Maths Extension 2 Thoughts and Feelings (1 Viewer)

[youssseeefffff-wahbbbaaa]

Hey guys, here is my solution to the very last question of the ext 2 hsc 2022. Though i wasnt able to fully complete it during the exam cuz it was too time pressured at least i came up with my own solution. Please correct me if theres any error. Good luck to everyone!

 

[idkkdi]

last years algebra bashing was way worse. this year you literally just had to integrate twice and you're done. not denying the difficulty but the mathematical component wasn't very demanding

compared to the rest of q16 parts this yr, it had a bit more algebra.

q16d should have been the favourite for the hardest q this yr. required a bit of creativity, while q16b looks like it came out of terry lee lmao.

 

[Daedalus13]

The James Ruse answer to Q 16 b I think is incorrect. The solution did not take into account that there are two equations of motion. One for the journey upwards where a = - g - 0.1v and for the journey downwards a = g - 0.1v.

This is what I thought in the exam (to my demise as I made some algebraic errors), but I think the change in sign of the velocity accounts for this change in direction. You end up with the same answer nonetheless (if you don't screw up like I did).

 

[okkk]

Well that's much better.

I've attached a screenshot again

View attachment 36704

I'm still confused by this solution lmao does anyone know how she's allowed to just use the upwards equation the whole way through and sub in x=0, seems to contradict everything I've learnt in mechanics

 

[notme123]

I'm still confused by this solution lmao does anyone know how she's allowed to just use the upwards equation the whole way through and sub in x=0, seems to contradict everything I've learnt in mechanics

Thats why i said that question felt illegal in one of my posts but this post below is the reason why.

I think the change in sign of the velocity accounts for this change in direction.

16 b and d are questions i feel people would get if they've seen similar versions or were taught those very specific details. can someone please prove I'm not crazy in saying I've seen 16 d somewhere before

 

[okkk]

compared to the rest of q16 parts this yr, it had a bit more algebra.

q16d should have been the favourite for the hardest q this yr. required a bit of creativity, while q16b looks like it came out of terry lee lmao.

Tbh the polynomial thing for 16d wasn't that necessary. Once you have that they're all magnitude 1, you can use z_1+z_2+z_3=1 to draw a rhombus (since their lengths are the same) and deduce one of them has to be 1 and the other two are of the form e^(itheta) and -e(itheta), then you can use the multiplying identity to get theta is pi/2, I think this is probably the most intuitive way cos I got up to that question with two minutes left cos I'd blown it all on the mechanics one

 

[Trebla]

I'm still confused by this solution lmao does anyone know how she's allowed to just use the upwards equation the whole way through and sub in x=0, seems to contradict everything I've learnt in mechanics

This is because v is defined as velocity in the solution (with upwards defined as positive) instead of speed (which is what most people do). When you define v as speed then you tend to define the direction of motion as positive. This usually makes adding/subtracting the vectors more intuitive as you can define the signs yourself using the magnitudes given - i.e. you don’t need to think about the sign of v.

Once you define v as velocity (which is a vector quantity), you have to take a bit more care on the signs because the vector can change direction.

Technically speaking though the question defined v as speed…

 

[idkkdi]

Tbh the polynomial thing for 16d wasn't that necessary. Once you have that they're all magnitude 1, you can use z_1+z_2+z_3=1 to draw a rhombus (since their lengths are the same) and deduce one of them has to be 1 and the other two are of the form e^(itheta) and -e(itheta), then you can use the multiplying identity to get theta is pi/2, I think this is probably the most intuitive way cos I got up to that question with two minutes left cos I'd blown it all on the mechanics one

nah you can't make that deduction that 'one of them has to be 1'.

 

[okkk]

nah you can't make that deduction that 'one of them has to be 1'.

It's impossible to construct unit vectors that add together to get one if one of the is not one. For example, construct the first one anywhere you want, then consider the second vector as one, if it shifts up or down, the distance between its tip and 1 would change and it no longer 1, so the third vector cannot get to one if the second vector is not 1. I think at least, I can't visualise a counter example

 

[youssseeefffff-wahbbbaaa]

q16d was a neat q.
You need to recognise that it was somewhat a polynomial question.

u get |z1| = |z2| = |z3| = 1.
z1+z2+z3 = 1, z1z2z3 = 1.
1/(z2z3) + 1/(z1z3) + 1/(z1z2) = 1.
=> z2z3 + z1z3 + z1z2 = 1 (noticing mod = 1, 1/z = z_conj, and then noticing rhs is real)
Notice that you get sum, product, sum of two roots formulas. Then you can make a cubic polynomial and solve that for the three complex numbers.

if this is not the correct method, well oops lol

Yo sorry i never read this, i just posted my solution to this (though i didnt actually have enough time to complete in the test) and coincidentally it was the exact same as urs. Congrats on solving it, just wanna make it clear that i didnt mean to plagiarise lol

 

[okkk]

This is because v is defined as velocity in the solution (with upwards defined as positive) instead of speed (which is what most people do). When you define v as speed then you tend to define the direction of motion as positive. This usually makes adding/subtracting the vectors more intuitive as you can define the signs yourself using the magnitudes given - i.e. you don’t need to think about the sign of v.

