A container in the form of a right circular cone of height 16 cm and base radius 4 cm is held vertex downward and filled with liquid. If the liquid leaks out from the vertex at a rate of 4 cm^3/s, find the rate of change of the depth of the liquid in the cone when half of the liquid has leaked out.

To solve this problem, we will need to use similar triangles and the volume of a cone formula.

Let's start by finding the volume of the cone when it is full. The volume of a cone is given by the formula:

V = (1/3)πr^2h

where V is the volume, r is the base radius, and h is the height.

Plugging in the given values, we have:

V_full = (1/3)π(4^2)(16)
= (1/3)π(16)(16)
= (1/3)π(256)
= 256π/3 cm^3

Now, let's find the rate of change of the depth of the liquid in the cone when half of the liquid has leaked out. Since the liquid leaks out of the vertex at a rate of 4 cm^3/s, the rate of change of the volume of the liquid can be written as -4 cm^3/s (negative because the volume is decreasing).

Let's assume the depth of the liquid in the cone when half of the liquid has leaked out is h_half.

By similar triangles, we know that the ratio of the remaining volume to the full volume is equal to the square of the ratio of the remaining height to the full height:

(V_liquid_remaining) / (V_full) = (h_remaining) ^ 2 / (h_full) ^ 2

Substituting in the given and assumed values, we have:

(V_liquid_remaining) / (256π/3) = (h_remaining)^2 / (16)^2

Simplifying, we get:

V_liquid_remaining = (h_remaining)^2 * (256π/3) / (16)^2

We can differentiate both sides of this equation with respect to time t to find the rate of change of V_liquid_remaining with respect to t:

d/dt(V_liquid_remaining) = d/dt[(h_remaining)^2 * (256π/3) / (16)^2]

Since we want to find the rate of change of the depth of the liquid in the cone, we can rewrite d/dt(V_liquid_remaining) as dh_remaining/dt:

dh_remaining/dt = d/dt[(h_remaining)^2 * (256π/3) / (16)^2]

Expand and differentiate each term separately:

dh_remaining/dt = (2h_remaining) * (256π/3) / (16)^2 * (dh_remaining/dt)

Now, let's substitute the given values:

V_liquid_remaining = (16/2)^2 * (256π/3) / (16)^2

Simplifying, we get:

V_liquid_remaining = 512π/3 cm^3

Substituting this value back into the equation and solving for dh_remaining/dt, we have:

dh_remaining/dt = (2h_remaining) * (256π/3) / (16)^2 * (-4)

dh_remaining/dt = -8h_remainingπ/12

Finally, when half of the liquid has leaked out, the remaining volume is half of the full volume:

512π/3 = (256π/3) / (16)^2 * (h_remaining)^2

Simplifying, we have:

(h_remaining)^2 = (16)^2 / 2
h_remaining = 8√2 cm

Now, let's substitute this value back into the previous equation to find dh_remaining/dt:

dh_remaining/dt = -8(8√2)π/12

Simplifying, we have:

dh_remaining/dt = -64√2π/12

Therefore, the rate of change of the depth of the liquid in the cone when half of the liquid has leaked out is approximately -5.31 cm/s.

To find the rate of change of the depth of the liquid in the cone when half of the liquid has leaked out, we need to use related rates.

Let's start by finding the volume of the cone. 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 equal to 3.14159, r is the base radius, and h is the height.

In this case, the height of the cone is 16 cm, and the base radius is 4 cm. Substituting these values into the formula, we get:

V = (1/3) * π * (4^2) * 16,
V = (1/3) * π * 64 * 16,
V = (1/3) * π * 1024.

Now, let's find the total volume of the liquid when half of it has leaked out. Since half of the liquid has leaked out, the remaining volume is half of the initial volume:

V_remaining = (1/2) * V,
V_remaining = (1/2) * (1/3) * π * 1024,
V_remaining = (1/6) * π * 1024.

Next, let's find the rate of change of the volume with respect to time. The volume is decreasing by 4 cm^3/s, so the rate of change of the volume is -4 cm^3/s.

Now, we can differentiate the volume equation with respect to time (t) to find the rate of change of the volume (dV/dt) with respect to time:

dV/dt = (1/6) * π * d(1024)/dt,

Since the radius and height of the cone are constant with respect to time, their partial derivatives are zero. Therefore, we only need to find d(1024)/dt:

d(1024)/dt = 0,

Thus, dV/dt = (1/6) * π * 0 = 0.

Since the rate of change of the volume is zero, it implies that the rate of change of the depth of the liquid in the cone is also zero. Therefore, the rate of change of the depth of the liquid in the cone when half of the liquid has leaked out is 0 cm/s.

When the water is x cm deep, and the surface is a circle of radius r

r/x = 4/16
r = x/4

v = pi/3 r^2 x = pi/3 x^2/16 x = pi/48 x^3

Now, what is x when half the liquid is gone?

pi/3 r^2 x = pi/3 * 16 * 16 / 2
x^3/4 = 128
x^3 = 2^11 = 2^9 * 4
x = 8 cbrt(4)

dv/dt = pi/16 x^2 dx/dt
4 = pi/16 * 64 cbrt(16) dx/dt
dx/dt = 64/pi / (128cbrt(2)) = .126 cm/s

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