A conical water tank with vertex down has a radius of 10 feet at the top and is 22 feet high. If water flows into the tank at a rate of 30 {\rm ft}^3{\rm /min}, how fast is the depth of the water increasing when the water is 14 feet deep?

I have tried several times but continue to mess up somewhere. Please help!

at a time of t min, height of water level = h, radius of water level = r

V = (1/3)πr^2 h
by ratios:
r/h = 10/22 = 5/11
11r = 5h
r = (5h/11)
V = (1/3)π(25h^2/121)h
= (25/363)π h^3
dV/dt = (25/121)π h^2 dh/dt

given: dV/dt = 30 ft^3/min
find: dh/dt when h = 14

30 = (25/121)π(196) dh/dt
dh/dt = appr .38197 ft/min

To solve this problem, we can use the concept of related rates and the formula for the volume of a cone.

The volume of a cone is given by the formula:

V = (1/3) * π * r^2 * h,

where V is the volume, r is the radius, h is the height, and π is approximately 3.14159.

In this problem, the water depth is increasing with time, so we are interested in finding the rate at which the depth (h) is changing (dh/dt) when the water is 14 feet deep.

First, let's write down the known values:
- Radius of the top of the cone (r) = 10 feet
- Height of the cone (H) = 22 feet
- Rate of change of volume (dV/dt) = 30 ft^3/min
- Depth of the water (h) = 14 feet

To find the rate at which the depth is increasing (dh/dt), let's differentiate the volume formula with respect to time:

dV/dt = (1/3) * π * 2 * r * dr/dt * h + (1/3) * π * r^2 * dh/dt.

Now substitute the given values:

30 = (1/3) * π * 2 * 10 * dr/dt * 14 + (1/3) * π * 10^2 * dh/dt.

Simplifying, we get:

30 = 280/3 * π * dr/dt + 100/3 * π * dh/dt.

Now, solve for dh/dt:

100/3 * π * dh/dt = 30 - 280/3 * π * dr/dt.

dh/dt = [30 - 280/3 * π * dr/dt] / (100/3 * π).

Now we need to find dr/dt, the rate at which the radius is changing. Since the water level is changing and the vertex of the cone is down, we can consider the similar triangles formed by the water level and the full cone.

Let's denote x as the distance from the top of the cone to the water level. Then, we can find the relationship between x, r, and H using similar triangles:

x / h = r / R,

where R is the radius of the bottom of the cone.

Substituting the given values, we have:

x / 14 = r / 10.

Simplifying, we get:

r = (10x) / 14.

Differentiating both sides with respect to time, we get:

dr/dt = (10/14) * dx/dt.

Now, we need to find dx/dt, the rate at which the water level is changing.
Given that the rate at which the depth of water is increasing (dh/dt) is the same as the rate at which the water level is increasing (dx/dt), we can substitute dh/dt = dx/dt into the equation:

dh/dt = [30 - 280/3 * π * dr/dt] / (100/3 * π).

Substitute dr/dt = (10/14) * dx/dt:

dx/dt = (14/10) * dh/dt = (7/5) * dh/dt.

Now, substitute dh/dt = dx/dt:

(7/5) * dh/dt = [30 - 280/3 * π * (10/14) * dx/dt] / (100/3 * π).

Simplifying, we get:

(7/5) * dh/dt = [30 - 20 * π * dx/dt] / (30/π).

Multiply both sides by 5/7 to isolate dx/dt:

dx/dt = 5/7 * [30 - 20 * π * dx/dt] / (30/π).

Now, we can solve for dx/dt:

7 * dx/dt = 5 * [30 - 20 * π * dx/dt] / (30/π).

Multiply both sides by (30/π) to isolate dx/dt:

7 * (30/π) * dx/dt = 5 * (30 - 20 * π * dx/dt).

Divide both sides by (7 * (30/π)):

dx/dt = 5 * (30 - 20 * π * dx/dt) / (7 * (30/π)).

Simplifying further, we get:

dx/dt = 150 / (7 - 20 * π).

Now, substitute the value of π (approximately 3.14159) to find the numerical value of dx/dt:

dx/dt = 150 / (7 - 20 * 3.14159).
dx/dt = 150 / (7 - 62.8318).
dx/dt = 150 / (-55.8318).
dx/dt ≈ -2.683 ft/min.

Therefore, the depth of the water is decreasing at a rate of approximately 2.683 ft/min when the water is 14 feet deep.

To find how fast the depth of the water is increasing, we need to find the rate of change of the depth of the water with respect to time.

Let's denote the height of the water as h feet and the radius of the water at height h as r feet.

We are given that the tank is conical with its vertex down, meaning that it forms a cone. The volume V of a cone can be calculated using the formula:

V = (1/3) * π * r^2 * h

We can rearrange this formula to solve for r^2:

r^2 = (3V) / (π * h)

Now, we can differentiate both sides of this equation with respect to time (t):

d(r^2)/dt = d((3V) / (π * h)) / dt

Using the chain rule, we can calculate each derivative separately. Let's do that now:

d(r^2)/dt = (d(3V/dt)/dt * (1/ (π * h))) - ((3V) / (π * h^2) * (dh/dt))

We are given that dV/dt = 30 ft^3/min (the rate at which water flows into the tank) and we need to find dh/dt (rate of change of the depth of the water with respect to time) when h = 14 ft.

Substituting the given values, we have:

d(r^2)/dt = (30 * (1/(π * 14))) - ((3V) / (π * 14^2) * (dh/dt))

Now, we need to find the value of V in terms of h:

V = (1/3) * π * 10^2 * h (since the radius at the top is 10 ft)

V = (1000/3) * π * h

Substituting this into the equation, we have:

d(r^2)/dt = (30 * (1/(π * 14))) - ((3(1000/3) * π * h) / (π * 14^2) * (dh/dt))

Simplifying, we get:

d(r^2)/dt = (30/14) - ((10 * h) / 196) * (dh/dt)

We are given that r = 10 ft when h = 14 ft, so we can substitute these values:

10^2 = (3V) / (π * 14)

100 = (3000/π) * (1/14)

Solving for π, we get:

π = (3000/1400)

Now we can substitute this value into the equation for d(r^2)/dt:

d(r^2)/dt = (30/14) - ((10 * h) / 196) * (dh/dt)

Finally, we can substitute the given values into this equation:

d(r^2)/dt = (30/14) - ((10 * 14) / 196) * (dh/dt)

Now we can solve for dh/dt:

dh/dt = ((30/14) - ((10 * 14) / 196)) / ((10 * 14) / 196)

dh/dt = (15/7) - (1/14)

dh/dt = 2 - (1/14)

dh/dt = 27/14 ft/min

Therefore, the depth of the water is increasing at a rate of 27/14 ft/min when the water is 14 ft deep.