spring balance shows the apparent
weight of an object to be 2.5 kg wt. What would be the reading if the lift were at rest?

2.5 = m(9.8+2)

m = 0.211864406 (try to get as many decimal places possible)

scale reading = m x (g+0)
= 2.1 kg wt.
Title: Re: Specialist 1/2 Question Thread!
Post by: Jakeybaby on September 07, 2016, 05:45:16 pm
2.5 = m(9.8+2)

m = 0.211864406 (try to get as many decimal places possible)

scale reading = m x (g+0)
= 2.1 kg wt.
I'd personally leave your value of m as a fraction, this would allow for more accurate results without having to waste time writing out 300 decimal places.

\begin{equation}{\frac{25}{118}}~.~{\left(9.81+0 \right)}\end{equation}

Title: Re: Specialist 1/2 Question Thread!
Post by: Gogo14 on September 08, 2016, 07:33:42 am
2.5 = m(9.8+2)

m = 0.211864406 (try to get as many decimal places possible)

scale reading = m x (g+0)
= 2.1 kg wt.
Why is it (9.8+2)? So lost
Title: Re: Specialist 1/2 Question Thread!
Post by: jamonwindeyer on September 08, 2016, 09:42:38 am
Why is it (9.8+2)? So lost

The formula used is \(W=mg\), with that additional 2\(ms^{-1}\) coming from the upwards acceleration of the lift! In the lifts frame of reference, this is experienced as a 2\(ms^{-1}\) acceleration downwards (think of being pushed to the floor on a rollercoaster that accelerates upwards, this is the same thing) ;D
Title: Re: Specialist 1/2 Question Thread!
Post by: Adequace on September 08, 2016, 08:02:53 pm
For this question http://m.imgur.com/3UDbKU3

So I found N = 8cos25, but the answer is resolving the normal force in to its components and then equating the vertical component to the weight. Isn't the reaction of the plane to the body, the entire normal force? They got 8/cos25
Title: Re: Specialist 1/2 Question Thread!
Post by: Syndicate on September 10, 2016, 08:58:28 pm
For this question http://m.imgur.com/3UDbKU3

So I found N = 8cos25, but the answer is resolving the normal force in to its components and then equating the vertical component to the weight. Isn't the reaction of the plane to the body, the entire normal force? They got 8/cos25

The answer given by the book is correct, as the Force F is perpendicular to the weight force not the normal force. Therefore, normal force = 8/ cos25.

(One of the ways to approach such a question is to draw a triangle (with all the forces). )
Title: Re: Specialist 1/2 Question Thread!
Post by: Gogo14 on September 11, 2016, 06:57:44 pm
A stone is projected upwards with a speed of 14 ms from a point O at the top of a mineshaft. Five seconds earlier, a lift began to descend the mineshaft from O with a constant speed of 3.5ms. Find the depth of the lift to the nearest metre at the instant when the stone falls on it. Neglect air resist
Ans: 33m
I had a look at the worked solutions, but that only made me more confused
Title: Re: Specialist 1/2 Question Thread!
Post by: ezferns on September 23, 2016, 02:49:21 pm
Hey, anyone know how to derive Lami's Rule using the Sine Rule and a triangle?
I can't really picture it in my head
This is for statics btw
Title: Re: Specialist 1/2 Question Thread!
Post by: Gogo14 on October 02, 2016, 03:32:12 pm
http://imgur.com/dhCbbsi
How do you do question 16b?
Ans for a is ]100/(9g)
Ans for b is ]2sqrt3
Title: Re: Specialist 1/2 Question Thread!
Post by: Syndicate on October 02, 2016, 07:01:25 pm
http://imgur.com/dhCbbsi
How do you do question 16b?
Ans for a is ]100/(9g)
Ans for b is ]2sqrt3

You can solve this question using the work-energy principle (I am not sure if it is part of the specialist curriculum).
Title: Re: Specialist 1/2 Question Thread!
Post by: Gogo14 on October 02, 2016, 08:09:01 pm
You can solve this question using the work-energy principle (I am not sure if it is part of the specialist curriculum).
Don't what that is lol.
But how do u do it by resolving vectors?
Title: Re: Specialist 1/2 Question Thread!
Post by: anotherworld2b on October 17, 2016, 07:59:23 pm
Could i please get help qith these 2 questions?
Title: Re: Specialist 1/2 Question Thread!
Post by: Syndicate on October 18, 2016, 07:27:32 pm
Could i please get help qith these 2 questions?

Have been really busy since the past week..  :-\ sorry about the late reply.

Q15)







Therefore it have been proven that x^5-x is divisible by 5, since x^5-x = 5n.

Q14 can be proved using a similar approach.
Title: Re: Specialist 1/2 Question Thread!
Post by: anotherworld2b on October 19, 2016, 12:33:59 am
Thank you for the help.
I was also windering how would yoj solve this question?
Have been really busy since the past week..  :-\ sorry about the late reply.

Q15)







Therefore it have been proven that x^5-x is divisible by 5, since x^5-x = 5n.

Q14 can be proved using a similar approach.
Title: Re: Specialist 1/2 Question Thread!
Post by: camrenis on October 19, 2016, 09:17:50 am
Thank you for the help.
I was also windering how would yoj solve this question?
Start with the formulas
\begin{align}
A&=2\pi rh+2\pi r^2\\
V&=\pi r^2 h
\end{align}
Substitute the volume in to \(V\) and solve for \(h\) in terms of \(r\).  Substitute this relationship into \(A\), and then you now should have \(A\) in terms of \(r\) only.  Derive \(A\) with respect to \(r\) and set to 0 to find the minimum value of \(r\).  The substitute this value of \(r\) into \(V\) with the volume to find \(h\).

Spoiler
\(h\) in terms of \(r\) is
\begin{align}
h=\frac{535}{\pi r^2}
\end{align}
Sub into \(A\) to get
\begin{align}
A&=2\pi r\frac{535}{\pi r^2}+2\pi r^2\\
&=\frac{1070}{r}+2\pi r^2
\end{align}
Derive with respect to \(r\)
\begin{align}
\frac{dA}{dr}&=-\frac{1070}{r^2}+4\pi r
\end{align}
Set to 0 to find where the minimum of \(A\) is (plotting \(A\) reveals that it is a local minimum and not a local maximum when \(\frac{dA}{dr}=0\)).
\begin{align}
0&=-\frac{1070}{r^2}+4\pi r\\
\implies r&=\left(\frac{535}{2\pi}\right)^\frac{1}{3}\\
&\approx 4.4
\end{align}
Sub into the volume equation to find \(h\):
\begin{align}
535 &= \pi \left(\frac{535}{2\pi}\right)^\frac{2}{3} \times h\\
\implies h&\approx 8.8
\end{align}
Title: Re: Specialist 1/2 Question Thread!
Post by: anotherworld2b on October 19, 2016, 10:41:58 pm
Thank you very much for your help
I was wondering would you do the same for this question?
Start with the formulas
\begin{align}
A&=2\pi rh+2\pi r^2\\
V&=\pi r^2 h
\end{align}
Substitute the volume in to \(V\) and solve for \(h\) in terms of \(r\).  Substitute this relationship into \(A\), and then you now should have \(A\) in terms of \(r\) only.  Derive \(A\) with respect to \(r\) and set to 0 to find the minimum value of \(r\).  The substitute this value of \(r\) into \(V\) with the volume to find \(h\).

Spoiler
\(h\) in terms of \(r\) is
\begin{align}
h=\frac{535}{\pi r^2}
\end{align}
Sub into \(A\) to get
\begin{align}
A&=2\pi r\frac{535}{\pi r^2}+2\pi r^2\\
&=\frac{1070}{r}+2\pi r^2
\end{align}
Derive with respect to \(r\)
\begin{align}
\frac{dA}{dr}&=-\frac{1070}{r^2}+4\pi r
\end{align}
Set to 0 to find where the minimum of \(A\) is (plotting \(A\) reveals that it is a local minimum and not a local maximum when \(\frac{dA}{dr}=0\)).
\begin{align}
0&=-\frac{1070}{r^2}+4\pi r\\
\implies r&=\left(\frac{535}{2\pi}\right)^\frac{1}{3}\\
&\approx 4.4
\end{align}
Sub into the volume equation to find \(h\):
\begin{align}
535 &= \pi \left(\frac{535}{2\pi}\right)^\frac{2}{3} \times h\\
\implies h&\approx 8.8
\end{align}
Title: Re: Specialist 1/2 Question Thread!
Post by: Gogo14 on October 22, 2016, 06:41:49 pm
Does sin(pi/2+pi/3) give cos(pi/3) or -cos(pi/3)
Title: Re: Specialist 1/2 Question Thread!
Post by: Syndicate on October 22, 2016, 09:04:57 pm
Does sin(pi/2+pi/3) give cos(pi/3) or -cos(pi/3)


Title: Re: Specialist 1/2 Question Thread!
Post by: Gogo14 on November 05, 2016, 11:45:40 pm
(Tech free)Determine the locus of points such that if P=(x,y) ,A=(-2,0) and B=(2,0) the PA+PB=4
Help! My exam is in 3 days!
Title: Re: Specialist 1/2 Question Thread!
Post by: jamonwindeyer on November 06, 2016, 12:15:25 am
(Tech free)Determine the locus of points such that if P=(x,y) ,A=(-2,0) and B=(2,0) the PA+PB=4
Help! My exam is in 3 days!

Disclaimer: No idea if this working out is VCE friendly. Just thought I'd lend a hand.

Hey Gogo! We'll take the points and substitute them into expressions for the distances PA and PB to start:




Now let's play with that relationship a bit so we can bring in the expressions without the square roots:



Notice we got a whole heap of cancellation into that last line. That's what we wanted to see. We can continue:



So this locus lies on the x-axis; however, it isn't the whole x-axis. We need to check something else to find the missing restriction, remember that for our condition to make sense, PA must be less than or equal to 4, and so must PB. Let's check that condition:




These two regions overlap only for \(-2\le x\le2\), so that's where our locus lies. Putting it together, this locus is just the line segment joining A and B.

Note that you could do this last bit just with some clever thinking. Clearly we can't venture beyond A or B with our locus, because that just doesn't make sense. As soon as we do, one of our distances is immediately, clearly, too large. You could probably not do any algebra for this question if you are good at picturing loci, but I thought I'd show everything :) sorry if this isn't VCE friendly, but I thought I'd lend a hand anyway! :)

Title: Re: Specialist 1/2 Question Thread!
Post by: Gogo14 on November 06, 2016, 12:29:29 pm
Disclaimer: No idea if this working out is VCE friendly. Just thought I'd lend a hand.

Hey Gogo! We'll take the points and substitute them into expressions for the distances PA and PB to start:




Now let's play with that relationship a bit so we can bring in the expressions without the square roots:



Notice we got a whole heap of cancellation into that last line. That's what we wanted to see. We can continue:



So this locus lies on the x-axis; however, it isn't the whole x-axis. We need to check something else to find the missing restriction, remember that for our condition to make sense, PA must be less than or equal to 4, and so must PB. Let's check that condition:




These two regions overlap only for \(-2\le x\le2\), so that's where our locus lies. Putting it together, this locus is just the line segment joining A and B.

Note that you could do this last bit just with some clever thinking. Clearly we can't venture beyond A or B with our locus, because that just doesn't make sense. As soon as we do, one of our distances is immediately, clearly, too large. You could probably not do any algebra for this question if you are good at picturing loci, but I thought I'd show everything :) sorry if this isn't VCE friendly, but I thought I'd lend a hand anyway! :)
Thanks!
Title: Re: Specialist 1/2 Question Thread!
Post by: samuelbeattie76 on February 02, 2017, 07:58:41 am
I have attached the question and its answer. What I do not understand is the 195 different sums possible. Isn't there 18 different possible sums from 1 to 18? Can some please explain this question I do not understand the answer given in the solutions manual.

Thanks a lot, any help is appreciated.
Title: Re: Specialist 1/2 Question Thread!
Post by: Shadowxo on February 03, 2017, 08:02:15 pm
I have attached the question and its answer. What I do not understand is the 195 different sums possible. Isn't there 18 different possible sums from 1 to 18? Can some please explain this question I do not understand the answer given in the solutions manual.

Thanks a lot, any help is appreciated.

I think you got a bit mixed up with the player and the number they were assigned
If each of the 35 players gets assigned a different number between 1 and 99, there  could be a player with the number 1 and another player with the number 2, so the sum of their numbers would be 3 (the lowest sum possible). There could also be a player with the number 99 and another player with the number 98, so the sum of their numbers would be 197 (the highest sum possible). So since the players could get any number between 1 and 99, the different sums could be anywhere between 3 (the lowest) and 197 (the highest), resulting in a possible 195 different sums
Does this help? :)
Title: Re: Specialist 1/2 Question Thread!
Post by: codebreaker1_91 on May 18, 2017, 06:34:14 pm
Hi!

A girl at the bottom of a 100m high cliff throws a tennis ball vertically upwards. At the same instant, a boy at the very top of the  cliff drops a golf ball so that it hits the tennis ball while both balls are still in the air. The acceleration of both balls can be taken as 10.0 m/s^2 downwards.

a. If balls collide when tennis ball is at top of path, what is position of the tennis ball when it strikes the golf ball?
b. With what speed is the tennis ball thrown for this to occur?
c. What is the speed of the golf ball when it strikes the tennis ball?
d. For how long has each ball been in motion when they collide?

Thankyou!
Title: Re: Specialist 1/2 Question Thread!
Post by: pha0015 on June 29, 2017, 09:47:13 pm
Dumb question alert!

I'm not too sure if I recall this question, but it's been baffling me:
If z=-4i what is the modulus-argument of Iz^3I

Wouldn't the modulus be 64 and the argument be 0?
However the teacher said that the argument was pi/2

Please help this poor soul
Title: Re: Specialist 1/2 Question Thread!
Post by: RuiAce on June 29, 2017, 10:22:27 pm
Dumb question alert!

I'm not too sure if I recall this question, but it's been baffling me:
If z=-4i what is the modulus-argument of Iz^3I

Wouldn't the modulus be 64 and the argument be 0?
However the teacher said that the argument was pi/2

Please help this poor soul


Edit: Just saw the bars around Iz^3I. Was that intended?
If yes then the argument is certainly 0 because the modulus is a real number
Title: Re: Specialist 1/2 Question Thread!
Post by: pha0015 on June 29, 2017, 10:30:03 pm


Edit: Just saw the bars around Iz^3I. Was that intended?
If yes then the argument is certainly 0 because the modulus is a real number


Edit: Just saw the bars around Iz^3I. Was that intended?
If yes then the argument is certainly 0 because the modulus is a real number

I do remember seeing the modulus signs, but I don't fully recall the wording of the question, so it may have been something like: if the modulus is Iz^3I then...

Thanks for the clarification anyways
Title: Re: Specialist 1/2 Question Thread!
Post by: keltingmeith on July 05, 2017, 02:46:41 pm
Dumb question alert!

I'm not too sure if I recall this question, but it's been baffling me:
If z=-4i what is the modulus-argument of Iz^3I

Wouldn't the modulus be 64 and the argument be 0?
However the teacher said that the argument was pi/2

Please help this poor soul

Interesting - not sure you've gone and calculated the argument for this, but there's two ways to look at it. The first is "what is the modulus and argument of the number |z^3| (the modulus of z^3)", in which case |z^3| is (by definition) a real number, and the argument MUST be 0, as you've initially suggested.

The second (and I imagine this is where the teacher was going), is where you ignore the modulus signs, and she actually intended you to find the modulus and argument of z^3 - the complex number. The modulus is, of course, done the same, but now for the argument, you'd need to use the formula arg(z^n)=n*arg(z), and then normalise it so it's in the domain (-pi,pi].
Title: Re: Specialist 1/2 Question Thread!
Post by: magicmania121 on July 07, 2017, 09:26:20 am
Hi can someone give me any good suggestions for a specialist tutor, I really wanna start now
Title: Re: Specialist 1/2 Question Thread!
Post by: Sine on July 07, 2017, 02:06:41 pm
Hi can someone give me any good suggestions for a specialist tutor, I really wanna start now
derrick ha, usually he only does yearly intakes and only at a unit 3/4 level. However you need to get in pretty early to secure a place. Otherwise check out anyone found HERE
Title: Re: Specialist 1/2 Question Thread!
Post by: atarmaths77 on July 16, 2017, 07:34:25 pm
Does anyone know how to model the heartbeat ekg using trigonometry, using e^(-x^2), where you multiply a sine or cosine function by another trigonometric function to look like this: https://malouffsaandpblog.files.wordpress.com/2013/03/trevor-ekg.jpg
Title: Re: Specialist 1/2 Question Thread!
Post by: magicmania121 on July 21, 2017, 07:52:45 am
Does anyon have the worked solutions for the book
Title: Re: Specialist 1/2 Question Thread!
Post by: keltingmeith on July 21, 2017, 08:04:45 am
Does anyone know how to model the heartbeat ekg using trigonometry, using e^(-x^2), where you multiply a sine or cosine function by another trigonometric function to look like this: https://malouffsaandpblog.files.wordpress.com/2013/03/trevor-ekg.jpg

Not in a way accessible to VCE, other than randomly guessing. Is this for an assignment?

Does anyon have the worked solutions for the book

Which book? You can probably purchase them directly from the publisher if you can't find them anywhere else. ;)
Title: Re: Specialist 1/2 Question Thread!
Post by: magicmania121 on July 21, 2017, 10:59:02 am
It's the specialist unit 1/2 book
Title: Re: Specialist 1/2 Question Thread!
Post by: keltingmeith on July 21, 2017, 01:23:07 pm
It's the specialist unit 1/2 book

I mean, that's cool, but there's a few of them. Which publisher is it?
Title: Re: Specialist 1/2 Question Thread!
Post by: magicmania121 on July 21, 2017, 07:15:34 pm
It's the Cambridge one like it is blue
Title: Re: Specialist 1/2 Question Thread!
Post by: Sine on July 21, 2017, 10:34:44 pm
It's the specialist unit 1/2 book
It's the Cambridge one like it is blue
Does anyon have the worked solutions for the book
Sorry unfortunately we cannot post copyright material on the forums

Your teacher should have access to the worked solutions so ask them.

Hopefully this helps
Title: Re: Specialist 1/2 Question Thread!
Post by: keltingmeith on July 21, 2017, 11:03:56 pm
Unfortunately, a quick Google search tells me that the worked solutions may not be available to purchase. I recommend following Sine's suggestion.
Title: Re: Specialist 1/2 Question Thread!
Post by: A TART on August 20, 2017, 06:40:00 pm
Hi :D I've been able to do this vector question and get the correct answer, however, I've been doing under the assumption the angle between the two is 0 (going in same direction). Can someone explain why this happens? Or am I wrong in my working out?
http://i.imgur.com/Q0ce9bp.jpg

Title: Re: Specialist 1/2 Question Thread!
Post by: VanillaRice on August 20, 2017, 07:04:01 pm
Hi :D I've been able to do this vector question and get the correct answer, however, I've been doing under the assumption the angle between the two is 0 (going in same direction). Can someone explain why this happens? Or am I wrong in my working out?
http://i.imgur.com/Q0ce9bp.jpg


You don't need to substitute a dot b in this case (you can, but a there is a simpler method). The dot product works very much like multiplication, so we can use that property in this question.
i.e.

With this in mind, have another go at the question. The most likely reason you are able to substitute theta = 0 is that when you do so, you are assuming the angle between a and b is 0, and also that the angle between a and (a+b) is 0 - note the two dot products are not the same.

Hope this helps :)
Title: Re: Specialist 1/2 Question Thread!
Post by: Shadowxo on August 20, 2017, 07:07:18 pm
Hi :D I've been able to do this vector question and get the correct answer, however, I've been doing under the assumption the angle between the two is 0 (going in same direction). Can someone explain why this happens? Or am I wrong in my working out?
http://i.imgur.com/Q0ce9bp.jpg


Great answer by VanillaRice (beaten by 1 minute haha)
a.(a+b) = a.a +a.b
It's one of the properties of the dot product :)
So then you end up with (a.a +a.b -a.b) / |a| = a.a / |a| which results in the answer - you don't need to assume theta = 0
Title: Re: Specialist 1/2 Question Thread!
Post by: Sine on August 20, 2017, 07:19:46 pm
Great answer by VanillaRice (beaten by 1 minute haha)
a.(a+b) = a.a +a.b
It's one of the properties of the cross product :)
So then you end up with (a.a +a.b -a.b) / |a| = a.a / |a| which results in the answer - you don't need to assume theta = 0
I really hate to do this but it's the dot product not cross. Cross product is not studied in spec maths. :) :)
Title: Re: Specialist 1/2 Question Thread!
Post by: VanillaRice on August 20, 2017, 07:32:33 pm
I really hate to do this but it's the dot product not cross. Cross product is not studied in spec maths. :) :)
Nevertheless, distributivity is still a property of both cross and dot products  ;D
Title: Re: Specialist 1/2 Question Thread!
Post by: Shadowxo on August 20, 2017, 07:36:34 pm
I really hate to do this but it's the dot product not cross. Cross product is not studied in spec maths. :) :)
Oops! Accidentally wrote cross instead of dot (been doing a lot of dot and cross product in maths lately). Applies to both but meant to say dot :)
Title: Re: Specialist 1/2 Question Thread!
Post by: A TART on August 20, 2017, 08:15:06 pm
You don't need to substitute a dot b in this case (you can, but a there is a simpler method). The dot product works very much like multiplication, so we can use that property in this question.
i.e.
With this in mind, have another go at the question. The most likely reason you are able to substitute theta = 0 is that when you do so, you are assuming the angle between a and b is 0, and also that the angle between a and (a+b) is 0 - note the two dot products are not the same.

