Sand falls from a conveyor belt at the rate of 20 m^3/min. The sand forms a conical pile where the height of the cone is always 4 times the radius. How fast is the radius changing when the pile is 4 meters high?
volume of cone
= V = (1/3)π r^2 h, but we know h = 4r or r = h/4
V = (1/3)π (h^2/16)(h)
= (1/48)π h^3
dV = (1/16)π r^2 dr/dt
when h = 4, r = 1
20 = (1/16)π(1) dr/dt
dr/dt = 320/π m/min
check my arithmetic, did it on the screen instead of writing it out first
To find the rate at which the radius is changing, we can use related rates.
Let's denote the height of the pile as h and the radius of the pile as r. We are given that the height of the cone is always 4 times the radius, so we have the equation:
h = 4r
Differentiating both sides with respect to time (t), we get:
dh/dt = 4(dr/dt)
We are also given that the sand falls from the conveyor belt at a rate of 20 m^3/min. This tells us that the rate of change of the volume of the cone is 20 m^3/min. The volume of a cone is given by:
V = (1/3)πr^2h
Substituting the relationship between h and r, we have:
V = (1/3)πr^2(4r)
V = (4/3)πr^3
Differentiating both sides with respect to time (t), we get:
dV/dt = (4/3)(3πr^2)(dr/dt)
20 = 4πr^2(dr/dt)
Now we have an equation relating the rate of change of volume (dV/dt) to the rate of change of radius (dr/dt), but we need to find dr/dt when h = 4 meters.
Since h = 4r, when h = 4, we have:
4 = 4r
r = 1
Substituting this value of r into our equation, we get:
20 = 4π(1)^2(dr/dt)
20 = 4π(dr/dt)
Simplifying further gives:
dr/dt = 5/π
Therefore, the radius is changing at a rate of 5/π m/min when the pile is 4 meters high.
To find how fast the radius is changing when the pile is 4 meters high, we need to use related rates.
Let's start by assigning variables to the quantities involved:
- Let h represent the height of the cone.
- Let r represent the radius of the cone.
- Let V represent the volume of the cone.
We know that the sand falls from the conveyor belt at a rate of 20 m^3/min. This tells us that dV/dt = 20 m^3/min.
The formula for the volume of a cone is V = (1/3)πr^2h. Since the height of the cone is always 4 times the radius, we can substitute h = 4r into the equation:
V = (1/3)πr^2(4r)
V = (4/3)πr^3
Differentiating both sides with respect to time (t), we get:
dV/dt = d((4/3)πr^3)/dt
To find dV/dt, we were given that dV/dt = 20 m^3/min.
Next, we need to find dr/dt, the rate at which the radius is changing with respect to time.
Now, we have:
20 = d((4/3)πr^3)/dt
Let's find d((4/3)πr^3)/dt:
d((4/3)πr^3)/dt = (4/3)π * 3r^2 * dr/dt
20 = (4/3)π * 3r^2 * dr/dt
Simplifying the equation:
20 = 4πr^2 * dr/dt
Next, we can rearrange the equation to solve for dr/dt:
dr/dt = 20 / (4πr^2)
Now, we need to find dr/dt when the pile is 4 meters high. Since the height (h) is always 4 times the radius (r), when the height is 4 meters, the radius will be 1 meter.
Substituting the values into the equation:
dr/dt = 20 / (4π(1^2))
dr/dt = 20 / (4π)
dr/dt = 5 / π
Therefore, the rate at which the radius is changing when the pile is 4 meters high is 5 / π meters per minute.