Rate of change - cones


by jsmith613
Tags: cones, rate
jsmith613
jsmith613 is offline
#1
Mar14-12, 04:00 PM
P: 614
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
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Mark44
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#2
Mar14-12, 04:20 PM
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Quote Quote by jsmith613 View Post
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
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
jsmith613 is offline
#3
Mar14-12, 04:37 PM
P: 614
Quote Quote by Mark44 View Post
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.

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
jsmith613 is offline
#4
Mar14-12, 04:45 PM
P: 614

Rate of change - cones


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
Mark44 is offline
#5
Mar14-12, 05:40 PM
Mentor
P: 21,059
Quote Quote by jsmith613 View Post
dv/dt = 4 cm3 min-1

tan(60) = r/h


r = √3* h


V = (1/3) π r2h



dv/dh = (1/3) π r2
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.
Quote Quote by jsmith613 View Post

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!


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