A filter filled with liquid is in the shape of a vertex-down cone with a height of 12 inches and a diameter of 18 inches at its open (upper) end. If the liquid drips out the bottom of the filter at the constant rate of 3 cubic inches per second, how fast is the level of the liquid dropping when the liquid is 2 inches deep?
Swagger
Oh, the liquid is probably desperate to find an exit and escape from its cone-straints! Let's calculate the rate at which it's getting out. We know the volume of the liquid that's leaving per second is 3 cubic inches. And since the shape of the filter is a cone, we can use its formula for volume:
V = (1/3)πr^2h
Where V is the volume, π is approximately 3.14159, r is the radius of the base of the cone, and h is the height of the liquid. In this case, we want to find the rate of change of the height, so let's differentiate the formula with respect to time:
dV/dt = (1/3)π(2r)(dh/dt)
Where dV/dt represents the rate of change of volume, and dh/dt represents the rate of change of height.
Now let's solve for dh/dt. We know the radius of the cone's base is half of its diameter, so r = 9 inches. And if the liquid is 2 inches deep, we can plug in these values:
3 = (1/3)π(2(9))(dh/dt)
3 = (6/3)π(dh/dt)
3 = 2π(dh/dt)
dh/dt = 3/(2π)
So, the rate at which the level of the liquid is dropping when it's 2 inches deep is approximately 3/(2π) inches per second. Keep a close eye on that liquid as it tries to find its way out!
To find the rate at which the level of the liquid is dropping, we can use related rates. Let's denote the rate at which the level of the liquid is dropping as dh/dt (in cubic inches per second).
Given information:
- The filter is in the shape of a vertex-down cone.
- The height of the cone is 12 inches.
- The diameter of the cone at its open (upper) end is 18 inches.
- The liquid drips out at a constant rate of 3 cubic inches per second.
We need to find dh/dt when the liquid is 2 inches deep.
To start, let's find an equation relating the height and radius of the cone.
Since the diameter at the open end is 18 inches and the radius is half the diameter, the radius of the cone can be found using the formula:
r = 18 / 2
r = 9 inches
The volume of a cone can be calculated using the formula:
V = (1/3) * π * r^2 * h
Differentiating both sides of the equation with respect to time (t), we get:
dV/dt = (1/3) * π * (2r * dr/dt) * h + (1/3) * π * r^2 * dh/dt
We know that dV/dt is the rate at which the liquid is flowing out, which is -3 cubic inches per second (negative sign indicates a decrease in volume).
Plugging in the values:
-3 = (1/3) * π * (2 * 9 * dr/dt) * 2 + (1/3) * π * 9^2 * dh/dt
Simplifying the equation:
-3 = 6π * dr/dt + 27π * dh/dt
We want to find dh/dt when the liquid is 2 inches deep, so we substitute h = 2 into the equation:
-3 = 6π * dr/dt + 27π * dh/dt
-3 = 6π * dr/dt + 27π * (dh/dt) at h = 2
Now we can solve for dh/dt:
dh/dt = (-3 - 6π * dr/dt) / (27π) at h = 2
To find dr/dt at h = 2, we can use similar triangles. We have a large cone with a height of 12 inches and a small cone with a height of 2 inches. The corresponding radii are 9 inches and R (unknown), respectively.
Using the similar triangles:
R / 2 = 9 / 12
R = (9 * 2) / 12
R = 3/2 inches
Now, we can differentiate the equation relating the height and radius of the cone with respect to time (t) to find dr/dt:
d(R^2)/dt = d((9/12) * h^2)/dt
2R * dR/dt = (9/12) * 2h * dh/dt
Substituting the values:
2 * (3/2) * dR/dt = (9/12) * 2 * 2 * dh/dt (since h = 2 when the liquid is 2 inches deep)
Simplifying:
3 * dR/dt = 3 * 2 * dh/dt
dR/dt = 2 * dh/dt
Substituting this value back into the earlier equation:
dh/dt = (-3 - 6π * (2 * dh/dt)) / (27π) at h = 2
Solving for dh/dt:
dh/dt = (-3 - 12π * dh/dt) / (27π) at h = 2
Multiplying through by 27π:
27π * dh/dt = -3 - 12π * dh/dt
Simplifying:
(27π + 12π) * dh/dt = -3
39π * dh/dt = -3
Finally, solving for dh/dt:
dh/dt = -3 / (39π)
Therefore, when the liquid is 2 inches deep, the rate at which the level of the liquid is dropping is approximately:
dh/dt ≈ -0.024 cubic inches per second
To solve this problem, we need to use the concept of related rates. We are given that the liquid is dripping out of the filter at a constant rate of 3 cubic inches per second. We need to find the rate at which the level of the liquid is dropping when the liquid is 2 inches deep.
Let's consider the volume of the liquid in the cone at a certain time. The volume V of a cone is given by the formula:
V = (1/3) * π * r^2 * h,
where r is the radius of the base of the cone and h is the height of the cone.
We are given that the diameter of the cone's open end is 18 inches, so the radius is half of that, which is 9 inches. The height of the cone is given as 12 inches.
Substituting the values into the formula, we have:
V = (1/3) * π * 9^2 * h.
The volume of the liquid is decreasing over time as it drips out at a rate of 3 cubic inches per second. Therefore, the rate at which the liquid volume is changing with respect to time is -3 cubic inches per second.
Now, let's find an expression for dh/dt, the rate at which the height of the liquid is changing with respect to time, when the liquid is 2 inches deep.
We know that V = (1/3) * π * 9^2 * h, so we can rewrite it as:
3V/π * 1/9^2 = h.
Differentiating both sides with respect to time, we get:
3 * dV/dt * 1/π * 1/9^2 = dh/dt.
We know that dV/dt = -3 since the volume is decreasing at a rate of 3 cubic inches per second.
Substituting the values, we have:
3 * (-3) * 1/π * 1/9^2 = dh/dt.
Simplifying the equation, we get:
-9/π * 1/81 = dh/dt.
Therefore, the rate at which the height of the liquid is dropping when the liquid is 2 inches deep is approximately -0.035 inches per second.
r=9
h=12
so, r = 3/4 h
v = 1/3 pi r^2 h
= 1/3 pi (3/4 h)^2 h
= 3/16 pi h^3
dv/dt = 9/16 pi h^2 dh/dt
so, just solve
-3 = 9/16 pi 2^2 dh/dt
for dh/dt
-3 = 4/3 pi (3/2)^2