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 5 cubic inches per second, how fast is the level of the liquid dropping when the liquid is 2 inches deep?
Well, well, well, looks like we have a leaky situation here! Let's dive into this conical filter problem.
First, let's find the volume of the liquid in the filter when it's 2 inches deep. We can use the formula for the volume of a cone:
V = (1/3) * π * r^2 * h
where V is the volume, π is approximately 3.14159, r is the radius, and h is the height.
Given that the diameter is 18 inches at the open end, the radius (r) would be half of that, which is 9 inches.
Now, let's plug in the values into the formula:
V = (1/3) * π * (9^2) * 2
V ≈ 1/3 * 3.14159 * 81 * 2
V ≈ 169.646 in³
Alrighty, now we know the volume of the liquid. Let's find the rate at which it's dropping. We're given that the liquid is leaking at a constant rate of 5 cubic inches per second. So:
dV/dt = -5 in³/s
Let's find dh/dt, which is the rate at which the height (level) of the liquid is dropping when the liquid is 2 inches deep.
We can use a little trick here. Since we know the volume of the liquid, we can now establish a relationship between the height and the volume of this cone:
V = (1/3) * π * r^2 * h
169.646 = (1/3) * π * (9^2) * h
h ≈ 1.268 inches
Using similar triangles, we can establish a relationship between the radius and the height:
r/h = 9/12
So, r = (3/4)h
Now, let's differentiate the equation V = (1/3) * π * r^2 * h with respect to t:
dV/dt = (1/3) * π * 2r * dr/dt * h + (1/3) * π * r^2 * dh/dt
-5 = (1/3) * π * 2(3/4h) * dr/dt * h + (1/3) * π * (3/4h)^2 * dh/dt
Phew, that was a mouthful! Let's simplify this equation a bit:
-5 = (1/3) * π * (3/2) * dr/dt * h^2 + (1/3) * π * (9/16) * h^2 * dh/dt
Since we're interested in finding dh/dt when h = 2, let's plug in the values:
-5 = (1/3) * π * (3/2) * dr/dt * 2^2 + (1/3) * π * (9/16) * 2^2 * dh/dt
-5 = (3/2) * π * dr/dt + (9/8) * π * dh/dt
Now, we can solve for dh/dt:
(9/8) * π * dh/dt = -5 - (3/2) * π * dr/dt
dh/dt = (-5 - (3/2) * π * dr/dt) / ((9/8) * π)
Let's plug in the given rate dr/dt = -5 in³/s:
dh/dt = (-5 - (3/2) * π * (-5)) / ((9/8) * π)
dh/dt = (-5 + (15/2) * π) / ((9/8) * π)
dh/dt ≈ -0.715 inches per second
So, the height of the liquid is dropping at a rate of approximately 0.715 inches per second when the liquid is 2 inches deep. Let's hope this clown show helped you understand the problem, and please don't take this seriously - it's just a joke!
To find the rate at which the level of the liquid is dropping when the liquid is 2 inches deep, we need to find the rate of change of the volume of the liquid with respect to time.
Given:
- Height of the cone (h) = 12 inches
- Diameter of the cone at the open end (d) = 18 inches
- Rate at which the liquid is dripping out (dr/dt) = 5 cubic inches per second
- Depth of the liquid (y) = 2 inches
First, let's find an expression for the volume of the liquid in terms of the depth y.
The volume of a cone can be calculated using the formula:
V = (1/3) * π * r^2 * h,
where V is the volume, π is a constant (approximately 3.14159), r is the radius of the cone, and h is the height of the cone.
Since the filter is initially vertex-down, we need to consider the inverted cone formed by the liquid. So, the height of the liquid in terms of y would be h - y.
Substituting the given values into the formula, we get:
V = (1/3) * π * [(18/2)^2] * (12 - y),
or simplifying:
V = (1/3) * π * 9^2 * (12 - y),
V = (1/3) * π * 81 * (12 - y),
V = 27π * (12 - y).
