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
LHS= y'(t1)/x'(t1)poWerdrY said:someone post the proof for second last part of q7
OMFG!!!!! i didnt think of implicit differentiation!!!!!!! omfg!!!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
8!/2 ( as john behind barabara is half of all cases. easy i know, but i didnt think of it in the exam)independantz said:Anyone got the answers to the perms and combs? and 3a(ii)
wow nice solution.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
yeww i had the same answerpoWerdrY said:8!/2 ( as john behind barabara is half of all cases. easy i know, but i didnt think of it in the exam)
i might be wrong but i got 0,pi/3,pi,5pi/3,2piscott.syd said:6(b): 0, 2pi/3, pi, 4pi/3, 2pi
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.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.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