Related Rates Calculus

Question

Water is leaking out of an inverted conical tank at a rate of 88000 cubic centimeters per min at the same time that water is being pumped into the tank at a constant rate. The tank has height 60 meters and the diameter at the top is 50 meters. If the water level is rising at a rate of 250 centimeters per minute when the height of the water is 20 meters, find the rate at which water is being pumped into the tank in cubic centimeters per minute

Answer 1

Place the origin at the bottom of the inverted cone.  Coordinate x is the radius of the present volume at any time t, and the height of the water is coordinate y.

 

By similar triangles x/y = 25/60 (when tank is at full capacity) so x = (5/12)y.

 

V = pi/3(x²y) = (pi/3)[(5/12)y]²y = 25(pi)y³/432

 

dV/dt = 25(pi)y²/144•dy/dt = C - 88,000

 

When y = 20 m & dy/dt =2.5 meters/min, we substitute these numbers to solve for C.

Answer 2

Hi Lilieth,

 

First, the total tank is an inverted cylinder of height H = 60 m and a diameter of 50 m, giving it a radius of R = 25 m.

 

At any time in the future when the tank is not full, the water in the tank forms a smaller cone of radius r < R and height h < H.

 

The volume of the water in the cone with radius r and height h is

 

V = (1/3)πr²h    (1)

 

We would like to find the rate of change of the volume by taking the derivative of this last equation, but, we have a problem. We have too many variables. We have r and h which are both functions of time. Since our final answer has to deal with height, we need to somehow eliminate r.

 

We can do this by considering similar triangles in the cone. The ratio of the radius of the liquid r to the height h is the same as the ratio of the total tank radius R and the total tank height H. In equation form,

 

r/R = h/H  (2)

 

We can solve Eq. (2) for r:

 

r = (R/H)h  (3)

 

Substitute Eq. (3) into Eq. (1) to eliminate r.

 

V = (1/3)π(R/H)²h²h, or

 

V = (1/3)π(R/H)²h³          (4)

 

Equation (4) is solely a function of h. Take the derivative of Eq. (4) with respect to time.

 

dV/dt = π(R/H)²h²dh/dt,   (5)

 

where we note that the 3 from the cube of h canceled the 1/3 in the derivative process, and the dh/dt came from the chain rule.

 

We know that dV/dt is having water pumped in at a constant rate C, which is not specified, and the tank is leaking at a constant rate of 88,000 cm³/min. So,

 

dV/dt = C - 88,000 cm³/min   (6)

 

Notice that R is positive since this represents water being added to the tank, and the 88,000 is negative since it represents water leaving the tank.

 

So, combining Eqs (5) and (6), we have

 

C - 88,000 cm³/min = π(R/H)²h²dh/dt    (7)

 

We want to solve Eq. (7) for the rate at which water is being pumped into the tank, C:

 

C = π(R/H)²h²dh/dt + 88,000 cm³/min    (8)

 

We are told that when h = 20 m = 2,000 cm, dh/dt = +250 cm/min, where the + sign indicates that the water is rising. We can now use Eq. (8) and solve for C.

 

C = π(25 m / 60 m)² (2000 cm)² (250 cm/min) + 88,000 cm³/min = 5.455 X 10⁸ cm³ / min