Hope this helps :)

Thanks :) Makes it a whole lot easier
Title: Re: Specialist 1/2 Question Thread!
Post by: lilhoo on August 25, 2017, 04:36:10 pm
Hey guys, I'm a year 11 student wondering if it is ok to do spesh 3/4 without 1/2.
I have been recommended to do spesh 3/4 on MANY occasions by my methods and recently, physics teacher (both also teach spesh 3/4).
If i can arrange "catch-up lessons" with either of the above teachers as well as studying by myself, what are my chances of surviving Spesh 3/4? Thanks
Title: Re: Specialist 1/2 Question Thread!
Post by: VanillaRice on August 25, 2017, 04:42:19 pm
Hey guys, I'm a year 11 student wondering if it is ok to do spesh 3/4 without 1/2.
I have been recommended to do spesh 3/4 on MANY occasions by my methods and recently, physics teacher (both also teach spesh 3/4).
If i can arrange "catch-up lessons" with either of the above teachers as well as studying by myself, what are my chances of surviving Spesh 3/4? Thanks
Your chances are good :)
I don't think that 1/2 is required to have a good basis to do 3/4 (although this may now be different with the new study design..). The most important thing is that you're willing to put in the work  ;)  You'll learn a few new concepts (complex numbers, vectors), so it might be worth focusing on these if you're wanting to do any work before next year. Given that it's been suggested to you many times by the teachers that actually teach the subject, I think it might be worth considering (especially if you like maths  ;D).
Title: Re: Specialist 1/2 Question Thread!
Post by: KiNSKi01 on October 11, 2017, 05:58:58 pm
These questions seem really straightforward but how do you figure out the different vectors and write them in correct format? :)
Title: Re: Specialist 1/2 Question Thread!
Post by: Eric11267 on October 11, 2017, 06:05:36 pm
These questions seem really straightforward but how do you figure out the different vectors and write them in correct format? :)
Ok so unit vector i represents one unit in the positive x direction, whilst the unit vector j represents one unit in the positive y direction. Moving from C to D is going 6 units in the negative x direction, so it is -6i. Moving from C to A is going 7 units in the negative x direction and 3 units in the positive y direction, so it is -7i+3j. Going from C to B is moving one unit in the negative x direction and three units in the positive y directoin, so it is -i+3j.
Title: Re: Specialist 1/2 Question Thread!
Post by: KiNSKi01 on October 11, 2017, 06:16:24 pm
Cheers Eric, I'm familiar with how to calculate the scalar product of vectors in two dimensions but I'm not completely sure on how to tackle questions with 3 dimensions. I assume its a similar process but I don't have answers to the questions  ;D
Title: Re: Specialist 1/2 Question Thread!
Post by: Eric11267 on October 11, 2017, 06:22:10 pm
Cheers Eric, I'm familiar with how to calculate the scalar product of vectors in two dimensions but I'm not completely sure on how to tackle questions with 3 dimensions. I assume its a similar process but I don't have answers to the questions  ;D
Yes its a similar process, just add the products of the corresponding coefficients for each unit vector. So for part a, it would ge -2 x 4 + -5 x -1 + 1 x 3 = 0
Title: Re: Specialist 1/2 Question Thread!
Post by: Shadowxo on October 11, 2017, 08:14:56 pm
Cheers Eric, I'm familiar with how to calculate the scalar product of vectors in two dimensions but I'm not completely sure on how to tackle questions with 3 dimensions. I assume its a similar process but I don't have answers to the questions  ;D
Yep, Eric's right. Just a bit of background knowledge:
We only multiply the parallel components (that is, i with i, j with j, k with k) because theta = 0 so cos(theta)=1. The perpendicular components (eg i and j) are perpendicular so cos(theta)=0. You should remember that a.b=|a||b|cos(theta) which is why we do this.
So (2i+3j).(2i+3j)=2i.2i+2*2i.3j+3j.3j = 4+0+9 = 13. (You don't have to do this in the exam or anything)
So just multiply the coefficients for each unit vector, as Eric said :)
You can do this for vectors in as many dimensions as you like, even higher than 3 :)
Title: Re: Specialist 1/2 Question Thread!
Post by: KiNSKi01 on October 11, 2017, 08:32:09 pm

So (2i+3j).(2i+3j)=2i.2i+2*2i.3j+3j.3j = 4+0+9 = 13. (You don't have to do this in the exam or anything)
So just multiply the coefficients for each unit vector, as Eric said :)


I'm not exactly sure what you have done here  ???  :-\ Don't u just multiply the coefficients in front of i and j so you get 2*2 + 3*3 = 14.
I know it is the same answer but where does the zero come from?
Title: Re: Specialist 1/2 Question Thread!
Post by: Shadowxo on October 11, 2017, 08:44:07 pm
I'm not exactly sure what you have done here  ???  :-\ Don't u just multiply the coefficients in front of i and j so you get 2*2 + 3*3 = 14.
I know it is the same answer but where does the zero come from?
Sorry :P it's not exactly necessary to show it just thought some background info would be nice
What I did was expand it out, and just wanted to show that i.j =0 as they're perpendicular (same with i.k and j.k) which is why you only multiply the coefficients in front of the same unit vectors, not the different ones.
Title: Re: Specialist 1/2 Question Thread!
Post by: Eric11267 on October 11, 2017, 08:50:50 pm
I'm not exactly sure what you have done here  ???  :-\ Don't u just multiply the coefficients in front of i and j so you get 2*2 + 3*3 = 14.
I know it is the same answer but where does the zero come from?
Adding up the products of the coefficients only works in unit vector notation because the vectors involved have magnitude 1 and each unit vector is perpendicular to the other two. Later you'll encounter questions that will involve the dot product of vectors which are not in unit vector notation (such as in vector geometry proof questions). Its important to know about the properties of the dot product.
For example if I have vectors a,b,c then:
 c· (a+b)= c·a+c·b   or
(c+a)·(c+b)= c·c+c·b+a·c+a·b or
c·c=|c|2   etc
I think its important you learn the definition of the dot product, because it will be useful in the future
Title: Re: Specialist 1/2 Question Thread!
Post by: KiNSKi01 on October 11, 2017, 09:18:49 pm

What I did was expand it out, and just wanted to show that i.j =0 as they're perpendicular (same with i.k and j.k)

Oh whoops sorry. Yeah I get ya know. Thanks Shadow and Eric for the help and advice!  ;)
Title: Re: Specialist 1/2 Question Thread!
Post by: KiNSKi01 on October 15, 2017, 12:32:58 pm
When calculating the unit vector do you need to first take out a common factor of the coefficients of 'i' and 'j'

e.g a= 6i + 4j goes to 1/root13 (3i + 2j)
Title: Re: Specialist 1/2 Question Thread!
Post by: LifeisaConstantStruggle on October 15, 2017, 12:43:15 pm
doesn't really matter :) you can take out the common factor at the end of your calculation and simplify it.
Title: Re: Specialist 1/2 Question Thread!
Post by: KiNSKi01 on October 15, 2017, 12:51:17 pm
How do you calculate the magnitude of vector AB when you don't know the individual 'i' and 'j' values of a and b?
Title: Re: Specialist 1/2 Question Thread!
Post by: LifeisaConstantStruggle on October 15, 2017, 12:55:37 pm
How do you calculate the magnitude of vector AB when you don't know the individual 'i' and 'j' values of a and b?

The dot product, a.b=0, which indicates that these two vectors are perpendicular. a (OA) and b (OB) would form the vertices of a right-angled triangle, and AB would be the hypotenuse of it. The magnitude of AB would be the length of the hypotenuse.
Title: Re: Specialist 1/2 Question Thread!
Post by: Sine on October 15, 2017, 12:56:54 pm
How do you calculate the magnitude of vector AB when you don't know the individual 'i' and 'j' values of a and b?
a.b = 0 means that vector a and b are perpendicular
we know the length of a is 40 and the length of b is 9
we can determine the length of AB via tthe pythagorean theorem (AB is just the pathway between A and B )

EDIT: didn't even realise someone posted the solution (must've missed that WARNING message) :P
Title: Re: Specialist 1/2 Question Thread!
Post by: KiNSKi01 on October 15, 2017, 01:14:44 pm
How do you approach a question like this. I have tried to split up A into its perpendicular and parallel components in relation to B but I don't know what to do from there  :-[
Title: Re: Specialist 1/2 Question Thread!
Post by: Sine on October 15, 2017, 01:27:15 pm
How do you approach a question like this. I have tried to split up A into its perpendicular and parallel components in relation to B but I don't know what to do from there  :-[
draw a large diagram to scale it may help you see things you haven't considered before
Title: Re: Specialist 1/2 Question Thread!
Post by: KiNSKi01 on October 15, 2017, 01:52:05 pm
I drew out a large diagram. Am I right in saying you want to calculate the component of (i +j) which is perpendicular to (3i + j); meaning you would find the dot product of these two position vectors divided by the magnitude of (3i +j)^2. When I do this I just get a fraction 2/5 instead of root17/5 (the answer to the question). What am I missing here?
Title: Re: Specialist 1/2 Question Thread!
Post by: KiNSKi01 on October 15, 2017, 08:02:47 pm
Any help would be great  ???
Title: Re: Specialist 1/2 Question Thread!
Post by: LifeisaConstantStruggle on October 15, 2017, 08:18:41 pm
Any help would be great  ???

Tbh I don't think the answers are correct? The shortest distance of the point (1,1) and the vector 3i+j would be the vector resolute of i+j in the direction perpendicular to 3i+j, but no answer seems to be correct.
Title: Re: Specialist 1/2 Question Thread!
Post by: VanillaRice on October 15, 2017, 08:26:57 pm
I drew out a large diagram. Am I right in saying you want to calculate the component of (i +j) which is perpendicular to (3i + j); meaning you would find the dot product of these two position vectors divided by the magnitude of (3i +j)^2. When I do this I just get a fraction 2/5 instead of root17/5 (the answer to the question). What am I missing here?
I agree with LifeisaConstantStruggle - I haven't actually been able to get the answer for this question either. However, I will show you my thought process:
You indeed looking for the vector component of OA which is perpendicular to OB. What you've calculated (2/5) is the scalar component of OA parallel to OB i.e. the vector projection of OA onto OB. Let us call the the component of OA parallel to OB a vector u.
u = 2/5<1,1>
So, the vector component of OA perpendicular to OB must be given by OA - u. The magnitude of this vector should give you the answer, however I get sqrt(10)/5.
Title: Re: Specialist 1/2 Question Thread!
Post by: KiNSKi01 on October 15, 2017, 08:27:31 pm
Wouldn't be the first time I've been given work with incorrect answers  :-X

What you said is exactly what I did and wouldn't the answer include 'i' and 'j' in it anyway since it doesn't ask for magnitude?
Title: Re: Specialist 1/2 Question Thread!
Post by: VanillaRice on October 15, 2017, 08:29:41 pm
Wouldn't be the first time I've been given work with incorrect answers  :-X

What you said is exactly what I did and wouldn't the answer include 'i' and 'j' in it anyway since it doesn't ask for magnitude?
If you think about it, the distance (or length) implies magnitude :)
Title: Re: Specialist 1/2 Question Thread!
Post by: KiNSKi01 on October 15, 2017, 08:31:51 pm
lol yeah true. Oh well it was an impossible question anyway  ;D
Title: Re: Specialist 1/2 Question Thread!
Post by: Sine on October 15, 2017, 09:12:47 pm
just did the question got same answer as VanillaRice - so there is no answer given ahah
you can also do this question (or check answer by) converting the vectors to lines in the cartesian plane to find the min distance.
Title: Re: Specialist 1/2 Question Thread!
Post by: KiNSKi01 on October 17, 2017, 06:49:35 pm
How do you answer a question like this? I know once I have values for cos u and sin u I just need to use one of the double angle formulas for cosine. But I have no idea how to find out what these values are
Title: Re: Specialist 1/2 Question Thread!
Post by: LifeisaConstantStruggle on October 17, 2017, 07:29:29 pm
I don't know if you've learnt the double angle formulas in spec 1/2, but the formula for this would be:

cos(2x)=1-2sin2(x)

http://www.vcaa.vic.edu.au/Documents/exams/mathematics/specmaths-formula-w.pdf

Here are all the trig identity formulas.
Title: Re: Specialist 1/2 Question Thread!
Post by: A TART on October 17, 2017, 07:43:18 pm
How do you answer a question like this? I know once I have values for cos u and sin u I just need to use one of the double angle formulas for cosine. But I have no idea how to find out what these values are

Hi!

If you look at your forumale sheet, the double angle forumula for cosine has 3 variations (couldn't think of a better word).

You are not limited to cos(2x)=(cos x)^2 - (sin x)^2 (see attached images)

Working:
[url=https://i.imgur.com/vNqbeiR.jpg]
[url=https://i.imgur.com/M7InnYt.jpg]
(I have a history of changing a 1 into a 2 and 2 into a 12 in an exam so take this with a grain of salt)
Title: Re: Specialist 1/2 Question Thread!
Post by: KiNSKi01 on October 17, 2017, 07:53:43 pm
Yeah I know the 3 different cosine double angle formulas but I was asking about what values you set sin/cos equal to. For this example, if you used a double angle formula with cos in it, would the value for cos u or cos v (Depending) be 4/5 because of pythag theorem?
Title: Re: Specialist 1/2 Question Thread!
Post by: VanillaRice on October 17, 2017, 08:23:04 pm
Yeah I know the 3 different cosine double angle formulas but I was asking about what values you set sin/cos equal to. For this example, if you used a double angle formula with cos in it, would the value for cos u or cos v (Depending) be 4/5 because of pythag theorem?
You can absolutely use the Pythagorean theorem to find the value of cos, however be sure to consider the domain of alpha and beta  (i.e. is it 4/5 or -4/5?).

Hope this helps :)
Title: Re: Specialist 1/2 Question Thread!
Post by: Shadowxo on October 18, 2017, 07:16:56 pm
Yeah I know the 3 different cosine double angle formulas but I was asking about what values you set sin/cos equal to. For this example, if you used a double angle formula with cos in it, would the value for cos u or cos v (Depending) be 4/5 because of pythag theorem?
You can find cos(alpha) or cos(beta) using pythag but you wouldn't need to for this question.

If they wanted you to find sin(2*alpha) however, you'd have to find out sin(alpha) as well, as sin(2*alpha)=2sin(alpha)cos(alpha)
Note that if you'd already found cos(2*alpha) you could use pythag / trig to find sin(2*alpha)
Hope this answers your question :)
Title: Re: Specialist 1/2 Question Thread!
Post by: KiNSKi01 on October 19, 2017, 09:14:17 pm
Thanks for the help!  ;D Circular functions are the worst   :-\
Title: Re: Specialist 1/2 Question Thread!
Post by: Dais_Deorum on October 20, 2017, 07:28:14 pm
Hi guys, first post here  :o,

Just a kinematics question that I want to make sure I've gotten right:

"Two particles, travelling in the same direction, pass a given point together at 8 m/s. Thereafter, one particle begins to slow down with a constant retardation of 2/7 m/s^2. The other continues for 6 seconds at 8 m/s and then slows down with a retardation of 1/9(t-6) m/s^2 where t is the time after the two particles were together. "

"Find how far each of the particles gave travelled when they come to rest."

The first particle was pretty simple; I got 112m. For the second one, I got an answer of 97m, which is the one I'm not quite sure about. Let me know! Cheers
Title: Re: Specialist 1/2 Question Thread!
Post by: Guideme on October 22, 2017, 06:30:51 pm
(https://uploads.tapatalk-cdn.com/20171022/999285985c57fb518f134294caf5cd3c.jpg)

Question 6 pls thank you
Title: Re: Specialist 1/2 Question Thread!
Post by: Dais_Deorum on October 22, 2017, 06:53:47 pm
Hey Guideme,

If the object being acted on by the force is in equilibrium (not moving at all), that force is balanced by an equal one acting in the opposite direction to it. So that force of 20kgwt acting down the plane happens to be a component of the gravity force, and is the result of mg * sin40.
So we know that:

mg * sin40 = 20

Therefore:

Mass = 20/sin40

In this case, the mass equals 31.1145kg

*edit*

To find the pressure this weight exerts on the surface, you must find the normal force, or the component of the force acting perpendicular to the plane. As the object is in equilibrium, the pressure acting on the surface will be the same as the normal force.  To find this we use the parallel force (20kgwt) and tan.

tan(theta) = opposite/adjacent

tan40 = 20/N

N = 20/tan40

N (pressure  on surface) = 26.8416kgwt
Title: Re: Specialist 1/2 Question Thread!
Post by: Shadowxo on October 22, 2017, 09:40:12 pm
Hi guys, first post here  :o,

Just a kinematics question that I want to make sure I've gotten right:

"Two particles, travelling in the same direction, pass a given point together at 8 m/s. Thereafter, one particle begins to slow down with a constant retardation of 2/7 m/s^2. The other continues for 6 seconds at 8 m/s and then slows down with a retardation of 1/9(t-6) m/s^2 where t is the time after the two particles were together. "

"Find how far each of the particles gave travelled when they come to rest."

The first particle was pretty simple; I got 112m. For the second one, I got an answer of 97m, which is the one I'm not quite sure about. Let me know! Cheers
Hey :)
So the kinematics formulas only work for constant acceleration, which the second one isn't, so we have to use integration.
The distance travelled during the first section (no acceleration) is v*t = 6*8=48m
The velocity in the second section (with acceleration) is the integral of acceleration with respect to time


Edit: Also note there are a couple ways to do this. One other way would be to find x as a function of t for t > 6 and sub in the value where v=0 (ie t=18), avoiding definite integrals
Title: Re: Specialist 1/2 Question Thread!
Post by: HighSchoolerRS on October 24, 2017, 06:22:03 pm
Hi guys.
I can't seem to figure out how to prove this and was hoping somebody could run me through it?
I am pretty sure that the question was the first one but there is a slight possibility that it is the second one.




Thank you!
Title: Re: Specialist 1/2 Question Thread!
Post by: VanillaRice on October 24, 2017, 06:50:52 pm
Hi guys.
I can't seem to figure out how to prove this and was hoping somebody could run me through it?
I am pretty sure that the question was the first one but there is a slight possibility that it is the second one.




Thank you!
It's the second one.
Beginning from the RHS and using the identity 1 + tan2(x) = sec2(x)
Spoiler


I'll let you finish it off :) Please post if you're having any issues.

Hope this helps :)
Title: Re: Specialist 1/2 Question Thread!
Post by: HighSchoolerRS on October 28, 2017, 04:49:21 pm
It's the second one.
Beginning from the RHS and using the identity 1 + tan2(x) = sec2(x)
Spoiler


I'll let you finish it off :) Please post if you're having any issues.

Hope this helps :)

Thank you! I actually got the answer now haha
Title: Re: Specialist 1/2 Question Thread!
Post by: KiNSKi01 on November 06, 2017, 08:10:14 pm
I had the exact same question in a recent test! I get to the stage cos^2(x/2) - sin^2(x/2) and I don't know what to do  :P Which identity applies to this part??
Title: Re: Specialist 1/2 Question Thread!
Post by: VanillaRice on November 06, 2017, 08:30:21 pm
I had the exact same question in a recent test! I get to the stage cos^2(x/2) - sin^2(x/2) and I don't know what to do  :P Which identity applies to this part??
Has your teacher talked about the formula sheet which you receive in both exam 1 and 2?
If not, here's a link. Have a look under the trigonometric functions section - you should be able to find your answer there :)

Hope this helps :)
Title: Re: Specialist 1/2 Question Thread!
Post by: KiNSKi01 on November 06, 2017, 08:50:03 pm
Yeh woops. forgot to make the connection that it was 1/2 theta and therefore I should be looking for a double angle identity  :P

I saw it instantly when I clicked on the link ahaha
Title: Re: Specialist 1/2 Question Thread!
Post by: KiNSKi01 on November 07, 2017, 07:15:06 pm
I can't get the answer 3 root 2 :-[

Can someone please tell me where im going wrong. To find vector ED I have been finding vector OE and OD. To find OE I subtracted vector OA from OE and to find vector OD i multiplied vector OB by 0.5. When I subtract vector OD from OE and and proceed to find the square root of the two coefficients of 'i' and 'j' squared; I always get an answer larger than 3 root 2!
Title: Re: Specialist 1/2 Question Thread!
Post by: Shadowxo on November 11, 2017, 09:31:08 pm
I can't get the answer 3 root 2 :-[

Can someone please tell me where im going wrong. To find vector ED I have been finding vector OE and OD. To find OE I subtracted vector OA from OE and to find vector OD i multiplied vector OB by 0.5. When I subtract vector OD from OE and and proceed to find the square root of the two coefficients of 'i' and 'j' squared; I always get an answer larger than 3 root 2!
So OD = -i +2j
AB = AO + OB = -OA + OB = -4i -7j -2i +4j = -6i -3j
So AE = 1/3 * AB = -2i -j
OE = OA +  AE = 4i +7j -2i -j = 2i +6j
ED = EO + OD = -OE + OD = -2i -6j -i +2j = -3i -4j

So the length would be √(9+16) = 5. I think it's an error with the answer :)
Title: Re: Specialist 1/2 Question Thread!
Post by: KiNSKi01 on November 11, 2017, 09:40:38 pm
Thank you Shadowxo!!

Thought I was going insane ahaha
Title: Re: Specialist 1/2 Question Thread!
Post by: Guideme on November 12, 2017, 07:49:46 pm
Hi everyone !

Can anyone explain question 8c to me please .

There isn't any examples for these type of questions :/
(https://uploads.tapatalk-cdn.com/20171112/193dc31a74588fd740ed6bdcd15901dd.jpg)

Thanks in advance !
Title: Re: Specialist 1/2 Question Thread!
Post by: VanillaRice on November 12, 2017, 08:13:56 pm
Hi everyone !

Can anyone explain question 8c to me please .

There isn't any examples for these type of questions :/
(https://uploads.tapatalk-cdn.com/20171112/193dc31a74588fd740ed6bdcd15901dd.jpg)

Thanks in advance !
Have a sketch of the vectors a and b. You can see that the shortest distance from point A to the line OB can be represented by a line which is perpendicular to OB. In other words, this line between the line OB and point A is actually the component of a perpendicular to b! If you can find the magnitude of this 'line', you can find the shortest distance.

Hope this helps :) Please post if you're still stuck.
Title: Re: Specialist 1/2 Question Thread!
Post by: Guideme on November 12, 2017, 08:51:52 pm
Have a sketch of the vectors a and b. You can see that the shortest distance from point A to the line OB can be represented by a line which is perpendicular to OB. In other words, this line between the line OB and point A is actually the component of a perpendicular to b! If you can find the magnitude of this 'line', you can find the shortest distance.

Hope this helps :) Please post if you're still stuck.

Thank you for your reply! I have drawn the vector graph but struggling to find the equation of the perpendicular line
Title: Re: Specialist 1/2 Question Thread!
Post by: KiNSKi01 on November 12, 2017, 08:53:59 pm
Hey I just did Guideme's question and got 12.5

Not sure if I got it right because of the double negative when you subtract the vector resolute of a in the direction b from a.
 My final line of working before calculating magnitude had 4i + j - (3i/2 - 3j/2)
Title: Re: Specialist 1/2 Question Thread!
Post by: Shadowxo on November 12, 2017, 09:06:41 pm
Hi everyone !

Can anyone explain question 8c to me please .

There isn't any examples for these type of questions :/
(https://uploads.tapatalk-cdn.com/20171112/193dc31a74588fd740ed6bdcd15901dd.jpg)

Thanks in advance !
So shortest vector from A to the line OB is the vector component of a perpendicular to b.
To find the distance, you simply need to find the magnitude of this vector, the magnitude of the answer to b)
Title: Re: Specialist 1/2 Question Thread!
Post by: KiNSKi01 on November 12, 2017, 09:32:10 pm
So shortest vector from A to the line OB is the vector component of a perpendicular to b.
To find the distance, you simply need to find the magnitude of this vector, the magnitude of the answer to b)

So if you aren't aware of how you calculate magnitude all you do is find the square root of the coeffecient of i squared + coeffecient of j sqaured

Title: Re: Specialist 1/2 Question Thread!
Post by: Guideme on November 12, 2017, 09:57:58 pm
Thank you shadowxo and kinski01!
So all you do is find the magnitude of the vector component -.-" .
I did all the hard work but didn't realise that it was just the magnitude of the last question lol.
Thank guys haha
Title: Re: Specialist 1/2 Question Thread!
Post by: KiNSKi01 on November 12, 2017, 10:00:08 pm
Can someone help me with this question?

