Expert: Paul Klarreich - 12/14/2006

Question
A conical tank (with vertex down) is 10 feet across the top and 12 feet deep.  If water is flowing into the tank at a rate of 10 cubic feet per minute, find the rate of change of the depth of the water when the water is 8 feet deep.

Answer
Questioner:  ridralev2
Category:  Calculus
 
Subject:  Related Rates
Question:  A conical tank (with vertex down) is 10 feet across the top and 12 feet deep.

If water is flowing into the tank at a rate of 10 cubic feet per minute, find the rate of change of the depth of the water when the water is 8 feet deep.
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Hi, Ridralev2, [Is that what your mother calls you?]

Is this your first attempt at R-R problems?  If so, the scheme is something like this:

1. Identify the variables in the problem -- the things that change.  Give them names.

2. Write their rates of change as derivatives WITH RESPECT TO time.  Note which are known and which is to be found.

3. Determine a relationship (yes, it is called 'related rates' for a reason) between the variables.  Use a diagram, use your life experience, use your general knowledge and brilliance, do whatever you have to.  This is the key step.

4. Now differentiate implicitly, then substitute the known quantities and rates, and solve for the unknown rate.

In this case,

Variables:
h = height (depth) of water in the tank
V = volume of water in the tank.

Rates:
dh/dt = rate of change of depth, TO BE FOUND
dV/dt = rate of increase of volume, given as 10 cu.ft/min

Relation:

The water in the tank is in the form of a cone, vertex down, and its volume is given by

V = 1/3 pi r^2 h

We have to get rid of the 'r' variable, and for that, we use similar triangles.  The water and tank, in cross section, look like this:

WARNING: THE FOLLOWING DISCUSSION MAY CONTAIN FRACTIONS AND OTHER MATERIAL INAPPROPRIATE FOR CERTAIN COMPUTING SYSTEMS.  VIEW IT IN A FIXED-SIZE FONT, SUCH AS COURIER.


10 feet across
---------------
\     |     /  ^
 \    | r  /   |
  \---|---/   12 feet deep.
    \ |h /     |
     \|/       |
      V        V
Proportions:

r   10
- = --
h   12

r = 5h/6

So V = 1/3 pi (5h/6)^2 h
   25 pi h^3
V = ---------
     108

Now we can go ahead:

dV   75 pi h^2 dh
-- = --------- --
dt     108     dt

dV   25 pi h^2 dh
-- = --------- --
dt     36      dt

Now we can put our values:

dV/dt = 10,  and  h = 8

    25 pi (8)^2 dh
10 = ----------- --
      36        dt

     5 pi(64)   dh
2 = ----------- --
      36        dt


     5 pi(64)   dh
2 = ----------- --
      36        dt

dh    72
-- = -------
dt   320 pi

dh    9
-- = -----
dt   40 pi