RELATED RATES HELP!!!

Question

Water is leaking out of an inverted conical tank at a rate of 11800.0 cubic centimeters per min at the same time that water is being pumped into the tank at a constant rate. The tank has height 110 meters and the diameter at the top is 55 meters. If the water level is rising at a rate of 220 centimeters per minute when the height of the water is 40 meters, find the rate at which water is being pumped into the tank in cubic centimeters per minute.

Answer 1

Start by listing what you are given. Draw a diagram if it helps to set up the problem. Make sure to convert given information into the same units. (1m = 100cm) Since the answer needs to be cm³·min^-1 you should convert meters to centimeters.

Given:

Rate water leaking = 11,800 cm³min^-1

 

water level is rising at a rate of 220 centimeters per minute
dh/dt = 220 cm·min^-1

when the height of the water is 40 meters
h = 40 m
h = 4,000 cm

total tank height = 110 m = 11,000 cm

tank diameter at top = 55 m
diameter = 5,500 cm
which means the radius, r = 2,750 cm

The tank is conical, so use the equation for volume of a cone. This relates the volume of the tank to the height.

V = (1/3)πr²h

As the tank drains/fills the radius is also changing but we only want to deal with two variables. Find an expression for r in terms of h.

The tank is an inverted cone so the radius meets the height at a right angle, forming a right triangle.

radius at top = 2,750 cm
total tank height = 11,000 cm

ratio: r/h = 2,750/11,000

r/h = 1/4
then
r = h/4

replace r with h/4 in the volume equation

V = (1/3)π(h/4)²h
V = (1/3)π(h²/16)h
V = (1/48)πh³

Now that Volume is in terms of only height, take the derivative with respect to time.

dV/dt = (1/48)π*(3)h² (dh/dt)

dV/dt = (3/48)πh² (dh/dt)

plug in the givens;
dh/dt = 220 cm·min^-1
h = 4,000 cm

dV/dt = (3/48)π(4,000cm)²*(220cm·min^-1)

dV/dt = 691,150,383 cm³·min^-1

dV/dt is the total change in volume. So since we need the rate at which water is being pumped in, we need to go one step further. Use the calculated total volume change (dV/dt) and the given rate of leaking 11,800 cm³·min^-1 to find the rate of incoming water.

dV/dt = (rate incoming water) - (rate leaking water)

 

691,150,383 cm³·min^-1 = (rate incoming water) - 11,800 cm³·min^-1

= 691,162,184 cm³·min^-1

 

* The answer seems really large but if the units are posted correctly in the question then it's a 110 meter tall tank with the rate of incoming water in cubic centimeters per minute. It would be more reasonable to request the rate in cubic meters per minute which would be 691 m³·min^-1