A machine starts dumping sand at the rate of 20 m3/min, forming a pile in the shape of a cone. The height of the pile is always twice the length of the base diameter. After 5 minutes, how fast is the base area of the base increasing?
I honestly have no idea how to start this question
Since the height is twice the diameter, the height is equal to the radius of the circular base.
Let the radius be r
V = (1/3)π r^2 h, but h = r
V = (1/3) π r^3
dV/dt = π r^2 dr/dt
dr/dt = 20/(πr^2) **
when t = 5 min, V = 5(20) = 100 m^3
100 = (1/3)πr^3
300 = πr^3
r^3 = 300/π
r = (300/π)^(1/3) ***
area of base = A = πr^2
dA/dt = 2π r dr/dt
= 2π(***)(**)
= ......
I will let you do the button-pushing
huh? height is twice the diameter, not half h = 4r
Make that change, Nicole, and redo the steps.
thanks Steve
To solve this problem, we need to find the rate at which the base area is increasing. Let's break down the problem step by step:
1. Visualize the situation: The machine is dumping sand, forming a pile in the shape of a cone. The height of the pile is always twice the length of the base diameter. After 5 minutes, we need to find how fast the base area is increasing.
2. Relate the given information: We know the rate at which the sand is being dumped, which is 20 m3/min. We also know the relationship between the height and the base diameter of the cone, i.e., the height is twice the length of the base diameter.
3. Identify the variables: Let's assign variables to the relevant quantities. Let h be the height of the cone (which is also twice the base diameter), and let r be the radius of the base. The rate at which the sand is being dumped is given as 20 m3/min, and we need to find the rate at which the base area is increasing.
4. Set up the equation: We know that the formula for the volume of a cone is V = (1/3)πr^2h. Rearranging this equation, we have h = 3V / (πr^2). As the height is always twice the base diameter, we have h = 2r. Combining these two equations, we get r = (3V / (2π))^(1/3).
5. Apply differentiation: We need to differentiate both sides of the equation with respect to time (t) to find the rates of change. Differentiating the equation r = (3V / (2π))^(1/3) with respect to t, we get dr/dt = [(dV/dt)*(3/2π)]^(1/3)
6. Plug in the values: We are given that dV/dt = 20 m3/min. Plugging this value into the equation, we have dr/dt = [(20)*(3/2π)]^(1/3).
7. Calculate the rate: Now, we can substitute the value of π and evaluate dr/dt. The final answer will give us the rate at which the base area is increasing after 5 minutes.
Following these steps, you can find the rate at which the base area of the pile is increasing.