A pile of sand in the shape of a cone whose base diameter is 4 times its height is being built up at 12 pi m3/min. How fast is the height of the pile increasing 3 min after the pouring begins.
follow the same method I just showed you in your other cone problem.
To find the rate at which the height of the sand pile is increasing, we can use derivatives and related rates.
Let's start by assigning some variables:
- Let h be the height of the cone (in meters)
- Let r be the radius of the cone's base (in meters)
- Let V be the volume of the cone (in cubic meters)
Given that the cone's base diameter is 4 times its height, we have r = 2h.
We know that the volume of a cone can be calculated using the formula:
V = (1/3)πr^2h
Since we have r = 2h, we can rewrite the volume formula using h:
V = (1/3)π(2h)^2h
V = (4/3)πh^3
Now, we can differentiate the equation with respect to time to find the rate of change of volume with respect to time:
dV/dt = (4/3)π(3h^2)(dh/dt)
Given that dV/dt (the rate at which the sand is being poured) is 12π m^3/min, we substitute that value into the equation:
12π = (4/3)π(3h^2)(dh/dt)
We can simplify the equation by canceling the π terms:
12 = 4(3h^2)(dh/dt)
Simplifying further:
12 = 12h^2(dh/dt)
Now, we need to find the value of dh/dt (the rate at which the height is increasing) at a specific time, which is 3 min after pouring begins.
At this time, we need to find the height of the pile. Since h represents the height at any given time, we can substitute 3 into the equation:
12 = 12(3^2)(dh/dt)
Simplifying further:
12 = 36(dh/dt)
Dividing both sides by 36:
dh/dt = 12/36
dh/dt = 1/3
Therefore, the height of the sand pile is increasing at a rate of 1/3 m/min, 3 minutes after pouring begins.