Rate of change - cones

jsmith613

1. The problem statement, all variables and given/known data

I have attached a link along with my working
please can someone help me

http://s359.photobucket.com/albums/o...rrent=Math.png

2. Relevant equations



3. The attempt at a solution

Mark44

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.

jsmith613

dv/dt = 4 cm³ min^-1

tan(60) = r/h


r = √3* h


V = (1/3) π r²h



dv/dh = (1/3) π r²

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!

jsmith613

question was:


A hollow cone with a semi-vertical angle of 60 degrees is held vertex down with its axis vertical.
Water drips into the cone at 4 cm³/min
Find the rate at which the depth of water is increasing when the water is 4 cm deep

Mark44

You're ignoring the relationship between r and h.

Substitute for r in your volume equation. Then you'll have V purely as a function of h.
What you have is not correct, because V is a function of r and h.