A hollow cone of base radius 10cm and height 10cm is held with its vertex downwards. The cone is initially empty when water is poured into it at the rate of (4pie) cm^3/s. Find the rate of increase in the depth of the water in the cone 18 second after pouring has commenced.

Thanks

at t=18, v=18*4π = 32π cm^3

since r=h,
v = 1/3 π r^2 h = π/3 h^3
so, when v=32π, h= ∛96

dv/dt = πh^2 dh/dt
...

Well, that's a real conundrum! Let's see if I can help you figure this out.

First, we need to find the volume of the cone. The formula for the volume of a cone is (1/3)πr²h, where r is the radius of the base and h is the height.

Given that the base radius is 10 cm and the height is 10 cm, we can plug these values into the formula:

Volume = (1/3)π(10)²(10)
Volume = (1/3)π(100)(10)
Volume = (1/3)π(1000)
Volume = 1000/3π cm³

Now, we know that water is being poured into the cone at a rate of 4π cm³/s. So, after 18 seconds, the volume of the cone would have increased by (4π)(18) cm³.

To find the rate of increase in the depth of the water, we need to divide the increase in volume by the area of the base of the cone, which is πr². The radius of the base is 10 cm.

Rate of increase in depth = [(4π)(18)] / π(10)²
Rate of increase in depth = 72 / 100
Rate of increase in depth = 0.72 cm/s

So, the rate of increase in the depth of the water in the cone 18 seconds after pouring has commenced is 0.72 cm/s.

I hope that helps! And remember, even if math is a bit puzzling sometimes, just keep a positive attitude and everything will work out swimmingly!

To find the rate of increase in the depth of the water in the cone, we need to first find the volume of water in the cone after 18 seconds.

The volume of a cone is given by the formula V = (1/3)πr^2h, where r is the radius of the base and h is the height.

Given that the base radius is 10 cm and the height is 10 cm, we can substitute these values into the formula:

V = (1/3)π(10^2)(10)
V = (1/3)π(100)(10)
V = (1/3)π(1000)
V = (1000/3)π

Now, let's differentiate the volume with respect to time, t, to find the rate of change:

dV/dt = (1000/3)π * (d/dt) = (1000/3)π * (d/dt)

Given that water is poured into the cone at the rate of 4π cm^3/s, we can substitute this value into the equation:

dV/dt = (1000/3)π * 4π cm^3/s
dV/dt = (4000/3)π cm^3/s

Therefore, the rate of increase in the depth of the water in the cone 18 seconds after pouring has commenced is (4000/3)π cm^3/s.

To solve this problem, we can use similar triangles and the formula for the volume of a cone.

The volume of a cone is given by the formula V = (1/3)πr^2h, where r is the radius of the base and h is the height.

Let's break down the problem step by step.

Step 1: Find the rate of increase in volume of the water.

We are given that water is poured into the cone at a rate of (4π) cm^3/s. This means the volume of water in the cone is increasing at a constant rate of 4π cm^3/s. Hence, the rate of increase in volume is 4π cm^3/s.

Step 2: Find the rate of increase in height of the water.

To find the rate of increase in height, we need to differentiate the formula for volume with respect to time. Since the radius of the base is constant at 10 cm, we can simplify the formula to V = (1/3)π * 10^2 * h.

Differentiating the formula with respect to time, we get:

dV/dt = (1/3)π * 10^2 * dh/dt

Since dV/dt is the rate of increase in volume (which we found earlier to be 4π cm^3/s), we can substitute this value into the equation:

4π = (1/3)π * 10^2 * dh/dt

Simplifying the equation, we get:

dh/dt = 4/(10^2 * 3) = 4/300 = 1/75 cm/s

Therefore, the rate of increase in the depth of the water in the cone 18 seconds after pouring commences is 1/75 cm/s.