Water is leaking out of an inverted conical tank at a rate of 12,000 cm3/min at the same time that water is being pumped into the tank at a constant rate. The tank has height 6 m and the diameter at the top is 4 m. If the water level is rising at a rate of 20 cm/min when the height of the water is 2 m, find the rate at which water is being pumped into the tank.
To solve this problem, we need to use related rates. Let's break down the problem into smaller steps:
Step 1: Find the rate at which the water is leaving the tank.
Given: The rate at which water is leaking out is 12,000 cm3/min.
The volume of a cone is given by the formula:
V = (1/3)πr^2h,
where V is the volume, π is the constant pi (approximately 3.14), r is the radius of the base of the cone, and h is the height of the cone.
In this case, we need to find the rate at which the water is leaving the tank, so we take the derivative of the volume with respect to time (t):
dV/dt = (1/3)π(2rh) * (dr/dt) + (1/3)π(r^2) * (dh/dt).
Here, dr/dt represents the rate at which the radius changes, and dh/dt represents the rate at which the height changes.
We are given that (dh/dt) = 20 cm/min (the water level is rising at a rate of 20 cm/min), and we need to find (dr/dt).
Step 2: Find the rate at which the radius of the water's surface is changing.
Given: The height of the water is 2 m.
Since the height of the cone is 6 m and the water level is at 2 m, the remaining height of the cone filled with air is 4 m.
To find the radius of the water's surface, we can use similar triangles:
h/r = (h + 4)/(4),
where h is the water's height and r is the radius of the water's surface.
Plugging in the given value, we have:
2/r = (2 + 4)/(4),
2/r = 6/4.
Solving for r, we find:
r = (4/6) * 2,
r = 8/6 m.
We need to find the rate at which the radius changes, so let's differentiate with respect to time (t):
d(2/r)/dt = d(6/4)/dt.
After simplifying, we get:
-2/dr/dt = 0,
dr/dt = 0.
So, the radius of the water's surface is not changing; it is constant.
Step 3: Substitute the values into the equation from Step 1 to find the rate at which water is being pumped into the tank.
Given: (dr/dt) = 0 (the rate at which the radius changes is 0).
Let's substitute this value into the equation from Step 1:
12,000 = (1/3)π(2r * 2) * 0 + (1/3)π(r^2) * 20.
Simplifying further, we have:
12,000 = 0 + (1/3)π(r^2) * 20,
12,000 = (2/3)π(r^2) * 20.
Dividing both sides by (2/3)π and 20, we get:
r^2 = 12,000 * (3/2) * (1/π) * (1/20),
r^2 = 180,
r = √180,
r ≈ 13.416 m.
So, the rate at which water is being pumped into the tank is the rate at which the volume increases. Let's substitute the values into the volume formula:
V = (1/3)π(13.416^2)(2) * dV/dt,
12,000 = (1/3)π(13.416^2)(2) * dV/dt.
Simplifying further, we have:
12,000 = 1.43 * dV/dt.
Finally, we can solve for dV/dt, which represents the rate at which water is being pumped into the tank:
dV/dt = 12,000 / 1.43,
dV/dt ≈ 8,391 cm3/min.
So, the rate at which water is being pumped into the tank is approximately 8,391 cm3/min.