Sand falls from a hopper into a conical pile with radius equal to half its height. The sand is poured at a rate of 3 m^3 each minute. How fast is the radius increasing after 10 minutes?
Need help with the set up and work through of the problem.
To solve this problem, we need to use related rates - a concept in calculus that deals with how the rates of change of quantities are related to each other. In this case, we are given the rate at which sand is poured (3 m³/minute) and we need to find the rate at which the radius is increasing.
Let's start by setting up some variables for the problem:
- Let V represent the volume of the pile at any given time (in cubic meters).
- Let r represent the radius of the pile at any given time (in meters).
- Let h represent the height of the pile at any given time (in meters).
We are given that the sand is poured at a rate of 3 m³/minute. This means that dV/dt (the rate of change of the volume with respect to time) is 3 m³/minute.
We are also given that the radius of the pile is equal to half its height, so we have the relationship r = h/2.
We can express the volume of a cone in terms of its radius and height using the formula V = (1/3)πr²h.
Now, let's differentiate the volume equation with respect to time (t) using the chain rule:
dV/dt = (1/3)π(2rh(dr/dt) + r²(dh/dt))
Since we are interested in finding the rate at which the radius is increasing after 10 minutes, we need to find dr/dt when t = 10.
We are given h = 2r, so substituting this into the volume equation and replacing dV/dt with 3 m³/minute, we have:
3 = (1/3)π(2r(2(dr/dt)) + r²(dh/dt))
Simplifying the equation further, we get:
1 = 4π(dr/dt) + 2πr(dh/dt)
Now, we have two unknowns in this equation: dr/dt and dh/dt. However, we can find dh/dt by using the fact that sand is pouring at a constant rate of 3 m³/minute. Since the height of the pile increases linearly with time, we can write dh/dt as (Δh)/(Δt), where Δh is the change in the height of the pile and Δt is the change in time. Substituting the given values, we have:
dh/dt = Δh/Δt = 3/1 = 3 m/minute
Substituting this value back into the equation, we have:
1 = 4π(dr/dt) + 2πr(3)
Simplifying further:
1 = 4π(dr/dt) + 6πr
Now, we have a single equation with one unknown (dr/dt). Rearranging the terms:
4π(dr/dt) = 1 - 6πr
dr/dt = (1 - 6πr)/(4π)
At t = 10 (after 10 minutes), we plug in the time value of t and solve for dr/dt:
dr/dt = (1 - 6πr)/(4π) = (1 - 6π(10))/(4π)
Now, we can calculate the value of dr/dt.
To solve this problem, we'll use related rates, which involves finding the rate at which one quantity is changing with respect to another. In this case, we need to find how fast the radius is increasing with respect to time.
Let's denote the height of the cone as h and the radius as r. We are given that the radius is equal to half the height, so we can write r = 1/2 * h.
Given that the sand is poured at a rate of 3 m^3 per minute, we need to find how fast the radius is changing (dr/dt) after 10 minutes.
First, we need to find an equation that relates the volume of the cone (V) to its height and radius. The formula for the volume of a cone is V = 1/3 * π * r^2 * h.
Substituting the value of r in terms of h, we have V = 1/3 * π * (1/2 * h)^2 * h.
Simplifying, V = 1/12 * π * h^3.
Next, we can find the rate at which the volume is changing with respect to time, which is dV/dt. We know that the volume is increasing at a rate of 3 m^3 per minute, so dV/dt = 3.
Differentiating V = 1/12 * π * h^3 with respect to time, we get dV/dt = 1/4 * π * h^2 * dh/dt.
We are interested in finding dh/dt, the rate at which the height is changing with respect to time.
Rearranging the equation, we have dh/dt = (4 / (π * h^2)) * dV/dt.
Substituting the given values of dV/dt = 3 and h = 10 (after 10 minutes), we can now solve for dh/dt.
dh/dt = (4 / (π * 10^2)) * 3.
Simplifying further, dh/dt = 4/π * 0.03 = 0.12/π.
Therefore, the rate at which the height is changing after 10 minutes is approximately 0.12/π meters per minute.