Once you define v as velocity (which is a vector quantity), you have to take a bit more care on the signs because the vector can change direction.

Technically speaking though the question defined v as speed…

It's just I graphed the curve she got for x on desmos, and it's symmetrical about the turning point implying that the time it takes to reach max height is the same as it takes to fall to the ground and that's not right right? It should take longer to fall down is my understanding of mechanics, I'm still struggling to understand how v is a vector resolves this asymmetry

 

[notme123]

It's just I graphed the curve she got for x on desmos, and it's symmetrical about the turning point implying that the time it takes to reach max height is the same as it takes to fall to the ground and that's not right right? It should take longer to fall down is my understanding of mechanics, I'm still struggling to understand how v is a vector resolves this asymmetry

conservation of energy is violated if the time going up and down is the same. should take longer to go down. that's why assuming impact speed is v0 is also completely wrong.

It's impossible to construct unit vectors that add together to get one if one of the is not one. For example, construct the first one anywhere you want, then consider the second vector as one, if it shifts up or down, the distance between its tip and 1 would change and it no longer 1, so the third vector cannot get to one if the second vector is not 1. I think at least, I can't visualise a counter example

ye i cant think of a good counter example but the method isn't rigorous or rather there isn't some established way to reach that conclusion

 

[okkk]

conservation of energy is violated if the time going up and down is the same. should take longer to go down. that's why assuming impact speed is v0 is also completely wrong.


ye i cant think of a good counter example but the method isn't rigorous or rather there isn't some established way to reach that conclusion

Ah I can say that cos our four vectors are z_1, z_2, z_3 and 1, and they all have modulus one, i.e 4 sides are equal, it has to be a rhombus which has parallel sides. so therefore one of them also has to be 1. That works right (hopefully lmao my yr 9 geometry is rusty)?

 

[notme123]

Ah I can say that cos our four vectors are z_1, z_2, z_3 and 1, and they all have modulus one, i.e 4 sides are equal, it has to be a rhombus which has parallel sides. so therefore one of them also has to be 1. That works right (hopefully lmao my yr 9 geometry is rusty)?

maybe if you had a diagram to explain it it would be easier but idk how you get a rhombus. did you do this in the actual exam?

 

[okkk]

maybe if you had a diagram to explain it it would be easier but idk how you get a rhombus. did you do this in the actual exam?

Yeah I did , it's a rhombus cos a rhombus is (well can be) defined as a four sided shape with equal side lengths and it's the only one with the definition (square is a rhombus) so that's the conclusion right?

 

[idkkdi]

It's impossible to construct unit vectors that add together to get one if one of the is not one. For example, construct the first one anywhere you want, then consider the second vector as one, if it shifts up or down, the distance between its tip and 1 would change and it no longer 1, so the third vector cannot get to one if the second vector is not 1. I think at least, I can't visualise a counter example

On second thought, your way works.

|z1|=|z2|=|z3|=1 and z1+z2+z3 = 1 does imply a rhombus with vectors z1,z2,z3, and (1,0).
Since a rhombus has parallel sides, one of z1/z2/z3 has to be the vector (1,0).
Not completely sure if you'll get the marks since it's HSC marking, but this method does work fine imo.

 

[Vall]

idk why smh articles make me cringe a little inside.

some friends and I were saying this on our discord. It feels like they have a bunch of template articles to release around HSC time.

 

[hogzillaAnson]

Typeset solutions. Hopefully not too many mistakes : )

 

[maths4dayz]

I probs got between 85 and 90 but pretty annoyed at how hard that was. I reckon an extra 20 or 30 minutes would've gone a long way. I got most of the vector, mechanics and proof questions but the complex number ones were random and I had a wtf moment. It was weird how they put in some easy questions though.

 

[ExtremelyBoredUser]

I probs got between 85 and 90 but pretty annoyed at how hard that was. I reckon an extra 20 or 30 minutes would've gone a long way. I got most of the vector, mechanics and proof questions but the complex number ones were random and I had a wtf moment. It was weird how they put in some easy questions though.

yep same assuming the solutions made are accurate, the paper felt inconsistent. MC was half hard, half easy. Q11-13 was really standard. Q14-15 was extremely slow pacing. Q16 was hard given the time crunch.