Sand is falling into a conical pile at the rate of 10 m3/sec such that the height of the pile is always half the diameter of the base of the pile. Find the rate at which the height of the pile is changing when the pile is 5 m. high.
h= r=5
V= (1/3)π(r)(r)(h)
dV/dt=3π(h)(h)(dh/dt)
10=45π(dh/dt)
The answer is supposed to be 2/5π m/s, but i don't see how.
from your
V= (1/3)π(r)(r)(h) , but r = h
V = (1/3)π h^3
dV/dt = π h^2 dh/dt
plug in the values ....
10 = π(25) dh/dt
dh/dt = 10/(25π) = 2/(5π)
I have no idea how you got your derivative.
Sand falls into a Konica on file at the rate of 10 ft^3/minsThe radius of the base of the fire is always equal to 1/2 of its altitude how fast is the altitude of the fighting greasing when it is 5th deep
Well, it seems like we have a "sandy" situation here! Let's see if we can decipher it with a little clownish humor.
First, let's break down the information we have:
- The rate at which sand is falling into the pile is 10 m3/sec.
- The height of the pile is always half the diameter of the base.
Now, let's find an equation that relates the height and the radius of the pile. Since the height is always half the diameter of the base, we can write h = r/2.
We also know that the volume V of a cone is given by V = (1/3)π(r^2)(h).
Now, let's differentiate both sides of this equation with respect to time t to find the rate of change of the volume:
dV/dt = (1/3)π * 2r * (dr/dt) * (h) + (1/3)π(r^2)(dh/dt).
Now, let's substitute the information we have:
dV/dt = (1/3)π * 2r * (dr/dt) * (h) + (1/3)π(r^2)(dh/dt).
10 = (1/3)π * 2r * (dr/dt) * (h) + (1/3)π(r^2)(dh/dt).
As we know that h = r/2, we can substitute this in the equation:
10 = (1/3)π * 2r * (dr/dt) * (r/2) + (1/3)π(r^2)(dh/dt).
Simplifying this equation, we get:
10 = (1/3)π * r * (dr/dt) * r + (1/3)π(r^2)(dh/dt).
10 = (1/3)π * r^2(dr/dt) + (1/3)π(r^2)(dh/dt).
Now we can factor out π(r^2):
10 = (1/3)π(r^2)(dr/dt + dh/dt).
Now let's solve for dh/dt:
dh/dt = (10 * 3) / (π(r^2)) - dr/dt.
But wait! We have a problem here. We don't have a given value for the rate at which the radius is changing (dr/dt). Therefore, we can't directly calculate dh/dt.
So unfortunately, my circular friend, it seems like we're stuck in the sand. Without knowing the rate at which the radius is changing, we can't find the rate at which the height is changing. Keep in mind that it's essential to have all the information at hand before attempting calculations.
Let's go through the steps to find the rate at which the height of the pile is changing.
Given:
The rate at which sand is falling into the pile = dV/dt = 10 m^3/sec
The height of the pile, h = 5 m (we are finding the rate when h = 5 m)
We are told that the height of the pile is always half the diameter of the base, so the radius, r = h/2 = 5/2 = 2.5 m
The volume of a cone is given by V = (1/3)π(r^2)(h).
Substituting the given values:
V = (1/3)π(2.5^2)(5)
V = (1/3)π(6.25)(5)
V = (1/3)(31.25)π
V = 10.4167π
We need to differentiate the volume equation with respect to time t:
dV/dt = (d/dt)[(1/3)(31.25)π]
Since the rate at which sand is falling into the pile is given as 10 m^3/sec:
10 = (d/dt)[(1/3)(31.25)π]
Now, let's find the rate at which the height of the pile is changing, which is dh/dt.
To do this, we rearrange the equation:
(d/dt)[(1/3)(31.25)π] = 10
Taking the derivative of V with respect to t, we get:
(31.25/3)π(dh/dt) = 10
Simplifying further:
31.25π(dh/dt) = 30
Now we can solve for dh/dt:
dh/dt = 30 / (31.25π)
Substituting the value of π as approximately 3.14:
dh/dt = 30 / (31.25 * 3.14)
Calculating this further:
dh/dt ≈ 0.96 m/s
Hence, the rate at which the height of the pile is changing when the pile is 5 m high is approximately 0.96 m/s, which is not the same as the given answer of 2/5π m/s. It seems there might be an error in the given solution or in the calculations.
To find the rate at which the height of the pile is changing when the pile is 5 m high, we need to differentiate the volume function with respect to time (t) using the chain rule.
Given:
- h = r = 5 (since the height of the pile is always half the diameter of the base)
- V = (1/3)π(r)(r)(h)
- dV/dt = 10 m^3/sec (rate at which sand is falling into the pile)
First, let's differentiate the volume equation with respect to time:
dV/dt = d/dt [(1/3)π(r)(r)(h)]
Using the product rule, we get:
dV/dt = (1/3)π[2r(dr/dt)(h) + r^2(dh/dt)]
Since h = r, the equation simplifies to:
dV/dt = (1/3)π[2r(dr/dt)(r) + r^2(dh/dt)]
dV/dt = (1/3)π[2r^2(dr/dt) + r^2(dh/dt)]
We can substitute the given values: r = 5, h = 5, and dV/dt = 10 into the equation:
10 = (1/3)π[2(5^2)(dr/dt) + 5^2(dh/dt)]
10 = (1/3)π[2(25)(dr/dt) + 25(dh/dt)]
10 = (1/3)(π/5)[2(25)(dr/dt) + 25(dh/dt)]
30 = 2π(5)(dr/dt) + 5π(dh/dt)
30 = 10π(dr/dt) + 5π(dh/dt)
6 = 2(dr/dt) + (dh/dt)
6 - 2(dr/dt) = (dh/dt)
We are looking for the rate at which the height (h) changes when h = 5 m, so let's substitute h = 5 into the equation:
dh/dt = 6 - 2(dr/dt)
To find dr/dt, we can use the given information that the rate at which the sand is falling into the pile (dV/dt) is 10 m^3/sec. Since the volume of the cone is given by V = (1/3)π(r^2)(h), 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)
Substituting the known values: dV/dt = 10, h = 5, and r = 5, we can solve for dr/dt:
10 = (1/3)π(2(5))(dr/dt)(5) + (1/3)π((5)^2)(dh/dt)
10 = (10/3)π(dr/dt) + (25/3)π(dh/dt)
30 = 10π(dr/dt) + 25π(dh/dt)
30 = (10π)(dr/dt) + (5π)(dh/dt)
dr/dt = (30 - 5π(dh/dt)) / (10π)
dr/dt = (6 - π(dh/dt)) / 2π
Now, substitute the expression for dr/dt back into the equation for dh/dt:
dh/dt = 6 - 2[(6 - π(dh/dt)) / 2π]
dh/dt = 6 - (6 - π(dh/dt)) / π
PI(dh/dt)= 6 - 6 + (dh/dt)
(dh/dt)(π-1)=0
Therefore, either (dh/dt) = 0 OR π = 1. However, since π ≠ 1, we assume (dh/dt) = 0.
Finally, when the pile is 5 m high, the rate at which the height of the pile is changing is zero, which means the height remains constant at that point. So, the rate is 0 m/s.