Wheat is poured through a chute at the rate of 10 ft^3/min and falls in a cone-shaped pile whose bottom radius is always half its height. How fast is the height of the cone increasing when the pile is 8 feet high?
Volume of a cone=1/3(pi)r^2h
R=1/2h Assume the water in cone is at height,h and radius is r.V=1/3Pi(1/2h)^2(h) V=1/12(Pi)h^3. by differentiating V:dV/dt=1/4(pi)h^2dh/dt 10=1/4*3.142*8^2dh/dt dh/dt=10/50.27=0.199ft/min
Well, well, well, looks like we've got some wheat woes going on here! Let's start by tackling this problem step by step.
We have a cone-shaped pile of wheat and we're interested in finding how fast the height of the cone is increasing. Perfecto! Now, we know that the wheat is being poured through a chute at a constant rate of 10 ft^3/min.
First things first, we can start by establishing a relationship between the various dimensions of our cone. We're told that the bottom radius of the cone is always half its height. So, let's call the height of the cone "h" and the bottom radius "r." That means our equation becomes r = h/2.
But hold on a sec, we need to find an equation that relates the rate of change of the height with respect to time (dh/dt) and the rate of change of the radius with respect to time (dr/dt).
What do we know? Well, the volume of a cone is given by V = (1/3) * π * r^2 * h. We know the rate of change of the volume, which is 10 ft^3/min. So we can differentiate this equation with respect to time:
dV/dt = (1/3) * π * (2r * dr/dt) * h + (1/3) * π * r^2 * dh/dt.
Now, we can plug in the values we know: dV/dt = 10 ft^3/min, r = h/2, and h = 8 ft. After some rearranging, our equation becomes:
10 = (4/3) * π * (h/2) * (h/2) * dh/dt + (1/3) * π * (h/2)^2 * dh/dt.
Let's simplify this further:
10 = (4/3) * π * (h^2/4) * dh/dt + (1/3) * π * (h^2/4) * dh/dt.
Don't you just love math and π? Now, we can factor out dh/dt and clean this up some more:
10 = π * (h^2/3) * dh/dt.
Finally, we can solve for dh/dt, which happens to be our goal. Dividing both sides by π * (h^2/3), we get:
dh/dt = 10 / (π * (h^2/3)).
Now all we have to do is plug in h = 8 ft and calculate:
dh/dt = 10 / (π * (8^2/3)).
Crunching the numbers, we find that dh/dt is approximately equal to 0.15 ft/min.
So there you have it: the height of the wheat cone is increasing at a rate of approximately 0.15 ft/min when the pile is 8 feet high. Now let's hope this explanation wasn't too corny for you!
Given:
Rate of wheat pouring: 10 ft³/min
Height of the cone: h (which is changing with time)
Bottom radius of the cone: r (which is always half the height)
We need to find the rate at which the height of the cone is changing when the pile is 8 feet high.
We can relate the height, radius, and volume of the cone using the formula for the volume of a cone:
Volume of a cone = (1/3)πr²h
Differentiating the volume formula with respect to time, we get:
d(Volume)/dt = d((1/3)πr²h) / dt
Since the rate of change of the volume is given by the rate at which the wheat is poured,
d(Volume)/dt = 10 ft³/min
We are required to find the rate at which the height is changing, so we need to differentiate the volume equation with respect to time and solve for dh/dt:
d(Volume)/dt = d((1/3)πr²h) / dt
10 = (1/3)πr² * dh/dt
Now we need to relate the height (h) and the radius (r) using the given information that the radius is always half the height:
r = h/2
Substitute this value of r into the equation:
10 = (1/3)π(h/2)² * dh/dt
Simplify further:
10 = (1/12)πh² * dh/dt
To find the value of dh/dt when the height is 8 feet (h = 8), substitute h = 8 into the equation:
10 = (1/12)π(8)² * dh/dt
Simplify and solve for dh/dt:
10 = (1/12)π * 64 * dh/dt
dh/dt = 10 * 12 / (π * 64)
Calculating the value of dh/dt:
dh/dt = 1.91 ft/min
Therefore, the height of the cone is increasing at a rate of 1.91 ft/min when the pile is 8 feet high.
To find the rate at which the height of the cone is increasing, we can use related rates. Related rates involve finding the rate at which one variable is changing with respect to another variable. In this case, we want to find how fast the height of the cone is changing with respect to time.
Let's denote the height of the cone as h, the radius of the cone as r, and the volume of the cone as V.
From the problem statement, we know that the cone is being filled by wheat at a constant rate of 10 ft^3/min. This means that the rate of change of the volume of the cone with respect to time is given by dV/dt = 10 ft^3/min.
The volume of a cone is given by the formula V = (1/3) * π * r^2 * h.
We are given that the bottom radius of the cone is always half its height, so we can write r = h/2.
Substituting this expression for r into the volume formula, we get V = (1/3) * π * (h/2)^2 * h.
Next, we can differentiate both sides of the volume equation with respect to time t:
dV/dt = (1/3) * π * d/dt((h/2)^2 * h)
Using the chain rule, we can differentiate each term:
dV/dt = (1/3) * π * (2(h/2) * (dh/dt) * h + (h/2)^2 * (dh/dt))
Simplifying this expression, we have:
10 ft^3/min = (1/3) * π * (h/2) * h * (dh/dt + (h/2) * (dh/dt))
Now, let's solve for dh/dt, the rate at which the height of the cone is changing:
Multiplying both sides by 3/π, we get:
(3/π) * 10 ft^3/min = (h/2) * h * (dh/dt + (h/2) * (dh/dt))
Simplifying further, we have:
30/π ft^3/min = (h^2/2) * (dh/dt + (h/2) * (dh/dt))
Now, we need to find the value of h when the pile is 8 feet high. Substituting h = 8 ft into the equation:
30/π ft^3/min = (8^2/2) * (dh/dt + (8/2) * (dh/dt))
Simplifying, we get:
30/π ft^3/min = 32 * (dh/dt + 4 * (dh/dt))
Dividing both sides by 32, we have:
15/π ft^3/min = dh/dt + 4 * dh/dt
Combining like terms, we get:
dh/dt = (15/π - 4) ft^3/min
Therefore, the height of the cone is increasing at a rate of (15/π - 4) ft^3/min when the pile is 8 feet high.