Expert: Paul Klarreich - 10/11/2008

Question
Hi im Suj,
Below is my question, hope you can solve it.

A water is in the shape of an inverted conical cone with top radius of 20m and depth of 15m. Water is flowing into the tank at a rate of 0.1 m^3/min

a) How fast is the depth of water in the tank increasing when the depth is 5m?

Water is now leaking from the tank at a rate that depends on the depth h(h=height of water in the tank) this rate is 0.1h^3/min.

a) How fast is the depth of water in the tank changing when the depth is 5m?
b) How full can the tank get  

Answer
Questioner:   Suj
Category:  Advanced Math
Private:  No
 
Subject:  Maths Question
Question:  Hi im Suj,
Below is my question, hope you can solve it.

A water is in the shape of an inverted conical cone with top radius of 20m and depth of 15m. Water is flowing into the tank at a rate of 0.1 m^3/min

a) How fast is the depth of water in the tank increasing when the depth is 5m?

Water is now leaking from the tank at a rate that depends on the depth h(h=height of water in the tank) this rate is 0.1h^3/min.

a) How fast is the depth of water in the tank changing when the depth is 5m?
b) How full can the tank get
-------------------------------------
Hi, Suj,

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.

AND, Please check the archives for other Related Rates examples.  There are a lot of them.  Click BROWSE PAST ANSWERS.
..................................................
Variables:

let  h = depth of water in tank, which is conical.
    r = radius of cone of water in tank.
    V = volume of water in tank.
Rates:

dV/dt given as 0.1

dh/dt to be found.

Keep in mind that
r    20
--- = ---
h    15

r    4
--- = ---
h    3

r = 4h/3

Part I:

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

V = (1/3) pi (4h/3)^2 h

V = (1/3) pi (16h^2/9) h

V = 16 pi h^3/27

Differentiate:

dV/dt = 48 pi h^2/27 dh/dt
dV/dt = 16 pi h^2/9 dh/dt
0.1   = 16 pi h^2/9 dh/dt


Substitute  h = 5, solve for dh/dt

You can do that part.
----------------------
Part II (water leaking at  0.1 h^3

Now dV/dt = input of water - output of water.

dV/dt = 0.1 - 0.1 h^3.

But dV/dt = 16 pi h^2/9 dh/dt

So:

0.1 - 0.1 h^3 = 16 pi h^2/9 dh/dt

Substitute  h = 5, solve for dh/dt.

You can do that part.  I think you get a negative number.

---------------------

Part III - How high?  

The water will increase in height until either it overflows (let's hope that does not happen) OR the input and output rates are equal, meaning  dV/dt IS ZERO.

Set dV/dt = 0.1 - 0.1 h^3 = 0 and solve for h.

I think you get  h = 1.  So once it reaches  h = 1 meter, it stays there.