Hsc 2002 (1 Viewer)

[Premus]

Hey

can u guys help me with Question 3 b) from the 2002 HSC paper
parts ii and iii


and question 7)
part a) i , ii, iii

thanks

 

[gman03]

part ii

tanget at P: x + p²y = 2cp
tanget at Q: x + q²y = 2cq

subtract bottom from top

y(p²-q²) = 2c(p-q)
so y (p-q)(p+q) = 2c(p-q)
so y = 2c / (p+q)

sub y in any eq:

x + p²y = 2cp
so x + 2cp²/ (p+q) = 2cp
so x = 2cp/(p+q) * ( [p+q] - p)
so x = 2cpq/(p+q)

 

[gman03]

Q3 iii:

tanget of P passes through (cq, 0):

tanget of P is x + p²y = 2cp
so (cq) + 0 = 2cp
so q = 2p

so parametric form of T becomes ( 4cp² / 3p, 2c / 3p)

so x_T = 4cp/ 3, y_T = 2c/3p

so xy = 8c²/ 9

you could work out the eccentricity yourself, I forgot how to

edit: i think is e = root 2

 

[gman03]

I have doubts I'm working on the question you are asking..... next time type up bits of the question too. Don't be lazy on maths.


Q7 part i
We know V = Ah
so V=Ay
differentiate it we get

dV/dt = A dy/dt (Noted that A is constant)

so dy/dt = (dV/dt) / A
= - k root y / A

 

[gman03]

Q7 part ii

dy/dt = - k root y / A
so dt/dy = -A/k * 1/(root y)

integrate we get t = -A/k * 2 (root y) + C

t = 0, y = y₀

so C = A/k * 2 (root y₀)

y=0, t=T, so

T = C = A/k * 2 (root y₀)

Now t = -A/k * 2 (root y) + C
so t/T = -A/k * 2 (root y)/C + 1
so A/k * 2 (root y)/ [A/k * 2 (root y₀)] = 1-t/T
so root (y/y₀) = 1-t/T
so y/y₀ = (1 - t/T)²
so y = y₀(1 - t/T)²

 

[gman03]

Q7 part iii

t = 10, y = y₀ / 2

and from above y = y₀(1 - t/T)²

so y₀ / 2= y₀(1 - 10/T)²

so 1 / root(2) = 1 - 10/T
so T = 10 (root 2 )/ [(root 2 ) - 1]

so it takes 10 (root 2 )/ [(root 2 ) - 1] seconds to empty the full cooler

 

[Premus]

hey thanks a lot!
and yeah rectangular hyperbolas always have an eccentricity of root 2.