The top half of a storage tank is a circular cylinder that is 5 meters tall and has a diameter 2 meters. The bottom half of the tank is shaped like an 8-meter inverted cone (pointed down). Let h represent the depth of the tank's contents.
At t = 0 minutes, a release valve at the bottom of the tank is opened and its contents flow out at a rate of 0.5 cubic meters per minute. Assuming the tank is completely full when the release valve is opened, answer the following:
a) Find the value of dh/dt when t = 30 minutes.
b) Find the value of dh/dt when h = 6 meters.
I am not going to do this because it is so long. However note that the change of volume per unit time is the surface area times the change of depth per unit time
dV/dt = pi r^2 dh/dt
so
5 = pi r^2 dh/dt
I mean
0.5 = pi r^2 dh/dt
To solve this problem, we need to find the rate of change of the depth of the tank, which is represented as dh/dt.
To solve part (a), we need to find the value of dh/dt at t = 30 minutes. To do this, we can calculate the rate at which the tank is being emptied at that specific time.
The tank is shaped like a combination of a cylinder and an inverted cone. Let's split the tank into two parts: the upper cylindrical part and the lower conical part.
Let's first find the volume V of the upper cylinder. We can use the formula for the volume of a cylinder, which is V = πr^2h, where r is the radius and h is the height.
Given that the diameter of the cylinder is 2 meters, the radius r is half of that, which is 1 meter. The height h is 5 meters, so the volume V is V = π * (1 meter)^2 * 5 meters = 5π cubic meters.
Now let's find the volume of the conical part. The volume V of a cone is given by V = (1/3)πr^2h, where r is the radius and h is the height.
Given that the cone has a height of 8 meters, we need to find the radius of the cone. Since the cone is shaped like an inverted cone, the radius of the base is half of the diameter of the top of the cone.
The diameter of the top of the cone is 2 meters, so the radius r is 1 meter. The height h is 8 meters, so the volume V of the cone is V = (1/3)π * (1 meter)^2 * 8 meters = (8/3)π cubic meters.
Now, let's consider the rate at which the tank is being emptied. It is given that the contents are flowing out at a rate of 0.5 cubic meters per minute.
This means that the total rate of change of the volume V with respect to time t is -0.5 cubic meters per minute, since the volume is decreasing. So, dV/dt = -0.5 cubic meters per minute.
Now, we can relate the rate of change of volume with respect to time (dV/dt) to the rate of change of height with respect to time (dh/dt) using the chain rule of differentiation.
The volume V of the entire tank is given by V = V_cylinder + V_cone = 5π + (8/3)π = (23/3)π cubic meters.
Differentiating both sides of the equation with respect to time t, we get dV/dt = (23/3)π * dh/dt.
Substituting the given value of dV/dt = -0.5, we can solve for dh/dt at t = 30 minutes:
-0.5 = (23/3)π * dh/dt
Simplifying the equation, we can solve for dh/dt:
dh/dt = -0.5 / ((23/3)π).
Calculate the value of -0.5 / ((23/3)π) to find the value of dh/dt at t = 30 minutes.
Now, let's solve part (b), which asks for the value of dh/dt when h = 6 meters.
To find dh/dt when h = 6 meters, we need to derive an equation that relates dh/dt with height h.
The volume of the tank can be expressed as V = V_cylinder + V_cone.
Using the volume formulas we derived earlier, this equation becomes:
V = 5π + (1/3)π * r^2 * h_cone.
Since the radius of the cone is related to the height, we can substitute r = (1 - h / 8) * 1, where h is the height of the tank.
Rewriting the equation in terms of h, we have:
V = 5π + (1/3)π * [(1 - h / 8) * 1]^2 * h.
To find the equation that relates the rate of change of the volume with respect to time (dV/dt) to the rate of change of height with respect to time (dh/dt), we differentiate both sides of the equation with respect to time t.
dV/dt = 0 + (1/3)π * [(1 - h / 8) * -1 * (1/8) * 2 * h + (1 - h / 8) * 1^2]
Note that the first term on the right-hand side (0) represents the derivative of 5π (constant term), and the second term represents the derivative of the cone volume equation.
Simplifying the derivative equation, we have:
dV/dt = (1/3)π * [(h / 8 - 1) * (-h / 4) + (h / 8 - 1)]
We know that dV/dt = -0.5 cubic meters per minute, so we can set the derived equation equal to -0.5 and solve for dh/dt:
-0.5 = (1/3)π * [(h / 8 - 1) * (-h / 4) + (h / 8 - 1)].
Solving this equation will give us the value of dh/dt when h = 6 meters.
I hope this explanation is helpful in understanding the process of solving the problem.