• Best of luck to the class of 2024 for their HSC exams. You got this!
    Let us know your thoughts on the HSC exams here
  • YOU can help the next generation of students in the community!
    Share your trial papers and notes on our Notes & Resources page
MedVision ad

General Thoughts: Mathematics Ext 1 (1 Viewer)

micuzzo

Member
Joined
Aug 31, 2008
Messages
489
Gender
Undisclosed
HSC
2009
how did ppl work out the Volume 1... i could not work it out!!!


oh no... dnt tell me it was as simple as [5h^2*root3]/2

plz dnt tell me i just worked it out

EDIT: no thats wrong... phew!
 
Last edited:

HailSatan

Member
Joined
Oct 17, 2009
Messages
94
Gender
Undisclosed
HSC
2009
Was question 2 (a) and 6 (b) (i) as easy as I thought they were?

What did other people get for them?
 

123chryssi123

New Member
Joined
Aug 24, 2008
Messages
1
Gender
Female
HSC
2009
I found it pretty hard even Q 4 and 5.
What the hell was the horizontal asymptote?
 

untouchablecuz

Active Member
Joined
Mar 25, 2008
Messages
1,693
Gender
Male
HSC
2009
7. b) ii)

did we have to check the nature of the point, becuase seriously

...

i differentiated implicitly and the 2nd derivative/1st derivative (checking each side) was a monster
 

6tgyuio

New Member
Joined
Sep 21, 2009
Messages
6
Gender
Male
HSC
2010
you shouldn't need to check its nature, it already gave when it was pos, it was max
 

study-freak

Bored of
Joined
Feb 8, 2008
Messages
1,133
Gender
Male
HSC
2009
7. b) ii)

did we have to check the nature of the point, becuase seriously

...

i differentiated implicitly and the 2nd derivative/1st derivative (checking each side) was a monster
Look at the wording of the question
"The maximum value occurs... dtheta/dx=0 and x is positive"
Thus you need to find x satisfying dtheta/dx=0 and x>0
no need to go further
 

untouchablecuz

Active Member
Joined
Mar 25, 2008
Messages
1,693
Gender
Male
HSC
2009
oh it did

thanks mate! :D

n2s: read questions more properly next time

...next time
 

elzy93

New Member
Joined
Apr 15, 2008
Messages
2
Location
Nowra
Gender
Female
HSC
2009
Q1-4 were fine, then I pretty much died. The supervisor said she saw my face fall when I started on Q5, haha.
 

AlexJB

Unmotivated
Joined
Jul 31, 2008
Messages
59
Gender
Male
HSC
2009
how did ppl work out the Volume 1... i could not work it out!!!


oh no... dnt tell me it was as simple as [5h^2*root3]/2

plz dnt tell me i just worked it out

EDIT: no thats wrong... phew!
You had to find the area of the triangle and then just multiply by 10.

This is what I did: was isoceles. Construct a line through the middle, form 2 equal right angled triangles. We know the height is h, and we know the angles of the triangle. Use trig to find out the length of the base of the triangle. That gives you 2 sides, which you can use to find the area, which you just multiply through by 10.
 

Stephie-Rae

Member
Joined
Mar 7, 2008
Messages
96
Gender
Female
HSC
2009
I was pretty happy with it. Loved Q's 1-4, 5 was good. The bloody binomials in 6 killed me, and i didn't get the last bit of 7 out. I'm just hoping i didn't make too many stupid errors.
 

micuzzo

Member
Joined
Aug 31, 2008
Messages
489
Gender
Undisclosed
HSC
2009
You had to find the area of the triangle and then just multiply by 10.

This is what I did: was isoceles. Construct a line through the middle, form 2 equal right angled triangles. We know the height is h, and we know the angles of the triangle. Use trig to find out the length of the base of the triangle. That gives you 2 sides, which you can use to find the area, which you just multiply through by 10.
yer i couldnt find the area... :(

did u keep it as a trig ratio but... coz from wat i remember htere the other sides were unkown
 

shady145

Banned
Joined
Dec 4, 2008
Messages
1,687
Gender
Female
HSC
2014
question 1-5=58/60
q6,7=10/24 ='(
thats wat i think i got
 

ShawnG

New Member
Joined
Jul 21, 2008
Messages
13
Gender
Male
HSC
2009
hey guys i got pi/4n for 5a)iv)
dx/dt = nroot(a^2-x^2)
then sin-1 (x/a) = nt
then x= asin(nt)
then particles speed is half maximum speed at (a/root2)
so sub a/root2 in then you get pi/4 = nt no?
 

untouchablecuz

Active Member
Joined
Mar 25, 2008
Messages
1,693
Gender
Male
HSC
2009
yer i couldnt find the area... :(

did u keep it as a trig ratio but... coz from wat i remember htere the other sides were unkown
yep

you had one angle 60, the adjacent side h

cos60 = h/hypotenosue

hypotenouse = h/cos60

A = (1/2)(h/cos60)(h/cos60)sin120
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top