A conical tank has height 3m and radius 2m at the top. Water flows in at a rate of 2m^3 /min. How fast is the water level rising when it is 2m.

9/8pi

Well, well, we have ourselves a conical tank in a tricky situation! Alright, let's dive into it.

We have a conical tank with a height of 3m and a top radius of 2m. Since the water is flowing in at a rate of 2m^3/min, we need to find how fast the water level is rising when it reaches a height of 2m. That's our target!

Now, let's make some jokes while solving this problem and remember to buckle up for some math fun!

First off, the volume of a cone can be found using the formula V = (1/3) * π * r^2 * h, where r is the radius and h is the height.

The height is decreasing at a rate of 2m/min. So dh/dt = -2m/min.

We want to find the rate at which the radius (r) is increasing when the water level is at 2m. Let's call this dr/dt.

Here comes the fun part. We need to relate the variables r, h, and V to find dr/dt when h = 2m.

Since the tank is conical and similar triangles are our best friends, we know that r/h = 2/3 because the top radius is 2m and the full height is 3m.

We also know that dh/dt = -2m/min, and we want to find dr/dt when h = 2m.

Using similar triangles, we can say that dr/dh = (2/3) * r/h. And we want to find dr/dt when h = 2m, so we need to find dr/dt when h = 2m.

Let's perform some mathematical acrobatics and rearrange the equation:

dr/dh = (2/3) * r/h
dr = (2/3) * (r/h) * dh
dr = (2/3) * (r/2) * (-2) [Substituting dh = -2m/min and h = 2m]
dr = -(2/3) * (r/2) * 2
dr = -2r/3

All right! We have our expression for dr. Now, let's substitute r = 2m (remember, this is the radius at a water level of 2m) and find dr/dt:

dr/dt = -2(2)/3
dr/dt = -4/3

So, when the water level is at 2m, the rate at which the radius is increasing is -4/3 m/min. Remember, the negative sign indicates a decrease in the radius. Oh, that cone just keeps getting spookier, doesn't it? 🤡

To find the rate at which the water level is rising when it is 2m, we need to use the related rates approach by considering the volume of water in the tank as a function of time.

Let's break down the problem step by step:

Step 1: Express the volume of the water in the tank as a function of its height.
In this case, since the shape of the tank is conical, the volume of the water can be expressed using the formula for the volume of a cone:
V = (1/3) * π * r^2 * h
where V is the volume, π is a mathematical constant (approximately equal to 3.14), r is the radius of the tank, and h is the height of the water in the tank.

Given that the height of the water is 2m, we can substitute the known values into the equation to find the volume at that height:
V = (1/3) * π * (2^2) * 2 = 8/3 * π

Step 2: Determine the rate at which the volume of water in the tank is changing with respect to time.
We are given that the water is flowing into the tank at a rate of 2m^3/min. Therefore, the rate of change of volume (dV/dt) with respect to time is 2m^3/min.

Step 3: Differentiate the volume equation with respect to time (t) using the chain rule.
dV/dt = (dV/dh) * (dh/dt)

We want to find dh/dt, which represents the rate at which the water level is rising when the height is 2m.

Step 4: Compute dh/dt.
From the formula for the volume of a cone, we differentiate both sides of the equation with respect to time (t):
dV/dt = (1/3) * π * (2r) * (dh/dt)
Given that the radius r is initially 2m, we can substitute the known values into the equation:
2 = (1/3) * π * 4 * (dh/dt)

Now, solve for dh/dt:
dh/dt = 2 / [(1/3) * π * 4] = 3 / (2π) = 3/6.28 ≈ 0.477 m/min

Therefore, when the water level is 2m, the water level is rising at a rate of approximately 0.477 m/min.

V=PIr^2*h

dV/dt=2rPIh dr/dt + PIr^2 dh/dt

you know dV/dt, dr/dt=0, and you are solving for dh/dt

Since it is a cone, and the height is 3 to a radius of 2, then no matter what the height is h = (3/2) * r and thus r = (2/3) * h. For the computations we want to get rid of the r by substituting (2/3) * h for r.


Now the capacity of the cone filled to the brim is (1/3) * h * pi * r^2. Substitute (2h / 3) for r, and (4h^2 / 9) for r^2.

v = (1/3) * h * pi * (4h^2 / 9) = (4pi / 27) * h^3

Therefore dv/dt = (4pi / 27) * 3h^2 dh/dt = (4 * pi / 9) * h^2 * dh/dt

dv/dt = (4 * pi / 9) * h^2 * dh/dt................. We want to find out what dh/dt is and we know that

dv/dt is 2m^3/min and that h = 2m. Substitute those into the equation above.

2 = (4 * pi / 9) * 4 * dh/dt = [(16 * pi / 9) * dh/dt]...........Divide both sides of this equation by [16 * pi / 9]

[2 / (16 * pi / 9) ] = dh/dt


So dh/dt = [2 / (16 * pi / 9) ] which is simplified to [ 1 / (16 * pi / 9) ]

But we aren't done. Multiply the numerator and denominator of that fraction by 9, getting:

(9 / (16 * pi))

So dh/dt = (9 / (16 * pi))

The water level is rising at (9 / 16 * pi) m / min. <--------------- Answer