I took t out as a common factor then divided both sides by (2cos2pi + 2sin2pi) then I get stuck   :(
Title: Re: Specialist 1/2 Question Thread!
Post by: Shadowxo on November 12, 2017, 10:59:53 pm
Can someone help me with this question?

I took t out as a common factor then divided both sides by (2cos2pi + 2sin2pi) then I get stuck   :(
In this case, you can't take out t as it's inside the function (although they could have been clearer). If you could, however, you should know that cos(2π)=1 and sin(2π)=0

I think this is a harder question than you'd normally expect.
Title: Re: Specialist 1/2 Question Thread!
Post by: Elsa//768 on November 27, 2017, 04:18:06 pm
Hey I just started Specialist today and i found it quite challenging... But the thing is that I actually got 95% as overall mark for my year 10 maths exams and i was in an accelerated maths class... Since im also doing two unit 3/4 subjects right now, should i give up specialist and switch into legal studies instead, or should i try to stick to specialist and see if things will work out?? THANKS A LOT!!
Title: Re: Specialist 1/2 Question Thread!
Post by: Sine on November 27, 2017, 04:41:38 pm
Hey I just started Specialist today and i found it quite challenging... But the thing is that I actually got 95% as overall mark for my year 10 maths exams and i was in an accelerated maths class... Since im also doing two unit 3/4 subjects right now, should i give up specialist and switch into legal studies instead, or should i try to stick to specialist and see if things will work out?? THANKS A LOT!!
Hey Elsa

I don't think anyone can say anything regarding the 95%/accelereated maths since it would be different at every school and generally years 7-10 and units 1/2 can't say anything about how well you will do for 3/4. I would stick with spec at least for 1 year and just see how it goes then if you don't like it / unhappy with it switch into legal. Which subject would you prefer spec or legal?
Title: Re: Specialist 1/2 Question Thread!
Post by: Elsa//768 on November 27, 2017, 06:00:39 pm
Hey Elsa

I don't think anyone can say anything regarding the 95%/accelereated maths since it would be different at every school and generally years 7-10 and units 1/2 can't say anything about how well you will do for 3/4. I would stick with spec at least for 1 year and just see how it goes then if you don't like it / unhappy with it switch into legal. Which subject would you prefer spec or legal?
I see... I'm not sure yet but I think im going to stick with specialist xD Thank u xD
Title: Re: Specialist 1/2 Question Thread!
Post by: RuiAce on December 04, 2017, 11:22:18 pm
In this case, you can't take out t as it's inside the function (although they could have been clearer). If you could, however, you should know that cos(2π)=1 and sin(2π)=0

I think this is a harder question than you'd normally expect.
Just saw this... ew why don't they teach you guys the auxiliary angle transform?
Title: Re: Specialist 1/2 Question Thread!
Post by: Syndicate on December 05, 2017, 12:14:29 am
Just saw this... ew why don't they teach you guys the auxiliary angle transform?

It's part of the course but nobody pays that much attention to it
Title: Re: Specialist 1/2 Question Thread!
Post by: Shadowxo on December 05, 2017, 11:40:42 am

It's part of the course but nobody pays that much attention to it
I don't recall ever being taught it, nor seeing a question like that  :-\
Just saw this... ew why don't they teach you guys the auxiliary angle transform?
They should  :P
Title: Re: Specialist 1/2 Question Thread!
Post by: keltingmeith on December 05, 2017, 01:17:11 pm

It's part of the course but nobody pays that much attention to it

Mirroring Shadowxo, auxiliary angle is not in the course. :P But also, fuq dat. We don't do enough trig to warrant learning it.
Title: Re: Specialist 1/2 Question Thread!
Post by: RuiAce on December 05, 2017, 01:27:29 pm
Mirroring Shadowxo, auxiliary angle is not in the course. :P But also, fuq dat. We don't do enough trig to warrant learning it.
<Solves asin(x)+bcos(x)=c equations>

<Not enough trig to warrant learning it>
Title: Re: Specialist 1/2 Question Thread!
Post by: keltingmeith on December 05, 2017, 01:34:32 pm
<Solves asin(x)+bcos(x)=c equations>

In a TEXTBOOK. Where they ask stupidly fucked questions that have never been asked before in exams. This is one example of one of those types of questions.
Title: Re: Specialist 1/2 Question Thread!
Post by: undefined on December 30, 2017, 08:07:50 pm
hi all how would you go by answering this question?
Title: Re: Specialist 1/2 Question Thread!
Post by: RuiAce on December 30, 2017, 08:30:57 pm
hi all how would you go by answering this question?

where \(a\) is the first term and \(d\) is the common difference.



Remark: Technically, the common difference is allowed to be 0; this is just the constant sequence; every number in the sequence is all the same. But I will assume that this can't happen purely because I don't know if this is of interest in the VCE course. I will only consider the other case to be correct.
Title: Re: Specialist 1/2 Question Thread!
Post by: pepper77 on February 04, 2018, 11:58:57 pm
Hello everyone.  :)

I just started Specialist Maths last week (we're doing complex numbers) and it's... really easy. Lots of fun, but easier than Methods easy. (off topic but even though I got an A+ on the final maths test last year I forgot just about everything, RIP.) Does Specialist pick up later?

Also, the teacher ran through partial fractions for like 40 minutes then moved on and I don't think I 100% understand them. There's nothing in my textbook about them, so if anyone knows where to find practice questions I would really appreciate it.
Title: Re: Specialist 1/2 Question Thread!
Post by: Sine on February 05, 2018, 12:09:58 am
Hello everyone.  :)

I just started Specialist Maths last week (we're doing complex numbers) and it's... really easy. Lots of fun, but easier than Methods easy. (off topic but even though I got an A+ on the final maths test last year I forgot just about everything, RIP.) Does Specialist pick up later?

Also, the teacher ran through partial fractions for like 40 minutes then moved on and I don't think I 100% understand them. There's nothing in my textbook about them, so if anyone knows where to find practice questions I would really appreciate it.
none of the maths is meant to be hard it's all about where you fall amongst everyone in the state which will depend on how many careless errors you make.

You should be able to make your own questions for partial fractions if not they are in the 1/2 spec book i think. Also you could also just google "Partial fractions questions"
Title: Re: Specialist 1/2 Question Thread!
Post by: PolySquared on February 24, 2018, 10:19:14 am
Hi, can I get some help on this question? I have no idea on how to start this question.

Find the value of sec θ if tanθ = 0.4 and θ is not in the 1st quadrant.
Title: Re: Specialist 1/2 Question Thread!
Post by: RuiAce on February 24, 2018, 10:24:28 am
Hi, can I get some help on this question? I have no idea on how to start this question.

Find the value of sec θ if tanθ = 0.4 and θ is not in the 1st quadrant.




______________________________________________________________________________


Alternate method using trigonometric identities
\begin{align*}\sec^2\theta &= 1+\tan^2\theta\\ &= 1 + \left(\frac25\right)^2\\ &= \frac{29}{25}\end{align*}

Title: Re: Specialist 1/2 Question Thread!
Post by: PolySquared on February 24, 2018, 05:20:22 pm
If

Then what is the value of 
Title: Re: Specialist 1/2 Question Thread!
Post by: TheAspiringDoc on February 24, 2018, 05:34:38 pm
If

Then what is the value of 
Just use the double angle formula twice. You should actually get 2 answers within the principal value of -Pi , pi
Title: Re: Specialist 1/2 Question Thread!
Post by: PolySquared on February 24, 2018, 05:56:29 pm
Just use the double angle formula twice. You should actually get 2 answers within the principal value of -Pi , pi

What do you mean by use it twice? Can you please demonstrate?
Title: Re: Specialist 1/2 Question Thread!
Post by: RuiAce on February 24, 2018, 05:57:09 pm
What do you mean by use it twice? Can you please demonstrate?
\begin{align*}\sin 4\theta &= 2\sin2\theta\cos2\theta\\ &= 4\sin\theta\cos\theta(\cos^2\theta-\sin^2\theta)\end{align*}
Title: Re: Specialist 1/2 Question Thread!
Post by: TheAspiringDoc on February 24, 2018, 06:25:41 pm
Hmm rui's method is correct but a bit tricky IMO. Just let 2theta = x and then find the double angle of sin(2x).
Title: Re: Specialist 1/2 Question Thread!
Post by: PolySquared on February 24, 2018, 06:53:45 pm
Can I please get some help on how to answer this?

√3 cos(2x) − sin(2x) in the form r cos(2x + α).
Title: Re: Specialist 1/2 Question Thread!
Post by: Sine on February 24, 2018, 07:09:56 pm
Can I please get some help on how to answer this?

√3 cos(2x) − sin(2x) in the form r cos(2x + α).
√3 cos(2x) − sin(2x) =  r cos(2x + α).
 RHS =  r cos(2x + α).
          = rcos(2x)cos( α) - r sin(2x)sin( α)
 
LHS = √3 cos(2x) − sin(2x)

LHS = RHS
√3 cos(2x) − sin(2x) = rcos(2x)cos( α) - r sin(2x)sin( α)
 From here we can see that
√3 cos(2x)  = rcos(2x)cos( α)
Hence √3 = rcos( α) 
√3/r = cos( α)
  α = cos-1(√3/r)

Also we can see
sin(2x) = r sin(2x)sin( α)
1 = rsin(α)     
1/r = sin(α) 
α = sin-1(1/r)

Therefore,   sin-1(1/r) = α = cos-1(√3/r)

sin-1(1/r) = cos-1(√3/r)

From here you should be able to eyeball that r = 2 and also that  α = π/6
So now we have
√3 cos(2x) − sin(2x) =  rcos(2x + α)
√3 cos(2x) − sin(2x) =  2cos(2x + π/6).
Title: Re: Specialist 1/2 Question Thread!
Post by: PolySquared on February 25, 2018, 10:56:48 am
Hi could I get some help on this question?

Write in the form

My answer is

but the textbook answer has 5pi/4 instead of pi/4.
Title: Re: Specialist 1/2 Question Thread!
Post by: jazzycab on February 25, 2018, 03:33:26 pm
Hi could I get some help on this question?

Write in the form

My answer is

but the textbook answer has 5pi/4 instead of pi/4.

Multiplying the entire expression by \(\frac{\sqrt{2}}{\sqrt{2}}\) we get:

If we then substitute for the exact values of \(\sin{\left(\frac{\pi}{4}\right)}=\frac{1}{\sqrt{2}}\) and \(\cos{\left(\frac{\pi}{4}\right)}=\frac{1}{\sqrt{2}}\), we get:

From here, we can recognise that what is in the brackets is simply the compound-angle formula for \(\sin{\left(a-b\right)}\):

But in this expression, our argument is not quite what we need (i.e. we need \(3x\) to be positive). We can use the fact that \(\sin{\left(-x\right)}=-\sin{\left(x\right)}\) to rectify this:

But now the sign of the entire expression is incorrect. Here we can use the fact that \(\cos{\left(-\pi\right)}=-1\) and \(\sin{\left(-\pi\right)}=0\) to force another compound angle formula for sine:


I'm certain that there is a more elegant way of doing this, but that's the way it fell out when I looked at it.
Title: Re: Specialist 1/2 Question Thread!
Post by: Yiyiyi on March 07, 2018, 10:22:05 am
Four teachers decide to swap desks at work. How many ways can this be done if no teacher is to sit at their previous desk?
Title: Re: Specialist 1/2 Question Thread!
Post by: PolySquared on March 11, 2018, 04:39:13 pm
Hi guys can I please get some help on this question.

Let A = (5,1), B = (0,4) and C = (-1,0). Find:

D such that vector AB = vector CD
Title: Re: Specialist 1/2 Question Thread!
Post by: RuiAce on March 11, 2018, 04:42:37 pm


This gives \(d_2 = 3\) and \(d_1 = -6\), assuming I did not mess up basic arithmetic
Title: Re: Specialist 1/2 Question Thread!
Post by: PolySquared on March 11, 2018, 04:52:23 pm


This gives \(d_2 = 3\) and \(d_1 = -6\), assuming I did not mess up basic arithmetic

Thank you so much
Title: Re: Specialist 1/2 Question Thread!
Post by: RuiAce on March 11, 2018, 04:56:08 pm
Four teachers decide to swap desks at work. How many ways can this be done if no teacher is to sit at their previous desk?

Let teacher 1 originally occupy desk 1, teacher 2 originally occupy desk 2 and etc.
There are 3 ways this can happen, since we have 3 other teachers.

Now, without loss of generality, assume that teacher 2 occupies desk 1.
Case 1: Teacher 1 occupies desk 2.
Then, teacher 3 must occupy desk 4 and vice versa, so this case can only happen 1 way.

Case 2: Either one of teachers 3 or 4 occupy desk 2. Note that there are 2 ways this can happen, since we have 2 other teachers.
Then, without further loss of generality, assume that teacher 3 occupies desk 2.

Subcase 2.1: Teacher 1 occupies desk 3.
This can't happen, because then teacher 4 occupies their own desk.

Subcase 2.2: Teacher 4 occupies desk 3.
Then teacher 1 must occupy desk 4. Only 1 way this can happen.
\[ \therefore \text{Ans: }3\big(1 + 2(0+1) \big) = 9.\]
____________________________________________________________

The total outcomes is just the number of ways the teachers can be assigned to sit anywhere, which is just 4! = 24.

The inclusion-exclusion principle becomes necessary because we double count the number of ways more than 1 teacher can sit at their own seat.
# of ways at least 1 sits at their own seat = 4 * 3! (choose 1 out of 4 teachers, force them to sit at their own desk, and then let the others sit anywhere)
# of ways at least 2 sit at their own seat = 6 * 2! (choose 2 out of 4 teachers, force them to sit at their own desk, and then let the others sit anywhere)
# of ways at least 3 sit in their own seat = 4 * 1! (choose 3 out of 4 teachers, force them to sit at their own desk, and let the last one sit anywhere. Of course, that happens to be their own seat anyway, but the inclusion/exclusion principle is just designed to work like this.)
# of ways all 4 sit in their own seat = 1.

\[ \therefore \text{Ans: }4! - 4\times3! + 6\times2! - 4\times1! + 1 =9.\]
Title: Re: Specialist 1/2 Question Thread!
Post by: PolySquared on March 11, 2018, 08:22:10 pm
Help please

Let C and D be points with position vectors c and d respectively. If |c| = 5 and |d| = 7  and c dot d = 4, find the absolute value of vector CD
Title: Re: Specialist 1/2 Question Thread!
Post by: RuiAce on March 11, 2018, 08:48:16 pm
Help please

Let C and D be points with position vectors c and d respectively. If |c| = 5 and |d| = 7  and c dot d = 4, find the absolute value of vector CD


Title: Re: Specialist 1/2 Question Thread!
Post by: PolySquared on March 15, 2018, 10:30:23 am
How do you do this question?

Let a =3i + 4j and b = i - j

Find the vector with the same magnitude of b with the same direction of a
Title: Re: Specialist 1/2 Question Thread!
Post by: jazzycab on March 15, 2018, 11:08:43 am
How do you do this question?

Let a =3i + 4j and b = i - j

Find the vector with the same magnitude of b with the same direction of a
Firstly, a vector in the direction of \(\textbf{a}\) with magnitude 1 is the unit vector, \(\hat{\textbf{a}}\):

We need to adjust the length of this vector so that it is the same as \(\textbf{b}\). That is, the required vector, \(\textbf{v}=\left|\textbf{b}\right|\hat{\textbf{a}}\):
Title: Re: Specialist 1/2 Question Thread!
Post by: randomnobody69420 on March 24, 2018, 12:05:23 pm
I feel like this question has already been asked a million times before but does anyone know where I can find worked solutions for Cambridge Specialist Maths 1&2 textbook?
Title: Re: Specialist 1/2 Question Thread!
Post by: K888 on March 24, 2018, 12:10:35 pm
I feel like this question has already been asked a million times before but does anyone know where I can find worked solutions for Cambridge Specialist Maths 1&2 textbook?
Hey there, it's best to ask your teacher about this. No-one can provide them to you here as they are a commercial product and thus subject to copyright. Your teacher will likely have a copy that your school has purchased, or will know where to get one.
Title: Re: Specialist 1/2 Question Thread!
Post by: PolySquared on April 09, 2018, 03:41:46 pm
Hi could I get some help on this question please.

In triangle OAB, let a = vector OA and b = vector OB. The point P is on AB such that vector AP = 2 times vector PB. express vector PB in terms of a and b
Title: Re: Specialist 1/2 Question Thread!
Post by: mzhao on April 09, 2018, 06:03:39 pm
Hi could I get some help on this question please.

In triangle OAB, let a = vector OA and b = vector OB. The point P is on AB such that vector AP = 2 times vector PB. express vector PB in terms of a and b

Hey PolySquared,

From the question, we have:
$$ \overrightarrow{AP} = 2\overrightarrow{PB} $$
and since
\begin{align*}
\overrightarrow{AB} &= \overrightarrow{AP} + \overrightarrow{PB}\\
\\
\implies \overrightarrow{AB} &= 2\overrightarrow{PB} + \overrightarrow{PB}\\
&= 3\overrightarrow{PB}\\
\implies \overrightarrow{PB} &= \frac{1}{3}\overrightarrow{AB}\\
&= \frac{1}{3}(\overrightarrow{OB} - \overrightarrow{OA})\\
\therefore \overrightarrow{PB} &= \frac{1}{3}(b-a)
\end{align*}
Title: Re: Specialist 1/2 Question Thread!
Post by: PolySquared on April 10, 2018, 01:03:20 pm
Could I get soem help on this question please.

Points A and B are defined by the position vectors a=3i+4j and b = 5i+12j

find the unit vector which bisects angle AOB
Title: Re: Specialist 1/2 Question Thread!
Post by: randomnobody69420 on April 11, 2018, 12:05:14 am
Is circle geometry supposed to be insanely hard? I feel like half of the questions in the cambridge 1/2 are impossible for me to do. I consider myself a decent maths student but I’m really struggling with this topic. I had no trouble with doing circle geometry in year 10 but this year...idk this questions seem much harder.

Is there any resources out there that can help me improve my problem solving skills when it comes to this topic? Or anything else that can help me solve these difficult questions (I would post them here but honestly there are way too many).

edit: I should specify: the main type of problems I’m having trouble with are the ones that ask you to prove something.
Title: Re: Specialist 1/2 Question Thread!
Post by: Sine on April 11, 2018, 12:35:03 am
Is circle geometry supposed to be insanely hard? I feel like half of the questions in the cambridge 1/2 are impossible for me to do. I consider myself a decent maths student but I’m really struggling with this topic. I had no trouble with doing circle geometry in year 10 but this year...idk this questions seem much harder.

Is there any resources out there that can help me improve my problem solving skills when it comes to this topic? Or anything else that can help me solve these difficult questions (I would post them here but honestly there are way too many).

edit: I should specify: the main type of problems I’m having trouble with are the ones that ask you to prove something.
use vectors for the proofs (if you have learnt it) and it also would be helpful to revise all the circle thoerems.

Also I found for circle gemotery proofs you sometimes had to be exposed to the solution in order to do the proof since the easiest proof counted in knowing really small details.
Title: Re: Specialist 1/2 Question Thread!
Post by: jazzycab on April 11, 2018, 07:45:41 am
Is circle geometry supposed to be insanely hard? I feel like half of the questions in the cambridge 1/2 are impossible for me to do. I consider myself a decent maths student but I’m really struggling with this topic. I had no trouble with doing circle geometry in year 10 but this year...idk this questions seem much harder.

Is there any resources out there that can help me improve my problem solving skills when it comes to this topic? Or anything else that can help me solve these difficult questions (I would post them here but honestly there are way too many).

edit: I should specify: the main type of problems I’m having trouble with are the ones that ask you to prove something.

Don't be afraid to try a few different things. It can be very difficult to identify how to approach a lot of these proofs. If you feel that something isn't working, try a different approach.
For these particular types of proofs, try and identify the geometric representation of the starting point, but more importantly, the end point. Then work backwards in terms of identifying all necessary information that would be required for that end point, hence, try and identify how to show that information from the start point
Title: Re: Specialist 1/2 Question Thread!
Post by: PolySquared on April 11, 2018, 12:18:05 pm
How do you prove that the diagonals of a square are of equal length and bisect eact other using vectors?
Title: Re: Specialist 1/2 Question Thread!
Post by: Sine on April 13, 2018, 12:23:30 pm
How do you prove that the diagonals of a square are of equal length and bisect eact other using vectors?
First define two sides of different "direction" as vectors a and b, now you should be able to express everything with only those two vectors.
For equal length: First how do we express the diagnol in terms of a and b then think about how you find the length of a vector?
For bisect: What properties inherently show bisecting? Either finding the lengths of each half of the diagnol or to find that both midpoints of the diagnol occur at the same place.
Title: Re: Specialist 1/2 Question Thread!
Post by: Sooomi on April 24, 2018, 06:37:35 pm
Hey all!!
So i'm currently doing Circle Geometry in Spesh 1/2 and I struggle with proofs and was wondering if I could get sure tips or some advice on improving my proofs and what to look out for and etc.
Just anything that will help or personally that you found useful!
Much appreciated!!
Title: Re: Specialist 1/2 Question Thread!
Post by: caffeine on April 25, 2018, 04:29:47 pm
Hey guys,

Sorry if this was the wrong area to post this, but I was given the cambridge senior maths chapter 18 (matrices) because my methods tutor (i'm in year 11) said it would give us a bigger overall picture on matrices. I'm having a little bit of trouble with exercise 18C. I know how to do the basic multiplying, but don't understand a few of the trickier questions, and I'd be super grateful if you guys could assist me with the following!!

Question 2: http://puu.sh/A9Yqb/283fa45c3e.png
Question 4 and Question 5: http://puu.sh/A9YrO/0bea0789de.png
Question 12: http://puu.sh/A9Yts/f21438a27d.png

Even if you guys can help me with even one of the questions it'd help a lot!!

Cheers!
Title: Re: Specialist 1/2 Question Thread!
Post by: jazzycab on April 25, 2018, 05:02:10 pm
Hey guys,

Sorry if this was the wrong area to post this, but I was given the cambridge senior maths chapter 18 (matrices) because my methods tutor (i'm in year 11) said it would give us a bigger overall picture on matrices. I'm having a little bit of trouble with exercise 18C. I know how to do the basic multiplying, but don't understand a few of the trickier questions, and I'd be super grateful if you guys could assist me with the following!!

Question 2: http://puu.sh/A9Yqb/283fa45c3e.png
Question 4 and Question 5: http://puu.sh/A9YrO/0bea0789de.png
Question 12: http://puu.sh/A9Yts/f21438a27d.png

Even if you guys can help me with even one of the questions it'd help a lot!!

