Sand falls from a conveyor belt onto a conical pile at a rate of 6ft/mi^3. The radius of the base is always equal to two-thirds of the pile’s height. At what rate is the height of the pile changing when the radius of the base of the pile is 2 ft?
I honestly have no idea how to do this. Please help.
sand falls at the rate of 6 ft/mi^3???
Do you mean 6 ft^3/min ?
r = 2h/3, so h = 3r/2
so, when r=2, h=3
v = 1/3 πr^2 h
= 1/3 π * (2/3 h)^2 * h
= 4/27 πh^3
dv/dt = 4/9 πh^2 dh/dt
You know that dv/dt = 6, so
4/9 π * 9 dh/dt = 6
dh/dt = 6/(4π) = 3/(2π) ft/min
Sweet! I managed to get as far as dv/dt = 4/9 πh^2 dh/dt , but couldn't figure it out from there, thanks!
To find the rate at which the height of the pile is changing, we need to use related rates, which involve finding the rates of change of various quantities with respect to time.
Let's assign variables to the quantities mentioned in the problem:
- Let V be the volume of sand in the pile (in cubic feet).
- Let r be the radius of the base of the pile (in feet).
- Let h be the height of the pile (in feet).
Now, we can establish the relationships between the variables:
1. The volume of a cone can be expressed in terms of the base radius and height using the formula V = (1/3)πr^2h.
2. The given condition states that the radius of the base is always equal to two-thirds of the height, which can be written as r = (2/3)h.
We are given the rate at which sand falls onto the pile, which is 6 ft^3/min. We need to find the rate at which the height of the pile is changing when the radius of its base is 2 ft. We need to find dh/dt (the rate of change of height) when dr/dt (the rate of change of radius) is known.
To find dh/dt, let's differentiate both sides of the equation r = (2/3)h with respect to time (t):
dr/dt = (2/3)dh/dt
We're given dr/dt = 0 ft/min because the problem states that the radius of the base of the pile is 2 ft. Therefore, we can substitute dr/dt = 0 ft/min into the equation above:
0 = (2/3)dh/dt
Now, we can solve for dh/dt:
dh/dt = 0 / (2/3) = 0 ft/min
Thus, the rate of change of the height of the pile when the radius of its base is 2 ft is 0 ft/min.