Water is leaking from the bottom of a tank in the shape of an inverted cone having an altitude of 12 feet and a radius of 2 feet. If the water is leaking at the rate of 0.25 cubic feet per minute, how

fast is the water level decreasing when the water is 4 feet deep?

v = 1/3 pi r^2 h

From the shape of the cone, you know that r = h/6
so, dr/dt = 1/6 dh/dt

dv/dt = pi/3 (2rh dr/dt + r^2 dh/dt)
= pi/3(2(h/6)h(1/6 dh/dt) + (h/6)^2 dh/dt)
= pi/18 h^2 dh/dt

Now just solve for dh/dt when
dv/dt = -1/4
h=4

To solve this problem, we can use related rates. Let's call the depth of the water in the tank "h" (in feet) and the radius of the water surface "r" (in feet).

We are given that the rate at which the water is leaking, dh/dt, is 0.25 cubic feet per minute. We need to find the rate at which the water level is decreasing, dh/dt, when the water is 4 feet deep.

First, let's express h and r in terms of each other. Since the shape of the tank is an inverted cone, we can use similar triangles to relate h and r.

From similar triangles, we have:

h / r = (altitude of the cone) / (radius of the base of the cone)

h / r = 12 / 2

h / r = 6

h = 6r

We can differentiate both sides of this equation with respect to time t:

dh/dt = 6(dr/dt)

Next, let's express r in terms of h using the formula for the volume of a cone:

V = (1/3) * π * r^2 * h

Since the radius of the base of the cone is twice the radius of the water surface, the radius at height h is given by:

r = (1/2) * h

Substituting this expression for r in the formula for the volume, we get:

V = (1/3) * π * ((1/2) * h)^2 * h

Simplifying further:

V = (1/3) * π * (1/4) * h^3

V = (1/12) * π * h^3

Now, let's differentiate both sides of this equation with respect to time t:

dV/dt = (1/12) * π * 3h^2 * dh/dt

But dV/dt is the rate at which the volume of water is changing, which is given as 0.25 cubic feet per minute. So we can substitute this value:

0.25 = (1/12) * π * 3h^2 * dh/dt

Simplifying further:

0.25 = (1/4) * π * h^2 * dh/dt

Now we can substitute the value of h = 4 into this equation to find dh/dt:

0.25 = (1/4) * π * (4^2) * dh/dt

0.25 = (1/4) * π * 16 * dh/dt

0.25 = 4π * dh/dt

0.25 / (4π) = dh/dt

0.0625 / π = dh/dt

Therefore, the rate at which the water level is decreasing when the water is 4 feet deep is approximately 0.0625 / π feet per minute.

To solve this problem, we need to use related rates. Related rates problems involve finding the rate at which one quantity is changing with respect to the rate at which another related quantity is changing.

Let's denote the radius of the water in the tank as "r" and the height of the water as "h". We are given that the water is leaking at a rate of 0.25 cubic feet per minute, which means the rate at which the volume of water is changing is -0.25 ft^3/min (negative because the volume is decreasing).

Given that the tank has the shape of an inverted cone, we can use similar triangles to relate the radius and the height of the cone. The height of the cone is given as 12 feet and the radius is given as 2 feet. Since the cone is inverted, the radius of the cone at any point is proportional to the height of the cone at that point.

Using similar triangles, we can set up the following proportion:

r / h = 2 / 12

Simplifying, we get:
r = (2/12) * h

Now, let's find an equation for the volume of water in the cone. The volume V of a cone is given by the formula:

V = (1/3) * pi * r^2 * h

Substituting the expression for r in terms of h that we found earlier:

V = (1/3) * pi * [(2/12) * h]^2 * h

Simplifying further:

V = (1/3) * pi * (4/144) * h^3

V = (1/108) * pi * h^3

To find the rate at which the water level is decreasing, we need to differentiate V with respect to t (time):

dV/dt = (1/108) * pi * 3h^2 * (dh/dt)

We know that dV/dt is -0.25 ft^3/min and we want to find dh/dt when the water is 4 feet deep, so we can substitute these values into the equation:

-0.25 = (1/108) * pi * 3(4^2) * (dh/dt)

Simplifying further:

-0.25 = (1/108) * pi * 3(16) * (dh/dt)

-0.25 = (48/108) * pi * (dh/dt)

Now, we can solve for dh/dt:

dh/dt = -0.25 * (108/48) * (1/pi)

Calculating this expression, we get:

dh/dt ≈ -0.135 ft/min

Therefore, the water level is decreasing at a rate of approximately -0.135 feet per minute when the water is 4 feet deep. The negative sign indicates that the water level is decreasing.