The discussion revolves around calculating the rate of change of water depth in a hollow cone with a semi-vertical angle of 60 degrees, as water drips into it at a rate of 4 cm³/min. Participants are examining the relationships between the volume of the cone, the radius, and the height of the water.
Some participants have provided guidance on how to approach the problem, suggesting that the original poster needs to consider the relationship between the radius and height more carefully. There is an ongoing exploration of different interpretations of the problem, particularly regarding the calculations leading to the rate of change of height.
Participants have noted the inconvenience of external links for sharing problems and solutions, indicating a preference for direct posting of the problem and work within the thread. There is also mention of a specific numerical answer that appears to differ from the calculations presented.
jsmith613 said:Homework Statement
I have attached a link along with my working
please can someone help me
http://s359.photobucket.com/albums/oo40/jsmith613/?action=view¤t=Math.png
Homework Equations
The Attempt at a Solution
dv/dt = 4 cm3 min-1Mark44 said:Please post the problem and your work here. IMO, it's a pain in the butt to have to open a web page to see the problem and the work, plus I can't insert a comment at the appropriate place where there's an error.
You're ignoring the relationship between r and h.jsmith613 said:dv/dt = 4 cm3 min-1
tan(60) = r/h
r = √3* h
V = (1/3) π r2h
dv/dh = (1/3) π r2
jsmith613 said:dh/dt = (dh/dv) * (dv/dt)
= 1/(1/3) π r2 * 4
= 12/(pi r2)
for h = 4
dh/dt = 0.079577
The answer given is 0.0265 cm/min. why?
I don't know how do use latex - see post for a clearer solution if you get lost!