Cheers!


Question 2 is testing your understanding of the matrix multiplication process, that is:
A product of 2 matrices is defined only if the number of columns in the first matrix is equal to the number of rows in the second matrix.
The matrices have the following dimensions:
\(\mathbf{X}:\ 2\times1,\ \mathbf{Y}:\ 2\times1,\ \mathbf{A}:\ 2\times2,\ \mathbf{B}:\ 2\times2,\ \mathbf{C}:\ 2\times2,\ \mathbf{I}:\ 2\times2\).
Note that the dimensions are listed as "number of rows"\(\times\)"number of columns"
Therefore the following matrix products are defined (of the ones listed):
\(\mathbf{A}\mathbf{Y},\ \mathbf{C}\mathbf{I}\).

For question 4, consider the matrix product \(\mathbf{AB}\) for two arbitrary \(2\times2\) matrices:

If \(\mathbf{AB}=\mathbf{O}\) then \(ae+bg=0,\ af+bh=0,\ ce+dg=0,\ cf+dh=0\).
All we need to determine is if there is any possible combination of these such that not all of \(a,\ b,\ c\text{ and }d\) are 0 and not all of \(e,\ f,\ g\text{ and }h\) are 0.
We have \(ae=-bg\Rightarrow\frac{a}{b}=-\frac{g}{e}\) and \(af=-bh\Rightarrow\frac{a}{b}=-\frac{h}{f}\). Therefore, \(\frac{g}{e}=\frac{h}{f}\).
We also have \(ce=-dg\Rightarrow\frac{c}{d}=-\frac{g}{e}\) and \(cf=-dh\Rightarrow\frac{c}{d}=-\frac{h}{f}\).
Combining all four of these gives us \(\frac{a}{b}=\frac{c}{d}=-\frac{g}{e}=-\frac{h}{f}\).
If we can find an example where this is true, for which none of \(a,\ b,\ c,\ d,\ e,\ f\text{ and }g\) are 0, then we have disproven the statement.
Consider the simple case where \(a=1,\ b=1,\ c=1,\ d=1,\ g=1,\ e=-1,\ h=1\text{ and }f=-1\). In this case, we get:

Thus, the statement is disproved (i.e. the null factor law is not true for matrices - although the product \(\mathbf{OX}=\mathbf{O}\), this doesn't necessarily mean that the product \(\mathbf{AB}=\mathbf{O}\) implies that \(\mathbf{A}=\mathbf{O}\) or \(\mathbf{B}=\mathbf{O}\) (or both).

Note that the way question 5 is asked gives a big hint as to the answer to question 4.
Firstly, if \(\mathbf{A}^2\) is defined, then \(\mathbf{A}\) must be square. Also, if the dimensions of \(\mathbf{A}\) are \(m\times m\), then the dimensions of \(\mathbf{A}^2\) and therefore \(\mathbf{O}\) are also \(m\times m\).
If \(\mathbf{A}\times\mathbf{A}=\mathbf{O}\), then \(\mathbf{A}^{-1}\mathbf{A}\mathbf{A}=\mathbf{A}^{-1}\mathbf{O}\), which gives \(\mathbf{I}\mathbf{A}=\mathbf{A}^{-1}\mathbf{O}\), finally giving \(\mathbf{A}=\mathbf{A}^{-1}\mathbf{O}\). Note that this seems to indicate that \(\mathbf{A}=\mathbf{O}\), however, this is ignoring the case where the inverse doesn't exist (i.e. when \(\text{det}\left(\mathbf{A}\right)=0\).)
Let's consider the \(2\times2\) case, to keep the arithmetic as simple as possible.

Now we have \(a^2+bc=0,\ ab+bd=0,\ ac+cd=0,\ bc+d^2=0\text{ and }ad=bc\), which gives \(a^2=-bc,\ ab=-bd,\ ac=-cd,\ bc=-d^2\text{ and }ad=bc\).
Consider the simple case where \(c=0\). This gives: \(a^2=-b\times0,\ ab=-bd,\ a\times0=-0\times d,\ b\times0=-d^2\text{ and }ad=b\times0\), which gives \(a^2=0,\ ab=-bd,\ 0=0,\ 0=-d^2\text{ and }ad=0\), finally giving \(a=0, c=0, d=0\). This will be true for ANY real value of \(b\), so simply choose a non-zero \(b\) and you have a matrix \(mathbf{A}\) for which this property is true (note that there are other solutions to)


For question 12a, note that the top row of the \(2\times2\) matrix contains the time, in minutes, that it takes John to drink a milkshake and eat a banana split, respectively and its bottom row contains the cost of a milkshake and a banana split respectively.
The matrix product gives us

The intermediate multiplication step gives us an insight into what the product represents. We have, in the top row, the time it takes John to drink a milkshake multiplied by 1, added to the time it takes John to eat a banana split multiplied by 2 and, in the bottom row, the cost of a milkshake multiplied 1, plus the cost of a banana split multiplied by 2.
Hence, the product tells us the time it would take John to drink a milkshake and eat 2 banana splits (row 1) and the cost of one milkshake and two banana splits (row 2).
The interpretation for part b, follows from that of part a:

Assuming that the speed that John's friends eat and drink at is exactly the same as John, we can interpret the information in this matrix as follows:
The time it takes John to drink a milkshake and eat 2 banana splits (row 1, column 1)
The cost of John's milkshake and 2 banana splits (row 2, column 1)
The time it takes one of John's friends to drink 2 milkshakes and eat a banana split (row 1, column 2)
The cost of John's first friend's banana split and 2 milkshakes (row 2, column 2)
The time it takes the second of John's friends to eat a banana split (row 1, column 3)
The cost of John's second friend's banana split (row 2, column 3)
Given that we weren't actually given any information about how fast John's friends eat and drink, you may interpret this matrix slightly differently (i.e. maybe John ordered different things on different days), but the information about his friends indicates that this is the logical way to answer it.

An additional note: The knowledge of these matrix ideas is well beyond what is required, even at Unit 3&4 Methods. A basic understanding of how matrices are multiplied is sufficient for what can be assessed on the VCAA exams.
Title: Re: Specialist 1/2 Question Thread!
Post by: jacquieg on September 01, 2018, 10:57:58 am
Hi everyone,
I've come across something strange in my antidiff homework, I've attached the question for those who want to check.
I know it's the integral of the equation between x=4 and x=5 so I long divided it then used partial fractions etc and i checked this part on the CAS so i'm sure it's all fine, and one part of equation that I had to anti differentiate was 1/3(x-3) so i just took out the 1/3 and differentiated it as usual, so I got 1/3loge|x-3|. But I got an incorrect answer for the question as a whole.
Then I thought if i just expand those brackets in the denominator I get 1/3x-9 and that antidifferentiated is 1/3loge|3x-9|.... so two different answers from the same equation.... how do I know which one is right and what's the difference between these 2? Is there only one way to correctly antidifferentiate 1/3x-9 or are both acceptable?
I'm so confused!
Title: Re: Specialist 1/2 Question Thread!
Post by: S_R_K on September 01, 2018, 03:44:49 pm
Hi everyone,
I've come across something strange in my antidiff homework, I've attached the question for those who want to check.
I know it's the integral of the equation between x=4 and x=5 so I long divided it then used partial fractions etc and i checked this part on the CAS so i'm sure it's all fine, and one part of equation that I had to anti differentiate was 1/3(x-3) so i just took out the 1/3 and differentiated it as usual, so I got 1/3loge|x-3|. But I got an incorrect answer for the question as a whole.
Then I thought if i just expand those brackets in the denominator I get 1/3x-9 and that antidifferentiated is 1/3loge|3x-9|.... so two different answers from the same equation.... how do I know which one is right and what's the difference between these 2? Is there only one way to correctly antidifferentiate 1/3x-9 or are both acceptable?
I'm so confused!

Note that:



Hence:



since ln(3) is a constant.
Title: Re: Specialist 1/2 Question Thread!
Post by: Purple_Mango on September 12, 2018, 05:20:07 pm
Hello everyone,
I don't know if this is the right place to post... but... it's relevant to Spesh 1&2, so it should be okay?
Anyway, I just had a SAC on vectors, and there was this one question I (and a majority of my classmates) had trouble with, and I would really like to know how one would answer it. Here it is, from the top of my memory, under spoiler in case some people don't want to read it.

Spoiler
Quadrilateral ABCD has the points E, F, G, H, P and Q as midpoints of AD, AB, BC, CD, AC, and BD respectively.
Show that EG, FH and PQ bisect each other at a point O.

7 Marks
I tried drawing a diagram of it - hopefully it's accurate !
I do not know how to use LaTex... so that is why it does not look right... I`m sorry
(https://i.imgur.com/vjTXA4s.png)

Title: Re: Specialist 1/2 Question Thread!
Post by: jacquieg on October 07, 2018, 08:44:39 pm
plz help question 8B, dont really get this topic either...
Title: Re: Specialist 1/2 Question Thread!
Post by: S_R_K on October 07, 2018, 10:21:24 pm
plz help question 8B, dont really get this topic either...

8b is asking for the second derivative of x with respect to t. You've already got the first derivative (it's given in the equation as dx/dt), so just differentiate again to get the second derivative.
Title: Re: Specialist 1/2 Question Thread!
Post by: anon101 on November 02, 2018, 09:33:34 pm
For learning:
Khan Academy (and practice)
Paul's Online Math Notes
Better Explained

If you're running out of fresh questions to do, maybe try a methods textbook? Obviously not all the content will be the same but there's plenty of relevant stuff in there. You should ask your teacher too, they might have more resources. Personally, I learned the most during practice (and actual ::) ) SACs. This might also be helpful :)
Is there anywhere I can find any year 11 practise exams free though?
Because as it's the end of the year, I feel what I need to do most is exam style questions.
Title: Re: Specialist 1/2 Question Thread!
Post by: Sine on November 03, 2018, 12:18:46 am
Is there anywhere I can find any year 11 practise exams free though?
Because as it's the end of the year, I feel what I need to do most is exam style questions.
As schools teach and assess different content it's quite difficult to get practice exams online that are indicative of what you will recieve from your teacher. Furthermore, commercial practice exams cannot be shared online due to copyright.
That is why it's recommended that you use the resources your teacher has provided you as that is probably what is going to be most like what you will get from the exam.
Title: Re: Specialist 1/2 Question Thread!
Post by: anon101 on November 10, 2018, 01:09:35 pm
The thing is.
Our teacher doesn't really give us anything at all!
I've just got the textbook, and it doesn't have exam-like questions.
Even for Methods 1&2 though, I got like 5 practise exams off a friend.
So why isn't it possible to find any for 1&2 Specialist. I know that the content's slightly different but still.
Shouldn't there be like topic tests for like Geometry and Circle Geometry with like exam styled questions?
Thanks a load dude
Title: Re: Specialist 1/2 Question Thread!
Post by: S_R_K on November 10, 2018, 02:38:07 pm
The thing is.
Our teacher doesn't really give us anything at all!
I've just got the textbook, and it doesn't have exam-like questions.
Even for Methods 1&2 though, I got like 5 practise exams off a friend.
So why isn't it possible to find any for 1&2 Specialist. I know that the content's slightly different but still.
Shouldn't there be like topic tests for like Geometry and Circle Geometry with like exam styled questions?
Thanks a load dude

The difference is that the Methods course is completely prescribed (ie. every school teaches exactly the same content). Specialist 1 & 2 is not; there is a lot of flexibility in what topics are chosen and how they are taught. So while there are commercial exams available for Specialist 1 & 2, schools may choose not to purchase them because the course they have run doesn't align with the questions on the commercial exams.

This also puts teachers in a difficult position because, typically, there is only one class and hence one teacher, so it's asking a lot for that one teacher to produce a whole lot of exam-like questions.

Don't worry, Units 3 & 4 is much better resourced.
Title: Re: Specialist 1/2 Question Thread!
Post by: anon101 on November 10, 2018, 07:19:32 pm
The difference is that the Methods course is completely prescribed (ie. every school teaches exactly the same content). Specialist 1 & 2 is not; there is a lot of flexibility in what topics are chosen and how they are taught. So while there are commercial exams available for Specialist 1 & 2, schools may choose not to purchase them because the course they have run doesn't align with the questions on the commercial exams.

This also puts teachers in a difficult position because, typically, there is only one class and hence one teacher, so it's asking a lot for that one teacher to produce a whole lot of exam-like questions.

Don't worry, Units 3 & 4 is much better resourced.

It's just that I need to do well in this exam if I'm going to do Specialist 3&4.
Because otherwise, I'm not quite sure how I'll go in year 12.
Thanks,
Kamron
Title: Re: Specialist 1/2 Question Thread!
Post by: accountingpro on November 21, 2018, 09:37:49 pm
(https://i.imgur.com/x7LXdmZ.png) Hey y'all can someone give us a hand with this one. Been trying for a while but I think I'm just missing 1 step repeatedly lol.
Title: Re: Specialist 1/2 Question Thread!
Post by: RuiAce on November 21, 2018, 10:22:53 pm
I've checked all 5 answers on WolframAlpha and none of them are true.
A
B
C
D
E

It may be worth mentioning that I got it down to \( \frac12 \tan x \cos 2x \), which Wolfram does agree with.
Title: Re: Specialist 1/2 Question Thread!
Post by: Sine on November 22, 2018, 12:06:52 am
(https://i.imgur.com/x7LXdmZ.png) Hey y'all can someone give us a hand with this one. Been trying for a while but I think I'm just missing 1 step repeatedly lol.
For these types of questions (only MCQ so you always have your CAS and don't need to show working out) if you are struggling to convert the question into an answer a not so eloquent solution would be to assume x = some random number (you can go to heaps of decimal places) and then substitute that same number for all 6 expressions (The question + 5 answers) then you can select the one that matches.

EDIT: the above is more of an exam technique when you can't do the question rather than something you should always rely on :)
Title: Re: Specialist 1/2 Question Thread!
Post by: accountingpro on November 22, 2018, 11:39:30 am
I've checked all 5 answers on WolframAlpha and none of them are true.
A
B
C
D
E

It may be worth mentioning that I got it down to \( \frac12 \tan x \cos 2x \), which Wolfram does agree with.

I think I got the same. Just was told today that none of these are correct lol.
Title: Re: Specialist 1/2 Question Thread!
Post by: S_R_K on November 22, 2018, 05:03:19 pm
For these types of questions (only MCQ so you always have your CAS and don't need to show working out) if you are struggling to convert the question into an answer a not so eloquent solution would be to assume x = some random number (you can go to heaps of decimal places) and then substitute that same number for all 6 expressions (The question + 5 answers) then you can select the one that matches.

EDIT: the above is more of an exam technique when you can't do the question rather than something you should always rely on :)

A similar technique is to try substituting in some numbers, and if the two expressions give you a different result, they can't be the same in general. This helps you rule out some of the options.
Title: Re: Specialist 1/2 Question Thread!
Post by: pm_me_ur_eggs on December 20, 2018, 03:04:39 pm
how do i even start this question

4. A woman has three children and two of them are teenagers, aged between 13 and 19.
The product of their three ages is 1050. How old is each child?
Title: Re: Specialist 1/2 Question Thread!
Post by: kiwikoala on December 20, 2018, 03:22:56 pm
4. A woman has three children and two of them are teenagers, aged between 13 and 19.
The product of their three ages is 1050. How old is each child?

I'd start by seeing which numbers between 13 and 19 can divide evenly into 1050.
Title: Re: Specialist 1/2 Question Thread!
Post by: AlphaZero on December 20, 2018, 03:42:46 pm
how do i even start this question

4. A woman has three children and two of them are teenagers, aged between 13 and 19.
The product of their three ages is 1050. How old is each child?

Just going to expand on what Steven said. Ages have to be natural numbers (positive whole numbers). So, we are looking for three natural numbers whose product is another natural number (mainly, 1050), with two of them being between 13 to 19. So, we should look for the factors of 1050 that are between 13 and 19.

I'll start you off. Using long division, or some other method, we find that 1050 is not divisible by 13, and so none of the children can be 13 years old. Now try 14, and see if it gives you progress :)
Title: Re: Specialist 1/2 Question Thread!
Post by: studyingg on December 20, 2018, 03:45:34 pm
how do i even start this question

4. A woman has three children and two of them are teenagers, aged between 13 and 19.
The product of their three ages is 1050. How old is each child?

is the answer 5, 14,15
Title: Re: Specialist 1/2 Question Thread!
Post by: S_R_K on December 20, 2018, 04:22:14 pm
The methods suggested by other people will work, but they are not efficient.

The best method is to use prime factorisation. The prime factorisation of 1050 is 2*3*5*5*7. Now just use these prime factors to find the ages of three children, two of which are between 13 and 19.
Title: Re: Specialist 1/2 Question Thread!
Post by: RuiAce on December 20, 2018, 04:25:50 pm
The methods suggested by other people will work, but they are not efficient.

The best method is to use prime factorisation. The prime factorisation of 1050 is 2*3*5*5*7. Now just use these prime factors to find the ages of three children, two of which are between 13 and 19.
I think, in this case they are more efficient. The efficiency would break down if the range of numbers were greater, but here you only had to guess and check a total of 7 numbers. Here, the prime factorisation is perhaps moderate in difficulty to compute, however having to pair things off can also take time. You could've started by doing, say, 2*3 instead of going to 3*5. (When I looked at those, I seemed to be drawn to the last 3 numbers first, making numbers exceeding 20 before I considered the whole bundle in the one go.)

On the other hand, your method would be more formal in my opinion here. It would be more efficient only after the amount of numbers you have to guess and check goes haywire, say, more than 15 of them.
Title: Re: Specialist 1/2 Question Thread!
Post by: S_R_K on December 20, 2018, 04:45:29 pm
Perhaps... I think the prime factorisation is pretty immediate, since 2*5 is clearly a factor, and there are very few ways to use the prime factors to get a product between 13 and 19...

Then again, once you find that 1050 is divisible by 14, the other two ages are pretty obvious.
Title: Re: Specialist 1/2 Question Thread!
Post by: pm_me_ur_eggs on December 20, 2018, 05:29:33 pm
Perhaps... I think the prime factorisation is pretty immediate, since 2*5 is clearly a factor, and there are very few ways to use the prime factors to get a product between 13 and 19...

Then again, once you find that 1050 is divisible by 14, the other two ages are pretty obvious.

sorry if this is obvious/assumed knowledge but what does this mean from the worked solutions?

7 × 2 = 14
5 × 3 = 15

does it mean that any of the factors multiplied together also equals another factor of the original number?
Title: Re: Specialist 1/2 Question Thread!
Post by: S_R_K on December 22, 2018, 10:13:35 am
sorry if this is obvious/assumed knowledge but what does this mean from the worked solutions?

7 × 2 = 14
5 × 3 = 15

does it mean that any of the factors multiplied together also equals another factor of the original number?

Not quite. Notice that both 525 and 210 are factors of 1050, but 525 * 210 is not a factor of 1050.

What is true is that any product of the prime factors of x will also be a factor of x. So, for example, since 2, 5, and 7 are prime factors of 1050, 2*5*7 = 70 is also a factor of 1050.
Title: Re: Specialist 1/2 Question Thread!
Post by: Scribe on December 22, 2018, 03:37:25 pm
When do you use tildes (~) and 'hats' (^) with regards to vectors, mechanics, etc? Also, with mechanics, do you write 'i-j direction', 'direction of motion', 'DOM', or 'positive direction', etc?

Thanks  :)
Title: Re: Specialist 1/2 Question Thread!
Post by: AlphaZero on December 22, 2018, 03:51:32 pm
When do you use tildes (~) and 'hats' (^) with regards to vectors, mechanics, etc? Also, with mechanics, do you write 'direction of motion', 'DOM', or 'positive direction', etc?

Thanks  :)

Tildes are used to denote any vector. For example: \[\underset{\sim}{\text{a}}=3\underset{\sim}{\text{i}}+4\underset{\sim}{\text{j}}.\]
Hats are used to denote unit vectors. For example: \[\underset{\sim}{\hat{\text{a}}}=\frac{1}{\sqrt{3^2+4^2}}\left(3\underset{\sim}{\text{i}}+4\underset{\sim}{\text{j}}\right)=\frac{3}{5}\underset{\sim}{\text{i}}+\frac{4}{5}\underset{\sim}{\text{j}}\]
As for wording in mechanics, any of those phrases are fine as long as you are consistent. Although, I will say, using some are more appropriate than others, but that depends on the context of the problem :)
Title: Re: Specialist 1/2 Question Thread!
Post by: Scribe on December 23, 2018, 05:05:45 pm
Consider the quadratic equation:
Find all values of a such that the equation will have two distinct real solutions.

Is the following answer that I found correct? Also, how would you write the answer (first or second one, or do they mean the same thing)?



Thanks  :)
Title: Re: Specialist 1/2 Question Thread!
Post by: RuiAce on December 23, 2018, 05:08:58 pm
Consider the quadratic equation:
Find all values of a such that the equation will have two distinct real solutions.

Is the following answer that I found correct? Also, how would you write the answer (first or second one, or do they mean the same thing)?



Thanks  :)
Your answer is correct. I can see that you solved \(a^2 - 4(4)(3) > 0\) because Wolfram gives the same answer.

They're both valid in theory because they do mean the same thing. But I wouldn't know if one is preferred in the VCE over the other.
Title: Re: Specialist 1/2 Question Thread!
Post by: Sine on December 23, 2018, 06:34:28 pm
Consider the quadratic equation:
Find all values of a such that the equation will have two distinct real solutions.

Is the following answer that I found correct? Also, how would you write the answer (first or second one, or do they mean the same thing)?



Thanks  :)
Both are fine in terms of vce

Although I prefer the first one since they answer is very clear and that is the usually notation vcaa will use in questions
Title: Re: Specialist 1/2 Question Thread!
Post by: Scribe on December 23, 2018, 08:44:54 pm
I don't know where to start with this question.

If the roots of are complex numbers and z is one of them, find:

i) Re(z) in terms of a
ii) The values of a such that Re(z) < 0.5

Thanks  :)
Title: Re: Specialist 1/2 Question Thread!
Post by: RuiAce on December 23, 2018, 09:08:59 pm
I don't know where to start with this question.

If the roots of are complex numbers and z is one of them, find:

i) Re(z) in terms of a
ii) The values of a such that Re(z) < 0.5

Thanks  :)



The next part is now easy - you're just solving \( -\frac{a}{6} < 0.5\)
Title: Re: Specialist 1/2 Question Thread!
Post by: unicornvce99.95 on January 11, 2019, 01:14:07 pm
HELP: 1. The number of ‘type A’ apple bugs present in an orchard is estimated to be 40 960, and the number is reducing by 50% each week. At the same time it is estimated that there are 40 ‘type B’ apple bugs, whose number is doubling each week. After how many weeks will there be the same number of each type of bug?

2.Consider the geometric sequence 1,a,a2,a3,.... Suppose that the sum of two consecutive terms in the sequence gives the next term in the sequence. Find a.

