Gravel is being dumped from a conveyor belt at a rate of 15 ft^3/hr and its coarseness is such that is forms a pile in the shape of an inverted right cone whose height is three times its base radius. How fast is the height of the pile increasing when the pile has a height of 10 feet?
Q: Gravel is being dumped from a conveyor belt at a rate of 25 ft3/min, and its coarseness is such that it forms a pile in the shape of a cone whose base diameter and height are always equal. How fast is the height of the pile increasing when the pile is 6 ft high? (Round your answer to two decimal places.)
A: since h = 2r, r = h/2
v = pi/3 r^2h = pi/12 h^3
So,
dv/dt = pi/4 h^2 dh/dt
When h=6, we have
25 = 9pi dh/dt
the units is a bit different but the same idea. Hope it helps
h=3r, so r = h/3
v = π/3 r^2 h = π/3 (h/3)^2 h = π/27 h^3
dv/dt = π/9 h^2 dh/dt
So, plug in your numbers to get dh/dt.
To solve this problem, we need to relate the changing height of the pile to its rate of change with respect to time.
Let's denote the height of the pile as h and the base radius as r. We're given that the height of the pile is three times its base radius, so we can write h = 3r.
We need to find the rate of change of the height (dh/dt) when the height is 10 feet.
First, let's establish the relationship between h and r in terms of time. Since the pile is formed by gravel being dumped from a conveyor belt, both h and r are changing with respect to time (t).
We can use similar triangles to relate the rates of change of h and r. Given that the cone is an inverted right cone, the similar triangles can be formed by connecting the tip of the cone, the center of the base, and a point on the circumference of the base.
According to the similar triangle relationship, we have:
(r + dr) / (h + dh) = r / h
Now let's differentiate this equation with respect to time (t) while assuming that r and h are functions of t.
Using the quotient rule, we get:
[(h + dh)(dr/dt) - (r + dr)(dh/dt)] / (h + dh)^2 = (dr/dt) / r
Next, we'll plug in the given values:
h = 10 ft
dh/dt = ?
dr/dt = 15 ft^3/hr (the rate at which gravel is being dumped)
Since we're looking for dh/dt when h = 10 ft, let's substitute these values into the equation and solve for dh/dt.
[(10 + dh)(15) - (r + dr)(dh/dt)] / (10 + dh)^2 = 15 / r
Simplifying the equation, we get:
150 + 15dh - 15r - dr(dh/dt) = 15(10 + dh) / r
Since the problem states that h = 3r, we can substitute that relationship into the equation:
150 + 15dh - 15r - dr(dh/dt) = 15(10 + dh) / (3r)
Simplifying further, we get:
150 + 15dh - 15r - dr(dh/dt) = 150 + 5dh
Rearranging the terms, we have:
15dh - dr(dh/dt) = 15dh - 15r
Canceling out the 15dh terms, we get:
-dr(dh/dt) = -15r
Dividing both sides by -dr and simplifying, we have:
dh/dt = 15r / dr
Now substitute h = 10 and r = h/3 into the equation:
dh/dt = 15(h/3) / dr
Simplifying, we get:
dh/dt = 15h / (3dr)
Since dr/dt = 15 ft^3/hr, we can substitute that into the equation:
dh/dt = 15(10) / (3 * 15)
Simplifying further, we get:
dh/dt = 5/3 ft/hr
Therefore, the height of the pile is increasing at a rate of 5/3 ft/hr when the pile has a height of 10 feet.