Q9&10, partial solutions (1 Viewer)

[myeewyee]

So this is what I got, no guarantees if its right. If someone could correct me on any I got wrong that would be sweet. Post any other answers here

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

 

[michellejai]

In 10b) iv) when you sub y=12 you end up getting x = 6, and when you sub y=13 then you end up getting 2 2/3 and 27/2, so yeah, x lies between 6 and 2 2/3.
v) when you did da/dx and found the values when A'=0, it lies outside of 2 2/3 and 6, so the min area of KLM is at one of the end points. and when you work that out it comes to be 56 1/3 cm sq, at y=13 and x=2 2/3..

 

[followme]

does it matter if i wrote 56.3? it's ridiculous if i lose a mark like that.

 

[fishy89sg]

omg so x=2 is wrong?

 

[xinxin89]

I got completely different answers:

Q10b)

(iv) X < 37/6

(v) Area = 32(3)^(1/2) with x

 

[michaeln36]

In 10b) iv) when you sub y=12 you end up getting x = 6, and when you sub y=13 then you end up getting 2 2/3 and 27/2, so yeah, x lies between 6 and 2 2/3.
v) when you did da/dx and found the values when A'=0, it lies outside of 2 2/3 and 6, so the min area of KLM is at one of the end points. and when you work that out it comes to be 56 1/3 cm sq, at y=13 and x=2 2/3..

meh...! im annoyed ! i started doing that for 10)iv) then stopped cause i didnt think it would work... now that i see the solns its not really that hard .. but alas.. 5 marks lost for me =(!

 

[rivergumrob]

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)

Oh man...i completed the square wrong...i ended up with a vertex of (3,1) and therefore a focus of (3,4)
grrr...
At least i got the next part... :mad1:
hmmm