A water tank has the shape of an inverted right circular cone of altitude 12 ft and
base radius of 6 ft. There is a hole in the bottom that is leaking at a rate of 0.3
gal/min. If water is being pumped into the tank at a rate of 10 gal/min,
a) Approximate the rate at which the water level is rising when the water is
3 ft deep. (1 gallon 1337 ft 2 ).
b) How long will it take for the tank to fill?
a) To find the rate at which the water level is rising when the water is 3 ft deep, we can use the concept of similar triangles and the formula for the volume of a cone.
First, we need to find the radius of the water level when it is 3 ft deep. We can use the ratio of the similar triangles: (radius of water level) / (radius of cone) = (height of water level) / (height of cone).
Let's assume the radius of the water level when it is 3 ft deep is r ft. So, we have (r/6) = (3/12).
Simplifying the equation, we get r = 1.5 ft.
Now, we can find the rate at which the water level is rising by taking the derivative of the volume with respect to time. The volume of a cone is given by V = (1/3) * π * r^2 * h, where r is the radius of the water level and h is the height of the water level. Since the height of the water level is changing with time, we can write h as a function of time: h(t) = 12 - t, where t is the time in minutes.
Now, let's differentiate the volume equation with respect to time using the chain rule:
dV/dt = (1/3) * π * (2r)(dr/dt) * h + (1/3) * π * r^2 * dh/dt
Since we are given that the water is being pumped into the tank at a rate of 10 gal/min and the leak is at a rate of 0.3 gal/min, the change in height of the water level with respect to time is dh/dt = 10 - 0.3 = 9.7 ft/min.
Substituting the values r = 1.5 ft, h = 12 - t, and dh/dt = 9.7 ft/min into the equation, we can calculate the rate at which the water level is rising:
dV/dt = (1/3) * π * (2*1.5)(dr/dt) * (12 - t) + (1/3) * π * (1.5)^2 * 9.7
Simplifying the equation, we get:
dV/dt = 15.3π - 1.5πt
Since the rate at which the water level is rising is given by dV/dt, when the water is 3 ft deep (t = 9), the rate would be:
dV/dt = 15.3π - 1.5π(9) = 15.3π - 13.5π = 1.8π ≈ 5.652 ft^3/min
Since 1 gallon is equal to 1337 ft^3, the rate in gallons would be approximately 5.652/1337 ≈ 0.004238 gal/min.
So, the approximate rate at which the water level is rising when the water is 3 ft deep is about 0.004238 gal/min.
b) To find how long it will take for the tank to fill, we need to determine the time it takes for the water level to reach the full height of 12 ft.
We know that dh/dt = 9.7 ft/min. Since the water level is increasing at a constant rate, we can use the equation h(t) = h0 + (dh/dt) * t, where h0 is the initial height of the water level and t is the time in minutes.
Setting h(t) = 12 ft and solving for t, we get:
12 = 12 - t * 9.7
t * 9.7 = 0
t = 0
Since the equation yields t = 0, it means that it will take 0 minutes for the tank to fill.
Therefore, the tank will fill instantly.