Related Rates in a water tank

  • Thread starter b.mueller5
  • Start date
  • #1

Homework Statement


A water tank has the shape of an inverted right-circular cone, with radius at the top 15 meters and depth 12 meters. Water is flowing into the tank at rate of 2 cublic meters per minute. How fast is the depth of water in the tank increasing at the instant when the depth is 8 meters


Homework Equations


V=(1/3)(pi)(r^2)(h)

The Attempt at a Solution


dv/dt=2 meter^3/min
dv/dt=(1/3)(pi)(2rh(dr/dt)+(r^2)(dh/dt))
but that has 2 unknow varibles in it.
 

Answers and Replies

  • #2
You need to relate the radius in terms of the height of the cone since both are changing with respect to time and because you only know the rate of change of the height. Once you do this, you will be able to differentiate it since you know the rate of change of the height. Try to find a way to relate the radius in terms of height. With this information can you figure out how to get rid of r and put it in terms of h?
 
  • #3
Yeah thanks that helps a lot
 

Related Threads on Related Rates in a water tank

Replies
1
Views
262
  • Last Post
Replies
0
Views
2K
  • Last Post
Replies
1
Views
12K
  • Last Post
Replies
1
Views
853
Replies
2
Views
3K
  • Last Post
Replies
1
Views
7K
Replies
1
Views
3K
  • Last Post
Replies
5
Views
18K
Replies
1
Views
3K
Top