Final answer:
The volume of the conical pile of wheat can be determined using the given formula v=1/3πr²h. Using this formula, the rate of change in volume can be calculated using the chain rule of differentiation. Similarly, the rate of change in the area of the base can be calculated using the formula A=πr² and the chain rule.
Step-by-step explanation:
Let's start by finding the relationship between the radius and the height of the cone. We know that the bottom radius is always half the altitude, so let's call the height h and the radius r. Therefore, r = h/2.
We also know the formula for the volume of a cone is V = (1/3)πr²h.
Substituting r = h/2 into the volume formula, we get V = (1/3)π(h/2)²h = (1/12)πh³.
Now, we can find the rate of change of the volume with respect to time using the chain rule of differentiation. We know that dV/dt = (dV/dh)(dh/dt). Since dh/dt is given as 20 ft³/min, we can solve for dV/dt when h = 6 ft.
Using dV/dh = (1/4)πh² and dh/dt = 20 ft³/min, we get dV/dt = (1/4)π(6)²(20) = 180π ft³/min.
This means that the volume is changing at a rate of 180π ft³/min when the pile is 6 ft high.
Next, let's find the rate of change of the area of the base. The formula for the area of the base is A = πr². Substituting r = h/2, we get A = π(h/2)² = (1/4)πh².
Now, we can find the rate of change of the area with respect to time using the chain rule. We know that dA/dt = (dA/dh)(dh/dt). Since dh/dt is given as 20 ft³/min, we can solve for dA/dt when h = 6 ft.
Using dA/dh = (1/2)πh and dh/dt = 20 ft³/min, we get dA/dt = (1/2)π(6)(20) = 60π ft²/min.
This means that the area of the base is changing at a rate of 60π ft²/min when the pile is 6 ft high.