sand is being poured onto the ground forming a conical pile whose height equals 1/4 of the diameter of the base. The sand is falling at a rate of 20cm^3/sec. How fast is the height of the sand pile increasing when it is 3cm high?

Why did the sand pile go to therapy? Because it had unresolved cone-plexes!

But in all seriousness, let's solve this problem. We have a conical pile of sand, where the height (h) is equal to 1/4 of the diameter (2r) of the base. We need to find how fast the height is increasing (dh/dt) when the height is 3 cm.

Let's use some geometry and calculus to find the solution.

First, let's find an expression for the height in terms of the radius (h(r)). Since the height is 1/4 of the diameter, we have h(r) = 1/4 * (2r) = r/2.

Now, we know that the volume of a cone is given by V = (1/3) * π * r^2 * h. Since the cone is being filled at a rate of 20 cm^3/sec, we have dV/dt = 20 cm^3/sec.

Let's substitute the expression for h into the volume formula and differentiate it with respect to time (t):

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

Now, we need to find dr/dt in terms of dh/dt. Differentiating both sides of the expression h = r/2 with respect to t:

dh/dt = (d/dt)(r/2)
= (1/2) * (dr/dt)

Substituting this into the previous equation:

20 = (1/3) * π * (2r * dr/dt + r^2 * (1/2) * (dr/dt))

We're given that the height is 3 cm, so we can substitute h = 3 and r = 2h into the equation:

20 = (1/3) * π * (2(2h) * dr/dt + (2h)^2 * (1/2) * (dr/dt))
20 = (1/3) * π * (4h * dr/dt + 2h^2 * dr/dt)
20 = (4/3) * π * h * (4 * dr/dt + 2h * dr/dt)

Now, we can solve for dr/dt:

dr/dt = (20 * 3) / ((4/3) * π * h * (4 + 2h))
dr/dt = (60) / (4h + 2h^2)

Substituting h = 3:

dr/dt = (60) / (4(3) + 2(3)^2)
dr/dt = (60) / (12 + 18)
dr/dt = (60) / (30)
dr/dt = 2 cm/sec

So, the height of the sand pile is increasing at a rate of 2 cm/sec when it is 3 cm high.

To solve this problem, we can use related rates, taking derivatives with respect to time.

Let's denote the height of the sand pile as h and the radius of the base as r. Since the height is 1/4 of the diameter, we have h = r/2.

We are given that the sand is falling at a rate of 20 cm^3/sec. This means that the volume of the pile is increasing at a rate of 20 cm^3/sec.

The volume of a cone can be calculated using the formula V = (1/3)πr^2h. In this case, the base is a circle, so the volume can be written as V = (1/3)πr^3/4.

Taking the derivative of both sides with respect to time (t), we can find the rate of change of volume with respect to time:

dV/dt = (1/3)(π/4)(3r^2(dr/dt) + r^3(dh/dt).

We are interested in finding how fast the height (h) is changing when it reaches 3 cm. This means we need to find dh/dt when h = 3, and we also know that r = 2h.

Substituting these values into the equation and rearranging, we get:

20 = (1/3)(π/4)(3(2h)^2(dr/dt) + (2h)^3(dh/dt).

Now, we can plug in the values we know: h = 3 and r = 2h = 6.

20 = (1/3)(π/4)(3(2(3))^2(dr/dt) + (2(3))^3(dh/dt).

Simplifying further:

20 = (1/3)(π/4)(3(6)^2(dr/dt) + (6)^3(dh/dt).

20 = (1/3)(π/4)(3(36)(dr/dt) + (216)(dh/dt).

20 = (1/3)(π/4)(108(dr/dt) + 216(dh/dt).

We are given that dr/dt = 0, since the sand is poured onto the ground without any change in the radius.

Plugging in this value and simplifying:

20 = (1/3)(π/4)(108(0) + 216(dh/dt).

20 = (1/3)(π/4)(0 + 216(dh/dt).

20 = (1/3)(π/4)(216(dh/dt).

To find dh/dt, we can solve for it:

dh/dt = (20 * 3 * 4) / (π * 216).

Simplifying further, we get:

dh/dt = 10 / (π * 9).

Therefore, the rate at which the height of the sand pile is increasing when it is 3 cm high is 10 / (π * 9) cm/sec.

To find the rate at which the height of the sand pile is increasing, we need to use related rates and the volume of a cone formula.

Let's denote the height of the cone as h, the radius of the base as r, and the volume of the cone as V.

We know that the height of the cone is 1/4 of the diameter of the base, which means h = 1/4 * d.

From the problem, we are given that the sand is falling at a rate of 20 cm^3/sec, which is the rate at which the volume of the cone is increasing (dV/dt = 20 cm^3/sec).

The volume of a cone is given by the formula V = (1/3) * π * r^2 * h.

Now, we need to find an equation that relates the height (h) and the radius (r) of the cone. Since h = 1/4 * d and the diameter is twice the radius, we can substitute d = 2r into the equation to express the volume in terms of only r and h: V = (1/3) * π * r^2 * (1/4 * d).

Simplifying this expression: V = (1/12) * π * r^2 * d.

Since r = d/2, we can substitute this in to express the volume in terms of only h: V = (1/12) * π * r^2 * (2h) = (1/6) * π * r^2 * h.

Now, differentiate both sides of the equation with respect to time (t) using the chain rule:
dV/dt = (1/6) * π * r^2 * dh/dt.

Substituting the given value of dV/dt (20 cm^3/sec) and rearranging the equation, we can solve for dh/dt, the rate at which the height of the sand pile is increasing:
dh/dt = (6/π * r^2) * (dV/dt).

To find the value of r when the height of the sand pile is 3 cm, we can use the relationship between the height (h) and the radius (r) in the cone: h = 1/4 * d = 1/4 * (2r) = r/2. Substituting h = 3 into this equation: 3 = r/2, which gives r = 6 cm.

Now, we can substitute the value of r into the expression for dh/dt to find the rate at which the height of the sand pile is increasing when it is 3 cm high:
dh/dt = (6/π * (6^2)) * 20 = (6/π * 36) * 20 = 3600/π cm/sec.

So, the height of the sand pile is increasing at a rate of 3600/π cm/sec when it is 3 cm high.

h = d/4 = r/2

So, r = 2h

v = π/3 r^2h = 4π/3 h^3
dv/dt = 4πh^2 dh/dt

Now just plug in your numbers.