Title: Re: Specialist 1/2 Question Thread!
Post by: S_R_K on January 11, 2019, 01:40:47 pm
HELP: 1. The number of ‘type A’ apple bugs present in an orchard is estimated to be 40 960, and the number is reducing by 50% each week. At the same time it is estimated that there are 40 ‘type B’ apple bugs, whose number is doubling each week. After how many weeks will there be the same number of each type of bug?

The numbers of Type A and Type B bugs each follows a geometric sequence. For Type A bugs, the initial value is 40960 and the ratio between consecutive terms is 0.5. Hence, the number of Type A bugs at week n is 40960*(0.5)^(n-1). Similarly, for Type B bugs, the initial value is 40 and the ratio between consecutive terms is 2. Hence the number of Type B bugs at week n is 40*(2)^(n-1). The number of each type of bug is the same when 40960*(0.5)^(n-1) = 40*(2)^(n-1).

Quote
2.Consider the geometric sequence 1,a,a2,a3,.... Suppose that the sum of two consecutive terms in the sequence gives the next term in the sequence. Find a.

If the sum of two consecutive terms gives the next term, then the sum of the first and second terms is equal to the third term. So 1 + a = a^2. (We could also use a + a^2 = a^3, and so on, but these all reduce to the first case).
Title: Re: Specialist 1/2 Question Thread!
Post by: unicornvce99.95 on January 11, 2019, 02:05:20 pm
The numbers of Type A and Type B bugs each follows a geometric sequence. For Type A bugs, the initial value is 40960 and the ratio between consecutive terms is 0.5. Hence, the number of Type A bugs at week n is 40960*(0.5)^(n-1). Similarly, for Type B bugs, the initial value is 40 and the ratio between consecutive terms is 2. Hence the number of Type B bugs at week n is 40*(2)^(n-1). The number of each type of bug is the same when 40960*(0.5)^(n-1) = 40*(2)^(n-1).

If the sum of two consecutive terms gives the next term, then the sum of the first and second terms is equal to the third term. So 1 + a = a^2. (We could also use a + a^2 = a^3, and so on, but these all reduce to the first case).


I got it...thankyou so much  :)
Title: Re: Specialist 1/2 Question Thread!
Post by: Gogurt on January 27, 2019, 09:42:50 pm
having trouble with this question. Any thoughts?
“If tanƟ=4/3 and 180˚<Ɵ<270˚, evaluate (sinƟ-2cos(-Ɵ))/(cotƟ-sinƟ)”
*no calculator

-thanks
Title: Re: Specialist 1/2 Question Thread!
Post by: mzhao on January 27, 2019, 10:04:42 pm
having trouble with this question. Any thoughts?
“If tanƟ=4/3 and 180˚<Ɵ<270˚, evaluate (sinƟ-2cos(-Ɵ))/(cotƟ-sinƟ)”
*no calculator

-thanks

Once you have the value of any one of sin x, cos x or tan x, as well as which quadrant x is in, you can work out the values of all of the others. For instance, you can draw a right angled triangle, plug in two of the edge lengths and find the length of the missing edge with Pythagoras. Alternatively, you can use any of the following formulae:

\begin{align}
\sin^2(x) + \cos^2(x) = 1\\
1+\tan^2(x) = \sec^2(x)\\
1+\cot^2(x) = \csc^2(x)\\
\end{align}

Spoiler: Values of sin θ and cos θ for this question
\begin{align}
\sin(\theta) = -\frac{4}{5}\\
\cos(\theta) = -\frac{3}{5}\\
\end{align}

With the value of sin θ and cos θ, you should be able to substitute them into the expression :)
Title: Re: Specialist 1/2 Question Thread!
Post by: Evolio on February 06, 2019, 12:06:39 pm
Hello guys.
I was wondering if anyone knew how to do  question b?
Title: Re: Specialist 1/2 Question Thread!
Post by: Jaijai on February 06, 2019, 11:05:12 pm
For number systems and sets, 2D question 11. Image attached.
How do you do this question? Thanks
Title: Re: Specialist 1/2 Question Thread!
Post by: DBA-144 on February 06, 2019, 11:11:51 pm
For number systems and sets, 2D question 11. Image attached.
How do you do this question? Obviously provide working out, thanks.

Just expand out the brackets. You should be able to recognise that this is the difference of two cubes formula. I don't know latex so sorry, can't provide a full solution for you.
Title: Re: Specialist 1/2 Question Thread!
Post by: Jaijai on February 06, 2019, 11:29:01 pm
Hello guys.
I was wondering if anyone knew how to do  question b?
To share the kindness DBA shared through helping me, in part a of this question you determined BD.
As it suggests to do, use the sine rule to work out CD.
Determine the angle opposite CD by determining the other unknown angle since this is possible using the known information. Once two angles are known, obviously the remaining angle can be found.
Refer to the attachment for the actual worked solution, I was merely explaining the process / what was done and why.
Title: Re: Specialist 1/2 Question Thread!
Post by: DBA-144 on February 06, 2019, 11:45:22 pm
To share the kindness DBA shared through helping me, in part a of this question you determined BD.
As it suggests to do, use the sine rule to work out CD.
Determine the angle opposite CD by determining the other unknown angle since this is possible using the known information. Once two angles are known, obviously the remaining angle can be found.
Refer to the attachment for the actual worked solution, I was merely explaining the process / what was done and why.

Ahahaha it would have been more kind if I gave you a fully worked solution but im too lazy to learn latex. Next time I might write it and attach a photo if time allows.

However, for this question, why disnt you consider the ambiguous case? Just something to ponder over. Not trying to sound smart or whatever, but this might help you with your understanding. :)
Title: Re: Specialist 1/2 Question Thread!
Post by: Jaijai on February 07, 2019, 12:06:32 am
Ahahaha it would have been more kind if I gave you a fully worked solution but im too lazy to learn latex. Next time I might write it and attach a photo if time allows.

However, for this question, why disnt you consider the ambiguous case? Just something to ponder over. Not trying to sound smart or whatever, but this might help you with your understanding. :)
The ambigiuous case? Hasn't done trigonometry for a while but wouldn't that break the triangle - sum of internal angles = 180° and if we considered the ambigiuous case, this angle plus the other known angle would already exceed 180°. Hence the ambiguous case isn't in this scenario as the angle has to be below 180 less the known angle (rough inequality)
Also, workings out for my question weren't actually necessary - I was attempting to turn a - b into the right hand side but I should have just done what you'd mentioned. Thanks for your support to the students (including myself) :)
Title: Re: Specialist 1/2 Question Thread!
Post by: Ionic Doc on February 24, 2019, 05:37:04 pm
So anyone learning sequences and series right now???

What chapter are you guys  ( and girls ) on
What topic are you studying
And What book are you using
My school uses Cambridge
Title: Re: Specialist 1/2 Question Thread!
Post by: Ionic Doc on February 25, 2019, 10:18:26 pm
When all eight factors of 30 are multiplied together, the product is 30k. What is the value of k?


someone help me ......................... :-[
Title: Re: Specialist 1/2 Question Thread!
Post by: AlphaZero on February 25, 2019, 11:03:10 pm
When all eight factors of 30 are multiplied together, the product is 30k. What is the value of k?


someone help me ......................... :-[
\begin{align*}30k&=(1\times 30)\times(2\times 15)\times (3\times 10)\times (5\times 6)\\
\implies 30k&=30^4\\
\implies \quad k&=30^3=27\,000\end{align*}
Title: Re: Specialist 1/2 Question Thread!
Post by: DBA-144 on February 25, 2019, 11:05:19 pm
\begin{align*}30k&=(1\times 30)\times(2\times 15)\times (3\times 10)\times (5\times 6)\\
\implies 30k&=30^4\\
\implies \quad k&=30^3=27\,000\end{align*}

Yeah I got the same answer, the user who posted the question said that it is 30^4,in spec 3/4 question thread. imo its an error on the question's/textbook's part- can you confirm?
Title: Re: Specialist 1/2 Question Thread!
Post by: AlphaZero on February 25, 2019, 11:09:09 pm
Yeah I got the same answer, the user who posted the question said that it is 30^4,in spec 3/4 question thread. imo its an error on the question's/textbook's part- can you confirm?

It's an error. If the question had instead been  "When all eight factors of 30 are multiplied together, the product is \(k\). What is the value of \(k\)?",  then  \(k=30^4\)  would be the correct answer.
Title: Re: Specialist 1/2 Question Thread!
Post by: lzxnl on February 26, 2019, 10:09:34 pm
I posted this in the 3/4 thread too, but it's useful to note that if there are only 8 factors of 30, and it's not a square number, this means you can pair the factors to multiply to give 30. Thus, without any multiplication or numerical arrangements at all, I know that the product must be 304.
Title: Re: Specialist 1/2 Question Thread!
Post by: Ionic Doc on March 13, 2019, 07:21:38 pm
For the sequence 12, 6, 3, ..., S8 is equal to what?

Thnx in advance
im legit dying cause I got a methods SAT and a specialist SAT tomorrow ...and I had a 3/4 Psych SAC yesterday ... :-[
Title: Re: Specialist 1/2 Question Thread!
Post by: AlphaZero on March 14, 2019, 09:17:50 am
For the sequence 12, 6, 3, ..., S8 is equal to what?

Thnx in advance
im legit dying cause I got a methods SAT and a specialist SAT tomorrow ...and I had a 3/4 Psych SAC yesterday ... :-[

This is a geometric series with first term  \(a=12\)  and common ratio  \(r=1/2\).  Thus, the sum of the first \(8\) terms is given by \[S_8=\frac{12\big((1/2)^8-1\big)}{1/2-1}=\frac{765}{32}\]
Title: Re: Specialist 1/2 Question Thread!
Post by: Evolio on March 14, 2019, 12:08:53 pm
Hi.
I was wondering how to find x when y=1/8
Title: Re: Specialist 1/2 Question Thread!
Post by: AlphaZero on March 14, 2019, 06:47:54 pm
Hi.
I was wondering how to find x when y=1/8

We have  \[y=5\!\times\!2^{2/3}x,\]and so substituting  \(y=1/8\)  yields \[x=\frac{1}{40\!\times\!2^{2/3}}=\frac{2^{1/3}}{80}\]
Title: Re: Specialist 1/2 Question Thread!
Post by: Evolio on March 18, 2019, 10:21:45 am
Okay, thanks!
 :)
Title: Re: Specialist 1/2 Question Thread!
Post by: Ionic Doc on March 18, 2019, 01:58:44 pm
hey is it me or is spec way more easier and interesting than methods
honestly methods is actually feeling more difficult than spec in year 11
I really  want to drop a math but I know I can't drop methods...so rip spec :-[
Title: Re: Specialist 1/2 Question Thread!
Post by: coldairballoon on March 30, 2019, 11:26:59 pm
Hi guys, first time posting on a forum! I feel like an old man with new technology, so if I've formatted anything wrong...oops.
Anyway, I just wanted to know if anybody could help me with this question? I don't know what I'm doing wrong, but I can't seem to come up with an answer (I sort of get it, but I need to show working/formal proof):

Prove that the recurrence relation xn+1 = (xn)2 + 2 does not converge to a single value.

If anyone could take a look at it it'd be so helpful for me! Thanks.
Title: Re: Specialist 1/2 Question Thread!
Post by: S_R_K on March 31, 2019, 09:38:11 am
hey is it me or is spec way more easier and interesting than methods
honestly methods is actually feeling more difficult than spec in year 11
I really  want to drop a math but I know I can't drop methods...so rip spec :-[

Spec is definitely more interesting, and it's pretty common to find Units 1 & 2 Spec easier than Units 1 & 2 Methods. A lot of that is due to the crowded curriculum in Methods, where you have to learn the content at a very fast pace, but in Spec you have more time to absorb the concepts and consolidate what you've learned. It's worth persevering with both, they complement each other very well in year 12, with quite a lot of overlapping content.

Hi guys, first time posting on a forum! I feel like an old man with new technology, so if I've formatted anything wrong...oops.
Anyway, I just wanted to know if anybody could help me with this question? I don't know what I'm doing wrong, but I can't seem to come up with an answer (I sort of get it, but I need to show working/formal proof):

Prove that the recurrence relation xn+1 = (xn)2 + 2 does not converge to a single value.

If anyone could take a look at it it'd be so helpful for me! Thanks.

Firstly, the recurrence relation is only well defined if we have a value for x0 (the initial value). But in this case, the proof of divergence can be done for any value of x0.

There are a couple of approaches. One is to simply prove that the sequence does not converge. We can do this by supposing that the sequence converges to L and deriving a contradiction. Given that xn+1 = (xn)^2 + 2, if we assume that the limit is L, then we'll have L = L^2 + 2 as n approaches infinity. This equation has no solution, hence the limit does not exist.

A stronger argument is to prove not only that the sequence does not converge, but that it is unbounded (ie. the terms approach infinity). This can be shown by noting that (xn)^2 > xn when xn > 1, but for this sequence xn > 1 for all n ≥ 1. Hence x(n+1) > xn + 2. So for any number L, we can find always find a term in the sequence that is larger than L.
Title: Re: Specialist 1/2 Question Thread!
Post by: RuiAce on March 31, 2019, 09:40:57 am
A stronger argument is to prove not only that the sequence does not converge, but that it is unbounded (ie. the terms approach infinity). This can be shown by noting that (xn)^2 > xn when n ≥ 1. Hence x(n+1) > xn + 2. So for any number L, we can find always find a term in the sequence that is larger than L.
Tiny subtlety. \((x_n)^2 > x_n\) only if we can assume that \( |x_n| > 1 \). Just need to add in that if \(|x_n| \leq 1\), then \( x_{n+1} = (x_n)^2 + 2 > 0+ 2 > 1 \geq x_n\).

Otherwise makes sense. Is this stuff actually in the spesh 1/2 course?
Title: Re: Specialist 1/2 Question Thread!
Post by: S_R_K on March 31, 2019, 09:47:00 am
Tiny subtlety. \((x_n)^2 > x_n\) only if we can assume that \( |x_n| > 1 \). Just need to add in that if \(|x_n| \leq 1\), then \( x_{n+1} = (x_n)^2 + 2 > 0+ 2 > 1 \geq x_n\).

Otherwise makes sense. Is this stuff actually in the spesh 1/2 course?

Thanks, I think I fixed up that in the edit – for any x0 we know that x1 = (x0)^2 + 2 > 1.

This is not really part of the course. There is a little bit on convergence / divergence of geometric series, but students aren't required to do any reasonably rigorous proofs. And in general, most of the sequences / series part of the course is on finite sequences / series.
Title: Re: Specialist 1/2 Question Thread!
Post by: Ash Grey on April 23, 2019, 06:16:49 pm
I'm confused with a few questions regarding sequences and series

1) I need to find the next 3 terms of,
    2,-1,4,3,6,-6

2) I'm so confused about the conversion from explicit to recursive relations, for example,
   how do you change u(n)=n^2-4n=7 into a recursive relation

3) find the value of a for which (3,a+2,a^2) defines an arithmetic sequence
Title: Re: Specialist 1/2 Question Thread!
Post by: Ionic Doc on April 24, 2019, 12:30:42 pm
During the first 8 months, baby Felix’s weight increased in the ratio 9 : 4. If his birth weight was 3.6 kg, then his weight after 8 months will be?

help
thnx
Title: Re: Specialist 1/2 Question Thread!
Post by: AlphaZero on April 24, 2019, 05:53:09 pm
I'm confused with a few questions regarding sequences and series

1) I need to find the next 3 terms of,
    2,-1,4,3,6,-6

2) I'm so confused about the conversion from explicit to recursive relations, for example,
   how do you change u(n)=n^2-4n=7 into a recursive relation

3) find the value of a for which (3,a+2,a^2) defines an arithmetic sequence

Hi there,  can I ask where you found Questions 1 and 2 from? You are only required in Specialist Maths to understand arithmetic and geometric sequences and series, and know some details of a few famous sequences, such as Euler's sequence and the Fibonacci sequence.

As for question 3, if  \(\{3,\ a+2,\ a^2\}\)  defines an arithmetic sequence, then the common difference between successive terms is constant.

Solution
So, we have \begin{align*}&(a+2)-3=a^2-(a+2)\\
\implies &a-1=a^2-a-2\\
\implies &a^2-2a-1=0\\
\implies &a=\frac{2\pm\sqrt{2^2-4(1)(-1)}}{2(1)}\\
\implies &a=1\pm\sqrt{2} \end{align*}
Title: Re: Specialist 1/2 Question Thread!
Post by: Evolio on July 09, 2019, 05:04:47 pm
Hi guys!

I was wondering how to do question 3 c (attached below)?
I'm pretty sure they're asking for the vector resolute. So, why are they giving us the magnitude of b and how can I use it to get the vector resolute? Wouldn't they have to give us the vector b?

Title: Re: Specialist 1/2 Question Thread!
Post by: S_R_K on July 09, 2019, 06:26:07 pm
Hi guys!

I was wondering how to do question 3 c (attached below)?
I'm pretty sure they're asking for the vector resolute. So, why are they giving us the magnitude of b and how can I use it to get the vector resolute? Wouldn't they have to give us the vector b?

A vector in the direction of a is ka where k is a real scalar.

In fact, given that you've already found the unit vector in the direction of a in the first part of the question, you should be able to just write down the answer to part (b)....(Hint: what is the magnitude of the vector you find in part (a)(ii)?) No need to use the formula for the vector resolute.
Title: Re: Specialist 1/2 Question Thread!
Post by: Evolio on July 10, 2019, 03:24:03 pm
Thank you for your help. I understand now!

I was also wondering how to do question 8 b (attached below)? I'm not sure how to approach it.
Title: Re: Specialist 1/2 Question Thread!
Post by: S_R_K on July 11, 2019, 01:13:26 am
Thank you for your help. I understand now!

I was also wondering how to do question 8 b (attached below)? I'm not sure how to approach it.

If v is the component of a in the direction of b, then av is the component perpendicular to b.
Title: Re: Specialist 1/2 Question Thread!
Post by: Evolio on July 11, 2019, 08:21:38 am
If v is the component of a in the direction of b, then av is the component perpendicular to b.
Yeah but how do you know that and how did you get there?
Title: Re: Specialist 1/2 Question Thread!
Post by: S_R_K on July 11, 2019, 09:25:20 am
Yeah but how do you know that and how did you get there?

Sorry my previous post was a bit imprecise.

There are infinitely many ways to resolve a into components such that one of them is parallel to b. We can always write a = kb + (a – kb), where k is any real scalar. Hence, if you have a vector v that is known to be a component of a parallel to b, then the other component must be av.

However, this other component is not necessarily perpendicular to b. That other component will only be perpendicular to b if v is the projection of a onto b – this is what the formula for the vector resolute gives you. Hence if v is known to be the vector resolute of a in the direction of b, then av must be perpendicular to b.


This image illustrates the idea: https://upload.wikimedia.org/wikipedia/commons/9/98/Projection_and_rejection.png
Title: Re: Specialist 1/2 Question Thread!
Post by: Evolio on July 11, 2019, 02:55:44 pm
Thank you for your help, S_R_K!
 ;D
Title: Re: Specialist 1/2 Question Thread!
Post by: ^^^111^^^ on July 13, 2019, 10:05:21 am
Hey there, I am usually never stuck with questions like this (sorry for the low difficulty I know it is kind of dumb  :P)

Here it is:

The distance travelled (s) by a particle varies partly with time and partly with the square of time. If it travels 142.5 m in 3 seconds and travels 262.5 m in 5 seconds, find:

a) how far it would travel in 6 seconds
b) how far it would travel during the sixth second.

I managed to do question a through part variation formula, but I don't understand the difference between question a and b. I checked the answer key and there were different answers for a and b

Thank you much appreciated  ;D
Title: Re: Specialist 1/2 Question Thread!
Post by: AlphaZero on July 13, 2019, 12:39:09 pm
Hey there, I am usually never stuck with questions like this (sorry for the low difficulty I know it is kind of dumb  :P)
...
I managed to do question a through part variation formula, but I don't understand the difference between question a and b. I checked the answer key and there were different answers for a and b
...

Never be afraid to ask questions regardless of the difficulty.

"During the sixth second" actually refers to the time between  \(t=5\)  and  \(t=6\).

Think about it this way. The first second is between  \(t=0\)  and  \(t=1\),  the second second is between  \(t=1\)  and  \(t=2\),  and so on.
Title: Re: Specialist 1/2 Question Thread!
Post by: Evolio on September 01, 2019, 08:52:49 am
Hi guys!
I was wondering how to do the question attached below?
I'm not sure how to go about working it out as the question just gave us the weights. Is this enough info to work this out?

Title: Re: Specialist 1/2 Question Thread!
Post by: S_R_K on September 01, 2019, 04:56:28 pm
Hi guys!
I was wondering how to do the question attached below?
I'm not sure how to go about working it out as the question just gave us the weights. Is this enough info to work this out?

Let a be the magnitude of acceleration of each mass; let T be the tension in the string. Writing an equation of motion (ie. using F = ma) for each mass we have:

1: 50a = T
2: 20a = 20g – T.

Adding equations gives 70a = 20g, so the acceleration of each mass is 2g/7. Hence the speed of each mass is 2gt/7, where t is in seconds.
Title: Re: Specialist 1/2 Question Thread!
Post by: Helish on September 01, 2019, 04:57:42 pm
Hey, guys,
Hi guys!
I was wondering how to do the question attached below?
I'm not sure how to go about working it out as the question just gave us the weights. Is this enough info to work this out?


Hi guys!
I was wondering how to do the question attached below?
I'm not sure how to go about working it out as the question just gave us the weights. Is this enough info to work this out?


Hi guys!
I was wondering how to do the question attached below?
I'm not sure how to go about working it out as the question just gave us the weights. Is this enough info to work this out?


Never be afraid to ask questions regardless of the difficulty.

"During the sixth second" actually refers to the time between  \(t=5\)  and  \(t=6\).

Think about it this way. The first second is between  \(t=0\)  and  \(t=1\),  the second second is between  \(t=1\)  and  \(t=2\),  and so on.
Hi guys!
I was wondering how to do the question attached below?
I'm not sure how to go about working it out as the question just gave us the weights. Is this enough info to work this out?


Hi guys!
I was wondering how to do the question attached below?
I'm not sure how to go about working it out as the question just gave us the weights. Is this enough info to work this out?


Never be afraid to ask questions regardless of the difficulty.

"During the sixth second" actually refers to the time between  \(t=5\)  and  \(t=6\).

Think about it this way. The first second is between  \(t=0\)  and  \(t=1\),  the second second is between  \(t=1\)  and  \(t=2\),  and so on.
Hey there, I am usually never stuck with questions like this (sorry for the low difficulty I know it is kind of dumb  :P)

Here it is:

The distance travelled (s) by a particle varies partly with time and partly with the square of time. If it travels 142.5 m in 3 seconds and travels 262.5 m in 5 seconds, find:

a) how far it would travel in 6 seconds
b) how far it would travel during the sixth second.

