An inverted conical tank (with vertex down) is 14 feet across the top and 24 feet deep. If water is flowing in at a rate of 12 ft3/min, find the rate of change of the depth of the water when the water is 10 feet deep.

sure

change in volume = surface area * change in depth
or
dV = pi r^2 dh
dV/dt = pi r^2 dh/dt = 12 ft^3/min
so
dh/dt = 6/(pi r^2)

if r = 7, h = 24
or
r = (7/24)h = (7/24)(10) = 70/24
so
dh/dt = 6 / [ pi (70/24)^2 ]

Hmmm. I get

r = 7/24 h
so, dr = 7/24 dh
r = 35/12 when h=10

v = 1/3 pi r^2 h
using the product rule,
dv = 1/3 pi (2rh dr + r^2 dh)
12 = pi/3 (2*(35/12)(10)(7/24 dh)+(35/12)^2 dh)
12 = 1225/144 pi dh
dh = 1728/1225 pi
= 12^3/35^2 pi

Maybe you can figure out whether one of us is correct.

Do you know what kind of formula I could use to solve this?

To find the rate of change of the depth of the water, we can use related rates, which involves finding the derivative of the variable with respect to time.

Let's start by assigning variables:
Let r be the radius of the water level in the tank (which will decrease as the water level decreases), and h be the depth of the water.

Given:
The inverted conical tank has a top diameter of 14 feet, which means the radius at the top is 7 feet (14/2 = 7).
The depth of the tank is 24 feet.
The rate of water flowing in is 12 ft3/min.

To find the rate of change of the depth (dh/dt), we need to find a relationship between the values.

The formula for the volume of a cone is V = (1/3)πr^2h.
Since the vertex of the cone is pointed down, the height of the cone is negative, so the expression for the volume in terms of r and h is:
V = -(1/3)πr^2h.

Now we can take the derivative of both sides with respect to time (t):
dV/dt = d(-(1/3)πr^2h)/dt.

We are given that dV/dt = 12 ft3/min, and we want to find dh/dt when h = 10 ft. We need to find dr/dt.

To do that, we can use similar triangles since the shape of the cone remains constant.
The ratio of the radius to the height is constant, so we can establish the following relationship:
r/h = 7/24.

To solve for r in terms of h:
r = (7/24)h.

Now, we can substitute this value into the equation for V:
V = -(1/3)π((7/24)h)^2h.

Next, we differentiate both sides with respect to t:
dV/dt = -(1/3)π * d((7/24)h)^2h/dt.

To find dh/dt when h = 10, let's substitute the known values into the equations:

dV/dt = 12 ft3/min
h = 10 ft

Now, let's differentiate the equation:
dV/dt = -(1/3)π * d((7/24)h)^2h/dt.

To differentiate the equation, we need to use the chain rule.
Let's break down the differentiation:

dV/dt = -(1/3)π * d((7/24)h)^2h/dh * dh/dt.

d((7/24)h)^2h/dh = (14/24)h(2h/h + (7/24)2).
d((7/24)h)^2h/dh = (7/12)h(2 + (49/24)).

Simplifying the expression:
d((7/24)h)^2h/dh = (7/12)h(2 + (49/24)).
d((7/24)h)^2h/dh = (7/12)h(24 + 49)/24.
d((7/24)h)^2h/dh = (7/12)h(73)/24.

Now we have:
12 = -(1/3)π * (7/12)h(73)/24 * dh/dt.

Simplifying further:
12 = -(1/3)π * (7/12)h(73)/24 * dh/dt.
12 = -(7/12)π * (73)/24 * dh/dt.

We can solve for dh/dt:
dh/dt = (12 * -24) / (-(7/12)π * 73).

Simplifying even further:
dh/dt = 288 / ((7/12)π * 73).

Now, substitute π as 3.14:
dh/dt = 288 / ((7/12) * 3.14 * 73).

Calculate the value:
dh/dt ≈ 288 / (2.92 * 73).
dh/dt ≈ 1.34 ft/min.

Therefore, the rate of change of the depth of the water when the water is 10 feet deep is approximately 1.34 ft/min.