Sand pouring from a conveyor belt forms a conical pile the radius of which is 3/4 of the height. If the sand is filling up at a constant rate of 1/2m^3/min, at what rate is the height of the pile is growing 3 min after the pouring starts?
since the radius is 3/4 the height,
The volume of the pile is thus
v = π/3 r^2 h = π/3 (3h/4)^2 h = 3π/16 h^3
dv/dt = 9π/16 h^2 dh/dt
now, plugging in your numbers,
at t=3, v = 3/2, so
3π/16 h^3 = 3/2
h = 2/∛π
so,
9π/16 4/∛π^2 dh/dt = 1/2
dh/dt = 2/9 ∛π m/min
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thank you steve
To find the rate at which the height of the pile is growing, we can use related rates.
Let's assign some variables:
- Let h be the height of the pile at time t (in minutes).
- Let r be the radius of the conical pile at time t.
Based on the given information, we know that the radius of the pile is 3/4 times the height, so we have: r = (3/4)h.
We also know that the sand is filling up the pile at a constant rate of 1/2 m^3/min. This means that the volume of the pile is increasing at a constant rate of 1/2 m^3/min.
The volume of a cone can be calculated using the formula: V = (1/3)πr^2h, where V is the volume, π is pi (approximately 3.14159), r is the radius, and h is the height.
To find the rate at which the height is growing, we need to find dh/dt, the derivative of the height with respect to time.
First, let's write the volume equation in terms of h and r:
V = (1/3)π[(3/4)h]^2h
V = (1/3)π(9/16)h^3
V = (3/16)πh^3
Now, we can differentiate both sides of the equation with respect to time (t):
dV/dt = d/dt[(3/16)πh^3]
The derivative of V with respect to t (dV/dt) is the rate at which the volume of the pile is changing, which is given as 1/2 m^3/min.
Therefore, we have: 1/2 = d/dt[(3/16)πh^3]
Now, we can solve for dh/dt, the rate at which the height of the pile is growing, by differentiating both sides of the equation with respect to time (t):
0 = d^2/dt^2[(3/16)πh^3]
Simplifying, we have:
0 = (9/16π)h^2(dh/dt)^2
To find dh/dt, we can rearrange the equation as follows:
(dh/dt)^2 = 0 / [(9/16π)h^2]
(dh/dt)^2 = 0
dh/dt = 0 (since the derivative squared is 0)
Therefore, the rate at which the height of the pile is growing is 0. The height remains constant after 3 minutes of pouring.
Note: This result might seem counterintuitive, but it is due to the specific conditions given in the problem.