Gravel is being dumped from a conveyor belt at a rate of 30 ft3/min. It forms a pile in the shape of a right circular cone whose base diameter and height are always the same. How fast is the height of the pile increasing when the pile is 21 ft high?
To find the rate at which the height of the pile is increasing, we can use related rates and apply the formula for the volume of a cone.
Let's denote the height of the cone as h and the radius of the base as r. Since the base diameter and height are always the same, the radius r is equal to the height h.
The volume of a cone can be expressed as V = (1/3)πr^2h.
Since the radius r is equal to the height h, we can rewrite the formula as V = (1/3)πh^3.
Now, we are given that the gravel is being dumped at a rate of 30 ft^3/min, meaning the volume of the pile is increasing at a rate of dV/dt = 30 ft^3/min.
To find how fast the height of the pile is increasing, we need to find dh/dt when h = 21 ft.
Differentiating the volume formula with respect to time (t) using implicit differentiation, we get:
dV/dt = (1/3)π(3h^2)(dh/dt)
Substituting the given values:
30 = (1/3)π(3(21^2))(dh/dt)
Simplifying, we have:
30 = 7π(441)(dh/dt)
Dividing both sides by 7π(441), we get:
dh/dt = 30 / (7π(441))
Evaluating this expression, we find:
dh/dt ≈ 0.00328 ft/min
Therefore, the height of the pile is increasing at a rate of approximately 0.00328 ft/min when the pile is 21 ft high.
To find how fast the height of the pile is increasing, we need to use related rates and the formula for the volume of a right circular cone.
The volume of a right circular cone can be calculated using the formula:
V = (1/3)πr^2h
Given that the base diameter and height are always the same, we can say that the radius (r) is half of the height (h).
Now, let's differentiate the volume equation with respect to time (t), and remember to use the chain rule since both the radius and height are changing with time.
dV/dt = (1/3)π[(2rh)(dh/dt) + (r^2)(dh/dt)]
Let's substitute the given values into the equation:
dV/dt = (1/3)π[(2(h/2)(dh/dt) + ((h/2)^2)(dh/dt)]
Simplifying the equation, we get:
dV/dt = (1/3)π[h(dh/dt) + (h^2/4)(dh/dt)]
Now, we can substitute the given values:
dV/dt = (1/3)π[21(dh/dt) + (21^2/4)(dh/dt)]
We are given that the volume is changing at a rate of 30 ft^3/min, so dV/dt = 30 ft^3/min:
30 = (1/3)π[21(dh/dt) + (21^2/4)(dh/dt)]
Now, let's solve for dh/dt by rearranging the equation:
30 = (1/3)π[21(dh/dt) + (441/4)(dh/dt)]
30 = (1/3)π[(84/4 + 441/4)(dh/dt)]
30 = (1/3)π[(525/4)(dh/dt)]
30 = (525/12)π(dh/dt)
10 = (525/12)π(dh/dt)
Finally, we can solve for dh/dt:
dh/dt = 10 / [(525/12)π]
dh/dt = 120 / (525π)
Therefore, the height of the pile is increasing at a rate of approximately 0.072 ft/min when the pile is 21 ft high.
so h = r
V = (1/3)πr^2 h ---- #1
V = (1/3)π h^3
dV/dt = π h^2 dh/dt ----- #2
you have all the values of #2 except dh/dt
Go for it