I managed to do question a through part variation formula, but I don't understand the difference between question a and b. I checked the answer key and there were different answers for a and b

Thank you much appreciated  ;D
If v is the component of a in the direction of b, then av is the component perpendicular to b.
Hi guys!

I was wondering how to do question 3 c (attached below)?
I'm pretty sure they're asking for the vector resolute. So, why are they giving us the magnitude of b and how can I use it to get the vector resolute? Wouldn't they have to give us the vector b?


I am currently in year 10 doing 1 and 2 business management. next year I am hoping to do eng 1&2 methodes1&2 chem1&2 business management 3&4 and specialist 1&2. I also want to do physics so I thought after I complete Business management, in year 12 I'll do physics 3&4 without 1&2. But teachers said that really hard, so now I am confused in doing physics 1&2 in year 11 and not do spec at all. Is it worth it or not. In my career I want to do medicine or astrophysicist, I really need help guys.
Title: Re: Specialist 1/2 Question Thread!
Post by: ^^^111^^^ on September 01, 2019, 08:41:05 pm
Hey, guys, I am currently in year 10 doing 1 and 2 business management. next year I am hoping to do eng 1&2 methodes1&2 chem1&2 business management 3&4 and specialist 1&2. I also want to do physics so I thought after I complete Business management, in year 12 I'll do physics 3&4 without 1&2. But teachers said that really hard, so now I am confused in doing physics 1&2 in year 11 and not do spec at all. Is it worth it or not. In my career I want to do medicine or astrophysicist, I really need help guys.
There are many factors that need to be taken into account. For example:
- Do you feel like Spec is right for you? I mean, are you capable of handling methods and spesh? If you feel like you enjoy maths in general and you are willing to put in enough hard work (I had to anyway) then from what I know you should be fine.
- You said you wanted to do medicine right, then have you ever thought of doing bio?
- Are specialist and/or methods prerequisites for your aspiring university course?
As for Physics units 3/4 without 1/2, I can't say much, but from what I know if you have enough determination it is possible to do it. But let me just warn you that you will be at a disadvantage (depends on your school as well tho), as some concepts of physics units 3/4 (like for example my textbook relies on prior knowledge to study the concepts like magnetic flux and stuff),need prior knowledge from units 1/2. but DW, the disadvantage is not relatively large.
Title: Re: Specialist 1/2 Question Thread!
Post by: Helish on September 01, 2019, 08:53:54 pm
I am not interested in bio, and my physics teacher has given me chapters to do from 1&2 physics which require the prior knowledge for 3&4, I am also in an advance maths class and currently receiving
There are many factors that need to be taken into account. For example:
- Do you feel like Spec is right for you? I mean, are you capable of handling methods and spesh? If you feel like you enjoy maths in general and you are willing to put in enough hard work (I had to anyway) then from what I know you should be fine.
- You said you wanted to do medicine right, then have you ever thought of doing bio?
- Are specialist and/or prerequisites for your aspiring university course?
As for Physics units 3/4 without 1/2, I can't say much, but from what I know if you have enough determination it is possible to do it. But let me just warn you that you will be at a disadvantage (depends on your school as well tho), as some concepts of physics units 3/4 (like for example my textbook relies on prior knowledge to study the concepts like magnetic flux and stuff), but DW, the disadvantage is not relatively large.
Hey, guys, I am currently in year 10 doing 1 and 2 business management. next year I am hoping to do eng 1&2 methodes1&2 chem1&2 business management 3&4 and specialist 1&2. I also want to do physics so I thought after I complete Business management, in year 12 I'll do physics 3&4 without 1&2. But teachers said that really hard, so now I am confused in doing physics 1&2 in year 11 and not do spec at all. Is it worth it or not. In my career I want to do medicine or astrophysicist, I really need help guys.
above 90, but also do I require Bio knowledge to pass the UCAT
Title: Re: Specialist 1/2 Question Thread!
Post by: ^^^111^^^ on September 01, 2019, 09:10:47 pm
I am not interested in bio, and my physics teacher has given me chapters to do from 1&2 physics which require the prior knowledge for 3&4, I am also in an advance maths class and currently receiving above 90, but also do I require Bio knowledge to pass the UCAT
Well if your physics teacher is giving you those chapters then I guess you should be fine. From what I can see, you should be able to easily survive a spesh class as well. As for UCAT, most likely not since UCAT should just be an aptitude test, but I would recommend bio bcos u said your career could be surrounding medicine as well. That's up to you anyway, since you also have chemistry as a VCE subject and given that you pursue it to units 3/4.
Title: Re: Specialist 1/2 Question Thread!
Post by: Helish on September 01, 2019, 09:56:01 pm
Thanks
Well if your physics teacher is giving you those chapters then I guess you should be fine. From what I can see, you should be able to easily survive a spesh class as well. As for UCAT, most likely not since UCAT should just be an aptitude test, but I would recommend bio bcos u said your career could be surrounding medicine as well. That's up to you anyway, since you also have chemistry as a VCE subject and given that you pursue it to units 3/4.
man for building up my willpower and helping me decide my subjects, I'll try my best to achieve the highest atar possible, Thanks again
Title: Re: Specialist 1/2 Question Thread!
Post by: caffinatedloz on September 01, 2019, 09:57:54 pm
Thanks man for building up my willpower and helping me decide my subjects, I'll try my best to achieve the highest atar possible, Thanks again

Great attitude! Best of luck! :D
Title: Re: Specialist 1/2 Question Thread!
Post by: ArtyDreams on September 08, 2019, 12:11:28 pm
Hey! Am doing further trigonmetry from the Cambridge textbook and am really stuck on proofs (I'm horrible at them!)

Can anyone possibly help me prove this identity?
sec (x) + cosec (x) cot (x) = sec (x) cosec^2 (x)

Also, when doing proofs llike this, do we tackle them as a show that question? Whats the best place to start?

I'd really appreciate any help.
Thanks!!!  :)
Title: Re: Specialist 1/2 Question Thread!
Post by: S_R_K on September 08, 2019, 12:20:45 pm
Hey! Am doing further trigonmetry from the Cambridge textbook and am really stuck on proofs (I'm horrible at them!)

Can anyone possibly help me prove this identity?
sec (x) + cosec (x) cot (x) = sec (x) cosec^2 (x)

Also, when doing proofs llike this, do we tackle them as a show that question? Whats the best place to start?

I'd really appreciate any help.
Thanks!!!  :)

Yes, a "show that" question is essentially asking you to give a proof.

For trig-identity questions like this, you generally have three methods. (1) Start with the LHS, and show, using legitimate steps, that it is equal to the RHS; (2) Start with the RHS and show, using legitimate steps, that it is equal to the LHS; (3) Work with both sides independently and show that they are each equal to some other expression. There is no foolproof method for knowing which is best in any given case. Improvement requires experience and practice; with that comes good judgement and intuition for how to get started.

For the question you've given, my first instinct is to multiply the LHS by cos(x), because that gives 1 + cot^2(x). That's useful, because I also know (from Pythgoras) that 1 + cot^2(x) = cosec^2(x). So, working backwards, now I consider what happens when I start with 1 + cot^2(x) = cosec^2(x), and divide both sides by cos(x)...
Title: Re: Specialist 1/2 Question Thread!
Post by: ArtyDreams on September 08, 2019, 12:31:41 pm
Yes, a "show that" question is essentially asking you to give a proof.

For trig-identity questions like this, you generally have three methods. (1) Start with the LHS, and show, using legitimate steps, that it is equal to the RHS; (2) Start with the RHS and show, using legitimate steps, that it is equal to the LHS; (3) Work with both sides independently and show that they are each equal to some other expression. There is no foolproof method for knowing which is best in any given case. Improvement requires experience and practice; with that comes good judgement and intuition for how to get started.

For the question you've given, my first instinct is to multiply the LHS by cos(x), because that gives 1 + cot^2(x). That's useful, because I also know (from Pythgoras) that 1 + cot^2(x) = cosec^2(x). So, working backwards, now I consider what happens when I start with 1 + cot^2(x) = cosec^2(x), and divide both sides by cos(x)...

Thank you so so much for answering - I really appreciate it! Your answer was really helpful. I do understand it a bit more now - just need to do more practice :D
Title: Re: Specialist 1/2 Question Thread!
Post by: Evolio on September 08, 2019, 01:11:54 pm
Let a be the magnitude of acceleration of each mass; let T be the tension in the string. Writing an equation of motion (ie. using F = ma) for each mass we have:

1: 50a = T
2: 20a = 20g – T.

Adding equations gives 70a = 20g, so the acceleration of each mass is 2g/7. Hence the speed of each mass is 2gt/7, where t is in seconds.
Thank you so much, S_R_K!
 :)
Title: Re: Specialist 1/2 Question Thread!
Post by: radiant roses on November 09, 2019, 04:28:02 pm
Hello,
Can I please have some help with question (b)
I attempted the question by making 3 cases:

1) x is less than or equal to 5/2
I got x = -9

2) x is greater than or equal to 4
I got x=11

3) x is between 5/2 and 4.
For this case, I got x=19/3 but the answers only say x=-9 and x=11
Why is this the case?
Title: Re: Specialist 1/2 Question Thread!
Post by: S_R_K on November 09, 2019, 04:49:46 pm
Hello,
Can I please have some help with question (b)
I attempted the question by making 3 cases:

1) x is less than or equal to 5/2
I got x = -9

2) x is greater than or equal to 4
I got x=11

3) x is between 5/2 and 4.
For this case, I got x=19/3 but the answers only say x=-9 and x=11
Why is this the case?

The answer is in your working out. x = 19/3 is not in the interval 5/2 < x < 4. It's a spurious solution that arises because of the way you remove the modulus signs to obtain a linear equation.
Title: Re: Specialist 1/2 Question Thread!
Post by: JimDADDY2009 on November 28, 2019, 08:25:15 pm
Hello everyone I am in year 9, and I have a question if you could help me answer it please thnx

The mathematician George enjoyed telling his friends he was x years old in the year x2. Find the year of Georges birth given he died in 1871 (rip George) 
Title: Re: Specialist 1/2 Question Thread!
Post by: keltingmeith on November 28, 2019, 10:01:16 pm
Hello everyone I am in year 9, and I have a question if you could help me answer it please thnx

The mathematician George enjoyed telling his friends he was x years old in the year x2. Find the year of Georges birth given he died in 1871 (rip George) 

(note: what you're about to see is my full approach at tackling the problem. This is to illustrate how I might approach this problem, which is far more valuable than me just giving you the answer)

So, we need to find a number such that when squared, it gives a number less than 1871. What's a number that's less than 1871? Well, 1871-y is. So, let's set up that first equation:

1871-y=x

Now, let's apply some logic. While we don't know what age George died at, in the 19th century, you weren't going to live for a hell of a long time - so let's just say he couldn't have been born before 1800. This gives us some room to work with.

And this is where I'm not sure what my next logical step would be - so, let's try playing around. What are the list of squares that'll get you between 1800 and 1871?

10^2=100
20^2=400
30^2=900
40^2=1600
45^2=2025

Too far, let's go back

42^2=1764
43^2=1849
44^2=1936

Brilliant, there's only one age that fits - being 43 in 1849.

Since 1849-43=1806, we can say our good pal' George was born in 1806, living to ripe old age of 65.
Title: Re: Specialist 1/2 Question Thread!
Post by: ErnieTheBirdi on December 17, 2019, 09:29:51 pm
Can someone please help me with this question, I don't really understand how to do it even after looking at the worked solutions. If a, 4a - 4 and 8a - 13 are successive terms of an arithmetic sequence, find a. Thankyou in advance
Title: Re: Specialist 1/2 Question Thread!
Post by: RuiAce on December 17, 2019, 09:51:50 pm
Can someone please help me with this question, I don't really understand how to do it even after looking at the worked solutions. If a, 4a - 4 and 8a - 13 are successive terms of an arithmetic sequence, find a. Thankyou in advance
In an arithmetic terms, successive terms must have the same common difference.

That means, the difference between the second and first term, should equal the difference between the third and second term.

So we end up with the equation
\[ (4a-4) - a = (8a-13) - (4a-4). \]
The solution to this equation is \(a=5\).
Title: Re: Specialist 1/2 Question Thread!
Post by: ErnieTheBirdi on December 18, 2019, 11:22:25 am
Could someone help me with this question? For the sequence 4, 8, 12, ... , find:
a) the sum of the first 9 terms
b) {n : Sn} = 180

mainly part B

P.S thank you to the lecturer who helped me with the previous question I had
Title: Re: Specialist 1/2 Question Thread!
Post by: ErnieTheBirdi on December 30, 2019, 11:48:34 pm
Show that the sum of the first 2n terms of an arithmetic sequence is n times the sum of the two middle terms.

Can someone please explain this question to me please, I don’t understand it even after looking at the worked solutions?
Title: Re: Specialist 1/2 Question Thread!
Post by: MoonChild1234 on January 23, 2020, 03:46:57 pm
If sin (x)=0.3, cos (a) =0.5, tan (b) =2.4, and x, a and b are in the first quadrant. Find the values of the following:

 sin (pi-a)


I got 1/2 but the answers say root 3/2

i just wanted some assistance on how to get to that answer.
 thanks in advance!
Title: Re: Specialist 1/2 Question Thread!
Post by: MoonChild1234 on January 23, 2020, 04:25:22 pm
Evaluate (sin a + cos a)^2 + (sin a - cos a)^2

I also wasn't really sure about this one, thanks again!
Title: Re: Specialist 1/2 Question Thread!
Post by: ^^^111^^^ on January 23, 2020, 09:14:30 pm
If sin (x)=0.3, cos (a) =0.5, tan (b) =2.4, and x, a and b are in the first quadrant. Find the values of the following:

 sin (pi-a)


I got 1/2 but the answers say root 3/2

i just wanted some assistance on how to get to that answer.
 thanks in advance!
(a) = cos-1(0.5)
sin(cos-1(0.5)) = squareroot(3)/2
Using the identity that sin(π-θ)=sinθ, we get that sin(π-a) is equal to sin (a) which is equal to squareroot(3)/2.
Evaluate (sin a + cos a)^2 + (sin a - cos a)^2

I also wasn't really sure about this one, thanks again!
Using the formulas (x+y)^2=x2+2xy+y2 and  (x-y)^2=x2-2xy+y2,
we get that sin2a+cos2a+2*sin(a)(cos)(a) and sin2a+cos2a-2*sin(a)(cos)(a).
Adding those two expressions, we get 2(sin2(a)) + 2(cos2(a)) = 2(sin2(a))+(cos2(a)).
Using the identity sin2(a)+cos2(a)=1, we get 2(1)=2. Therefore, your answer should be 2.
Title: Re: Specialist 1/2 Question Thread!
Post by: MoonChild1234 on March 20, 2020, 09:23:44 pm
hi! i was wondering how to do these questions:

|x-4|-|x-2|=6

|2x-5|-|4-x|=10
Title: Re: Specialist 1/2 Question Thread!
Post by: S_R_K on March 21, 2020, 01:46:22 pm
hi! i was wondering how to do these questions:

|x-4|-|x-2|=6

|2x-5|-|4-x|=10

Use the definition of |x| to break into cases and solve, making sure to check your solutions against the domain. eg:

If x > 4, then x – 4 - (x – 2) = 6, gives no solution.
If 4 > x > 2, then –(x – 4) – (x – 2) = 6 gives x = 0. This solution is rejected because it is outside the domain.
If 2 > x, then –(x – 4) – (–(x – 2)) = 6 gives no solution.

Hence there is no solution to |x – 4| – |x – 2| = 6.
Title: Re: Specialist 1/2 Question Thread!
Post by: MoonChild1234 on March 21, 2020, 04:23:59 pm
thank you! but how would i know what cases to break it into?
Title: Re: Specialist 1/2 Question Thread!
Post by: fun_jirachi on March 21, 2020, 04:58:17 pm
thank you! but how would i know what cases to break it into?

In general, for some equation \(|x-a| + |x-b| = c \ (a < b)\), you're going to have three cases - \(x < a, a \leq x \leq b, x > b\). In most cases of equations in this form, you'll have what we call a bucket curve (try graphing one of your examples on Desmos or GeoGebra! it'll make sense :) ). Each of the three cases represents each 'part' of the bucket. Muck around a bit with the equations, and hopefully it'll start making a bit more sense as to where the cases come from :)

Hope this helps :)
Title: Re: Specialist 1/2 Question Thread!
Post by: xpikachuzz on March 23, 2020, 11:22:29 pm
https://imgur.com/etBnp6h . Find angle COD where O is the center and abcd lie on the circumference. context: my school is closing tomorrow and i had a sac today, got stumped on this question.
If it helps this topic was on congruence (of triangles), similarity (of triangles) and circle theorems.
Title: Re: Specialist 1/2 Question Thread!
Post by: S_R_K on March 24, 2020, 03:23:02 pm
https://imgur.com/etBnp6h . Find angle COD where O is the center and abcd lie on the circumference. context: my school is closing tomorrow and i had a sac today, got stumped on this question.
If it helps this topic was on congruence (of triangles), similarity (of triangles) and circle theorems.

∠BCA = 40° (half the angle at the centre subtended by the same arc).
∠APC = 128°
Hence, ∠CAD = 12°.
Hence ∠COD = 24° (using the same theorem as above).
Title: Re: Specialist 1/2 Question Thread!
Post by: xpikachuzz on March 24, 2020, 03:39:10 pm
∠BCA = 40° (half the angle at the centre subtended by the same arc).
∠APC = 128°
Hence, ∠CAD = 12°.
Hence ∠COD = 24° (using the same theorem as above).

Thank you i feel so dumb for not getting this.
Title: Re: Specialist 1/2 Question Thread!
Post by: S_R_K on March 27, 2020, 04:39:04 pm
Hi

I really need help with this question:

You just need to find the value of x

Thanks

A couple of bits of circle geometry that are useful: (1) The point where the tangents intersect and the centres of the two circles are collinear. (2) The line passing through these points bisects the angle between the tangents.
Title: Re: Specialist 1/2 Question Thread!
Post by: SS1314 on March 27, 2020, 05:19:06 pm
Hey

I needed help with another question:

A tunnel 300 m long has a semicircular cross-section. It requires temporary supports, AB and BC, as shown.
Point B is vertically above D. If AB = 15 m and BC = 8 m, find:
(a)   the diameter of the tunnel

(b)   the height of B above the floor of the tunnel

(c)   the distance D is from the wall at C

(d)   the maximum height a 3-metre-wide truck could be and still pass under the supports.

I just needed help with question d


Thanks  :)
Title: Re: Specialist 1/2 Question Thread!
Post by: fun_jirachi on March 27, 2020, 06:14:26 pm
Hey there!

Here's a diagram of the situation we have in part (d):
(https://i.imgur.com/1sJbZk2.png)

Basically, the max height we have is if the truck passes right through the middle. We have that the tunnel is 17m wide, which is where all the numbers come from. Thus we can also deduce the radius - which I haven't labelled, then use Pythagoras' theorem to find the max height h as labelled on the diagram.

Hope this helps :)
Title: Re: Specialist 1/2 Question Thread!
Post by: S_R_K on March 27, 2020, 06:21:20 pm
Hey there!

Here's a diagram of the situation we have in part (d):
(https://i.imgur.com/1sJbZk2.png)

Basically, the max height we have is if the truck passes right through the middle. We have that the tunnel is 17m wide, which is where all the numbers come from. Thus we can also deduce the radius - which I haven't labelled, then use Pythagoras' theorem to find the max height h as labelled on the diagram.

Hope this helps :)

You are assuming that the truck drives down the middle of the tunnel. The truck can be taller if it doesn't drive down the middle (because of the location of the supports AB and BC).
Title: Re: Specialist 1/2 Question Thread!
Post by: S_R_K on March 27, 2020, 06:48:22 pm
Hey

I needed help with another question:

A tunnel 300 m long has a semicircular cross-section. It requires temporary supports, AB and BC, as shown.
Point B is vertically above D. If AB = 15 m and BC = 8 m, find:
(a)   the diameter of the tunnel

(b)   the height of B above the floor of the tunnel

(c)   the distance D is from the wall at C

(d)   the maximum height a 3-metre-wide truck could be and still pass under the supports.

I just needed help with question d


Thanks  :)

For (d), without assuming that the truck drives down the middle, here's a method to find the maximum height:

Let X and Y be points on AB and BC, respectively, such that XY is 3 m. Then BXY is similar to BAC. From this you can deduce the distance from X to AC (or from Y to AC).
Title: Re: Specialist 1/2 Question Thread!
Post by: SS1314 on March 27, 2020, 08:14:27 pm
For (d), without assuming that the truck drives down the middle, here's a method to find the maximum height:

Let X and Y be points on AB and BC, respectively, such that XY is 3 m. Then BXY is similar to BAC. From this you can deduce the distance from X to AC (or from Y to AC).

Thank you  :)
Title: Re: Specialist 1/2 Question Thread!
Post by: v.l.o.n.e. on April 15, 2020, 12:08:03 pm
ummmm hi guys i'm a year 10 and I have a question on the subject. I was wondering whether it is necessary to know the proofs for all the formulae which are in the book. For example, there's this formula which I saw which was like:

HCF(a,b)=HCF(a,r) if b=aq+r where like all the numbers are integers or something

So for a formula like this, would I need to know the proof for it? I am aware that there is a proofs chapter, but that is more general stuff like "proof that x^4 is greater than y^4 is greater than blah blah".

IN PERORATION, do I need to learn the proofs of formulae? Are they tested on exams or sacs? (I go to mhs so I expect the worst)
Title: Re: Specialist 1/2 Question Thread!
Post by: S_R_K on April 15, 2020, 08:03:29 pm
You're doing Specialist Maths - if you want to learn maths properly, learn the proofs. Whether you will be asked to reproduce them on an assessment is beside the point. The Cambridge book does not include proofs that are beyond the ability of any reasonably capable student.
Title: Re: Specialist 1/2 Question Thread!
Post by: ShirinKothari on May 03, 2020, 12:18:42 pm
Hi all,
I was wondering why there isn't a specialist math 1/2 ATARNOTES complete course notes book?
It would be very helpful if there were to be one, as I am taking specialist maths units 1 & 2 next year.

Kind Regards,
Shirin.
Title: Re: Specialist 1/2 Question Thread!
Post by: Sine on May 03, 2020, 12:28:46 pm
Hi all,
I was wondering why there isn't a specialist math 1/2 ATARNOTES complete course notes book?
It would be very helpful if there were to be one, as I am taking specialist maths units 1 & 2 next year.

Kind Regards,
Shirin.
Not an official response from ATAR Notes but it could possibly be because of the content for Specialist Maths 1/2 differs quite a lot between schools in Victoria. Teachers take a lot of control of this subject and sometimes skip heaps of parts of the study design and add new topics. The reason they do this is because the study design of Spec 1/2 is not a great precursor for preparing for specialist maths 3/4.
Title: Re: Specialist 1/2 Question Thread!
Post by: bluechai on May 15, 2020, 05:14:55 pm
1) the first 3 terms in geometric sequence are 1,p,q. If the first 3 terms of an arithmetic sequence are 1,q,p.
Show that 2p^2 - p - 1 = 0

2) given that geometric series 1+p+q+... converges, find
i) p
i) 1+ p+ q +...

