A water tank has the shape of an inverted right circular cone whose base radius is equal to half its height. Water is being pumped in at 2 cubic meters per second. How fast is the water rising when the water is 3 meters deep?

radius = (1/2)h

r = (1/2)h
r = h/2

V = (1/3)πr^2 h
= (1/3)π(h^2/4)h
= (1/12)πh^3
dV/dt = (1/4)πh^2 dh/dt

when h = 3, dV/dt = 2
2 = (1/4)π(9) dh/dt
dh/dt = 8/(9π) m/s

Check my calculations, I did not write them out first.

To solve this problem, we need to apply related rates.

Let's denote the height of the water in the tank as h (in meters) at a given time. We are given that the base radius is equal to half its height, so the radius of the cone at any height is r = (h/2).

We are asked to find how fast the water level is rising (dh/dt) when the water is 3 meters deep. We are given that the water is being pumped in at a rate of 2 cubic meters per second, which means dV/dt = 2 m^3/s.

To relate the variables, we need the formula for the volume of a cone:

V = (1/3)πr^2h

Differentiating both sides with respect to time (t) using the chain rule, we get:

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

Since r = (h/2), we can substitute this expression into the equation:

dV/dt = (1/3)π(2h/2)(dh/dt + (h^2/4) * (dh/dt))

Now, we can substitute the given values:

2 = (1/3)π(2*3/2)(dh/dt + (3^2/4) * (dh/dt))

Simplifying the equation:

2 = (1/3)π(3)(dh/dt + (9/4) * (dh/dt))

2 = (9/4)π(dh/dt + (9/4) * (dh/dt))

Multiplying through by (4/9π), we get:

(8/9π) = dh/dt + (81/16) * dh/dt

Combining like terms:

(8/9π) = (1 + 81/16) * dh/dt

(8/9π) = (97/16) * dh/dt

dh/dt = (8/9π) * (16/97)

Simplifying:

dh/dt ≈ 1.465 m/s

Therefore, when the water is 3 meters deep, the water level is rising at a rate of approximately 1.465 meters per second.

To solve this problem, we can use related rates, a technique in calculus that allows us to find the rate at which one quantity is changing given the rate at which another quantity is changing.

Let's denote the height of the water in the tank as h (in meters) and the radius of the water surface as r (in meters).

We are given that the water tank has the shape of an inverted right circular cone, and its base radius is equal to half its height. This implies that the radius of the cone at any height h is r = h/2.

We are also given that water is being pumped into the tank at a rate of 2 cubic meters per second. This means that the volume V of water in the tank is changing at a rate of dV/dt = 2 m³/s.

To find how fast the water is rising, we need to find dh/dt, the rate at which the height of the water is changing when the water is 3 meters deep.

We know that the volume of a cone is given by the formula V = (1/3)πr²h, where π is a constant.

Differentiating both sides of this equation with respect to time t using the chain rule, we get:

dV/dt = (1/3)((d/dt)(πr²h) + (d/dt)(r²)(dh/dt))

Substituting r = h/2, we can rewrite this as:

2 = (1/3)((d/dt)(π(h/2)²h) + (d/dt)((h/2)²)(dh/dt))

Simplifying further, we get:

2 = (1/3)((π/4)(4h³) + (1/4)(2h)(dh/dt))

2 = (1/3)(πh³ + h(dh/dt))

2 = (1/3)πh³ + (1/3)h(dh/dt)

Now, let's solve for dh/dt.

Rearranging the equation, we get:

(1/3)h(dh/dt) = 2 - (1/3)πh³

Multiplying both sides by 3/h, we have:

dh/dt = 6/h - πh²

We are given that the water is 3 meters deep, so we can substitute h = 3 into the equation:

dh/dt = 6/3 - π(3)²
= 2 - 9π

Therefore, when the water depth is 3 meters, the water is rising at a rate of 2 - 9π meters per second.