MedVision ad

Mathematics help please! (1 Viewer)

Drongoski

Well-Known Member
Joined
Feb 22, 2009
Messages
4,255
Gender
Male
HSC
N/A
Hey Guys! Id really appreciate it if someone could help me out as soon as possible.

I have a question from the 2001 HSC paper, Mathematics Advanced.

Question 8 c)

http://community.boredofstudies.org/images/attach/pdf.gif

please help me understand

Thankyou :)
dv/dt max near the bottom (indicated with line) where bottle is narrowest

and min where bottle is "fattest" (also marked byline)

As time t increases, from the bottom of bottle, dv/dt increases until it hits a max and then very gently decreases to a min and then increases again until it fills up. Need a diagram to explain more fully, I'm sorry.
 

ktl_edwards

Member
Joined
Mar 27, 2009
Messages
35
Gender
Female
HSC
2009
Oh thanks for replying...a little bit hard to understand though.. :( :uhoh:
 

ktl_edwards

Member
Joined
Mar 27, 2009
Messages
35
Gender
Female
HSC
2009
hmm...i might be able to, but not today.

I appreciate your help! Its so hard to understand math stuff on here since most of understanding math is by watching another person...uno?

Thank you so much though! :)

Hopefully my brother will help me tonight...
 

boxhunter91

Member
Joined
Nov 16, 2007
Messages
736
Gender
Male
HSC
2009
When i done that question i just based it upon the curve y=x^3
Success one did as follows:
dy/dx will be maximum at a depth where glass is narrowest, ie at y=2
dy/dx will be minimum where glass is widest ie at y=7.5
 

gurmies

Drover
Joined
Mar 20, 2008
Messages
1,209
Location
North Bondi
Gender
Male
HSC
2009
Well, if you think about it:

The narrower the segment, the less "radius" I suppose you could call it, you have to till for an increase in height. Recognise that before height can increase, the entire surface has to be filled with water (surface at any height). As a result, with a wide segment, it takes times to fill the outer part and hence takes more time to increase height.

Is it an awful question? Yes.
 

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

Top