Expert: Paul Klarreich - 3/31/2008

Question
Could you please help me with this question.  I do not understand how to answer it.  I am currently studdying this type of question but we have only completed examples with conical tanks.  I know that the formula for the volume of the tank is V=1/3Bh where B=area of the base and h=height

A tank has the shape of an inverted pyramid of height 50 ft whose base is a square measuring 40 ft by 40 ft.  The liquid in the tank is being drained at the rate of 8(square root)x ft^3/min, where x represents the depth of the liquid in feet.
a. Write an expression for the volume of the liquid in the tank as a function of x.
b. Let t represent the time in minutes.  Using x and t, write a differential equation that describes this situation.
c. Find a general solution to the differential equation you wrote in part (b).
d. Assume the tank is full at time t=0.  Give an expression for the depth x as a function of t.
e. How long does it take to empty the tank?

Thank you very much for your assistance.

Answer
Questioner:   Aly
Category:  Calculus
Private:  No
 
Subject:  A drained inverted pyramid question
Question:  Could you please help me with this question.  I do not understand how to answer it.  I am currently studdying this type of question but we have only completed examples with conical tanks.  I know that the formula for the volume of the tank is V=1/3Bh where B=area of the base and h=height

A tank has the shape of an inverted pyramid of height 50 ft whose base is a square measuring 40 ft by 40 ft.  The liquid in the tank is being drained at the rate of 8(square root)x ft^3/min, where x represents the depth of the liquid in feet.
a. Write an expression for the volume of the liquid in the tank as a function of x.
b. Let t represent the time in minutes.  Using x and t, write a differential equation that describes this situation.
c. Find a general solution to the differential equation you wrote in part (b).
d. Assume the tank is full at time t=0.  Give an expression for the depth x as a function of t.
e. How long does it take to empty the tank?

Thank you very much for your assistance.
....................................................
Hi, Aly,

In your tank, since the full tank has  side of base = 4/5 height, that will be true of any partially full tank.  You will always have:

s = 4h/5, and you will have:

V = (1/3) s^2 h

V = (1/3) 16h^3/25

V = 16h^3/75

[Using h instead of x.]

Differentiate:

dV/dt = 48h^2/75  dh/dt

dV/dt = 16h^2/25  dh/dt

But dV/dt = 8 sqrt(h), you said.

So your equation is:

8 sqrt(h) = 16h^2/25  dh/dt

sqrt(h) = 2h^2/25  dh/dt

To integrate, separate the variables:

dt = 2h^(3/2)/25 dh

t = 2h^(5/2)/(5/2)25  + C

t = 4h^(5/2)/125  + C

If h = 50 when t = 0, substitute and find C.

0 = 4(50)^(5/2)/125  + C

0 = 4(50)^2 sqrt(50)/125  + C

You can do the rest.