Now, we can find the rate of change of volume with respect to time by differentiating V with respect to t (time):
dV/dt = d/dt (27π * (12 - y)),
dV/dt = -27π * (dy/dt).
The negative sign arises because y and V are inversely proportional.
We are given the rate at which the liquid is dripping out, dr/dt = 5 cubic inches per second.
Setting dV/dt equal to dr/dt and solving for dy/dt (which represents the rate at which the liquid level is dropping), we get:
-27π * (dy/dt) = 5,
dy/dt = -5 / (-27π),
dy/dt = 5 / (27π).
Approximating the value, we have:
dy/dt ≈ 0.058 cubic inches per second.
Therefore, the rate at which the level of the liquid is dropping when the liquid is 2 inches deep is approximately 0.058 cubic inches per second.
To solve this problem, we need to find the rate at which the level of the liquid is dropping when it is 2 inches deep. We can do this by applying the concept of related rates.
Let's denote the depth of the liquid in the cone as "h" (in inches) and the radius of the circular cross-section at height "h" as "r" (in inches). We know that the diameter of the cone at its open end is 18 inches, so the radius will be half that, or 9 inches.
Since the liquid is dripping out at a constant rate of 5 cubic inches per second, we can determine the rate at which the volume of the liquid is decreasing with respect to time. The volume of a cone is given by the formula V = (1/3)πr^2h. Taking the derivative of both sides with respect to time (t), we get dV/dt = (1/3)π(2rh(dr/dt) + r^2(dh/dt)).
Now, we need to find the values of r and dr/dt (the rate at which the radius is changing) in terms of h and dh/dt (the rate at which the depth is changing). Since the cone is vertex-down, we can use similar triangles to find r in terms of h. The depth of the liquid and the height of the cone are in a proportion, so h/12 = r/9. Solving for r, we get r = (9h)/12 = (3h)/4.
Next, let's find dr/dt in terms of h and dh/dt. Taking the derivative of both sides of the equation h/12 = r/9 with respect to time, we get (1/12)(dh/dt) = (1/9)(r(dh/dr)(dr/dt)). Rearranging and substituting the value of r, we have dh/dt = (9/4)(dr/dt).
Now we can substitute the values of r and dh/dt in terms of h into the expression for dV/dt that we derived earlier. Using r = (3h)/4 and dh/dt = (9/4)(dr/dt), we get dV/dt = (1/3)π[(2(3h)/4((9/4)(dr/dt))(dr/dt) + ((3h)/4)^2((9/4)(dr/dt))]. Simplifying this, we have dV/dt = (1/3)π[(27/8)(dr/dt)(2h + (81/16)h^2)].
Now, we know that dV/dt = -5 (since the liquid is draining out) and we need to find dh/dt when h = 2. Substituting these values into the equation, we have -5 = (1/3)π[(27/8)(dr/dt)(4 + (81/16)(2)^2)]. Simplifying this equation, we get -5 = (1/3)π[(27/8)(dr/dt)(4 + (81/16)(4))]. Solving for dr/dt, we have dr/dt = -15/(π(27/8)(5 + 81/4)).
To find the value of dh/dt, we can substitute dr/dt = -15/(π(27/8)(5 + 81/4)) and h = 2 into dh/dt = (9/4)(dr/dt). Evaluating this expression, we can find the rate at which the level of the liquid is dropping when it is 2 inches deep.
Please note that this calculation will require mathematical operations that are best performed using a calculator or a software tool such as Python or MATLAB.
at a time of t seconds ...
let the radius of the circular water level be r
let the height of the water be h
12/9 = h/r
12r = 9h
4r = 3h ----> r = 3h/4
V = (1/3)π r^2 h
= (1/3)π (9h^2/4)(h)
= (3/4)π h^3
dV/dt = (9/4) h^2 dh/dt
5 = (9/4)π (4) dh/dt
solve for dh/dt