Calculus HELP!!!!!?

Question

Water is leaking out of an inverted conical tank at a rate of 7800.0 cubic centimeters per min at the same time that water is being pumped into the tank at a constant rate. The tank has height 11.0 meters and the diameter at the top is 6.0 meters. If the water level is rising at a rate of 21.0 centimeters per minute when the height of the water is 2.5 meters, find the rate at which water is being pumped into the tank in cubic centimeters per minute.

7830.669 is incorrect!!!

Answer 1

First of all, you need to determine the volumne equation of the tank, which is inverted cone,

V = (1/3)πr^2 h

Since you know the relationship between the radius (r) and the height (h):

r / h = 3 / 11
r = 3h/11

(note that the unit conversion is not needed above, because both would cancel with each other)

Now, sub the above equation into the volume equation (the reason you do this is to get rid of the variable "r", since when you take the derivative, you can get dh/dt)

V = (π/3) * (3h/11)^2 *h
V = 3πh^3 / 121

Now, you can take the derivative because on the left side, you can get dV/dt (which is what you want to find) and on the right side, you can get dh/dt (which you are given):

dV/dt = (3π/121) * 3h^2 * dh/dt

Since you know the rate of change of the height (dh/dt) is 21 cm/min at height 250 cm (just to convert back to cm for consistency because the rate is in cm^3/min), sub these information into the above equation yields:

dV/dt = (3π/121) * 3 * (250)^2 * 21
dV/dt ~ 306694.74 cm^3/min

Remember, this is NET rate. In other words, (flow in - flow out). Otherwise, if flow out > flow in, the water level will not rise!! So,

dV/dt = flow in - flow out = 306694.74
flow in - 7800 = 306694.74
flow in = 314494.74 cm^3/min

(sorry, I mis-read the height as cm instead of m before!! The above should be the right answer)

Answer 2

Let R be the constant rate the water is being pumped in, expressed in cubic cm per minute. The net flow into the tank is:

(1) (R - 7800) cm^3 / min

The volume of the cone of water at time t is:

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

where r(t) and h(t) are the radius of the surface area of the water, and height of the water at time t.

You are given that the height of the conical tank is 11 m and the radius of the top of the tank = 6/2 = 3m

We have r(t) / h(t) = 3/11

r(t) = (3/11) h(t)

V(t) = (pi/3) (3/11)^2 x h(t)^3 = (3pi/121) h(t)^3

d V(t)/dt = (3pi/121) x 3 h(t)^2 x d h(t)/dt

We are given that when h(t) =2.5m = 250cm, d h(t)/dt = 21 cm/min. So when h(t) = 250cm

d V(t)/dt = (9pi/121) (250cm)^2 x 21cm/min

= 306695 cm^3 / min

So from (1) we have:

(R-7800) cm^3/min = 306695 cm^3/min

So R = 314495 cm^3/min

Answer 3

2.5m / 11.0m = .227273 = height of the water as a fraction of the height of the cone.

.227273 x 6.0 m = 1.3636 = diameter at surface of water

pi x (1.3636 / 2)^2 = 5.8418 m^2 = area of surface of water

net flow rate (in m^3 / min) / area of surface of water (in m^2) =
Instantaneous rate that water level is rising (in m / min)

((x - 7800) cm^3 / min) x (1 m / 100 cm)^3 / (5.8418 m^2) =
21 cm / min x (1 m / 100 cm)

(x - 7800) / (1,000,000 x 5.8418) m / min = (21 / 100) m / min

x / 5,841,800 = 21 / 100 + 7800 / 5,841,800

x = 21 x (5,841,800 / 100) + 7800

x = 1,218,982 cm^3 / min = 1.219 m^3 / min

Not sure I did all the calculations just right.
But if you redo the above on paper, you should catch any oopses.