Question 9:

A)

12y = x^2 - 6x - 3

= (x-3)^2 - 12

12y + 12 = (x-3)^2

(x-3)^2 = 4.3(y+1)

a = 3, vertex (3,-1)

.: focus (3,2)

Bi)

sub t=0 --> dv/dt = 120

so find t when dv/dt = 240:

240 = 120 + 26t - t^2

t^2 - 26t + 120 = 0

t = 20, 6

ii)

V = S 120 + 26t - t^2 dt = 120t + 13t^2 - t^3/3 + C

iii)

t = 0, V = 1500 ---> C = 1500

.: V = 120t + 13t^2 - t^3/3 + 1500

t = 30 ---> V = something.

.: water lost = 7000 - that something

I think I got that one wrong in the exam, I remember getting some huge number.

Ci) V = 1/3*pi*r^2*h where h = x (given)

The triangle below O has sides a,r,(x-a)

.: r^2 = a^2 - (x-a)^2 = 2ax - x^2 (pythagoras' theorem)

.: V = 1/3*pi*x*(2ax - x^2) = 1/3*pi*(2ax^2 - x^3)

ii) simple differentiation. I got x = 4a/3 or something, which hopefully is right.

Question 10:

i) KQ = 12 - 6 - x = 6 - x

KL = 6 + x (its PK folded)

.: Ql^2 = (6+x)^2 - (6-x)^2 (pythagoras' in KQL)

= 24x

ii) LMS = QLM (alternate angles between parallel lines)

LMS = LMN + NMS = LMN + 90

similarly QLM = QLK + 90

.: LMN + 90 = QLK + 90

.: LMN = QLK

LNM = QKL = 90 (given)

.: triangles are equiangular & similar

ML/KL = MN/QL (corresponding sides in similar triangles are in proportion)

.: y/(6+x) = 12/2rt(6x)

y = 6(6+x)/rt(6x)

= (6+x)rt6/rtx

iii) simply A = 1/2.b.h where h = (6+x) and base = y given above

iv) Screwed this up! Solved for 12<=y and 13=>y , getting 8/3<=x<=27/2 -- but 27/2 is larger than the side so thats wrong

v) Also screwed this up! Found it to be x=2, calculated it all. Then did part iv) and realised it was outside the domain.

so i just found the area when x = 8/3.

If anyone is confident they got the last 2 parts right, i'd like to see your working. cheers