Thanks in advance
Title: Re: Specialist 1/2 Question Thread!
Post by: Opengangs on May 15, 2020, 05:46:05 pm
1) the first 3 terms in geometric sequence are 1,p,q. If the first 3 terms of an arithmetic sequence are 1,q,p.
Show that 2p^2 - p - 1 = 0

2) given that geometric series 1+p+q+... converges, find
i) p
i) 1+ p+ q +...

Thanks in advance
1) On the one hand, we have \(q - 1 = p - q\). On the other hand, we have \(\displaystyle \frac{p}{1} = \frac{q}{p}\). So \(p^2 = q\). Then substituting that into our first equation, we get
\begin{align*}
p^2 - 1 &= p - p^2 \\
2p^2 - p - 1 &= 0.
\end{align*}

2)
Again, we use the fact that \(1 + p + q + \dots\) is a geometric series to get
\begin{align*}
\frac{p}{1} = \frac{q}{p} \implies p^2 = q \implies p = \pm \sqrt{q}.
\end{align*}
Since the series converges, then it will converge to \(\displaystyle \frac{a}{1 - r}\) where \(a\) is the first term and \(r\) is the common ratio \(p\). Hence, the series converges to \(\displaystyle \frac{1}{1 - p}\). This assumes that \(|p| < 1\).
Lmk if there are any errors or questions! :))
Title: Re: Specialist 1/2 Question Thread!
Post by: ErnieTheBirdi on May 19, 2020, 09:56:47 am
I have an upcoming SAC for Unit 1 for Spesh and its my first Spesh SAC. I was just wondering if someone could give me a heads up or pointers to what to look out for or study to do well on it?
Title: Re: Specialist 1/2 Question Thread!
Post by: Sine on May 19, 2020, 02:01:16 pm
I have an upcoming SAC for Unit 1 for Spesh and its my first Spesh SAC. I was just wondering if someone could give me a heads up or pointers to what to look out for or study to do well on it?
What topics have you  learnt so far this year and what will be tested on the sac?
Title: Re: Specialist 1/2 Question Thread!
Post by: ErnieTheBirdi on May 20, 2020, 01:19:18 am
What topics have you learnt so far this year and what will be tested on the sac?
I'm not too sure what will be tested, mainly because we've been at home and all? We've done Sequences and series, Circle geometry, Further Trig, Complex and now we are doing Trigonometric ratios and applications. I think our teacher said when we return to school we may be doing Stats
Title: Re: Specialist 1/2 Question Thread!
Post by: jessieleung on June 01, 2020, 08:23:08 am
I struggle with geometric proofs. Any good websites with videos and questions that would help me with geometric proofs? Thanks
Title: Re: Specialist 1/2 Question Thread!
Post by: keltingmeith on June 05, 2020, 05:44:19 pm
I'm not too sure what will be tested, mainly because we've been at home and all? We've done Sequences and series, Circle geometry, Further Trig, Complex and now we are doing Trigonometric ratios and applications. I think our teacher said when we return to school we may be doing Stats

Cool - study those, that'll help.

I struggle with geometric proofs. Any good websites with videos and questions that would help me with geometric proofs? Thanks

Here's a good video with tips. Yes, it's an hour long, but it's REALLY REALLY good, I recommend watching it instead of doing maths study one night. Otherwise, exposure is the best. Just do every question you can find. Look up "examples of geomteric proofs", and instead of reading through, stop at the bit where they say what they're going to prove, and try it yourself. The more tricks you pick up, the easier it'll become. But starting with Grant's 10 tips in that video will help a lot - and he'll even show you how to apply them to a geometric proof! You may have seen the proof before, but still watch - it'll give you an idea of how to approach things.

Another thing you can do: in a lot of your classes, you're given different formulae and asked to take them for granted. They only work because you CAN prove them. So, you should try and prove them - because it's honestly very possible to come up with a convincing argument for all of them even if not a full proof. Here's some properties that I find really fun to prove geometrically:

a) Prove that (x+y)^2=x^2+2xy+y^2
b) Prove (x-y)^2 = x^2-2xy-y^2
c) Prove x^2-y^2=(x-y)(x+y)

It'll help if you think of x as a line with length x, which means that x^2 is a square with side-lengths x - that way these random algebraic letters become shapes that you can play around with. Also, don't be scared to cut these shapes up - actually get a pair of scissors and cut things up, then play around with them! Hands-on learning can often work wonders
Title: Re: Specialist 1/2 Question Thread!
Post by: SmartWorker on July 26, 2020, 09:05:47 pm
Hey All!

When doing the quotient & chain rule for example for this function


Then is this the proper working out? Like am I using the proper notation? Sorry I am new to Latex. Please tell me if there is a quicker accepted way. My teacher makes us write it out using composite functions (she marks whether we have included it), So idk if this is the right way?









Thank you!😀
Title: Re: Specialist 1/2 Question Thread!
Post by: keltingmeith on July 27, 2020, 12:28:52 am
Hey All!

When doing the quotient & chain rule for example for this function


Then is this the proper working out? Like am I using the proper notation? Sorry I am new to Latex. Please tell me if there is a quicker accepted way. My teacher makes us write it out using composite functions (she marks whether we have included it), So idk if this is the right way?









Thank you!😀


If you're required to show all steps, then yes that's what it should look like. However, it's totally considered as fine to skip the notation for the chain rule and to just do it. Eg, instead of writing:

tan^2(x)=z^2
.: 2z*z'

To just write:

d/dx (tan^2(x))=2tan(x)*(d/dx)(tan(x))=2tan(x)sec^2(x)
Title: Re: Specialist 1/2 Question Thread!
Post by: SmartWorker on August 04, 2020, 01:18:53 pm
Hey,

How would you do this question? (attached it below). I can do a and c once I have looked at a graph of it on desmos (attached below). But I would have not know how to do it otherwise.

What is the transformations that takes f(x) = arccos (x) to f(x) = arccos (6/x) (I was thinking maybe its a dilation by a factor of x^2/6 from the y-axis - but idk how this works?)

Thank you!
Title: Re: Specialist 1/2 Question Thread!
Post by: keltingmeith on August 04, 2020, 01:34:09 pm
Hey,

How would you do this question? (attached it below). I can do a and c once I have looked at a graph of it on desmos (attached below). But I would have not know how to do it otherwise.

Super important - there's nothing wrong with this. Don't forget that you have your calculator available to you in about 2/3 of your exam time, so don't ever feel bad about the quicker method being by calculator - if not your only method. Tbh, I think this question would be quite mean to ask in non-calc exam, but definitely doable.

We know that the domain for arcos(x) is [-1,1]. This means that for any function f(x), arcos(f(x)) only works when -1<=f(x)<=1. So, if f(x) is 6/x, then we know that we MUST have:



Alright - so here's where things get tricky. If we want to make x the numerator, we can't do this in our current form - we need to do this by splitting up the inequality. So, we need to satisfy the following two equations:



Do you think you'd be able to take over from here?

b should be straightforward, it's just plugging into the differentiation rule for arcos(f(x)). Sketching the graph is straight up outside the study design, I wouldn't worry about it (unless someone wants to show me otherwise, but I don't remember ever seeing it)

What is the transformations that takes f(x) = arccos (x) to f(x) = arccos (6/x) (I was thinking maybe its a dilation by a factor of x^2/6 from the y-axis - but idk how this works?)

Thank you!


AFAIK, what you've said is right - I don't think the typical DRT transformation scheme really has a point for reciprocal or composition of functions.

In fact, checking the study design AGAIN, it only mentions linear transformations - this function is not a linear transformation of arcos(x) OR of 1/x, so you don't need to worry about specifying the transformations of this type.
Title: Re: Specialist 1/2 Question Thread!
Post by: schoolstudent115 on August 04, 2020, 01:49:02 pm
1. Domain of arccos(x) = [-1,1]
    Domain of Arccos(6/x):
    If you imagine a graph of you will realise that this means that: .

2. F’(x): use chain rule and distributing any powers along the way ~
.  Now that fraction is only positive if the denominator is positive:


Well x^2>0 for all x except 0.

So this means that for the whole thing to be positive, then (+ * + == +).

Rearranging and simplifying: .

So now we have two intersecting conditions: x is any number except for 0, and x>6. Intersecting the sets and this means that x>6. Done.

3. As for the graph, we know 3 things:
- It must be strictly increasing over x>6, but it increases slower as x—>infty (the derivative’s denominator will grow due to the x^2 term)
- we know the domain from (1)
- we know that on the LHS of the domain (-inf,-6] is mapped from (-1,0] (from the original domain of arccos(x)). And on arccos(x), over x: (-1,0] it is decreasing, so since we are using 6/x, then over (-inf,-6) it is increasing, or likewise if we flip the domain (-6,-inf) it is decreasing (from right to left).

This is enough info to graph.


But I’m not sure that is precisely how they’d do it.
Title: Re: Specialist 1/2 Question Thread!
Post by: S_R_K on August 04, 2020, 02:53:15 pm
Note that arccos(1/x) = arcsec(x), so you can graph arccos(1/x) by graphing sec(x) for 0 ≤ x ≤ pi, and then reflecting about y = x.
Title: Re: Specialist 1/2 Question Thread!
Post by: schoolstudent115 on August 04, 2020, 03:30:15 pm
F=Arccos(1/x)
G=arcsec(x)
X = sec(G)=1/cos(G)
1/X = cos(F)
X = 1/cos(F) hence cos(F)=cos(G) ==~== F =~= G.

I didn’t think of it like that.
I’d graph cos(x)—>sec(x)—>arcsec(x)—>~—>arccos(1/x)—>dilate
The relation didn’t seem so obvious as the 1/x was wrapped up in the function.
Title: Re: Specialist 1/2 Question Thread!
Post by: S_R_K on August 04, 2020, 03:56:34 pm
F=Arccos(1/x)
G=arcsec(x)
X = sec(G)=1/cos(G)
1/X = cos(F)
X = 1/cos(F) hence cos(F)=cos(G) ==~== F =~= G.

In general this reasoning is a bit suspicious. It only works if you make some assumptions about F and G being one-to-one functions of x. (Which is true in this case, but be careful in general).
Title: Re: Specialist 1/2 Question Thread!
Post by: aspiringantelope on August 08, 2020, 12:56:01 pm
Hello,
Do Trig Sum/Difference formulae work for sin^2,cos^2 and tan^2?
Like does this work?

\(sin^2\left(x\right)cos^2\left(a\right)-cos^2\left(x\right)sin^2\left(a\right)\) = sin^2(x-a)

Thank you!
Title: Re: Specialist 1/2 Question Thread!
Post by: S_R_K on August 08, 2020, 02:50:44 pm
Hello,
Do Trig Sum/Difference formulae work for sin^2,cos^2 and tan^2?
Like does this work?

\(sin^2\left(x\right)cos^2\left(a\right)-cos^2\left(x\right)sin^2\left(a\right)\) = sin^2(x-a)

Thank you!

No. \(\sin^2(u-v) = (\sin(u)\cos(v)-\sin(v)\cos(u))^2 \neq (\sin(u)\cos(v))^2-(\sin(v)\cos(u))^2\)

The equation you wrote above is just the common howler \((x - a)^2 = x^2 - a^2\).
Title: Re: Specialist 1/2 Question Thread!
Post by: redset8 on August 11, 2020, 04:54:24 pm
Hi,
Encountered this question, but don't know how to go about proving it. If anyone could help, that would be much appreciated!
Title: Re: Specialist 1/2 Question Thread!
Post by: keltingmeith on August 11, 2020, 06:09:32 pm
Hi,
Encountered this question, but don't know how to go about proving it. If anyone could help, that would be much appreciated!

Oh boy, love a good circle proof. Everything I'm spitting out is literally me working through this question in real time - that means at some point I may hit a block or have to disregard something entirely. I think it's important to SHOW this process, because a lot of students will often ask, "but how did you know to try x?" - the true, honest to goodness, answer is - I didn't know to try x, I just did it, and it happened to work out.

Okay, so the question is - how can we prove this? Well, we know that all tangents to a circle meet the radius at 90 degrees - so maybe we can prove that each PQ and PR are parrallel to the respective tangents at the points PQ and PR cross on the circle edge. BUT, that would only work if P is the middle of the circle - which isn't necessarily true.

Okay, instead of questioning how to answer the question, let's instead just figure out what we know and work from there. So, <ADC and <ABC must add to 180 degrees, and the same for <DAB and <DCB. But, I'm not sure how we could use this... But it does mean we know the angles <QBC and <RDC. It's worth noting down, even if we never use it. Let's say <ADC=<QBC=d, and <ABC=<RDC=b.

In fact, we can use this to figure out all the angles in the triangle. Using the same logic, <RCD=<QCD=<DAB=a (again, I've just decided that it should be called a, because that makes things easier). So that means that we now know all the angles in the two triangles they made up, RDC and QBC. In fact, now we know:

<DRC=180-b-a
<BQC=180-d-a

So, how does that help us? Well, let's call the intersection between QD and RP, M, and the intersection between RB and QP, N (I hope you're labelling this on your own circle), then all we need to prove is that either <MQP + <QMP = 90 degrees OR <NRP + <RNP = 90 degrees. And, using the above, we know that:

<DRC = 2*<NRP = 180 - b - a, <NRP = 90 - (b+a)/2
<BQC = 2*<MQP = 180 - d - a, <MQP = 90 - (d+a)/2

So for simplicity, I'm just going to work on <PNR. So I know for <NRP + <PNR = 90, then I need <PNR = (b+a)/2 - the question is, how can I get there? Well, I know that <PNR=<BNQ=180-<CNQ=180-<PNB. Can I figure any of those out?

The problem with any of these triangles is that N doesn't lie on the circle - it's inside the circle, and we don't have many proofs that rely on working INSIDE the circle - so frustratingly, knowing we're so close, this approach looks doomed to fail.



Okay, new idea. What if we made a circle using the points P, Q, and R. Then, all we'd need to do is prove that the length QR passes through the centre of that circle. Then the angle <QPR HAS to be 90 degrees.

... Except I'm not sure how we could use any of the information we've been given to inform ourselves of this circle we just made. Okay, maybe I'll go back to the (a+b)/2 thing



Okay, so let's try a slightly different approach, and we're going to extend QP so that it intersects with AR - and we'll call that point Z because I'm running out of letters that make sense. Since RP bisects the triangle ZRN, if PQ and PR are perpendicular, then <RZN=<RNZ. No, I'm stopping this train of thought, things are just going to get worse before they get better. There's gotta be a simpler way.



Okay, I think I've got it. If I go back to the (a+b)/2 thing, I know that <QNB=<RNP, and I want to show that <RNP=(a+b)/2. I may have realised this earlier if I labelled myself (oops), but I just realised that I CAN figure out <QNB because I know all the other angles in the triangle QNB:

<BQN=<MQP=90-(d+a)/2
<NBQ=d
<QNB=180-<BQN-<NBQ=180-(90-d/2-a/2)-d=90+a/2-d/2=90+(a-d)/2

Okay, not looking good. But, we know that d=180-b, so we then get:

<RNP=<QNB=90+a/2 - (180-b)/2=90 + a/2 - 180/2 + b/2 = 90 + a/2 - 90 + b/2 = (a+b)/2

Which means that:

<RPQ=<RPN=180-<RNP-<NRP=180-(a+b)/2-(90-(a+b)/2)=90

Fuck, this was a work-out. I'm sure this got confusing along the way, so try re-doing this yourself to see if it makes more sense


EDIT: After giving up I hit a break-through, so hopefully you've noticed this and come back after I made my initial post? Oops
Title: Re: Specialist 1/2 Question Thread!
Post by: redset8 on August 11, 2020, 08:57:30 pm
Wow. Just wow. I started out like you, thinking about cyclic quadrilaterals, and then couldn't really progress.
I've worked through your solution in my book. I'll now try it again by myself.

Which means that:

<RPQ=<RPN=180-<RNP-<NRP=180-(a+b)/2-(90-(a+b)/2)=90

Just the very last point (I was so close...), how does <NRP = (90-(a+b)/2). Maybe I'm just not seeing it tonight...
Title: Re: Specialist 1/2 Question Thread!
Post by: keltingmeith on August 12, 2020, 08:34:22 am
Just the very last point (I was so close...), how does <NRP = (90-(a+b)/2). Maybe I'm just not seeing it tonight...

So that was something I proved much earlier in the process. Remember that the line BP bisects the angle <BRA, and so the angles <DRP and <PRC (which is the same as <PRN, since R, N, and C are collinear) are exactly half the angle <BRA. If the question is what's the angle <BRA - since we decided that the angles <RAB=a, and <ABR=b, then angle <BRA = 180-a-b. Cut that in half, and you get <NRP
Title: Re: Specialist 1/2 Question Thread!
Post by: redset8 on August 13, 2020, 04:48:04 pm
Ah, thanks heaps! Geometry practice for me...
Title: Re: Specialist 1/2 Question Thread!
Post by: sogreatsosad on August 13, 2020, 05:19:16 pm
Can someone please show me step by step how I'd solve this? Thanks
Title: Re: Specialist 1/2 Question Thread!
Post by: keltingmeith on August 13, 2020, 05:24:05 pm
Can someone please show me step by step how I'd solve this? Thanks

What are you struggling with? What have you tried yourself?
Title: Re: Specialist 1/2 Question Thread!
Post by: sogreatsosad on August 13, 2020, 05:34:33 pm
What are you struggling with? What have you tried yourself?

I tried multiplying it repeatedly, utilising the fact that i^2 is a real number. Can't get more minute than that, need help
Title: Re: Specialist 1/2 Question Thread!
Post by: keltingmeith on August 13, 2020, 05:37:44 pm
I tried multiplying it repeatedly, utilising the fact that i^2 is a real number. Can't get more minute than that, need help

Yeah, so as you've seen, that'll quickly become an issue because it only works for when n is an integer - which it might not be. Have you learned about using De Moivre's theorem? Because that'll be integral to solving this question
Title: Re: Specialist 1/2 Question Thread!
Post by: sogreatsosad on August 13, 2020, 05:39:58 pm
Yeah, so as you've seen, that'll quickly become an issue because it only works for when n is an integer - which it might not be. Have you learned about using De Moivre's theorem? Because that'll be integral to solving this question

I haven't learn that, no. I'll learn out the theorem then try the problem again, if that doesn't work then I'll come back here. Thanks for telling me about De Moivre's theorem!
Title: Re: Specialist 1/2 Question Thread!
Post by: keltingmeith on August 13, 2020, 05:43:11 pm
I haven't learn that, no. I'll learn out the theorem then try the problem again, if that doesn't work then I'll come back here. Thanks for telling me about De Moivre's theorem!

All good - information on the internet can be confusing, but there should be information in your textbook - and here's a quick summary:

Once you've converted your complex number, z, to polar form, then it has a modulus (r) and an argument (theta), and has the form:



Then, the form of z^n is:

Title: Re: Specialist 1/2 Question Thread!
Post by: sogreatsosad on August 13, 2020, 07:31:56 pm
All good - information on the internet can be confusing, but there should be information in your textbook - and here's a quick summary:

Once you've converted your complex number, z, to polar form, then it has a modulus (r) and an argument (theta), and has the form:



Then, the form of z^n is:



Hi, I tried using DM's theorem. I converted the given problem to polar form, and expanded n. My approach has been that basically sin((-pi*n)/3 cannot equal i, as if it did then i times i means a real number. However, I tried to use the CAS to solve for n, but it's simply returning "false". Where to now?
Title: Re: Specialist 1/2 Question Thread!
Post by: S_R_K on August 13, 2020, 08:38:29 pm
Hi, I tried using DM's theorem. I converted the given problem to polar form, and expanded n. My approach has been that basically sin((-pi*n)/3 cannot equal i, as if it did then i times i means a real number. However, I tried to use the CAS to solve for n, but it's simply returning "false". Where to now?

The argument of an imaginary number is \(\frac{\pi}{2} + \pi k\) where k is an integer. Equate that to the argument of \((2-2\sqrt{3}i)^n\) and solve for the smallest positive integer n.
Title: Re: Specialist 1/2 Question Thread!
Post by: M-D on August 30, 2020, 08:15:30 pm
Hi,
I need help with the following question:

Find the values of p and q such that is parallel to the x-axis.

I have worked out that:



I don't know how to get specific values for p and q so is parallel to the x-axis.

Any assistance will be much appreciated.
Title: Re: Specialist 1/2 Question Thread!
Post by: keltingmeith on August 30, 2020, 08:19:21 pm
Hi,
I need help with the following question:

Find the values of p and q such that is parallel to the x-axis.

I have worked out that:



I don't know how to get specific values for p and q so is parallel to the x-axis.

Any assistance will be much appreciated.

Well, if the vector is parallel to the x-axis, then it should be equal to some multiple of the i unit vector - does that help?
Title: Re: Specialist 1/2 Question Thread!
Post by: redset8 on November 04, 2020, 08:08:31 pm
Hey,
A little guidance on the attached question would be much appreciated! I'm definitely missing something here...
Title: Re: Specialist 1/2 Question Thread!
Post by: keltingmeith on November 04, 2020, 08:13:00 pm
Hey,
A little guidance on the attached question would be much appreciated! I'm definitely missing something here.

Absolutely - before I help, I'm curious. What do you notice in this picture? Is there anything that jumps out at you? The 'trick' with circle theorem questions is to not focus on what you need to find, and to instead think what you notice about everything else in the question, then slowly build through smaller links until you eventually find one which tells you an unknown. So, what smaller things (maybe related to the pronumerals - maybe not) have you noticed about this particular circle?
Title: Re: Specialist 1/2 Question Thread!
Post by: redset8 on November 04, 2020, 08:22:16 pm
Absolutely - before I help, I'm curious. What do you notice in this picture? Is there anything that jumps out at you? The 'trick' with circle theorem questions is to not focus on what you need to find, and to instead think what you notice about everything else in the question, then slowly build through smaller links until you eventually find one which tells you an unknown. So, what smaller things (maybe related to the pronumerals - maybe not) have you noticed about this particular circle?

Uh all I notice is that it's not stated that BC is a straight line. Oh and triangle DOC is isosceles. So is triangle BOD. (due to radius) If, however, BC is diameter, then <BDC is 90, but I don't know if that helps.
Title: Re: Specialist 1/2 Question Thread!
Post by: keltingmeith on November 04, 2020, 08:29:04 pm
Uh all I notice is that it's not stated that BC is a straight line. Oh and triangle DOC is isosceles. So is triangle BOD. (due to radius) If, however, BC is diameter, then <BDC is 90, but I don't know if that helps.

That's awesome! So, I think you can assume BC is a straight line - and if it is, that also means that ABCD is a cyclic quadrilateral. So, this should tell you what the value of a is - and hopefully using the facts you've already noticed, you should be able to get the rest.

