• 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

ROFLQUESTION rates of change ROFLQUESTION (1 Viewer)

menty

Member
Joined
Feb 11, 2004
Messages
119
Location
riding the roflcopter made by loldude
Gender
Male
HSC
2010
Water is being pumped into a pool at 2metres cubed per minute. The pool is 20m long and 5m wide. Itts depth decreases uniformly from 5m to 1m at the other end.
A) show the volume of water at depth h is 12.5h^2. Hence find he rate of increase of i) depth and ii) SA of the water when the water depth at the deepest end is on metre
 

justchillin

Member
Joined
Jun 11, 2005
Messages
210
Gender
Male
HSC
2005
Area trap = 1/2xhx5(h) at depth = h
therefore volume = 25/2h^2 (width = 5)
U can do the rest...just chain rule, use ur head lol:)
 

justchillin

Member
Joined
Jun 11, 2005
Messages
210
Gender
Male
HSC
2005
Draw a diagram..id put mine up but im spastic with computers...gl tomorrow mate
 

Stefano

Sexiest Member
Joined
Sep 27, 2004
Messages
296
Location
Sydney, Australia
Gender
Male
HSC
2005
justchillin said:
Area trap = 1/2xhx5(h) at depth = h
Could you explain that? are you using the fact that the Area of a RHS triangle is ab/2 ? If so where did you get 5h from ?

menty said:
ta lollerchillin. Just couldnt work out the roflarea.
lmao, you sound like someone who plays too much Counter-Strike :D
 

justchillin

Member
Joined
Jun 11, 2005
Messages
210
Gender
Male
HSC
2005
K at x=h draw a horizontal line going 2 the deep end. Then use pythagoras on the top triangle to work out perpendicular heights...comes out at 5h for me...fuck i need 2 put my diagram up but i cant seriously...its like a trap face with depth = 5 on left and depth = 1 on right...
 

Stefano

Sexiest Member
Joined
Sep 27, 2004
Messages
296
Location
Sydney, Australia
Gender
Male
HSC
2005
justchillin said:
K at x=h draw a horizontal line going 2 the deep end. Then use pythagoras on the top triangle to work out perpendicular heights...comes out at 5h for me...fuck i need 2 put my diagram up but i cant seriously...its like a trap face with depth = 5 on left and depth = 1 on right...
Don't fret about a diagram. I'm pretty sure mine is correct.

What's the length of the hypotenuse??
 

justchillin

Member
Joined
Jun 11, 2005
Messages
210
Gender
Male
HSC
2005
sorry my internet has been down...I dont really know how 2 explain it apart from what I've been saying...what bit are u unsure about? And stefano what was that link: i dont get it?
Edit: you dont need the length of the hypotenuse... the 2 sides are h and 5 and the perpindicular height is found using right angled triangle...
 
Last edited:

Stefano

Sexiest Member
Joined
Sep 27, 2004
Messages
296
Location
Sydney, Australia
Gender
Male
HSC
2005
The link was a...link... lol.. to a thread of mine entitled 'Quick Polynomial Question' in the 3u forums. Check it out. No-one has been able to help yet. I thought maybe you could?
 

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

Top