ITT: We go through the 3U paper and post up answers. (1 Viewer)

Scissors

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I'm stupid so I won't post up any answers.

Anyway, go!
 

scott.syd

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i can try:
1(a) : -27
1(b): -3/(1-9x^2)^1/2
1(c): pi/3
1(d): 10264320
1(e): ((2)^1/2)/12
1(f): 3<x<5

2(a): 1/2
2(b): (11)^1/2
2(c): -5
2(d): 4.11
 

elliots

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scott.syd said:
i can try:
1(a) : -27
1(b): -3/(1-9x^2)^1/2
1(c): pi/3
1(d): 10264320
1(e): ((2)^1/2)/12
1(f): 3<x<5

2(a): 1/2
2(b): (11)^1/2
2(c): -5
2(d): 4.11


got all of those :)
 

scott.syd

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poWerdrY said:
someone post the proof for second last part of q7
LHS= y'(t1)/x'(t1)
=(Vsin-gt1)/Vcos
RHS=tan a - tan b
=h(t2-t1)/Vt1t2cos
sub t2= 2Vsin/g - t1
and t1t2= 2h/g

therefore: RHS= h (2Vsin/g -2t1)/V(2h/g)cos
=(Vsin-gt1)/Vcos
=LHS
 

poWerdrY

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scott.syd said:
LHS= y'(t1)/x'(t1)
=(Vsin-gt1)/Vcos
RHS=tan a - tan b
=h(t2-t1)/Vt1t2cos
sub t2= 2Vsin/g - t1
and t1t2= 2h/g

therefore: RHS= h (2Vsin/g -2t1)/V(2h/g)cos
=(Vsin-gt1)/Vcos
=LHS
OMFG!!!!! i didnt think of implicit differentiation!!!!!!! omfg!!!
 

poWerdrY

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independantz said:
Anyone got the answers to the perms and combs? and 3a(ii)
8!/2 ( as john behind barabara is half of all cases. easy i know, but i didnt think of it in the exam)
 

u-borat

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scott.syd said:
LHS= y'(t1)/x'(t1)
=(Vsin-gt1)/Vcos
RHS=tan a - tan b
=h(t2-t1)/Vt1t2cos
sub t2= 2Vsin/g - t1
and t1t2= 2h/g

therefore: RHS= h (2Vsin/g -2t1)/V(2h/g)cos
=(Vsin-gt1)/Vcos
=LHS
wow nice solution.

i didn't quite get it, but i think the way you were meant to (cos implicit diff is only 4unit amirite?) was to get the general equation of the paraboal by getting rid of t in your general motion equations, then differentiate that, subbing in x=Hcotalpha then equate that to tan wierd squiggly thing.

didn't quite get the answer though :(
 
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poWerdrY said:
8!/2 ( as john behind barabara is half of all cases. easy i know, but i didnt think of it in the exam)
yeww i had the same answer
but i had a mind blank with the question and considered cases. took a long time
but same result =]
 

scott.syd

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i'll put the other answers i got, not proofs though, they are too long:
3(a)(ii): -2=<x=<4/3
4(a)(ii): 12.44pm
4(b)(i): 7!= 5040 ways
5(a)(ii): y= 1-(1-2x)^1/2
5(a)(iii): 1/2
5(b): a=2/3 metres T=2pi/3 seconds
6(a)(ii): 96.0m
6(b): 0, 2pi/3, pi, 4pi/3, 2pi
6(c)(i): (p+q)!/(p!q!)
6(c)(ii): (p+q)!/p!q!
 

independantz

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Anyone got answers to 3a(ii) and the height of the tower for the 3d trig 6a(ii)
 

conics2008

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srrhy guys I'm using itouch so it's hard but I'm confident with mine
 

poWerdrY

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u-borat said:
wow nice solution.

i didn't quite get it, but i think the way you were meant to (cos implicit diff is only 4unit amirite?) was to get the general equation of the paraboal by getting rid of t in your general motion equations, then differentiate that, subbing in x=Hcotalpha then equate that to tan wierd squiggly thing.

didn't quite get the answer though :(
thats what i thought, but then the equation of the parabola would be y= -ax(x-r). and i couldnt think of a way to get rid of a.
 

scott.syd

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u-borat said:
wow nice solution.

i didn't quite get it, but i think the way you were meant to (cos implicit diff is only 4unit amirite?) was to get the general equation of the paraboal by getting rid of t in your general motion equations, then differentiate that, subbing in x=Hcotalpha then equate that to tan wierd squiggly thing.

didn't quite get the answer though :(
but i didn't do it by implicit diff, i just did normal differentiation coz the tan of the angle at the point was the vertical velocity/horizontal velocity. so ya. i wouldn't how else to do it.
 

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