The County Roads Department is stockpiling salt for use next winter. The salt is being unloaded from
railroad cars by a conveyor belt at a rate of 20 cubic feet per minute. The salt falls into a conical pile
whose diameter at the base is four times the height of the pile. At what rate is the diameter of the pile
changing when the pile is 12 feet high?
Show steps please!
Thank you!
let the height be h
then the diameter of the base is 4h, thus the radius is 2h
given: dV = 20 ft^3/min
find: d(2h)/dt , when h = 12
V = (1/3)π r^2 h
= (1/3) π (4h^2)(h)
= (4/3) π h^3
dV/dt = 4π h^2 dh/dt
20 = 4π((144) dh/dt
dh/dt = 5/(144π)
so
d(2h)/dt = 10/(144π)
= 5/(72π)
= appr .0221 ft/min
I was just reading the last solution and I was wondering why he had the 1/3 before the pir^2h.
To find the rate at which the diameter of the pile is changing when the pile is 12 feet high, we can use related rates.
Given:
- The rate at which salt is being unloaded is 20 cubic feet per minute.
- The diameter at the base of the pile is four times the height of the pile.
- We need to find the rate at which the diameter is changing when the pile is 12 feet high.
Let's assign variables to the quantities involved in the problem:
- Let V be the volume of the cone at any time t (in cubic feet).
- Let r be the radius of the base of the cone at any time t (in feet).
- Let h be the height of the pile at any time t (in feet).
We know that the volume of a cone can be calculated using the formula V = (1/3)πr^2h.
To find the relationship between r and h, we are given that the diameter at the base of the pile is four times the height of the pile. Since the diameter is twice the radius, we can write: r = 2h.
We need to find the rate at which the diameter (2r) is changing when the pile is 12 feet high. This means we need to find dr/dt when h = 12.
To solve the problem, we will use implicit differentiation. Differentiate both sides of the equation r = 2h with respect to t (time):
dr/dt = 2(dh/dt)
Now let's find an expression for dh/dt, the rate at which the height of the pile is changing at any given time. We are given that salt is being unloaded from the conveyor belt at a rate of 20 cubic feet per minute. This means that dV/dt = 20.
To find an expression for dV/dt, we can differentiate the volume formula V = (1/3)πr^2h with respect to t:
dV/dt = (1/3)π(2r)(dh/dt) + (1/3)πr^2(dh/dt)
Simplifying this expression, we have:
20 = (2/3)πr(dh/dt) + (1/3)πr^2(dh/dt)
Now, substitute the value of r in terms of h (r = 2h):
20 = (2/3)π(2h)(dh/dt) + (1/3)π(4h^2)(dh/dt)
Combine the terms:
20 = (4/3)πh(dh/dt) + (4/3)πh^2(dh/dt)
Now, let's solve for dh/dt:
dh/dt = 20 / [(4/3)πh + (4/3)πh^2]
= 15 / [πh + πh^2]
Finally, substitute h = 12 into the expression for dh/dt:
dh/dt = 15 / [π(12) + π(12^2)]
= 15 / [12π + 144π]
= 15 / (156π)
≈ 0.0303 ft/min
Therefore, when the pile is 12 feet high, the diameter of the pile is changing at a rate of approximately 0.0303 feet per minute.