Wheat is falling from a chute onto a level floor at a rate of 8pi ft3/min to form a conical pile. If the height of the pile is always equal to the radius of its base, at what rate is the radius increasing when the pile is 8 ft deep?
V = (1/3)πr^2h
but h=r
V= (1/3)πr^3
dV/dt = πr^2 dr/dt
so when h = 8 , r = 8 and dV/dt = 8π
8π = π(8)^2 dr/dt
dr/dt = 8π/(64π) = 1/8 ft/min
THANKS!
Why did the wheat go on strike?
Because it felt like it was being taken for granite!
Let's start by visualizing the situation. We have a conical pile forming on the floor, with wheat falling from a chute onto it. The height of the pile is always equal to the radius of its base.
We need to find the rate at which the radius is increasing when the pile is 8 ft deep.
Let's call the height of the pile h and the radius of its base r. According to the problem, h = r.
Now, we need to find an equation relating the rate at which the radius is changing (dr/dt) with the given rate at which the wheat is falling (dV/dt), where V represents the volume of the pile.
The volume of a cone is given by the formula V = (1/3)πr^2h.
Differentiating both sides of the equation with respect to time t, we get dV/dt = (1/3)(2πrh)(dh/dt) + (1/3)πr^2(dr/dt).
We are given that dV/dt = 8π ft^3/min (the rate at which the wheat is falling). We also know that h = 8 ft when we want to find dr/dt.
Substituting the given values into the equation and simplifying, we get 8π = (16/3)πr(dr/dt) + (1/3)πr^2(dr/dt).
We can further simplify this equation as 24 = 16r(dr/dt) + r^2(dr/dt).
Factoring out dr/dt, we have 24 = (16r + r^2)(dr/dt).
To find dr/dt at h = 8 ft, we substitute r = h = 8 ft into the equation and solve for dr/dt.
24 = (16 * 8 + 8^2)(dr/dt).
24 = (128 + 64)(dr/dt).
24 = 192(dr/dt).
Dividing both sides by 192, we get dr/dt = 24/192 = 1/8 ft/min.
So, at the moment when the pile is 8 ft deep, the radius is increasing at a rate of 1/8 ft/min.
I hope that helps! If you have any more questions, feel free to ask.
To find the rate at which the radius is increasing, we need to relate the height and radius of the cone and then differentiate with respect to time. Let's use the similar triangles in this problem.
Let's assume that at any given time, the height of the cone is h and the radius is r. Since the height of the pile is always equal to the radius of its base, h = r.
We know that the volume of a cone is given by: V = (1/3)πr²h.
Given that the rate at which the wheat is falling is 8π ft³/min, we can also say that dh/dt = 8π ft³/min.
Now, let's differentiate the volume of the cone equation with respect to time to relate the rates of change:
dV/dt = (1/3)π(2r)(dh/dt)
Since h = r, we can simplify the equation as follows:
dV/dt = (2/3)πr(dh/dt)
We want to find the rate at which the radius is increasing, so we need to differentiate with respect to time again:
(dV/dt) / (dr/dt) = (2/3)πr(dh/dt) / (dr/dt)
The left-hand side of the equation represents the rate at which the volume is changing with respect to the radius, and the right-hand side represents the rate at which the height is changing with respect to time.
Since we know the height at a particular time (when the pile is 8 ft deep), we can substitute h = r = 8 into the equation:
(dV/dt) / (dr/dt) = (2/3)π(8)(8) / (8π)
(dV/dt) / (dr/dt) = 16/3
Therefore, the rate at which the radius is increasing when the pile is 8 ft deep is 16/3 ft/min.
To find the rate at which the radius is increasing, we can use related rates, which involves applying the formulas for volume and surface area of a cone.
Let's first define the variables:
- V: Volume of the cone (in this case, the wheat pile)
- r: Radius of the base of the cone (which would also be the height of the pile)
- h: Height of the cone (which would also be the radius of the base of the pile)
Given that the height of the pile (h) is always equal to the radius of its base (r), we can write h = r.
We are told that the volume of the pile is increasing at a rate of 8π ft^3/min. Therefore, dV/dt = 8π, where t is the time.
The volume of a cone is given by the formula V = (1/3)πr^2h. Substituting h = r, we have V = (1/3)πr^3.
Differentiating both sides with respect to time (t), we get dV/dt = (1/3)π * 3r^2(dr/dt).
Now, we know that dV/dt = 8π, so we can substitute that information into the equation:
8π = (1/3)π * 3r^2(dr/dt).
Simplifying, we have:
8 = r^2(dr/dt).
We want to find the rate at which the radius is increasing, so we need to find dr/dt when the pile is 8 ft deep. Since h = r, when the height of the pile is 8 ft, the radius is also 8 ft.
Plugging in r = 8 into the equation, we have:
8 = (8^2)(dr/dt).
Simplifying further:
8 = 64(dr/dt).
Now we can solve for dr/dt:
dr/dt = 8/64 = 1/8 ft/min.
Therefore, the rate at which the radius is increasing when the pile is 8 ft deep is 1/8 ft/min.