EDIT: Also, that picture is so not to scale it hurts lmao. I don't think that shape can actually exist if the angle BAD is 65 degrees
Title: Re: Specialist 1/2 Question Thread!
Post by: redset8 on November 04, 2020, 08:36:28 pm
That's awesome! So, I think you can assume BC is a straight line - and if it is, that also means that ABCD is a cyclic quadrilateral. So, this should tell you what the value of a is - and hopefully using the facts you've already noticed, you should be able to get the rest.

So if ABCD is a cyclic quadrilateral (opposites sum to 180), then a=180-65=115. But that would mean b = 180 -2a = -50...? If triangle DOC is isosceles.
Or have I managed to stuff up the angles of a triangle?
Title: Re: Specialist 1/2 Question Thread!
Post by: keltingmeith on November 04, 2020, 08:42:32 pm
So if ABCD is a cyclic quadrilateral (opposites sum to 180), then a=180-65=115. But that would mean b = 180 -2a = -50...? If triangle DOC is isosceles.
Or have I managed to stuff up the angles of a triangle?

Yeah - the fact that this shape CANNOT exist makes me think it's just a dud question. Notice how a looks acute, but the angle BAD looks obtuse? That's not just because it looks that way, it's because the angles HAVE to be that way for this specific shape to exist. I'm curious if you have answers for this question? If not, why don't you assume that a=65 degrees. In the spoiler, I'll tell you what b and c are, so that you can work it out yourself and have something to check against:

Spoiler
b=180-2a=180-130=50
2c+(180-b)=180 <==> c=(1/2)b=25
Alternatively,
a+c+90=180 <==> c=90-a=25
Title: Re: Specialist 1/2 Question Thread!
Post by: redset8 on November 04, 2020, 08:47:48 pm
Yeah - the fact that this shape CANNOT exist makes me think it's just a dud question. Notice how a looks acute, but the angle BAD looks obtuse? That's not just because it looks that way, it's because the angles HAVE to be that way for this specific shape to exist. I'm curious if you have answers for this question? If not, why don't you assume that a=65 degrees. In the spoiler, I'll tell you what b and c are, so that you can work it out yourself and have something to check against:
Unfortunately I don't have answers; it's an old school 1/2 Specialist exam, but the answers are just missing for this section A.
If we let a=65, I got the same answers as you.
Side note, how does your 'alternatively' section work? As in how can we say a+c+90=180? Is it still to do with the opposite angles of cyclic quadrilateral sum to 180?
Title: Re: Specialist 1/2 Question Thread!
Post by: keltingmeith on November 04, 2020, 09:06:17 pm
Unfortunately I don't have answers; it's an old school 1/2 Specialist exam, but the answers are just missing for this section A.
If we let a=65, I got the same answers as you.
Side note, how does your 'alternatively' section work? As in how can we say a+c+90=180? Is it still to do with the opposite angles of cyclic quadrilateral sum to 180?

That's using the theorem that an angle subtended on a circle from the diameter is always 90 degrees
Title: Re: Specialist 1/2 Question Thread!
Post by: redset8 on November 04, 2020, 09:12:20 pm
That's using the theorem that an angle subtended on a circle from the diameter is always 90 degrees

Ah yeah now I see it.
Thanks heaps for your help!!!
Title: Re: Specialist 1/2 Question Thread!
Post by: xpikachuzz on November 23, 2020, 03:01:53 pm
this one was one my year 11 exam
if i have the values of sin(x) and cos(x) then what is sin(x/2).
im guesing i need to expand it but i dont know how to... on my exam i guessed sin(x/2) = sin(x)cos(x)
Title: Re: Specialist 1/2 Question Thread!
Post by: keltingmeith on November 23, 2020, 04:23:54 pm
this one was one my year 11 exam
if i have the values of sin(x) and cos(x) then what is sin(x/2).
im guesing i need to expand it but i dont know how to... on my exam i guessed sin(x/2) = sin(x)cos(x)

Yeah, not quite! You've mixed up the double-angle formula \(\sin(2u)=2\sin(u)\cos(u)\). I could re-phrase this using u=x/2, and get:

\[
\sin(x)=2\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right)
\]

Which, as you can see, is very different to what you did! You'd need to be looking into some other double angle formula - but here's a hint. If you know that \(\sin(\alpha)=\beta\), then you also know that \(\cos(\alpha)=\sqrt{1-\beta^2}\). You can also do this in reverse, and if \(\cos(\alpha)=\beta\), then \(\sin(\alpha)=\sqrt{1-\beta^2}\) You may not need to use my hint, either. There is a method that doesn't need that hint, but hopefully it's stimulating some thought
Title: Re: Specialist 1/2 Question Thread!
Post by: Ruchir on December 07, 2020, 09:10:26 am
Hello I am in year 11 and I am doing Spesh in the holidays , to get a head start.
I was stuck at this series question in the text book Ex 4E Q12
Find    1−x^2 +x^4 −x^6 +x^8 −···+x^2m   ,where m is even.
Also please explain the answer.
Thanks :'(
Title: Re: Specialist 1/2 Question Thread!
Post by: Bri MT on December 07, 2020, 12:00:28 pm
Hello I am in year 11 and I am doing Spesh in the holidays , to get a head start.
I was stuck at this series question in the text book Ex 4E Q12
Find    1−x^2 +x^4 −x^6 +x^8 −···+x^2m   ,where m is even.
Also please explain the answer.
Thanks :'(


Hey, I haven't done spec so I'm not 100% on how they would expect you to find it but here are some thoughts that may help:

This is an alternating series (it's a series where we see it alternating b/w + and -)
This then leads to thinking, well how do they get that alternation? We can do this by having (-1)^n and then when n is an odd number it will be (-1)^n = -1 but if n is an even number (-1)^2 = 1. (When you apply this double check if you start at positive or negative, sometimes you might do n + 1 rather than n as the power.)

Let's move on from the signs for a moment, we can easily see that we are incrementing the power of x by 2 each time. Since we start at a_0 = 1, we know that the power is starting at 0. We can represent this information using x^2n

Let's put this together:

a_n is the sum of (-1)^n x^2n

we are told that m is even (so that gives us a plus for our last term)


Hope this helps!
Title: Re: Specialist 1/2 Question Thread!
Post by: Ruchir on December 07, 2020, 12:49:20 pm
Hey, I haven't done spec so I'm not 100% on how they would expect you to find it but here are some thoughts that may help:

This is an alternating series (it's a series where we see it alternating b/w + and -)
This then leads to thinking, well how do they get that alternation? We can do this by having (-1)^n and then when n is an odd number it will be (-1)^n = -1 but if n is an even number (-1)^2 = 1. (When you apply this double check if you start at positive or negative, sometimes you might do n + 1 rather than n as the power.)

Let's move on from the signs for a moment, we can easily see that we are incrementing the power of x by 2 each time. Since we start at a_0 = 1, we know that the power is starting at 0. We can represent this information using x^2n

Let's put this together:

a_n is the sum of (-1)^n x^2n

we are told that m is even (so that gives us a plus for our last term)


Hope this helps!



I appreciate the effort you took to solve this question but the answer is a bit different
This is the answer is a bit different.
Title: Re: Specialist 1/2 Question Thread!
Post by: keltingmeith on December 07, 2020, 01:35:03 pm
Hello I am in year 11 and I am doing Spesh in the holidays , to get a head start.
I was stuck at this series question in the text book Ex 4E Q12
Find    1−x^2 +x^4 −x^6 +x^8 −···+x^2m   ,where m is even.
Also please explain the answer.
Thanks :'(



Have you covered the geometric series at all? If not, I reckon you just need to read up on that, and you'll probably be fine
Title: Re: Specialist 1/2 Question Thread!
Post by: Ruchir on December 07, 2020, 01:49:49 pm

Have you covered the geometric series at all? If not, I reckon you just need to read up on that, and you'll probably be fine
Yes I have but this question is particular is baffling me, and I can’t see why there are supposed to be m+1 terms
Title: Re: Specialist 1/2 Question Thread!
Post by: keltingmeith on December 07, 2020, 02:07:38 pm
Yes I have but this question is particular is baffling me, and I can’t see why there are supposed to be m+1 terms

Okay, well try breaking the series up and see if that makes things easier. Eg,
When m=1, you go up to x^(2m)=x^2: 1 - x^2
When m=4, you go up to x^(2m)=x^8: 1 - x^2 + x^4 - x^6 + x^8

Try a few more, fill out the series, count up the terms, and see what you notice
Title: Re: Specialist 1/2 Question Thread!
Post by: Ruchir on December 07, 2020, 02:24:54 pm
Okay, well try breaking the series up and see if that makes things easier. Eg,
When m=1, you go up to x^(2m)=x^2: 1 - x^2
When m=4, you go up to x^(2m)=x^8: 1 - x^2 + x^4 - x^6 + x^8

Try a few more, fill out the series, count up the terms, and see what you notice

Thanks I got it. :)
Just 1 quick question
When they say m is even, do they mean all values of m are supposed to be even?
Title: Re: Specialist 1/2 Question Thread!
Post by: Ruchir on December 19, 2020, 04:13:56 pm
I don’t understand when for 9 a and b.
Here’s what I did
for A. 6*6*5*3 = 540 Pr(even) = 540/1470
For B just Pr(odd) = 930/1470
But the answer if for B is 720/1470
And for A is 750/1470
Please help and explain what I did wrong
Title: Re: Specialist 1/2 Question Thread!
Post by: cloudyy on January 21, 2021, 11:16:32 pm
Could someone help me with this q?

From the top of a cliff 160m high, two buoys are observed. Their bearings are 337 and 308. Their respective angles of depression are 3 and 5. Calculate the distance between the buoys.

I've tried drawing a diagram to figure it out, but I'm way too confused on how to draw the diagram with bearing included. Any tips?
Title: Re: Specialist 1/2 Question Thread!
Post by: fun_jirachi on January 22, 2021, 01:07:50 am
Hey there!

Open for a diagram
(https://i.imgur.com/L0ckaKE.png)

If you open the above spoiler, I've drawn a diagram on paint. Hopefully it's clear enough, the green triangle corresponds to the buoy with bearing 337 and the red triangle corresponds to the buoy with bearing 308.

Try using different colours and angle your North-South axis a bit so you leave space for the vertical axis for 3D diagrams like this, and feel free to exaggerate the angles a bit as diagrams don't really need to be accurate as they only serve as an interpretive tool.

Further help in the spoiler below if you need it, just in case of followups:
Spoiler
Find the distance from the cliff to each of the buoys using trig ratios, then use the cosine rule to find the distance between the buoys.

Hope this helps :)
Title: Re: Specialist 1/2 Question Thread!
Post by: cloudyy on January 22, 2021, 05:08:37 pm
Hey there!

Open for a diagram
(https://i.imgur.com/L0ckaKE.png)

If you open the above spoiler, I've drawn a diagram on paint. Hopefully it's clear enough, the green triangle corresponds to the buoy with bearing 337 and the red triangle corresponds to the buoy with bearing 308.

Try using different colours and angle your North-South axis a bit so you leave space for the vertical axis for 3D diagrams like this, and feel free to exaggerate the angles a bit as diagrams don't really need to be accurate as they only serve as an interpretive tool.

Further help in the spoiler below if you need it, just in case of followups:
Spoiler
Find the distance from the cliff to each of the buoys using trig ratios, then use the cosine rule to find the distance between the buoys.

Hope this helps :)

You're my lifesaver!! Can't thank you enough, thank you!
Title: Re: Specialist 1/2 Question Thread!
Post by: pans on February 01, 2021, 09:01:52 pm
Can someone please explain CHAPTER 4A Q1E.

Title: Re: Specialist 1/2 Question Thread!
Post by: Ruchir on February 01, 2021, 09:18:48 pm
Can someone please explain CHAPTER 4A Q1E.

It is saying that tn+1 which is the next term is equal to twice the nth term term or current term (tn) adding the previous term tn-1
In context with the question the current term is   t2 = 3 and the previous term is t1=1

There fore t3 = 2*3(or t2) + 1(or tn-1)
Therefore t3 = 7
The follows on with the next term being t4 , current term being t3 and previous term being t2.
Title: Re: Specialist 1/2 Question Thread!
Post by: pans on February 01, 2021, 09:37:43 pm
omg DUDE THANK YOU I GET IT NOW
Title: Re: Specialist 1/2 Question Thread!
Post by: pans on February 01, 2021, 09:40:22 pm
Can someone please explain 4A Q7C.
Title: Re: Specialist 1/2 Question Thread!
Post by: fun_jirachi on February 01, 2021, 10:32:08 pm
You should use the recurrence relation found in part b).

Given that \(t_0 = 94.3s\), you need to be finding \(t_8\) using the relation as this indicates 8 weekly 4% reductions from her time.
Title: Re: Specialist 1/2 Question Thread!
Post by: pans on February 02, 2021, 04:16:25 pm
pls explain 4C Q12
Title: Re: Specialist 1/2 Question Thread!
Post by: pans on February 02, 2021, 04:16:53 pm
You should use the recurrence relation found in part b).

Given that \(t_0 = 94.3s\), you need to be finding \(t_8\) using the relation as this indicates 8 weekly 4% reductions from her time.

Thank you!
Title: Re: Specialist 1/2 Question Thread!
Post by: fun_jirachi on February 02, 2021, 06:15:15 pm
pls explain 4C Q12

What exactly do you not understand? Is it perhaps setting up the arithmetic sequence/arithmetic sequences altogether (or something completely different)? Your question is quite vague - we ideally want to explain arithmetic sequences (and other sequences too!) so you understand them completely and can apply said explanations to every form of question :D
Title: Re: Specialist 1/2 Question Thread!
Post by: Cate_m on February 13, 2021, 04:10:12 pm
Hey I've got a Spesh sac next week and I would really appreciate some help on this question

Express the following as a partial fraction


Title: Re: Specialist 1/2 Question Thread!
Post by: fun_jirachi on February 13, 2021, 04:12:17 pm
Hi there!

I'd really love to help, but without further context it's impossible to answer this question! Is there any other information that you could give us?

Title: Re: Specialist 1/2 Question Thread!
Post by: Cate_m on February 13, 2021, 05:01:52 pm
Yep, this is the summary in the textbook of what they want done
Title: Re: Specialist 1/2 Question Thread!
Post by: fun_jirachi on February 13, 2021, 05:29:21 pm
I'm sorry, I didn't read the text above the question! My bad :(

Firstly, note that \((x-2)(x^2-4) = x^3-2x^2-4x+8 = x^3-2x^2-3x+9 - (x+1)\).

Hence, we can do the following:
.

We can then apply the method of partial fractions to the remaining fractional part.



Hence, \(\frac{x^3-2x^2-3x+9}{x^2-4} = x - 2 + \frac{3}{4(x-2)} + \frac{1}{4(x+2)}\). Try this with a few other examples and see how you go :)
Title: Re: Specialist 1/2 Question Thread!
Post by: Cate_m on February 13, 2021, 10:55:56 pm
Hey, it's me again. Could somebody please explain 11b
Title: Re: Specialist 1/2 Question Thread!
Post by: fun_jirachi on February 14, 2021, 12:34:00 am
You're given that \(d = k\sqrt{h}\) for some constant \(k\). You can find said constant by substituting \(d=4.8, \ h = 1.8\). Note also that height is in metres while distance is in kilometres - so you to make it easier, you can make the necessary conversion. You don't have to, but it just means you have to remember your units really well. Also, it is implied that you should assume the person in the question is standing up at sea level.

Thus, we have that \(k = \frac{4800}{\sqrt{1.8}} \implies \ d = \frac{4800\sqrt{h}}{\sqrt{1.8}}\).

The first question basically tells you that at the top of the tower, they can see \(\frac{4800\sqrt{5.8}}{\sqrt{1.8}} m\) into the distance. This is implied to be the same distance from the person to the top of the mast. A diagram drawn with this information should look something like the one below.

Spoiler
(https://i.imgur.com/1d8sTQo.png)

From here, you can use Pythagoras' Theorem to find the distance from the yacht to the tower (which is exactly on the coast).

Totally understand your confusion as there are a lot of assumptions you have to make - not the best question in the world imo

Hope this helps :)

Title: Re: Specialist 1/2 Question Thread!
Post by: Cate_m on February 14, 2021, 11:15:11 am
Thank you, you're a life saver
Title: Re: Specialist 1/2 Question Thread!
Post by: pans on February 16, 2021, 08:46:52 pm
Hello,
Can someone pls explain 4E Q12, 13
Title: Re: Specialist 1/2 Question Thread!
Post by: fun_jirachi on February 16, 2021, 10:25:04 pm
Hey!

Q12 was answered here.

Q13a - each time you tear the pieces of paper, you double the number of pieces of paper. You start with 1, then after the first step you have 2. After two steps, you have 4. Is there a geometric series you can construct that will tell you how many will you have after 40 steps? If you stack each piece of 0.05mm paper, the stack should just be the number of pieces you have after 40 steps * 0.05mm.

Q13b - essentially, 384400 km = 3.844 x 1011 mm. At what step does the number of pieces of paper exceed this number? use the same geometric series as part a).

Try this again and get back to us if you're stuck.
Title: Re: Specialist 1/2 Question Thread!
Post by: pans on February 20, 2021, 07:58:52 pm
can anyone help me with sequence and series hard Q!
Title: Re: Specialist 1/2 Question Thread!
Post by: pans on February 20, 2021, 08:18:04 pm
can someone also pls help me with these 3 sequence and series Q
Title: Re: Specialist 1/2 Question Thread!
Post by: fun_jirachi on February 21, 2021, 12:48:03 am
Hey there!

Please try not to double post - you can edit your additional questions into the previous one :)

What have you tried? Ideally, I'd love to help you by giving you more hints, but at some point, it does become a bit counterintuitive as you won't be getting hints at school and in exams - it would help you learn more if you told us what you've tried, then in response, we point you in a better direction. Hopefully, this makes sense!
Title: Re: Specialist 1/2 Question Thread!
Post by: Rachelrachel on March 20, 2021, 11:10:33 am
I've been stuck on this for a while:

If z1 and z2 are the two complex solutions to the quadratic equation ax2 + bx + c = 0, and P1 and P2 are the corresponding points on an Argand diagram, what is the cosine of the angle between P1, P2, and the origin in terms of a b and c?
Title: Re: Specialist 1/2 Question Thread!
Post by: fun_jirachi on March 20, 2021, 03:04:30 pm
A few hints:
- If \(z_1, z_2\) are complex roots to a quadratic equation, how are they related?
- Can you express the points \(P_1, P_2\) in terms of \(a, b, c\)?
- If possible, graph the origin and these two points on the Argand plane (does require you get the first hint) - what do you notice about the position of these points?
- Connect each point to the origin so you have segments \(OP_1\) and \(OP_2\). Is there a relationship between these segments and the angles they make with the coordinate axes?
- You can use the fact that \(\cos (\angle P_1OP_2) = \frac{P_1 \ \bullet \ P_2}{|P_1||P_2|}\) - but is there an easier method?  (Think geometrically rather than algebraically)

Hope this helps :)
Title: Re: Specialist 1/2 Question Thread!
Post by: Rachelrachel on March 20, 2021, 08:58:30 pm
Thank you very much for the help, fun_jirachi.

To be honest, it is the geometric part that I'm having difficulty with. I know that the angle between P1 and P2 would be twice the argument of z1, but I don't understand how to find the cosine of an angle in geometric terms.

(https://i.imgur.com/YWm6sQs.jpg)
Title: Re: Specialist 1/2 Question Thread!
Post by: fun_jirachi on March 20, 2021, 09:12:16 pm
Got everything apart from that, which is great! If you have the coordinates of a point on the Argand plane, how would you usually calculate the argument? You're right in that \(\cos (P_1OP_2) = \cos (2\text{arg}(z_1))\). Right-angled trig should be handy (which is what I meant when I said think geometrically (awful wording, in hindsight - sorry! that one's on me :( )).
Title: Re: Specialist 1/2 Question Thread!
Post by: Ruchir on May 01, 2021, 09:28:23 pm
Hey guys need help with this question
I got part a, but I am stuck with part b.


Let A and B be the points defined by the position vectors a=i+3j   and b=i+j
respectively. Find:
a the vector resolute of a in the direction of b
b a unit vector through A perpendicular to OB
Title: Re: Specialist 1/2 Question Thread!
Post by: S_R_K on May 02, 2021, 07:31:20 pm
Hey guys need help with this question
I got part a, but I am stuck with part b.


Let A and B be the points defined by the position vectors a=i+3j   and b=i+j
respectively. Find:
a the vector resolute of a in the direction of b
b a unit vector through A perpendicular to OB

1. The vector resolute of \(\mathbf{a}\) perpendicular to \(\mathbf{b}\) is given by: \(\mathbf{a}\) - the vector resolute of \(\mathbf{a}\) in the direction of \(\mathbf{b}\).

2. The unit vector in the direction of a vector v is given by \(\frac{1}{\lvert \mathbf{v} \rvert} \mathbf{v}\)
Title: Re: Specialist 1/2 Question Thread!
Post by: mabajas76 on August 18, 2021, 02:17:01 pm
Hi hope all is well.
I have really been struggling with chapter 23 statics in the cambridge text book...my teacher didn't give the best explanation and home learning has really got me down.
Could somebody please explain how this question is answered carfully? I am just kinda confused on why they do what they do.
Cheers.
Title: Re: Specialist 1/2 Question Thread!
Post by: fun_jirachi on August 18, 2021, 04:25:08 pm
Here's a relevant diagram:
(https://i.imgur.com/YKS6xJ2.png)

There are a fair amount of assumptions in that diagram, but logically it makes sense. The most important one that you should note is that the block does not move ie. the net horizontal force is zero, and the net vertical force is also zero.

Guiding questions:
- First, determine the angle denoted x in the diagram. You already know how much force rope one exerts horizontally. Is there a trigonometric ratio that could help you here?
- Then, determine the downwards force of the block by recalling that the net vertical force is zero. This will be equal to the combined upwards force exerted by the two men. Is there another trigonometric ratio that could help you? Don't forget that you want the mass of the block at the end, not the downwards force of the block.

Hope this helps :)

Title: Re: Specialist 1/2 Question Thread!
Post by: Ryan Heng on January 17, 2022, 04:33:58 pm
Hey guys I need help with this question (vectors)

The position vectors of three points A, B and C relative to an origin O are a, b and ka respectively.

The point P lies on AB and is such that AP = 2PB.The point Q lies on BC and is such that CQ = 6QB.

a) Find in terms of a and b,the position vectors of P and Q
Title: Re: Specialist 1/2 Question Thread!
Post by: fun_jirachi on January 18, 2022, 10:23:30 pm
Here are some hints to help you complete the question
\(\vec{OA} + \vec{AB} = \vec{OB}\).
\(\vec{AP} = \frac{1}{3}\vec{AB}\).
\(\vec{OA} + \vec{AP} = \vec{OP}\).