Grain is being dropped into a silo from a tall conveyor belt forming a right circular cone with diameter at the base equal to the height. The grain is being added at the rate of 10 cubic inches per second. How fast is the height changing when the height is 80 inches
To find how fast the height is changing when the height is 80 inches, we need to use related rates.
Let's denote the height of the cone as h inches and the radius (half of the diameter) as r inches. Since the diameter at the base of the cone is equal to the height, we can write h = 2r.
The volume of a cone is given by the formula V = (1/3)πr^2h.
We are given that the volume is changing at a rate of 10 cubic inches per second, which means dV/dt = 10.
We want to find dh/dt, the rate at which the height is changing when the height is 80 inches. To do this, we need to find a relationship between dh/dt and dV/dt.
We can start by differentiating the equation V = (1/3)πr^2h with respect to time t:
dV/dt = (1/3)π(2rh)(dh/dt) + (1/3)πr^2(dh/dt)
Simplifying this expression, we get:
10 = (2/3)πr(h)(dh/dt) + (1/3)πr^2(dh/dt)
Since h = 2r, we can substitute this into the equation:
10 = (2/3)πr(2r)(dh/dt) + (1/3)πr^2(dh/dt)
Simplifying further, we have:
10 = (4/3)πr^2(dh/dt) + (1/3)πr^2(dh/dt)
Combining like terms, we get:
10 = (5/3)πr^2(dh/dt)
Now, we can solve for dh/dt:
dh/dt = (10 / ((5/3)πr^2))
To find the height, we substitute r = h/2:
dh/dt = (10 / ((5/3)π(h/2)^2))
dh/dt = (20 / ((5/3)π(h^2/4)))
dh/dt = (20 * (4/3)π) / h^2
Now we can find dh/dt when the height is 80 inches:
dh/dt = (20 * (4/3)π) / (80^2)
= (20 * (4/3)π) / 6400
≈ 0.0007854 in/sec
Therefore, when the height is 80 inches, the height is changing at a rate of approximately 0.0007854 inches per second.