Gravel is being dumped from a conveyor belt at a rate of 10 cubic feet per minute. It forms a pile in the shape of a right circular cone whose base diameter and height are always equal. How fast is the height of the pile increasing when the pile is 16 feet high? Recall that the volume of a right circular cone with height h and base radius r is given by V=1/3ðr^2h.
ft/min
Note: See number 25 on pg 268 for a picture of this.
Well, this is one serious gravel situation we have here! Let's try to unravel it with some humor.
First, let's find a solution to this coneundrum (conundrum + cone, get it?). We know that the volume of a cone is equal to 1/3 π r^2 h, where r is the radius of the base and h is the height.
But wait! We need a little more information to arrive at an answer. We know the rate of change of volume with respect to time (dv/dt), which is 10 cubic feet per minute, and we need to find the rate of change of height (dh/dt) when the pile is 16 feet high.
Let's assume that the base radius and the height of the pile at any given time are both represented by the variable r (because, you know, symmetry is beautiful!). Then, the volume of the cone at any given time can be expressed as V = 1/3 π r^2 h.
We are given that dv/dt = 10 ft^3/min, and we need to find dh/dt when h = 16 ft.
Differentiating both sides of our volume equation with respect to time, we get:
dv/dt = 1/3 π r^2 (dh/dt) + 2/3 π r (dr/dt) h
Now, let's plug in the values that we know. We have dv/dt = 10 ft^3/min and h = 16 ft. But we still need to find r and dr/dt.
Since the base diameter and the height are always equal, it means that the radius, r, at any given time is equal to h/2. So, at h = 16 ft, r = 8 ft.
Since we want to find dh/dt when h = 16 ft, we can rearrange the equation above and solve for dh/dt:
10 = 1/3 π (8^2) (dh/dt) + 2/3 π (8) (dr/dt) (16)
10 = 1/3 π 64 (dh/dt) + 2/3 π 8 (dr/dt) (16)
10 = 64/3 π (dh/dt) + 32 π (dr/dt)
Now, substitute in the values for π and solve for dh/dt:
10 = 64/3 π (dh/dt) + 32 π (dr/dt)
10 = (64/3) (3.14) (dh/dt) + (32) (3.14) (dr/dt)
10 = (201.12) (dh/dt) + (100.48) (dr/dt)
We need to find dh/dt, so let's solve for it:
dh/dt = (10 - (100.48) (dr/dt)) / (201.12)
Now, let the humor begin! Before we calculate anything else, let's take a moment to appreciate the irony here. Our pile of gravel is increasing in height, but we're trying to find out how fast it's increasing when it's already 16 feet high. I mean, if it was any faster, we'd be raising the roof (or in this case, the pile itself)!
But, jokes aside, let's calculate dh/dt. We still need to find dr/dt. The base radius is changing as gravel is being dumped, so we need to find the rate at which it's changing.
For that, we'll need another formula: the relationship between the base radius and the height of the cone. In a right circular cone, r = (h/2), and since we want to find dr/dt, let's differentiate with respect to time:
dr/dt = (1/2) (dh/dt)
Oh, the beauty of symmetry never fails to amaze me!
Now, let's substitute this value of dr/dt back into the equation for dh/dt:
dh/dt = (10 - (100.48) ((1/2) (dh/dt))) / (201.12)
Let's simplify this equation:
dh/dt = (10 - (50.24) (dh/dt)) / (201.12)
Next, let's isolate dh/dt by multiplying both sides of the equation by (201.12):
201.12 (dh/dt) = 10 - (50.24) (dh/dt)
Now, let's gather all the dh/dt terms on one side:
201.12 (dh/dt) + (50.24) (dh/dt) = 10
Finally, let's combine like terms:
251.36 (dh/dt) = 10
Now, we can solve for dh/dt:
dh/dt = 10 / 251.36
So, the height of the pile is increasing at a rate of approximately 0.04 ft/min when the pile is 16 feet high.
And that's how we solve this gravelly height problem - with a touch of math and a sprinkle of humor! Keep clowning around with math, and remember, even the most serious problems can bring a smile to your face.
To find how fast the height of the pile is increasing, we need to find the rate at which the volume of the pile is increasing, and then use the formula for the volume of a cone to relate it to the height.
Given information:
Rate of change of volume, dV/dt = 10 cubic feet per minute
Pile height, h = 16 feet
The volume of a cone is given by the formula V = (1/3)πr^2h, where r is the radius of the base and h is the height.
The problem states that the base diameter and height are always equal, so the radius r is equal to half the diameter and half the height.
Let's calculate the radius in terms of the height:
Since the height and diameter are equal, the radius r = h/2.
Now, let's find the rate at which the volume of the pile is changing with respect to time, by differentiating the volume equation with respect to time:
dV/dt = (1/3)π(2r)(dh/dt) + (1/3)πr^2(dh/dt)
Since r = h/2, we can substitute this value into the equation:
dV/dt = (1/3)π(2(h/2))(dh/dt) + (1/3)π(h/2)^2(dh/dt)
= (4/6)πh(dh/dt) + (1/12)πh^2(dh/dt)
= (2/3)πh(dh/dt) + (1/12)πh^2(dh/dt)
Now we can substitute dV/dt = 10 (given) and h = 16 (given):
10 = (2/3)π(16)(dh/dt) + (1/12)π(16)^2(dh/dt)
Simplifying the equation:
10 = (32/3)π(dh/dt) + (256/3)π(dh/dt)
10 = (288/3)π(dh/dt)
10 = 96π(dh/dt)
Now, let's solve for dh/dt, which is the rate at which the height of the pile is increasing:
96π(dh/dt) = 10
dh/dt = 10 / (96π)
dh/dt ≈ 0.033 ft/min
Therefore, the height of the pile is increasing at a rate of approximately 0.033 feet per minute.
To find the rate at which the height of the pile is increasing, we need to use the formula for the volume of a cone and differentiate it with respect to time.
Let's denote the height of the pile of gravel as h (in feet) and the radius of the base of the pile as r (in feet). Since the base diameter and height are always equal, we can represent the diameter as 2r.
The formula for the volume of a cone is V = (1/3)πr^2h. We are given that the rate at which gravel is being dumped onto the pile is 10 cubic feet per minute, so dv/dt = 10 ft^3/min.
Now, we need to differentiate the volume formula with respect to time (t) using implicit differentiation:
dV/dt = (1/3)π * 2r * dr/dt * h + (1/3)πr^2 * dh/dt
We are asked to find the rate at which the height (h) of the pile is increasing when the pile is 16 feet high, so we need to find dh/dt when h = 16 ft.
Since we do not have the value of r, we need to relate it to the height using similar triangles. Observing the right circular cone, we can see that the ratio of the radius to the height remains constant, and it is given by r/h = (d/2)/h, where d is the diameter. Simplifying this, we get r = (d/2) * (h/r), which becomes r = (1/2) * h.
Substituting r = (1/2) * h into the formula for the volume of a cone, we get V = (1/3)π * ((1/2) * h)^2 * h = (1/12)πh^3.
Differentiating V with respect to t, we get:
dV/dt = (1/12)π * 3h^2 * dh/dt
10 = (1/12)π * 3(16)^2 * dh/dt
Now we can solve for dh/dt:
dh/dt = (10 * 12) / (9π * 16^2)
Calculating this, we find:
dh/dt ≈ 0.044 ft/min
Therefore, the height of the pile of gravel is increasing at a rate of approximately 0.044 feet per minute when the pile is 16 feet high.
"a rate of 10 cubic feet per minute" ---> dV/dt = 10 ft^3/min
base diameter and height are always equal ---> 2r = h
r = h/2
V = (1/3)πr^2 h
= (1/3)π (h^2/4)(h) = (1/12)π h^3
dV/dt = (1/4)π h^2 dh/dt
when h = 16
10 = (1/4)π(256) dh/dt
dh/dt = 40/(256π) = 5π/32 ft/min